Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Liang C++ 5E Revel Assigned Programming Quiz and Programming Project Solution Note: For every question you entered, test it using the solution.

Chapter 1 Programming Quiz 1.6: Question 1: (00000-10501) Write a statement that prints Hello World to the screen. cout << "Hello World";

Programming Quiz 1.6: Question 2: (00000-10502) Write a complete program that prints Hello World to the screen. #include <iostream> using namespace std; int main() { cout << "Hello World"; return 0; } Programming Quiz 1.6: Question 3: (00000-10503) Suppose your name was Alan Turing. Write a statement that would print your last name, followed by a comma, followed by your first name.

Do not print anything else (that includes blanks). cout << "Turing,Alan"; Chapter 1: Programming Project 1: (345021-00007) (Display three messages) Write a program that displays Welcome to C++ Welcome to Computer Science Programming is fun For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; int main() { cout << "Welcome to C++" << endl; cout << "Welcome to Computer Science" << endl; Solutions Manual for Introduction to C++ Programming and Data cout << "Programming is fun" << endl;

Structures 5th edition by Daniel Liang return 0;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } Chapter 1: Programming Project 2: (345021-00008) (Compute expressions) Write a program that displays the result of (9.5 x 4.5 - 2.5 x 3) / (45.5 - 3.5) For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; int main() { cout << (9.5 * 4.5 - 2.5 * 3) / (45.5 - 3.5) << endl; return 0; }

Chapter 1: Programming Project 3: (345021-00009) (Population projection) The U.S. Census Bureau projects population based on the following assumptions: One birth every 7 seconds.

One death every 13 seconds.

One new immigrant every 45 seconds Write a program that displays the population for each of the next five years. Assume the current population is 312,032,486 and one year has 365 days.

Hint: In C++, if two integers perform division, the result is the quotient. The fractional part is truncated.

For example, 5 / 4 is 1 (not 1.25) and 10 / 4 is 2 (not 2.5). To get an accuate result with the fractional part, one of the values involved in the division must be a number with a decimal point. For example, 5.0 / 4 is 1.25 and 10 / 4.0 is 2.5.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; int main() { cout << 312032486 + 365 * 24 * 60 * 60 / 7.0 - 365 * 24 * 60 * 60 / 13.0 + 365 * 24 * 60 * 60 / 45.0 << endl; cout << 312032486 + 2 * 365 * 24 * 60 * 60 / 7.0 - 2 * 365 * 24 * 60 * 60 / 13.0 + 2 * 365 * 24 * 60 * 60 / 45.0 << endl; cout << 312032486 + 3 * 365 * 24 * 60 * 60 / 7.0 - 3 * 365 * 24 * 60 * 60 / 13.0 + 3 * 365 * 24 * 60 * 60 / 45.0 << endl; cout << 312032486 + 4Introduction * 365 * 24 * to 60 * 60 Programming / 7.0 - 4 * 365 24 * 60 * 60 Solutions Manual for C++ and* Data / 13.0 + 4 * 365 * 24 * 60 * 60 / 45.0 << endl;

Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << 312032486 + 5 * 365 * 24 * 60 * 60 / 7.0 - 5 * 365 * 24 * 60 * 60 / 13.0 + 5 * 365 * 24 * 60 * 60 / 45.0 << endl; return 0; } Chapter 1: Programming Project 4: (345021-00153) (Simple computation) The formula for computing the discriminant of a quadratic equation ax^2 + bx + c = 0 is b^2 – 4ac. Write a program that computes the discriminant for the equation 3x^2 + 4x + 5 = 0.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise01_01Extra using namespace std; int main() { cout << 4 * 4 - 4 * 3 * 5 << endl; return 0; } Chapter 1: Programming Project 5: (345021-00154) (Physics: acceleration) Average acceleration is defined as the change of velocity divided by the time taken to make the change, as shown in the following formula: a = (v1 - v0) / t Here, v0 is the starting velocity in meters/second, v1 is the ending velocity in meters/second, and t is the time span in seconds. Assume v0 is 5.6, v1 is 10.5, and t is 0.5, and write the code to display the average acceleration.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise01_02Extra using namespace std; int main() { cout << (10.5 - 5.6) / 0.5 << endl; return 0; } Chapter 2

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 2.2: Question 1: (00000-10982) Declare a double variable named distance.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang double distance;

Programming Quiz 2.2: Question 2: (00000-10983) Write a statement that declares a double variable named dosage. double dosage;

Programming Quiz 2.3: Question 1: (00000-10968) Write a statement that reads a floating point (real) value from standard input into temperature.

Assume that temperature has already been declared as a double variable. cin >> temperature; Programming Quiz 2.5: Question 1: (00000-10514) Write a complete program that •

declares an int variable,

•

reads a value from the keyboard into that variable, and

•

writes to standard output the square of the variable's value.

Besides the number, nothing else should be written to standard output. #include <iostream> using namespace std; int main() { int k; cin >> k; cout << k * k; return 0; }

Programming Quiz 2.5: Question 2: (00000-10515) Write a complete program that •

declares an int variable,

•

reads a value from the keyboard into that variable, and

•

writes to standard output the variable's value, twice the value, and the square of the value, separated by spaces.

Besides the numbers, nothing else should be written to standard output except for spaces separating the values. #include <iostream> using namespace std; int main() { int k; Solutions cin >> k;Manual for Introduction to C++ Programming and Data Structures Liang cout << k 5th << "edition " << k *by2 Daniel << " " << k * k;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return 0; } Programming Quiz 2.6: Question 1: (00000-10539) Given two int variables oldRecord and newRecord, write a statement that gives newRecord the same value that oldRecord has. newRecord = oldRecord; Programming Quiz 2.6: Question 2: (00000-10538) Given an int variable drivingAge that has already been declared, write a statement that assigns the value 17 to drivingAge. drivingAge = 17; Programming Quiz 2.6: Question 3: (00000-10947) Write a statement to set the value of num to 4 (num is a variable that has already been declared). num = 4;

Programming Quiz 2.7: Question 1: (00000-10519) Declare an int constant, MonthsInYear, whose value is 12.

Note: The quizzes are a third-party product. The name for the constant does not follow the convention used in the text. It would be named MONTHS_IN_YEAR if the convention in the text is used. This is the only place we alert our users. We will not explicitly point it out in other places. const int MonthsInYear = 12;

Programming Quiz 2.7: Question 2: (00000-10520) Declare an int constant MonthsInDecade whose value is the value of the constant MonthsInYear (already declared) multiplied by 10. const int MonthsInDecade = MonthsInYear * 10;

Programming Quiz 2.8: Question 1: (00000-10505) Assume that an int variable x that has already been declared and initialized.

Write an expression whose value is 1 more than x. x + 1 Programming Quiz 2.8: Question 2: (00000-10526)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Assume that variables verbalScore and mathScore are already declared and assigned values.

Write an expression that computes the sum of the two variables verbalScore and mathScore.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang verbalScore + mathScore Programming Quiz 2.8: Question 3: (00000-10528) Assume two variables endingTime and startingTime are already declared and assigned values.

Write an expression that computes the difference of the variables endingTime and startingTime. endingTime - startingTime Programming Quiz 2.8: Question 4: (00000-10530) Given the variable pricePerCase, write an expression corresponding to the price of a dozen cases. pricePerCase * 12 Programming Quiz 2.8: Question 5: (00000-10932) Write an expression that computes the remainder of the variable principal when divided by the variable divisor. (Assume both are type int.) principal % divisor

Programming Quiz 2.9: Question 1: (00000-10533) Write an expression that computes the average of the values 12 and 40.

(12 + 40) / 2 Programming Quiz 2.9: Question 2: (00000-11013) The dimensions (width and length) of room1 have been read into two variables: width1 and length1.

The dimensions of room2 have been read into two other variables: width2 and length2. Write a single expression whose value is the total area of the two rooms. length1 * width1 + length2 * width2 Programming Quiz 2.11: Question 1: (00000-10541) Given an int variable bridgePlayers, write a statement that increases the value of that variable by 4. bridgePlayers += 4; Programming Quiz 2.11: Question 2: (00000-10542) Given an int variable profits, write a statement that increases the value of that variable by a factor of 10. profits *= 10;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 2.11: Question 3: (00000-10961)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write a statement using a compound assignment operator to subtract 10 from minutes_left (an int variable that has already been declared and initialized). minutes_left -= 10; Programming Quiz 2.11: Question 4: (00000-10962) Write a statement using a compound assignment operator to cut the value of pay in half (pay is an integer variable that has already been declared and initialized). pay /= 2;

Programming Quiz 2.11: Question 5: (00000-10964) Write a statement using a compound assignment operator to change val to the remainder when val is divided by 16 (val has already been declared and initialized) val %= 16; Programming Quiz 2.12: Question 1: (00000-10958) Write a statement using the increment operator or combined assignment operator to increase the value of num_items (an already declared integer variable) by 1. num_items++; Programming Quiz 2.12: Question 2: (00000-10544) Given an int variable timer, write a statement that uses the decrement operator to decrease the value of that variable by 1. timer--; Programming Quiz 2.12: Question 3: (00000-10959) Write a statement using the decrement operator to decrease the value of count (an already declared integer variable) by 1. count--; Programming Quiz 2.13: Question 1: (00000-10608) Given two int variables distance and speed, write an expression that divides distance by speed using floating point arithmetic, i.e., a fractional result should be produced. static_cast<double>(distance) / speed Programming Quiz 2.13: Question 2: (00000-11017) In mathematics, the Nth harmonic number is defined to be 1 + 1/2 + 1/3 + 1/4 + ... + 1/N. So, the first harmonic number is 1, the second is 1.5, the third is 1.83333... and so on.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write an expression whose value is the 8th harmonic number.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang 1 + 1.0 / 2 + 1.0 / 3 + 1.0 / 4 + 1.0 / 5 + 1.0 / 6 + 1.0 /7 + 1.0 / 8

Chapter 2: Programming Project 1: (345021-00010) (Financial application: calculate tips) Write a program that reads the subtotal and the gratuity rate, then computes the gratuity and total.

For example, if the user enters 10 for subtotal and 15% for gratuity rate, the program displays $1.5 as gratuity and $11.5 as total.

Sample Run Enter the subtotal and a gratuity rate: 10 15 The gratuity is $1.5 and total is $11.5 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; int main() { cout << "Enter the subtotal and a gratuity rate: "; double subtotal, gratuityRate; cin >> subtotal >> gratuityRate; double gratuity = subtotal * gratuityRate / 100; double total = subtotal + gratuity; cout << "The gratuity is " << gratuity << " and total is " << total << endl; return 0; }

Chapter 2: Programming Project 2: (345021-00011) (Find the number of years) Write a program that prompts the user to enter the minutes (e.g., 1 billion) as an integer and displays the number of years and days for the minutes. For simplicity, assume a year has 365 days.

Sample Run Enter the number of minutes: 1000000000 1000000000 minutes is approximately 1902 years and 214 days For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std;

Solutions int main() Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { // Prompt the user to enter the number of minutes

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "Enter the number of minutes: "; long minutes; cin >> minutes; long totalNumberOfDays = minutes / (24 * 60); long numberOfYears = totalNumberOfDays / 365; long remainingNumberOfDays = totalNumberOfDays % 365; // Display results cout << minutes << " minutes is approximately " << numberOfYears << " years and " << remainingNumberOfDays << " days." << endl; return 0; }

Chapter 2: Programming Project 3: (345021-00012) (Geometry: distance of two points) Write a program that prompts the user to enter two points (x1, y1) and (x2, y2) and displays their distance between them.

The formula for computing the distance is: Square root of ((x2 - x1) squared + (y2 - y1) squared) Note that you can use pow(a, 0.5) to compute square root of a.

Sample Run Enter x1 and y1: 1.5 -3.4 Enter x2 and y2: 4 5 The distance between the two points is 8.764131445842194 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <cmath> using namespace std; int main() { // Enter the first point with two double values cout << "Enter x1 and y1: "; double x1, y1; cin >> x1 >> y1; // Enter the second point with two double values cout << "Enter x2 and y2: "; double x2, y2; cin >> x2 >> y2; // Compute the distance double distance = pow((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2), 0.5);

Solutions Manual for Introduction to C++ and Data cout << "The distance of the two points is Programming " << distance; Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return 0; }

Chapter 2: Programming Project 4: (345021-00013) (Geometry: area of a triangle) Write a program that prompts the user to enter three points (x1, y1), (x2, y2), (x3, y3) of a triangle and displays its area.

The formula for computing the area of a triangle is: s = (side1 + side2 + side3) / 2 area = square root of s(s - side1)(s - side2)(s - side3) Sample Run Enter three points for a triangle: 1.5 -3.4 4.6 5 9.5 -3.4 The area of the triangle is 33.6 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise02_19 #include <cmath> using namespace std; int main() { // Enter three points for a triangle double x1, y1, x2, y2, x3, y3; cout << "Enter three points for a triangle: "; cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3; // Compute the length of the three sides double side1 = pow((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2), 0.5); double side2 = pow((x1 - x3) * (x1 - x3) + (y1 - y3) * (y1 - y3), 0.5); double side3 = pow((x3 - x2) * (x3 - x2) + (y3 - y2) * (y3 - y2), 0.5); double s = (side1 + side2 + side3) / 2; double area = pow(s * (s - side1) * (s - side2) * (s - side3), 0.5); cout << "The area of the triangle is " << area << endl; return 0; }

Chapter 2: Programming Project 5: (345021-00014) (Financial application: future investment value) Write a program that reads in investment amount, annual interest rate, and number of years, and

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang displays the future investment value using the following formula: futureInvestmentValue =

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang investmentAmount x (1 + monthlyInterestRate)^(numberOfYears x 12) For example, if you enter amount 1000.56, annual interest rate 4.25%, and number of years 1, the future investment value is $1043.92.

Sample Run Enter investment amount: 1000.56 Enter annual interest rate in percentage: 4.25 Enter number of years: 1 Accumulated value is $1043.92 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html.

#include <iostream> #include <cmath> using namespace std; int main() { double investmentAmount, annualInterestRate, numberOfYears; cout << "Enter investment amount: "; cin >> investmentAmount; cout << "Enter annual interest rate: "; cin >> annualInterestRate; cout << "Enter number of years: "; cin >> numberOfYears; double monthlyInterestRate = annualInterestRate / 1200; double accumulatedValue = investmentAmount * pow(1 + monthlyInterestRate, numberOfYears * 12); cout << "Future value is $" << accumulatedValue << endl; return 0; }

Chapter 3 Programming Quiz 3.2: Question 1: (00000-10597) Working overtime is defined as having worked more than 40 hours during the week.

Given the variable hoursWorked, write an expression that evaluates to true if the employee worked overtime. hoursWorked > 40

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Programming Quiz 3.2: Question 2: (00000-10599)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Given the variables numberOfMen and numberOfWomen, write an expression that evaluates to true if the number of men is greater than or equal to the number of women. numberOfMen >= numberOfWomen

Programming Quiz 3.2: Question 3: (00000-10907) Write an expression that evaluates to true if the value of the int variable x is divisible (with no remainder) by the int variable y. (Assume that y is not zero.) All the variables used in the programming quizzes are assumed to be declared already unless explicitly specified otherwise.

This is the only place we alert our users. We will not explicitly point it out in other places. x % y == 0

Programming Quiz 3.2: Question 4: (00000-10547) Write an expression that evaluates to true if the int variable x contains an even value, and false if it contains an odd value. x % 2 == 0

Programming Quiz 3.3: Question 1: (00000-10561) Write an if statement that assigns 10,000 to the variable bonus if the value of the variable goodsSold is greater than 500,000. Assume bonus and goodsSold have been declared and initialized. if (goodsSold > 500000) bonus = 10000; Programming Quiz 3.3: Question 2: (00000-10562) Write an if statement that decreases the variable shelfLife by 4 if the variable outsideTemperature is greater than 90.

Assume shelfLife and outsideTemperature have been declared and initialized. if (outsideTemperature > 90) shelfLife -= 4;

Programming Quiz 3.3: Question 3: (00000-11060) Assume that the variables gpa, deansList, and studentName have been declared and initialized.

Write a statement that both adds 1 to deansList and prints studentName to standard output if gpa exceeds 3.5. if (gpa > 3.5) { deansList++; cout << studentName << endl; Solutions Manual for Introduction to C++ Programming and Data }

Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 3.4 : Question 1: (00000-10564) Write an if/else statement that compares the variable age with 65, adds 1 to the variable seniorCitizens if age is greater than or equal to 65, and adds 1 to the variable nonSeniors otherwise. Assume all variables have been declared and initialized. if (age >= 65) seniorCitizens++; else nonSeniors++;

Programming Quiz 3.4: Question 2: (00000-11119) NOTE: in mathematics, the square root of a negative number is not real; in C++ therefore, passing such a value to the square root function is an error.

Given a double variable named areaOfSquare, write the code to read in a value, the area of some square, into areaOfSquare and print out the length of the side of that square.

HOWEVER: if any value read in is not valid input, just print the message "INVALID". cin >> areaOfSquare; if (areaOfSquare >= 0.0) cout << pow(areaOfSquare, 0.5); else cout << "INVALID"; Programming Quiz 3.5: Question 1: (00000-10567) Write a multi-way if statement that adds 1 to the variable minors if the variable age is less than 18, adds 1 to the variable adults if age is 18 through 64, and adds 1 to the variable seniors if age is 65 or older. if (age < 18) minors++; else if (age < 65) adults++; else seniors++;

Programming Quiz 3.5: Question 2: (00000-10568) Write a multi-way if statement that compares the double variable pH with 7.0 and makes the following assignments to the bool variables neutral, base, and acid: •

false, false, true if pH is less than 7

•

false, true, false if pH is greater than 7

•

true, false, false if pH is equal to 7

if (pH < 7) { neutral = false; base = false; acid = true; Solutions Manual for Introduction to C++ Programming and Data } Structures 5th edition by Daniel Liang else if (pH > 7)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { neutral = false; base = true; acid = false; } else { neutral = true; base = false; acid = false; }

Programming Quiz 3.5: Question 3: (00000-11064) Online Book Merchants offers premium customers 1 free book with every purchase of 5 or more books and offers 2 free books with every purchase of 8 or more books. It offers regular customers 1 free book with every purchase of 7 or more books, and offers 2 free books with every purchase of 12 or more books.

Write a statement that assigns freeBooks the appropriate value based on the values of the bool variable isPremiumCustomer and the int variable nbooksPurchased. Assign 0 to freeBooks if no free books are offered. Assume the variables freeBooks, isPremiumCustomer, and nbooksPurchased are already declared. if (isPremiumCustomer) { if (nbooksPurchased >= 8) freeBooks = 2; else if (nbooksPurchased >= 5) freeBooks = 1; else freeBooks = 0; } else { if (nbooksPurchased >= 12) freeBooks = 2; else if (nbooksPurchased >= 7) freeBooks = 1; else freeBooks = 0; }

Programming Quiz 3.10: Question 1: (00000-10556) Given the variables temperature and humidity, write an expression that evaluates to true if and only if the temperature is greater than 90 and the humidity is less than 10. temperature > 90 && humidity < 10

Programming Quiz 3.10: Question 2: (00000-10557) Given the int variables yearsWithCompany and department, write an expression that evaluates to true if yearsWithCompany is less than 5 and department is not equal to 99. yearsWithCompany < 5 && department != 99

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 3.10: Question 3: (00000-10558)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Given the bool variable isFullTimeStudent and the int variable age, write an expression that evaluates to true if age is less than 19 or isFullTimeStudent is true. age < 19 || isFullTimeStudent

Programming Quiz 3.10: Question 4: (00000-11067) Clunker Motors Inc. is recalling all vehicles from model years 2001-2006. A bool variable named recalled has been declared.

Given an int variable modelYear, write a statement that assigns true to recalled if the value of modelYear falls within the recall range and assigns false otherwise.

Do not use an if statement in this exercise! recalled = 2001 <= modelYear && modelYear <= 2006; Programming Quiz 3.11: Question 1: (00000-11118) Clunker Motors Inc. is recalling all vehicles from model years 1995-1998 and 2004-2006.

A bool variable named norecall has been declared.

Given an int variable modelYear, write a statement that assigns true to norecall if the value of modelYear does not fall within the two recall ranges and assigns false otherwise.

Do not use an if statement in this exercise! norecall = !(1995 <= modelYear && modelYear <= 1998 || 2004 <= modelYear && modelYear <= 2006); Programming Quiz 3.11: Question 2: (00000-11116) Clunker Motors Inc. is recalling all vehicles from model years 2001-2006.

Given an int variable modelYear, write a statement that prints the message "NO RECALL" to standard output if the value of modelYear does not fall within that range. if (modelYear < 2001 || modelYear > 2006) cout << "NO RECALL";

Programming Quiz 3.11: Question 3: (00000-11066) Clunker Motors Inc. is recalling all vehicles from model years 2001-2006.

Given a variable modelYear, write a statement that prints the message "RECALL" to standard output if the value of modelYear falls within that range. if (2001 <= modelYear && modelYear <= 2006) cout << "RECALL";

Programming Quiz 3.13: Question 1: (00000-11074)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang HTTP is the protocol that governs communications between Web servers and Web clients (i.e. browsers). Part of the protocol includes a status code returned by the server to tell the browser the status of its most recent page request. Some of the codes and their meanings are listed below:

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang •

200, OK (fulfilled)

•

403, forbidden

•

404, not found

•

500, server error

Given an int variable status, write a switch statement that prints out the appropriate label from the above list based on status. switch (status) { case 200: cout << "OK (fulfilled)"; break; case 403: cout << "forbidden"; break; case 404: cout << "not found"; break; case 500: cout << "server error"; break; } Programming Quiz 3.13: Question 2: (00000-11073) In the Happy Valley School System, children are classified by age as follows: •

less than 2, ineligible

•

2, toddler

•

3-5, early childhood

•

6-7, young reader

•

8-10, elementary

•

11 and 12, middle

•

13, impossible

•

14-16, high school

•

17-18, scholar

•

greater than 18, ineligible

Given an int variable age, write a switch statement that prints out the appropriate label from the above list based on age. switch (age) { case 2: cout << "toddler"; break; case 3: case 4: case 5: cout << "early childhood"; break; case 6: case 7: cout << "young reader"; break; case 8: case 9: case 10: cout << "elementary"; break; case 11: case 12: cout << "middle"; break; case 13: cout << "impossible"; break; case 14: case 15: case 16: cout << "high school"; break; case 17: case 18: cout << "scholar"; break; default: cout << "ineligible"; break; } Solutions Manual for Introduction to C++ Programming and Data

Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 3.14: Question 1: (00000-10569)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write an expression using the conditional operator (? :) that compares the value of the variable x to 5 and results in: •

x if x is greater than or equal to 5

•

-x if x is less than 5

x >= 5 ? x : -x

Programming Quiz 3.14: Question 2: (00000-10777) Assume that day is an int variable whose value is 1 or 2 or ... or 7.

Write an expression whose value is "sun" or "mon" or "tue" or "wed" or "thu" or "fri" or "sat" based on the value of day. day == 1 ? "sun" : day == 2 ? "mon" : day == 3 ? "tue" : day == 4 ? "wed":day == 5 ? "thu" : day == 6 ? "fri" : "sat" Programming Quiz 3.14: Question 3: (00000-10806) Assume that credits is an int variable whose value is 0 or positive.

Write an expression whose value is "freshman" or "sophomore" or "junior" or "senior" based on the value of credits.

In particular: if the value of credits is less than 30 the expression's value is "freshman"; 30-59 would be a "sophomore"; 60-89 would be "junior" and 90 or more would be a "senior". credits < 30 ? "freshman" : credits <= 59 ? "sophomore" : credits <= 89 ? "junior" : "senior"

Chapter 3: Programming Project 1: (345021-00015) The description of this project is given in Programming Exercise 3.1 in the Chapter 3 Programming Exercise from the Book section.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <cmath> using namespace std; int main() { cout << "Enter a, b, c: "; double a, b, c, discriminant, r1, r2; cin >> a >> b >> c; discriminant = b * b - 4 * a * c;

Solutions Manual for if (discriminant < 0)Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "The equation has no real roots" << endl; } else if (discriminant == 0) { r1 = (-b + pow(discriminant, 0.5)) / (2 * a); cout << "The root is " << r1 << endl; } else // (discriminant >= 0) { r1 = (-b + pow(discriminant, 0.5)) / (2 * a); r2 = (-b - pow(discriminant, 0.5)) / (2 * a); cout << "The roots are " << r1 << " and " << r2 << endl; } return 0; } Chapter 3: Programming Project 2: (345021-00016) (Algebra: solve 2 × 2 linear equations) You can use Cramer’s rule to solve the following 2 × 2 system of linear equation: ax + by = e cx + dy = f x = (ed - bf) / (ad - bc) y = (af - ec) / (ad - bc) Write a program that prompts the user to enter a, b, c, d, e and f and displays the result.

If ad - bc is 0, report that “The equation has no solution”.

Sample Run 1 Enter a, b, c, d, e, f: 9.0 4.0 3.0 -5.0 -6.0 -21.0 x is -2 and y is 3 Sample Run 2 Enter a, b, c, d, e, f: 1.0 2.0 2.0 4.0 4.0 5.0 The equation has no solution For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; int main() { cout << "Enter a, b, c, d, e, f: "; double a, b, c, d, e, f; cin >> a >> b >> c >> d >> e >> f;

Solutions Manual for Introduction to C++ Programming and Data double detA a * d - bby* Daniel c; Structures 5th= edition Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang if (detA == 0) cout << "The equation has no solution" << endl; else { double x = (e * d - b * f) / detA; double y = (a * f - e * c) / detA; cout << "x is " << x << " and y is " << y << endl; } return 0; } Chapter 3: Programming Project 3: (345021-00017) (Find the number of days in a month) Write a program that prompts the user to enter the month and year and displays the number of days in the month.

For example, if the user entered month 2 and year 2012, the program should display February 2012 has 29 days If the user entered month 3 and year 2015, the program should display March 2015 has 31 days For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; int main() { cout << "Enter a month in the year (e.g., 1 for Jan): "; int month; cin >> month; cout << "Enter a year: "; int year; cin >> year; int numberOfDaysInMonth = 0; switch (month) { case 1: cout << "January " << year; numberOfDaysInMonth = 31; break; case 2: cout << "February " << year; if (year % 400 == 0 || (year % 4 == 0 && year % 100 != 0)) numberOfDaysInMonth = 29; else numberOfDaysInMonth = 28; Solutions break; Manual for Introduction to C++ Programming and Data case 3: 5th edition by Daniel Liang Structures

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "March " << year; numberOfDaysInMonth = 31; break; case 4: cout << "April " << year; numberOfDaysInMonth = 30; break; case 5: cout << "May " << year; numberOfDaysInMonth = 31; break; case 6: cout << "June " << year; numberOfDaysInMonth = 30; break; case 7: cout << "July " << year; numberOfDaysInMonth = 31; break; case 8: cout << "August " << year; numberOfDaysInMonth = 31; break; case 9: cout << "September " << year; numberOfDaysInMonth = 30; break; case 10: cout << "October " << year; numberOfDaysInMonth = 31; break; case 11: cout << "November " << year; numberOfDaysInMonth = 30; break; case 12: cout << "December " << year; numberOfDaysInMonth = 31; break; } cout << " has " << numberOfDaysInMonth << " days"; return 0; } Chapter 3: Programming Project 4: (345021-00018) (Science: day of the week) Zeller’s congruence is an algorithm developed by Christian Zeller to calculate the day of the week.

The formula is: h = (q + 26(m + 1)/10 + k + k/4 + j/4 + 5j) % 7

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang where - h is the day of the week (0: Saturday, 1: Sunday, 2: Monday, 3: Tuesday, 4: Wednesday, 5: Thursday, 6: Friday).

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang - q is the day of the month.

- m is the month (3: March, 4: April, . . . , 12: December). January and February are counted as months 13 and 14 of the previous year.

- j is year/100 - k is the year of the century (i.e., year % 100)

Note that all divisions in this exercise perform an integer division.

Write a program that prompts the user to enter a year, month, and day of the month, and displays the name of the day of the week.

Sample Run 1 Enter year: (e.g., 2012): 2015 Enter month: 1-12: 1 Enter the day of the month: 1-31: 25 Day of the week is Sunday Sample Run 2 Enter year: (e.g., 2012): 2012 Enter month: 1-12: 5 Enter the day of the month: 1-31: 12 Day of the week is Saturday (Hint: January and February are counted as 13 and 14 in the formula, so you need to convert the user input 1 to 13 and 2 to 14 for the month and change the year to the previous year.)

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; int main() { cout << "Enter year: (e.g., 2008): "; int year; cin >> year; cout << "Enter month: 1-12: "; int month; cin >> month; if (month == 1) { month = 13; year--; } else if (month == 2) { month =Manual 14; Solutions for Introduction to C++ Programming and Data year--; Structures 5th edition by Daniel Liang }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "Enter the day of the month: 1-31: "; int dayOfMonth; cin >> dayOfMonth; int k = year % 100; int j = (int)(year / 100); int dayOfWeek = (int)(dayOfMonth + (int)((month + 1) * 26.0 / 10) + k + (int)(k / 4.0) + (int)(j / 4.0) + 5 * j) % 7; if (dayOfWeek == 0) cout << "Day of the week is Saturday" << endl; else if (dayOfWeek == 1) cout << "Day of the week is Sunday" << endl; else if (dayOfWeek == 2) cout << "Day of the week is Monday" << endl; else if (dayOfWeek == 3) cout << "Day of the week is Tuesday" << endl; else if (dayOfWeek == 4) cout << "Day of the week is Wednesday" << endl; else if (dayOfWeek == 5) cout << "Day of the week is Thursday" << endl; else if (dayOfWeek == 6) cout << "Day of the week is Friday" << endl; return 0; }

Chapter 3: Programming Project 5: (345021-00019) The description of this project is given in Programming Exercise 3.19 in the Chapter 3 Programming Exercise from the Book section.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <cmath> using namespace std; int main() { // Enter a point with two double values cout << "Enter a point with two coordinates: "; double x, y; cin >> x >> y; // Compute the distance double distance = pow(x * x +

y * y, 0.5);

if (distance <= 10) cout << "Point (" << x << ", " << y << ") is in the circle" << endl; else cout << "Point (" << x << ", " << y << ") is not in the circle" << endl; return 0;Manual for Introduction to C++ Programming and Data Solutions } Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Chapter 4 Programming Quiz 4.2: Question 1: (00000-11034) The area of a square is stored in a double variable named area. Write an expression whose value is length of one side of the square. sqrt(area)

Programming Quiz 4.2: Question 2: (00000-11035) Write the necessary preprocessor directive to enable the use of functions like sqrt and sin. #include <cmath>

Programming Quiz 4.2: Question 3: (00000-10804) If a right triangle has sides of length a, b and c and if c is the largest, then it is called the "hypotenuse" and its length is the square root of the sum of the squares of the lengths of the shorter sides (a and b).

Assume that variables a and b have been declared as double and that a and b contain the lengths of the shorter sides of a right triangle. Write an expression for the length of the hypotenuse. sqrt(a * a + b * b)

Programming Quiz 4.3: Question 1: (00000-11101) Assume that x is a char variable that has been declared and already given a value. Write an expression whose value is true if and only if x is an uppercase letter.

'A' <= x && x <= 'Z'

Programming Quiz 4.3: Question 2: (00000-11102) Assume that x is a char variable that has been declared and already given a value. Write an expression whose value is true if and only if x is a lowercase letter. 'a' <= x && x <= 'z'

Programming Quiz 4.3: Question 3: (00000-11103) Assume that x is a char variable that has been declared and already given a value. Write an expression whose value is true if and only if x is a decimal digit (0-9).

'0' <= x && x <= '9' Programming Quiz 4.3: Question 4: (00000-11105) Assume that x is a char variable that has been declared and already given a value. Write an expression whose value is true if and only if x is a letter.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang 'A' <= x && x <= 'Z' || 'a' <= x && x <= 'z'

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 4.3: Question 5: (00000-11106) Assume that x is a char variable that has been declared and already given a value. Write an expression whose value is true if and only if x is alphanumeric, that is either a letter or a decimal digit.

'A' <= x && x <= 'Z' || 'a' <= x && x <= 'z' || '0' <= x && x <= '9'

Programming Quiz 4.4: Question 1: (00000-11038) Write the necessary preprocessor directive to enable the use of the rand function. #include <cstdlib> Programming Quiz 4.8: Question 1: (00000-10845) Given a string variable address, write a string expression consisting of the string "http://" concatenated with the variable's string value. So, if the value of the variable were "www.turingscraft.com", the value of the expression would be "http://www.turingscraft.com". "http://" + address Programming Quiz 4.8: Question 2: (00000-11069) Clunker Motors Inc. is recalling all vehicles in its Extravagant line from model years 1999-2002.

Given an int variable modelYear and a string modelName, write a statement that prints the message "RECALL" to standard output if the values of modelYear and modelName match the recall details. if (modelName == "Extravagant" && 1999 <= modelYear && modelYear <= 2002) cout << "RECALL" << endl; Programming Quiz 4.8: Question 3: (00000-11071) Clunker Motors Inc. is recalling all vehicles in its Extravagant line from model years 1999-2002 as well all vehicles in its Guzzler line from model years 2004-2007.

Given an int variable modelYear and a string modelName, write a statement that prints the message "RECALL" to standard output if the values of modelYear and modelName match the recall details. if (modelName == "Extravagant" && 1999 <= modelYear && modelYear <= 2002 || modelName == "Guzzler" && 2004 <= modelYear && modelYear <= 2007) cout << "RECALL" << endl; Programming Quiz 4.8: Question 4: (00000-11098) Assume that message is a string variable.

Write a statement to display its value on standard output. cout << message;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 4.8: Question 5: (00000-11008)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Declare k, d, and s so that they can store an integer, a real number, and a small word (without spaces).

Use these variables to first read in an integer, a real number, and a small word and print them out in reverse order (i.e., the word, the real, and then the integer) all on the same line, separated by exactly one space from each other.

Then, on a second line, print them out in the original order (the integer, the real, and the word), separated again by exactly one space from each other. string s; int k; double d; cin >> k >> d >> s; cout << s << " " << d << " " << k << "\n"; cout << k << " " << d << " " << s << "\n"; Programming Quiz 4.9: Question 1: (00000-11039) Write the necessary preprocessor directive to enable the use of the time function. #include <ctime> Programming Quiz 4.10: Question 1: (00000-11024) Write the necessary preprocessor directive to enable the use of the stream manipulators like setw and setprecision. #include <iomanip> Programming Quiz 4.10: Question 2: (00000-11025) Assume that m is an int variable that has been given a value. Write a statement that prints it out in a print field of 10 positions. cout << setw(10) << m << endl; Programming Quiz 4.10: Question 3: (00000-11026) Assume that x is a double variable that has been given a value. Write a statement that prints it out with exactly three digits to the right of the decimal point no matter what how big or miniscule its value is. cout << setprecision(3) << fixed << x; Programming Quiz 4.10: Question 4: (00000-11027) Assume that x is a double variable that has been initialized. Write a statement that prints it out, guaranteed to have a decimal point, but without forcing scientific (also known as exponential or enotation). cout << showpoint << x; Programming Quiz 4.10: Question 5: (00000-11307)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Given three variables, k, m, n, of type int that have already been declared and initialized, write some code that prints each of them in a 9 position field on the same line.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang For example, if their values were 27, 987654321, -4321, the output would be: |xxxxxxx27987654321xxxx-4321 NOTE: The vertical bar, |, on the left above represents the left edge of the print area; it is not to be printed out. Also, we show x in the output above to represent spaces-- your output should not actually have xs! cout << setw(9) << k << setw(9) << m << setw(9) << n << endl; Programming Quiz 4.11: Question 1: (00000-11045) Given an ifstream object named input1, associate it with a file named winterdata.txt by opening the file for reading. input1.open("winterdata.txt");

Programming Quiz 4.11: Question 2: (00000-11048) Given an ofstream object named output, associate it with a file named yearsummary.txt by opening the file for writing. output.open("yearsummary.txt");

Programming Quiz 4.11: Question 3: (00000-11049) Given that corpdata is an ifstream object that has been used for reading data and that there is no more data to be read, write the necessary code to complete your use of this object. corpdata.close();

Programming Quiz 4.11: Question 4: (00000-11050) Given the availability of an ofstream object named output, write the code to output the string "3.14159" into a file called pi.

(Do not define a main function.) output.open("pi"); output << "3.14159"; output.close();

Programming Quiz 4.11: Question 5: (00000-11051) Given the availability of an ifstream object named input, write the code to read an integer into a variable datum that has already been declared from a file called rawdata.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Assume that reading that one integer is the only operation you will carry out with this file.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang (Note: write just the statements, do not define a main function.) input.open("rawdata"); input >> datum; input.close();

Programming Quiz 4.11: Question 6: (00000-11052) Given the availability of an ifstream object named indata and an ofstream object named outdata, write the code to read an integer from a file called currentsales and write twice its value into a file called projectedsales.

Assume that this is the extent of the input and output that this program will do. int sales; indata.open("currentsales"); outdata.open("projectedsales"); indata >> sales; outdata << 2 * sales; indata.close(); outdata.close();

Chapter 4: Programming Project 1: (345021-00020) (Geometry: area of a regular polygon) A regular polygon is an n-sided polygon in which all sides are of the same length and all angles have the same degree (i.e., the polygon is both equilateral and equiangular).

The formula for computing the area of a regular polygon is: area = n * s^2 / (4 * tan (pi/n)) s is the length of a side. Write a program that prompts the user to enter the number of sides and their length of a regular polygon and displays its area.

Sample Run Enter the number of sides: 5 Enter the length of a side: 6.5 The area of the polygon is 72.69 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <cmath> using namespace std; double area(int n, double side); int main() { // Enter the number of sides cout << "Enter of sides:to "; C++ Programming and Data Solutions Manual the fornumber Introduction int numberOfSides; Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cin >> numberOfSides; cout << "Enter the length of a side: "; double side; cin >> side; double area = numberOfSides * side * side / tan(3.14159 / numberOfSides) / 4; cout << "The area of the polygon is " << area << endl; return 0; } Chapter 4: Programming Project 2: (345021-00021) (Convert an uppercase letter to lowercase) Write a program that prompts the user to enter an uppercase letter and converts it to a lowercase letter.

For an incorrect input, display invalid input.

Sample Run 1 Enter an uppercase letter: T The lowercase letter is t Sample Run 2 Enter an uppercase letter: 5 5 is an invalid input For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise04_11 using namespace std; int main() { cout << "Enter an uppercase letter: "; char uppercase; cin >> uppercase; if ('A' <= uppercase && uppercase <= 'Z') { int offset = 'a' - 'A'; char lowercase = uppercase + offset; cout << "The lowercase letter is " << lowercase << endl; } else { cout << uppercase << " is an invalid input" << endl; } return 0; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Chapter 4: Programming Project 3: (345021-00022)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang (Hex to binary) Write a program that prompts the user to enter a hex digit and displays its corresponding binary number in four digits.

Sample Run 1 Enter a hex digit: B The binary value is 1011 Sample Run 2 Enter a hex digit: 7 The binary value is 0111 Sample Run 3 Enter a hex digit: G G is an invalid input For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <string> #include <cctype> using namespace std; int main() { cout << "Enter a hex character: "; char hex; cin >> hex; string result = ""; switch (toupper(hex)) { case '0': result = "0000"; break; case '1': result = "0001"; break; case '2': result = "0010"; break; case '3': result = "0011"; break; case '4': result = "0100"; break; case '5': result = "0101"; break; case '6': result = "0110"; break; case '7': result = "0111"; break; case '8': result = "1000"; break; case '9': result = "1001"; break; case 'A': result = "1010"; break; case 'B': result = "1011"; break; case 'C': result = "1100"; break; case 'D': result = "1101"; break; case 'E': result = "1110"; break; case 'F': result = "1111"; break; default: cout Manual << hex for << " Introduction is an invalid to input" endl; Solutions C++ << Programming and Data return 0; Structures 5th edition by Daniel Liang }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "The binary value is " << result << endl; return 0; } Chapter 4: Programming Project 4: (345021-00023) (Order three cities) Write a program that prompts the user to enter three cities and displays them in ascending order. City names may contain spaces.

Sample Run Enter the first city: Chicago Enter the second city: New York Enter the third city: Atlanta The three cities in alphabetical order are Atlanta Chicago New York For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <string> using namespace std; int main() { cout << "Enter the first city: "; string s1, s2, s3; getline(cin, s1); cout << "Enter the second city: "; getline(cin, s2); cout << "Enter the third city: "; getline(cin, s3); if (s1 > s2) { // Swap s1 with s2 string temp = s1; s1 = s2; s2 = temp; } if (s2 > s3) { // Swap s2 with s3 string temp = s2; s2 = s3; s3 = temp; }

Solutions Manual for Introduction to C++ Programming and Data if (s1 > s2) Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { // Swap s1 with s2 string temp = s1; s1 = s2; s2 = temp; } cout << "The three cities in alphabetical order are "; cout << s1 << " " << s2 << " " << s3 << endl; return 0; } Chapter 4: Programming Project 5: (345021-00024) (Check SSN) Write a program that prompts the user to enter a Social Security number in the format ddd-dd-dddd, where d is a digit. Your program should check whether the input is valid.

Sample Run 1 Enter a SSN: 232-23-5435 232-23-5435 is a valid social security number Sample Run 2 Enter a SSN: 23-23-5435 23-23-5435 is an invalid social security number For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <string> #include <cctype> using namespace std; int main() { cout << "Enter a string for SSN: "; string ssn; cin >> ssn; bool isValid = ssn.length() == 11 && isdigit(ssn[0]) && isdigit(ssn[1]) && isdigit(ssn[2]) && ssn[3] == '-' && isdigit(ssn[4]) && isdigit(ssn[5]) && ssn[6] == '-' && isdigit(ssn[7]) && isdigit(ssn[8]) && isdigit(ssn[9]) && isdigit(ssn[10]); if (isValid) cout << ssn << " is a valid social security number" << endl; else Solutions to C++ Programming Data cout <<Manual ssn <<for " isIntroduction an invalid social security number"and << endl;

Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return 0; }

Chapter 5 Programming Quiz 5.2: Question 1: (00000-10576) Given an int variable k that has already been declared, use a while loop to print a single line consisting of 97 asterisks.

Use no variables other than k. Don't forget to initialize k with an appropriate value. k = 0; while (k < 97) { cout << '*'; k++; }

Programming Quiz 5.2: Question 2: (00000-11127) Given an int variable n that has already been declared, write some code that repeatedly reads a value into n until at last a number between 1 and 10 (inclusive) has been entered. Don't forget to initialize n with an appropriate value. n = -1; while (n < 1 || n > 10) { cin >> n; }

Programming Quiz 5.2: Question 3: (00000-11131) Write a loop that reads positive integers from console input, printing out those values that are greater than 100, and that terminates when it reads an integer that is not positive. The printed values should be separated by single blank spaces. Declare any variables that are needed. int x; cin >> x; while (x > 0) { if (x > 100) cout << x << " "; cin >> x; } Programming Quiz 5.3: Question 1: (00000-11132) Write a loop that reads positive integers from console input, printing out those values that are even, separating them with spaces, and that terminates when it reads an integer that is not positive.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Declare any variables that are needed. int x;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cin >> x; while (x > 0) { if (x % 2 == 0) cout << x << " "; cin >> x; }

Programming Quiz 5.3: Question 2: (00000-11133) Write a loop that reads positive integers from console input and that terminates when it reads an integer that is not positive. After the loop terminates, it prints out the sum of all the even integers read.

Declare any variables that are needed. int x; int sum = 0; cin >> x; while (x > 0) { if (x % 2 == 0) sum += x; cin >> x; } cout << sum; Programming Quiz 5.4: Question 1: (00000-10579) Given an int variable n that has been initialized to a positive value and, in addition, int variables k and total that have already been declared, use a while loop to compute the sum of the cubes of the first n counting numbers, and store this value in total.

Thus if n equals 4, your code should put 1*1*1 + 2*2*2 + 3*3*3 + 4*4*4 into total.

Use no variables other than n, k, and total.

Do not modify n. Don't forget to initialize k and total with appropriate values. total = 0; k = 1; while (k <= n) { total += k * k * k; k++; }

Programming Quiz 5.4: Question 2: (00000-10578) Given int variables k and total that have already been declared, use a while loop to compute the sum of the squares of the first 50 counting numbers, and store this value in total.

Thus your code should put 1*1 + 2*2 + 3*3 +... + 49*49 + 50*50 into total.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Use no variables other than k and total. Don't forget to initialize k and total with appropriate values.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang total = 0; k = 1; while (k <= 50) { total += k * k; k++; } Programming Quiz 5.4: Question 3: (00000-10589) Given that two int variables, total and amount, have been declared, write a loop that reads integers into amount and adds all the non-negative values into total. The loop terminates when a value less than 0 is read into amount. Don't forget to initialize total to 0. total = 0; cin >> amount; while (amount >= 0) { total += amount; cin >> amount; } Programming Quiz 5.5: Question 1: (00000-11128) Given a char variable c that has already been declared, write some code that repeatedly reads a value from console input into c until at last a 'Y' or 'y' or 'N' or 'n' has been entered. cin >> c; while (c != 'y' && c != 'Y' && c != 'N' && c != 'n') cin >> c; Programming Quiz 5.5: Question 2: (00000-11129) Given a string variable s that has already been declared, write some code that repeatedly reads a value from console input into s until at last a "Y" or "y" or "N" or "n" has been entered. s = ""; while (s != "Y" && s != "y" && s != "N" && s != "n") cin >> s; Programming Quiz 5.5: Question 3: (00000-11136) Write a loop that reads strings from console input where the string is either "duck" or "goose". The loop terminates when "goose" is read in. After the loop, your code should print out the number of "duck" strings that were read. string word; int n = 0; cin >> word; while (word == "duck") { n++; cin >> word; } cout << n << endl;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 5.7: Question 1: (00000-10580) Given an int variable k that has already been declared, use a do-while loop to print a single line consisting of 97 asterisks.

Use no variables other than k. Don’t forget to initialize k with an appropriate value. k = 97; do { cout << '*'; k--; } while (k > 0); Programming Quiz 5.7: Question 2: (00000-10582) Given int variables k and total that have already been declared, use a do-while loop to compute the sum of the squares of the first 50 counting numbers, and store this value in total.

Thus your code should put 1*1 + 2*2 + 3*3 +... + 49*49 + 50*50 into total.

Use no variables other than k and total. Don’t forget to initialize k and total with appropriate values. total = 0; k = 1; do { total += k * k; k++; } while (k <= 50); Programming Quiz 5.7: Question 3: (00000-10583) Given an int variable n that has been initialized to a positive value and, in addition, int variables k and total that have already been declared, use a do-while loop to compute the sum of the cubes of the first n whole numbers, and store this value in total.

Thus if n equals 4, your code should put 1*1*1 + 2*2*2 + 3*3*3 + 4*4*4 into total.

Use no variables other than n, k, and total. Don’t forget to initialize k and total with appropriate values. total = 0; k = 1; do { total += k * k * k; k++; } while (k <= n); Programming Quiz 5.8: Question 1: (00000-10572) Given an int variable k that has already been declared, use a for loop to print a single line consisting of 97 asterisks.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Use no variables other than k.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang for (k = 0; k < 97; k++) cout << '*'; Programming Quiz 5.8: Question 2: (00000-10574) Given int variables k and total that have already been declared, use a for loop to compute the sum of the squares of the first 50 counting numbers, and store this value in total. Thus your code should put 1*1 + 2*2 + 3*3 +... + 49*49 + 50*50 into total. Use no variables other than k and total. total = 0; for (k = 1; k <= 50; k++) total += k * k; Programming Quiz 5.8: Question 3: (00000-10575) Given an int variable n that has been initialized to a positive value and, in addition, int variables k and total that have already been declared, use a for loop to compute the sum of the cubes of the first n whole numbers, and store this value in total.

Thus if n equals 4, your code should put 1*1*1 + 2*2*2 + 3*3*3 + 4*4*4 into total.

Use no variables other than n, k, and total. total = 0; for (k = 1; k <= n; k++) total += k * k * k;

Programming Quiz 5.9: Question 1: (00000-10971) Write a for loop that prints the integers 1 through 40, separated by spaces or new lines. You may use only one variable, count, which has already been declared as an int. for (count = 1; count <= 40; count++) cout << count << endl; Programming Quiz 5.9: Question 2: (00000-10808) Given an int variable i that has already been declared, write a for loop that prints the integers 0 through 39, separated by spaces.

Use no variables other than i. for (i = 0; i < 40; i++) cout << i << " "; Programming Quiz 5.9: Question 3: (00000-10811) Write a for loop that prints the odd integers 11 through 121 inclusive, separated by spaces. for (int k = 11; k <= 121; k += 2) cout << k << " ";

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 5.10: Question 1: (00000-10586)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Assume that the int variables i, j and n have been declared, and n has been initialized. Write code that causes a "triangle" of asterisks to be output to the screen, i.e., n lines should be printed out, the first consisting of a single asterisk, the second consisting of two asterisks, the third consisting of three, etc.

The last line should consist of n asterisks.

Thus, for example, if n has value 3, the output of your code should be: * ** *** for (i = 1; i <= n; i++) { for (j = 1; j <= i; j++) cout << "*"; cout << "\n"; } Programming Quiz 5.10: Question 2: (00000-11142) Write a loop that displays all possible combinations of two letters where the letters are 'a', or 'b', or 'c', or 'd', or 'e'.

The combinations should be displayed in ascending alphabetical order: aa ab ac ad ae ba bb ... ee for (char i = 'a'; i <= 'e'; i++) for (char j = 'a'; j <= 'e'; j++) cout << i << j << endl;

Programming Quiz 5.10: Question 3: (00000-11143) Given a char variable last that has been initialized to a lowercase letter, write a loop that displays all possible combinations of two letters in the range 'a' through last.

The combinations should be displayed in ascending alphabetical order: For example, if last was initialized to 'c', the output should be: aa ab ac ba bb bc

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang ca cb cc

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang for (char c1 = 'a'; c1 <= last; c1++) for (char c2 = 'a'; c2 <= last; c2++) cout << c1 << c2 << endl; Chapter 5: Programming Project 1: (345021-00001) (Count positive and negative numbers and compute the average of numbers) Write a program that reads an unspecified number of integers, determines how many positive and negative values have been read, and computes the total and average of the input values (not counting zeros). Your program ends with the input 0. Display the average as a floating-point number.

If you entire input is 0, display 'No numbers are entered except 0.'

Sample Run 1 Enter an integer, the input ends if it is 0: 1 2 -1 3 0 The number of positives is 3 The number of negatives is 1 The total is 5 The average is 1.25 Sample Run 2 Enter an integer, the input ends if it is 0: 0 No numbers are entered except 0 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; int main() { int countPositive = 0, countNegative = 0; int count = 0, total = 0, num; cout << "Enter an int value, the program exits if the input is 0: "; cin >> num; while (num != 0) { if (num > 0) countPositive++; else if (num < 0) countNegative++; total += num; count++; // Read the next data cin >> num; } if (count == 0) cout << "No numbers are entered except 0"; else { Solutions Manual for Introduction to "C++ Programming and Data cout << "The number of postives is << countPositive << endl; cout << "The number of negatives is " << countNegative << endl; Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "The total is " << total << endl; cout << "The average is " << (total * 1.0 / count) << endl; } return 0; }

Chapter 5: Programming Project 2: (345021-00151) (Occurrence of max numbers) Write a program that reads integers, finds the largest of them, and counts its occurrences. Assume that the input ends with number 0. Suppose that you entered 3 5 2 5 5 5 0; the program finds that the largest is 5 and the occurrence count for 5 is 4.

Hint: Maintain two variables, max and count. max stores the current max number and count stores its occurrences. Initially, assign the first number to max and 1 to count. Compare each subsequent number with max. If the number is greater than max, assign it to max and reset count to 1. If the number is equal to max, increment count by 1.

Sample Run 1 Enter an integer (0: for end of input): 3 Enter an integer (0: for end of input): 5 Enter an integer (0: for end of input): 2 Enter an integer (0: for end of input): 5 Enter an integer (0: for end of input): 5 Enter an integer (0: for end of input): 5 Enter an integer (0: for end of input): 0 The largest number is 5 The occurrence count of the largest number is 4

Sample Run 2 Enter an integer (0: for end of input): 0 No numbers are entered except 0 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise05_41 using namespace std; int main() { // Prompt the user to enter the first number int number; cout << "Enter an integer (0: for end of input): "; cin >> number; if (number == 0) { cout << "No numbers are entered except 0" << endl; return 0; } int max = number; int count = 1; // Prompt the user to enter the remaining numbers

Solutions Manual!=for while (number 0) Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "Enter an integer (0: for end of input): "; cin >> number; // Read the next number if (number > max) { max = number; count = 1; } else if (number == max) count++; } cout << "\n" << "The largest number is " << max << "\n" << "The occurrence count of the largest number is " << count << endl; return 0; } Chapter 5: Programming Project 3: (345021-00003) (Find the two highest scores) Write a program that prompts the user to enter the number of students and each student’s name and score, and displays the name and score of the student with the highest score and the student with the second-highest score.

Sample Run Enter the number of students: 5 Enter a student name: Smith Enter a student score: 60 Enter a student name: Jones Enter a student score: 96 Enter a student name: Peterson Enter a student score: 85 Enter a student name: Greenlaw Enter a student score: 98 Enter a student name: Zhang Enter a student score: 95 Top two students: Greenlaw's score is 98.0 Jones's score is 96.0

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <string> using namespace std; int main() { // Prompt the user to enter the number of students cout << "Enter the number of students: "; int numberOfStudents; cin >> numberOfStudents;

Solutions Manual for Introduction to C++ Programming and Data double score1 = -1; Structures 5th edition

by Daniel Liang double score2 = -2;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang string name1; string name2; for (int i = 0; i < numberOfStudents; i++) { cout << "Enter a student name: "; string name; cin >> name; cout << "Enter a student score: "; double score; cin >> score; if (score > score1) { score2 = score1; score1 = score; name2 = name1; name1 = name; } else if (score > score2) { score2 = score; name2 = name; } } cout << "Student " << name1 << " has the highest score " << score1 << endl; cout << "Student " << name2 << " has the second highest score " << score2 << endl; return 0; } Chapter 5: Programming Project 4: (345021-00004) (Financial application: compare loans with various interest rates) Write a program that lets the user enter the loan amount and loan period in number of years and displays the monthly and total payments for each interest rate starting from 5% to 8%, with an increment of 1/8.

Sample Run Loan Amount: 10000 Number of Years: 5 Interest Rate Monthly Payment Total Payment 5.000% 188.71 11322.74 5.125% 189.28 11357.13 5.250% 189.85 11391.59 ...

7.875% 202.17 12129.97 8.000% 202.76 12165.83

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang For the formula to compute monthly payment, see LiveExample 2.11, ComputeLoan.cpp. For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang #include <iostream> #include <iomanip> #include <cmath> using namespace std; int main() { int numberOfYears; double loanAmount; // Enter loan amount cout << "Enter loan amount, for example 120000.95: "; cin >> loanAmount; // Enter number of years cout << "Enter number of years as an integer, \nfor example 5: "; cin >> numberOfYears; // Display the header cout << setw(13) << "Interest Rate" << setw(20) << "Monthly Payment" << setw(20) << "Total Payment" << endl; double monthlyInterestRate; double monthlyPayment; double totalPayment; for (double annualInterestRate = 5.0; annualInterestRate <= 8.0; annualInterestRate += 1.0 / 8) { // Obtain monthly interest rate monthlyInterestRate = annualInterestRate / 1200; // Compute mortgage monthlyPayment = loanAmount * monthlyInterestRate / (1 - (pow(1 / (1 + monthlyInterestRate), numberOfYears * 12))); totalPayment = monthlyPayment * numberOfYears * 12; // Display results cout << setw(12) << fixed << showpoint << setprecision(3) << annualInterestRate << "%"; cout << setw(20) << fixed << showpoint << setprecision(2) << monthlyPayment; cout << setw(20) << showpoint << setprecision(2) << totalPayment << endl; } return 0; }

Chapter 5: Programming Project 5: (345021-00005) (Count vowels and consonants) Assume letters A/a, E/e, I/i, O/o, and U/u as the vowels.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write a program that prompts the user to enter a string and displays the number of vowels and consonants in the string.

Sample Run Enter a string: Programming is fun The number of vowels is 5 The number of consonants is 11 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html.

#include <iostream> #include <string> #include <cctype> using namespace std; int main() { cout << "Enter a string: "; string s; getline(cin, s); int countVowels = 0; int countConsonants = 0; for (int i = 0; i < s.size(); i++) { char temp = toupper(s[i]); if (temp == 'A' || temp == 'E' || temp == 'I' || temp == 'O' || temp == 'U') countVowels++; else if (isalpha(temp)) countConsonants++; } cout << "The number of vowels is " << countVowels << endl; cout << "The number of consonants is " << countConsonants << endl; return 0; } Chapter 5: Programming Project 6: (345021-00006) (Check password) Some websites impose certain rules for passwords. Suppose the password rules are as follows: - A password must have at least eight characters.

- A password must consist of only letters and digits.

- A password must contain at least two digits.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write a program that prompts the user to enter a password and displays valid password if the rules are followed or invalid password otherwise.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <string> #include <cctype> using namespace std; int main() { cout << "Enter a string for password: "; string s; cin >> s; // or getline(cin, s); bool isValid = true; // Only letters and digits? for (unsigned i = 0; i < s.size(); i++) if (!isalnum(s[i])) isValid = false; // Check length if (s.size() < 8) isValid = false; // Count the number of digits int count = 0; for (unsigned i = 0; i < s.size(); i++) if (isdigit(s[i])) count += 1; if (count < 2) isValid = false; if (isValid) cout << "valid password" << endl; else cout << "invalid password" << endl; return 0; }

Chapter 6 Programming Quiz 6.3: Question 1: (00000-10658) Write the definition of a function twice, that receives an int parameter and returns an int that is twice the value of that parameter. int twice(int x) { return 2 * x; } Programming Quiz 6.3: Question 2: (00000-10909)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write the definition of a function oneLess, which receives an int parameter and returns an int that is one less than the value of the parameter. So if the parameter's value is 7, the function returns the value 6. If the parameter's value happens to be 44, the functions returns the value 43. int oneLess(int x) { return x - 1; } Programming Quiz 6.3: Question 3: (00000-10913) Write the definition of a function signOf that receives an int parameter and returns -1 if the parameter is negative, returns 0 if the parameter is 0 and returns 1 if the parameter is positive.

So, if the parameter's value is 7 or 803 or 141 the function returns 1.

But if the parameter's value is -22 or -57, the function returns -1.

And if the parameter's value is 0, the function returns 0. int signOf(int x) { return x < 0 ? -1 : (x > 0 ? 1 : 0); } Programming Quiz 6.4: Question 1: (00000-10640) printErrorDescription is a function that accepts one int parameter and returns no value.

Write a statement that calls the function printErrorDescription, passing it the value 14. printErrorDescription(14);

Programming Quiz 6.4: Question 2: (00000-10654) Write the definition of a function printDottedLine, which has no parameters and doesn't return anything.

The function prints a single line consisting of 5 periods (terminated by a new line character) . void printDottedLine() { cout << ". ....\n"; }

Programming Quiz 6.4: Question 3: (00000-10657) Write the definition of a function printLarger, which has two int parameters and returns nothing.

The function prints the larger value of the two parameters on a single line by itself. (For purposes of this exercise, the "larger" means "not the smaller".) void printLarger(int x, int y) { Solutions Manual to C++ Programming and Data cout << (x > y ?for x : Introduction y) << endl; } Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 6.5: Question 1: (00000-10918) Assume that printStars is a function that takes one argument and returns no value. It prints a line of n stars (followed by a newline character ) where n is the value of the argument received.

Write a statement that invokes printStars to make a line of 35 stars. printStars(35);

Programming Quiz 6.5: Question 2: (00000-10642) add is a function that accepts two int parameters and returns their sum.

Two int variables, euroSales and asiaSales, have already been declared and initialized.

Another int variable, eurasiaSales, has already been declared.

Write a statement that calls add to compute the sum of euroSales and asiaSales and stores this value in eurasiaSales. eurasiaSales = add(euroSales, asiaSales); Programming Quiz 6.5: Question 3: (00000-10643) toThePowerOf is a function that accepts two int parameters and returns the value of the first parameter raised to the power of the second.

An int variable cubeSide has already been declared and initialized.

Another int variable, cubeVolume, has already been declared.

Write a statement that calls toThePowerOf to compute the value of cubeSide raised to the power of 3 and stores this value in cubeVolume. cubeVolume = toThePowerOf(cubeSide, 3);

Programming Quiz 6.7: Question 1: (00000-11093) Write the definitions for three functions named max.

Each receives two parameters of the same type and returns the larger of the two values.

Define one of these functions to apply to type double, another to type int, and a third to type char. double max(double a, double b) { return a > b ? a : b; } int max(int a, int b) { return a > b ? a : b; } char max(char a, char b) { return a > b ? a : b; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 6.8: Question 1: (00000-10651)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write a statement that declares a prototype for a function powerTo, which has two parameters.

The first is a double and the second is an int. The function returns a double. double powerTo(double, int); Programming Quiz 6.8: Question 2: (00000-10653) Write a statement that declares a prototype for a function printTodaysDate, which has no parameters and doesn't return anything. void printTodaysDate(); Programming Quiz 6.8: Question 3: (00000-10647) Write a statement that declares a prototype for a function printErrorDescription, which has an int parameter and returns nothing. void printErrorDescription(int); Programming Quiz 6.9: Question 1: (00000-11083) Write the definition of a function named timeOnHighway that receives three parameters, all of type double: mileEndingPoint, mileStartingPoint, and speed.

The first parameter indicates the mile marker on an interstate at which a vehicle goes to.

The second parameter indicates the mile marker on an interstate at which a vehicle starts at.

The third parameter indicates the speed of the vehicle in miles per hour.

The function returns the number of hours it takes a vehicle to go from the starting mile marker to the ending one.

The function has a default value for the speed: 55 miles per hour, and a default value for mileStartingPoint: 0.0. double timeOnHighway(double mileEndingPoint, double mileStartingPoint = 0.0, double speed = 55.0) { return (mileEndingPoint - mileStartingPoint) / speed; }

Programming Quiz 6.11: Question 1: (00000-11084) Write the definition of a function named newbie that receives no parameters and returns true the first time it is invoked (when it is a "newbie"), and that returns false every time that it is invoked after that. bool newbie() { static bool everCalled = false; bool firstTime = !everCalled; everCalled = true; return firstTime; Solutions Manual for Introduction to C++ Programming and Data }

Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 6.11: Question 2: (00000-11085) Write the definition of a function named max that receives an int parameter and returns the largest value that it has been called with so far.

So, if you make these calls to max, max(5), max(3), max(12), max(4), the values returned will be (respectively): 5, 5, 12, and 12.

Hint: use a static local variable to track the largest number that has ever been called. int max(int v) { static bool everCalled = false; static int vmax; if (!everCalled) { everCalled = true; vmax = v; } else if (v > vmax) vmax = v; return vmax; }

Programming Quiz 6.11: Question 3: (00000-11087) Write the definition of a function named averager that receives a double parameter and return-- as a double -- the average value that it has been passed so far.

So, if you make these calls to average using averager(5.0), averager(15.0), averager(4.3), the values returned will be (respectively): 5.0, 10.0, 8.1.

Hint: use a static local variable to track the average of all the numbers been called. double averager(double v) { static double total = 0.0; static int n = 0; n++; total += v; return total / n; }

Programming Quiz 6.12: Question 1: (00000-10663) Given an int variable i, declare a reference variable r that is a reference to i. int& r = i; Programming Quiz 6.12: Question 2: (00000-10665) Write the header for a function addOne that accepts a single int reference parameter and returns nothing.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Name the parameter x.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang void addOne(int& x) Programming Quiz 6.12: Question 3: (00000-10666) Write a function addOne that adds 1 to its int reference parameter. The function returns nothing. void addOne(int& x) { x++; } Programming Quiz 6.12: Question 4: (00000-11091) Write the definition of a function named swapints that is passed two int variables. The function returns nothing but exchanges the values of the two variables. So, if j and k have (respectively) the values 15 and 23, and the invocation swapints(j, k) is made, then upon return, the values of j and k will be 23 and 15 respectively. void swapints(int& x, int& y) { int t = x; x = y; y = t; } Chapter 6: Programming Project 1: (345021-00025) (Financial application: compute the future investment value) Write a function that computes future investment value at a given interest rate for a specified number of years. The future investment is determined using the formula in Programming Exercise 2.23.

Use the following function header: double futureInvestmentValue( double investmentAmount, double monthlyInterestRate, int years) For example, futureInvestmentValue(10000, 0.05/12, 5) returns 12833.59.

Write a test program that prompts the user to enter the investment amount (e.g., 1000) and the interest rate (e.g., 9%) and prints a table that displays future value for the years from 1 to 30, as shown below: The amount invested: 1000 Annual interest rate: 9 Years Future Value 1 1093.81 2 1196.41 ...

29 13467.25 30 14730.58 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <iomanip> Solutions Manual for Introduction to C++ Programming and Data #include <cmath> using namespace std; Structures 5th edition

by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang double futureInvestmentValue(double investmentAmount, double monthlyInterestRate, int numOfYears) { return investmentAmount * pow(1 + monthlyInterestRate, numOfYears * 12); } int main() { cout << "The amount invested: "; double loanAmount; cin >> loanAmount; cout << "Annual interest rate: "; double annualInterestRate; cin >> annualInterestRate; cout << setw(10) << left << "Years" << setw(10) << "Future Value" << endl; for (int i = 1; i <= 30; i++) { double futureValue = futureInvestmentValue(loanAmount, annualInterestRate / 1200, i); cout << setw(10) << left << i << fixed << setprecision(2) << setw(10) << futureValue << endl; } return 0; } Chapter 6: Programming Project 2: (345021-00026) The description of this project is given in Programming Exercise 6.13 in the Chapter 6 Programming Exercise from the Book section.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <iomanip> #include <cmath> using namespace std; double m(int i) { double pi = 0 int sign = 1; for (int k = 1; k <= i; k++) { pi += sign / (2 * k - 1.0); sign = -1 * sign; } return 4 * pi; } int main() { cout << left << for setw(10) << "i" << to left << Programming setw(10) << "m(i)" << endl; Solutions Manual Introduction C++ and Data for (int i = 1; i <= 1000; i = i + 100) Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << left << setw(10) << i << fixed << setprecision(4) << m(i) << endl; return 0; } Chapter 6: Programming Project 3: (345021-00027) (Algebra: solve 2 × 2 linear equations) You can use Cramer’s rule to solve the following 2 × 2 system of linear equation: ax + by = e cx + dy = f x = (ed - bf) / (ad - bc) y = (af - ec) / (ad - bc)

Write a function with the following header: void solveEquation(double a, double b, double c, double d, double e, double f, double& x, double& y, bool& isSolvable) If ad - bc is 0, the equation has no solution and isSolvable should be false.

Write a program that prompts the user to enter a, b, c, d, e, and f and displays the result.

If ad - bc is 0, report that “The equation has no solution.”

See Programming Exercise 3.3 for sample runs.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; void solveEquation(double a, double b, double c, double d, double e, double f, double& x, double& y, bool& isSolvable) { double detA = a * d - b * c; if (detA == 0) isSolvable = false; else { x = (e * d - b * f) / detA; y = (a * f - e * c) / detA; isSolvable = true; } } int main()

Solutions Manual for Introduction to C++ Programming and Data { Structures edition by x, Daniel Liang double a, 5th b, c, d, e, f, y;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang bool isSolvable; cout << "Enter a, b, c, d, e, f: "; cin >> a >> b >> c >> d >> e >> f; solveEquation(a, b, c, d, e, f, x, y, isSolvable); if (isSolvable) cout << "x is " << x << " and y is " << y << endl; else cout << "The equation has no solution" << endl; return 0; } Chapter 6: Programming Project 4:

(345021-00028)

(Binary to decimal) Write a function that returns a decimal number from a binary string. The function header is as follows: int bin2Dec(const string& binaryString) For example, binaryString 10001 is 17 (1*2^4+0*2^3+0*2^2+0*2+1=17 ).

So, bin2Dec("10001") returns 17. Write a test program that prompts the user to enter a binary number as a string and displays its decimal equivalent value.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise06_39 #include <string> using namespace std; int bin2Dec(const string& binaryString) { int value = 0; for (int i = 0; i < binaryString.length(); i++) { value = value * 2 + binaryString[i] - '0'; } return value; } int main() { cout << "Enter a bianry number: "; string bianryString; cin >> bianryString; cout << bin2Dec(bianryString) << endl; return 0; } Chapter 6: Programming Project 5:

(345021-00029)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang (Decimal to binary)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write a function that parses a decimal number into a binary number as a string. The function header is as follows: string dec2Bin(int value) See Appendix D, “Number Systems,” for converting a decimal into a binary.

Write a test program that prompts the user to enter a decimal number and displays its equivalent binary value.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise06_41 #include <string> using namespace std; string dec2Bin(int value) { // Convert decimal to binary string binary = ""; while (value != 0) { int binaryValue = value % 2; // Convert a decimal value to a binary digit char binaryDigit = static_cast<char>(binaryValue + '0'); binary = binaryDigit + binary; value = value / 2; } return binary; } int main() { cout << "Enter a decimal number: "; int number; cin >> number; cout << dec2Bin(number) << endl; return 0; } Chapter 7 Programming Quiz 7.2: Question 1: (00000-10630) We informally define the term "corresponding element" as follows: The first element in an array and the last element of the array are corresponding elements.

Similarly, the second element and the second last element are corresponding elements.

The third element and the third last element are corresponding elements -- and so on.

Given an array a and a variable n that contains the number of elements in a, write an expression for the

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang corresponding element of a[i].

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang a[n-1-i]

Programming Quiz 7.2: Question 2:

(00000-10631)

Reversing the elements of an array involves swapping the corresponding elements of the array: the first with the last, the second with the next to the last, and so on, all the way to the middle of the array.

Given an array a, an int variable n containing the number of elements in a, and two other int variables, k and temp, write a loop that reverses the elements of the array.

Do not use any other variables besides a, n, k, and temp. for (k = 0; k < n / 2; k++) { temp = a[k]; a[k] = a[n-1-k]; a[n-1-k] = temp; } Programming Quiz 7.2: Question 3: (00000-10635) Given •

an int variable k,

•

an int array incompletes that has been declared and initialized,

•

an int variable nIncompletes that contains the number of elements in the array,

•

an int variable studentID that has been initialized, and

•

an int variable numberOfIncompletes,

write code that counts the number of times the value of studentID appears in incompletes and assigns this value to numberOfIncompletes.

You may use only k, incompletes, nIncompletes, studentID, and numberOfIncompletes. for (numberOfIncompletes = 0, k = 0; k < nIncompletes; k++) if (incompletes[k] == studentID) numberOfIncompletes++;

Programming Quiz 7.5: Question 1: (00000-10646) printArray is a function that has two parameters. The first parameter is an array of element type int and the second is an int, the number of elements in the array. The function prints the contents of the array parameter; it does not return a value. inventory is an array of ints that has been already declared and filled with values. n is an int variable that holds the number of elements in the array.

Write a statement that prints the contents of the array inventory by calling the function printArray. printArray(inventory, n);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 7.5: Question 2: (00000-10652)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write a statement that declares a prototype for a function printArray, which has two parameters.

The first parameter is an array of element type int and the second is an int, the number of elements in the array.

The function does not return a value. void printArray(int[], int); Programming Quiz 7.6: Question 1: (00000-10662) Write the definition of a function printArray, which has two parameters.

The first parameter is an array of ints and the second is an int, the number of elements in the array.

The function does not return a value.

The function prints out each element of the array, on a line by itself, in the order the elements appear in the array, and does not print anything else. void printArray(const int a[], int n) { for (int i = 0; i < n; i++) cout << a[i] << endl; }

Programming Quiz 7.6: Question 2: (00000-10669) Write the definition of a function named sumArray that receives two parameters: an array of element type int and an int that contains the number of elements of the array.

The function returns the sum of the elements of the array as an int. int sumArray(const int a[], int n) { int total = 0; while (n > 0) total += a[--n]; return total; } Programming Quiz 7.6: Question 3: (00000-10670) Write the definition of a function, isReverse, whose first two parameters are arrays of integers of equal size, and whose third parameter is an integer indicating the size of each array.

The function returns true if and only if one array is the reverse of the other. ("Reverse" here means same elements but in reverse order.) bool isReverse(const int a[], const int b[], int n) { int k; Manual for Introduction to C++ Programming and Data Solutions for (k = 0; < n && a[k] b[n-1-k]; Structures 5thk edition by == Daniel Liangk++); return k == n;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } Programming Quiz 7.11: Question 1: (00000-11122) Declare a char array named line of size 50, and write a statement that reads in the next line of console input into this array. (Assume the line may contain whitespace characters and the total number of the characters in the line is less than 50.) char line[50]; cin.getline(line, 50); Programming Quiz 7.11: Question 2: (00000-10708) Write an expression that evaluates to true if and only if the C-string s equals the C-string "end". strcmp(s, "end") == 0 Programming Quiz 7.11: Question 3: (00000-10709) Write an expression that evaluates to true if the value of the C-string s1 is greater than the value of Cstring s2. strcmp(s1, s2) > 0 Programming Quiz 7.11: Question 4: (00000-10710 10713) Write an expression that evaluates to true if the value of C-string lastName is greater than the Cstring "Dexter". strcmp(lastName, "Dexter") > 0 Chapter 7: Programming Project 1:

(345021-00030)

(Assign grades) Write a program that reads student scores, gets the best score, and then assigns grades based on the following scheme: Grade is A if score is >= best – 10; Grade is B if score is >= best – 20; Grade is C if score is >= best – 30; Grade is D if score is >= best – 40; Grade is F otherwise.

The program prompts the user to enter the total number of students, then prompts the user to enter all of the scores, and concludes by displaying the grades.

Sample Run Enter the number of students: 4

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Enter 4 scores: 40 55 70 58

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Student 0 score is 40 and grade is C Student 1 score is 55 and grade is B Student 2 score is 70 and grade is A Student 3 score is 58 and grade is B For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; int main() { // Maximum number of students const int MAX_NUMBER_OF_STUDENTS = 100; int scores[MAX_NUMBER_OF_STUDENTS]; // Array scores int best = 0; // The best score char grade; // The grade int numberOfScores; cout << "Enter the number of students: "; cin >> numberOfScores; cout << "Enter " << numberOfScores << " score: "; // Read scores and find the best score for (int i = 0; i < numberOfScores; i++) { cin >> scores[i]; if (scores[i] > best) best = scores[i]; } // Assign and display grades for (int i = 0; i < numberOfScores; i++) { if (scores[i] >= best - 10) grade = 'A'; else if (scores[i] >= best - 20) grade = 'B'; else if (scores[i] >= best - 30) grade = 'C'; else if (scores[i] >= best - 40) grade = 'D'; else grade = 'F'; cout << "Student " << (i + 1) << " score is " << scores[i] << " and grade is " << grade << "\n"; } return 0; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Chapter 7: Programming Project 2: (345021-00031)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang (Count occurrence of numbers) Write a program that reads at most 100 integers between 1 and 100 and counts the occurrence of each number. Assume the input ends with 0.

Sample Run Enter the integers between 1 and 100: 2 5 6 5 4 3 23 43 2 0 2 occurs 2 times 3 occurs 1 time 4 occurs 1 time 5 occurs 2 times 6 occurs 1 time 23 occurs 1 time 43 occurs 1 time Note that if a number occurs more than once, the plural word “times” is used in the output. Numbers are displayed in an increasing order.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; int main() { int counts[100] = {0}; // counts[0] is to count the number of occurrence of number 1 // counts[99] is to count the number of occurrence of number 100 cout << "Enter the numbers between 1 and 100 ending with 0: "; // Read all numbers int number; // number read from a file cin >> number; while (number != 0) { counts[number - 1]++; cin >> number; } // Display result for (int i = 0; i < 100; i++) { if (counts[i] > 0) cout << (i + 1) << " occurs " << counts[i] << (counts[i] == 1 ? " time" : " times") << endl; } return 0;Manual for Introduction to C++ Programming and Data Solutions } Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Chapter 7: Programming Project 3: (345021-00032) (Print distinct numbers) Write a program that reads in 10 numbers and displays distinct numbers (i.e., if a number appears multiple times, it is displayed only once). The numbers are displayed in the order of their input and separated by exactly one space.

(Hint: Read a number and store it to an array if it is new. If the number is already in the array, discard it.

After the input, the array contains the distinct numbers.)

Sample Run Enter ten numbers: 1 2 3 2 1 6 3 4 5 2 The distinct numbers are: 1 2 3 6 4 5 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html.

#include <iostream> using namespace std; int main() { int numbers[10]; // Store distinct values int size = 0; // Indicate how many distict values are in the array numbers cout << "Enter ten integers: "; for (int i = 0; i < 10; i++) { // Read and store numbers in an array if it is new int value; cin >> value; bool isInArray = false; for (int j = 0; j < size; j++) { if (numbers[j] == value) { isInArray = true; break; } } if (!isInArray) { numbers[size] = value; size++; } } cout << "The number of distinct numbers is " << size << endl; cout << "The distinct numbers are: "; for (int i = 0; i < size; i++) cout << numbers[i] << " "; return 0;Manual for Introduction to C++ Programming and Data Solutions } Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Chapter 7: Programming Project 4:

(345021-00033)

(Statistics: compute deviation) The description of this project is given in Programming Exercise 7.11 in the Chapter 7 Programming Exercise from the Book section. For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <cmath> using namespace std; /** Function for computing mean of an array of double values */ double mean(const double x[], int size) { double sum = 0; for (int i = 0; i < size; i++) sum += x[i]; return sum * 1.0 / size; } /** Function for computing mean of an array of int values */ double mean(const int x[], int size) { int sum = 0; for (int i = 0; i < size; i++) sum += x[i]; return sum / size; } /** Function for computing deviation of double values */ double deviation(const double x[], int size) { double mean1 = mean(x, size); double squareSum = 0; for (int i = 0; i < size; i++) { squareSum += pow(x[i] - mean1, 2); } return sqrt(squareSum / (size - 1)); } /** Function for computing deviation of int values */ double deviation(const int x[], int size) { double mean1 = mean(x, size); double squareSum = 0; for (int i = 0; i < size; i++) {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang squareSum += pow(x[i] - mean1, 2); }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return sqrt(squareSum / (size - 1)); } /** Function for printing array */ void printArray(const double x[], int size) { for (int i = 0; i < size; i++) cout << x[i] << " "; cout << endl; } int main() { const int SIZE = 10; // Declare and create an array for 10 numbers double numbers[SIZE]; cout << "Enter ten numbers: "; for (int i = 0; i < SIZE; i++) { cin >> numbers[i]; } // Print numbers // printArray(numbers, 10); // Display mean and deviation cout << "The mean is " << mean(numbers, 10) << endl; cout << "The standard deviation is " << deviation(numbers, 10) << endl; return 0; } Chapter 7: Programming Project 5: (345021-00034) (Sorted?)

Write the following function that returns true if the list is already sorted in increasing order: bool isSorted(const int list[], int size) Write a test program that prompts the user to enter a list and displays whether the list is sorted or not.

Assume that the maximum number of integers in the list is 100. So declare array of size 100.

Sample Run 1 Enter the size of the list: 8 Enter list: 10 1 5 16 61 9 11 1 The list is not sorted Sample Run 2 Enter the size of the list: 10

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Enter list: 1 1 3 4 4 5 7 9 11 21

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang The list is already sorted

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; bool isSorted(const int list[], int size) { for (int i = 0; i < size - 1; i++) if (list[i] > list[i + 1]) return false; return true; } int main() { const int MAX_SIZE = 80; // Enter list size cout << "Enter the size of the list: "; int size; cin >> size; int list[MAX_SIZE]; cout << "Enter the list: "; for (int i = 0; i < size; i++) cin >> list[i]; if (isSorted(list, size)) cout << "The list is already sorted" << endl; else cout << "The list is not sorted" << endl; return 0; } Chapter 8 Programming Quiz 8.2: Question 1: (00000-10735) Declare a 3x3 two-dimensional array of int values named tictactoe. int tictactoe[3][3];

Programming Quiz 8.2: Question 2: (00000-10736) Declare an 8x8 two-dimensional array of strings named chessboard. string chessboard[8][8];

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 8.2: Question 3: (00000-10737)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Declare a two-dimensional array of int values named arr with 3 rows and 10 columns. int arr[3][10];

Programming Quiz 8.3: Question 1: (00000-10743) Given two 3x3 arrays of integers, x1 and x2, write the code needed to copy every value from x1 to its corresponding element in x2. for (int i = 0; i < 3; i++) for (int j = 0; j < 3; j++) x2[i][j] = x1[i][j]; Programming Quiz 8.3: Question 2:

(00000-10744)

You are given a 6x8 (6 rows, 8 columns) array of int, x, already initialized and three int variables: max, i, and j.

Write the code so that max will have the largest value in the array x. Please consider the case that all the numbers in the matrix may be negative when writing your code. max = x[0][0]; for (i = 0; i < 6; i++) for (j = 0; j < 8; j++) if (max < x[i][j]) max = x[i][j]; Programming Quiz 8.3: Question 3:

(00000-10745)

Given two 2x3 (2 rows, 3 columns) arrays of int, x1 and x2, both already initialized, two int variables, i and j, and a bool variable x1Rules.

Write the code that is necessary for x1Rules to be true if and only if every element in x1 is bigger than its corresponding element in x2 and is false otherwise. x1Rules = true; for (i = 0; i < 2; i++) for (j = 0; j < 3; j++) if (x1[i][j] <= x2[i][j]) x1Rules = false; Programming Quiz 8.8: Question 1: (00000-10985) Declare an array of int values westboundHollandTunnelTraffic that can store the number of vehicles going westbound through the Holland Tunnel - on a particular hour (numbered 0 through 23) - on a particular day (numbered 0 through 6) - on a particular week (numbered 0 through 51) - over the last ten years (numbered 0 through 9).

The innermost dimension should be years, with the next being weeks, and so on.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang int westboundHollandTunnelTraffic[10][52][7][24];

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Chapter 8: Programming Project 1: (345021-00035) (Sum elements in each column) Write a function that returns the sum of all the elements in a specified column in a matrix using the following header: const int SIZE = 4; double sumColumn(const double m[][SIZE], int rowSize, int columnIndex); Write a test program that reads a 3-by-4 matrix and displays the sum of each column.

Sample Run Enter a 3-by-4 matrix row by row: 1.5 2 3 4 5.5 6 7 8 9.5 1 3 1 Sum of the elements at column 0 is 16.5 Sum of the elements at column 1 is 9 Sum of the elements at column 2 is 13 Sum of the elements at column 3 is 13 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; const int SIZE = 4; double sumColumn(const double m[][SIZE], int rowSize, int columnIndex) { double total = 0; for (int i = 0; i < rowSize; i++) total += m[i][columnIndex]; return total; } int main() { double matrix[3][SIZE]; cout << "Enter a 3-by-4 matrix row by row: " << endl; for (int i = 0; i < 3; i++) for (int j = 0; j <Introduction SIZE; j++) Solutions Manual for to C++ Programming and Data cin >> matrix[i][j]; Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang for (int j = 0; j < SIZE; j++) cout << "Sum of the elements at column " << j << " is " << sumColumn(matrix, 3, j) << endl; return 0; }

Chapter 8: Programming Project 2: (345021-00036) (Algebra: add two matrices) The description of this project is given in Programming Exercise 8.5 in the Chapter 8 Programming Exercise from the Book section. For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; const int N = 3; /** The method for adding two matrices */ void addMatrix(const double a[][N], const double b[][N], double c[][N]) { for (int i = 0; i < N; i++) for (int j = 0; j < N; j++) c[i][j] = a[i][j] + b[i][j]; } /** Print result */ void printResult(const double m1[][N], const double m2[][N], const double m3[][N], char op) { for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) cout << " " << m1[i][j]; if (i == N / 2) cout << " " << op << " else cout << " ";

";

for (int j = 0; j < N; j++) cout << " " << m2[i][j]; if (i == N / 2) cout << " = "; else cout << " "; for (int j = 0; j < N; j++) cout << " " << m3[i][j]; cout << endl; } Solutions Manual for Introduction to C++ Programming and Data } Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang int main() { // Create two matrices as two dimensional arrays double matrix1[N][N]; double matrix2[N][N]; double result[N][N]; // Assign random values to matrix1 and matrix2 cout << "Enter matrix1: "; for (int i = 0; i < N; i++) for (int j = 0; j < N; j++) { cin >> matrix1[i][j]; } cout << "Enter matrix2: "; for (int i = 0; i < N; i++) for (int j = 0; j < N; j++) { cin >> matrix2[i][j]; } // Add two matrices and print the result addMatrix(matrix1, matrix2, result); cout << "The addition of the matrices is " << endl; printResult(matrix1, matrix2, result, '+'); }

Chapter 8: Programming Project 3: (345021-00150) (Markov matrix) An n by n matrix is called a positive Markov matrix if each element is positive and the sum of the elements in each column is 1. Write the following function to check whether a matrix is a Markov matrix. const int SIZE = 3; bool isMarkovMatrix(const double m[][SIZE]); Write a test program that prompts the user to enter a 3 by 3 matrix of double values and tests whether it is a Markov matrix. Here are sample runs: Sample Run 1 Enter a 3-by-3 matrix row by row: 0.15 0.875 0.375 0.55 0.005 0.225 0.30 0.12 0.4 It is a Markov matrix Sample Run 2 Enter a 3-by-3 matrix row by row: 0.95 -0.875 0.375 0.65 0.005 0.225 0.30 0.22 -0.4 It is not a Markov matrix If you get a logical or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

#include <iostream> // Exercise08_25 using namespace std;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang const int SIZE = 3; bool isMarkovMatrix(const double m[][SIZE]); int main() { cout << "Enter a 3 by 3 matrix row by row: "; double m[SIZE][SIZE]; for (int i = 0; i < SIZE; i++) for (int j = 0; j < SIZE; j++) cin >> m[i][j]; if (isMarkovMatrix(m)) cout << "It is a Markov matrix" << endl; else cout << "It is not a Markov matrix" << endl; return 0; } bool isMarkovMatrix(const double m[][SIZE]) { // Check positive for (int i = 0; i < SIZE; i++) for (int j = 0; j < SIZE; j++) if (m[i][j] < 0) return false; // Check the sum of each column for (int j = 0; j < SIZE; j++) { double sum = 0; for (int i = 0; i < SIZE; i++) sum += m[i][j]; if (abs(sum - 1) >= 0.0000000001) // sum is not 1 return false; } return true; }

Chapter 8: Programming Project 4:

(345021-00038)

(Central city) Given a set of cities, the central point is the city that has the shortest total distance to all other cities.

Write a program that prompts the user to enter the number of cities and the locations of the cities (coordinates), and finds the central city and its total distance to all other cities. Assume that the maximum number of cities is 20.

Sample Run Enter the number of cities: 5

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Enter the coordinates of the cities: 2.5 5 5.1 3 1 9 5.4 54 5.5 2.1 The central city is at (2.5, 5.0)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang The total distance to all other cities is 60.81 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <cmath> using namespace std; const int MAX_NUMBER_OF_CITIES = 20; double distance(const double c1[], const double c2[]) { return sqrt((c1[0] - c2[0]) * (c1[0] - c2[0]) + (c1[1] - c2[1]) * (c1[1] - c2[1])); } // Return the total distance from i to all other cities double totalDistance(const double cities[][2], int numberOfCities, int i) { double total = 0; for (int j = 0; j < numberOfCities; j++) total += distance(cities[i], cities[j]); return total; } int main() { cout << "Enter the number of cities: "; int numberOfCities; cin >> numberOfCities; cout << "Enter the coordinates of the cities: "; double cities[MAX_NUMBER_OF_CITIES][2]; for (int i = 0; i < numberOfCities; i++) { cin >> cities[i][0] >> cities[i][1]; } double minTotal = totalDistance(cities, numberOfCities, 0); int minIndex = 0; for (int i = 1; i < numberOfCities; i++) { double total = totalDistance(cities, numberOfCities, i); if (minTotal > total) { minTotal = total; minIndex = i; } } cout << "The central city is at (" << cities[minIndex][0] << ", " << cities[minIndex][1] << ")" << endl; cout << "The total distance to all other cities is " << minTotal << endl;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return 0; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Chapter 9 Programming Quiz 9.3: Question 1: (00000-11195) Objects of the Calculator class require no additional information when created.

Declare an object named calc of type Calculator using the no-arg constructor.

Calculator calc;

Programming Quiz 9.3: Question 2: (00000-11196) Objects of the BankAccount class require a name (string) and a social security number (string) be specified (in that order) upon creation.

Declare an object named account, of type BankAccount, using the values "John Smith" and "12345-6789" as the name and social security number respectively.

BankAccount account("John Smith", "123-45-6789");

Programming Quiz 9.3: Question 3: (00000-11197) Objects of the BankAccount class require a name (string) and a social security number (string) be specified (in that order) upon creation.

Declare two strings corresponding to a name and a social security number and read values into them from standard input (in that order) using cin >>.

Use these values to declare an object of type BankAccount named newAccount. string name; string ssnum; cin >> name; cin >> ssnum; BankAccount newAccount(name, ssnum);

Programming Quiz 9.3: Question 4: (00000-11198) Objects of the Window class require a width (int) and a height (int) be specified (in that order) upon creation.

Declare an object named window, of type Window, corresponding to an 80 x 20 window.

Window window(80, 20); Programming Quiz 9.3: Question 5: (00000-11199) Objects of the Window class require a width (int) and a height (int) be specified (in that order) upon creation.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Declare an object named window, of type Window, whose width and length are obtained from the (already declared and initialized) variables, w and l.

Window window(w, l); Programming Quiz 9.3: Question 6: (00000-11200) Objects of the Window class require a width (int) and a height (int) be specified (in that order) upon creation.

Declare two integers corresponding to a width and a length and read values into them from standard input (in that order).

Use these values to declare an object of type Window named newWindow. int width; int height; cin >> width; cin >> height; Window newWindow(width, height);

Programming Quiz 9.4: Question 1: (00000-11201) Creating objects of the Currency class requires a name (string), a currency symbol (string), and the number of decimal places (int) usually associated with the currency (in that order).

Declare an object named curr of type Currency corresponding to "US Dollar" with the symbol "$" and 2 decimal places.

Currency curr("US Dollar", "$", 2);

Programming Quiz 9.4: Question 2: (00000-11204) Creating objects of the Currency class requires a name (string), a currency symbol (string), and the number of decimal places (int) usually associated with the currency (in that order).

Creating objects of the Money class requires a Currency object and the actual amount (int) (in that order).

Declare an object of type Currency named canadianDollars and corresponding to the name "Canadian Dollar", the symbol "C$", and 2 decimal places.

Declare an object of type Money named valueInCanadians corresponding to 230 units of canadianDollars.

Currency canadianDollars("Canadian Dollar", "C$", 2); Money valueInCanadians(canadianDollars, 230); Programming Quiz 9.4: Question 3: (00000-11205) Objects of class Name are defined by supplying last name (string) and a first name (string) in this order.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Declare an object of type Name named name and corresponding to a last name of "Black" and a first name of "Susan".

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Name name("Black", "Susan"); Programming Quiz 9.9: Question 1: (00000-10874) Write the header file (.h file) of a class Counter containing: •

A data member counter of type int.

•

A data member named limit of type int.

•

A constructor that takes two int arguments.

•

A function called increment that accepts no parameters and returns no value.

•

A function called decrement that accepts no parameters and returns no value.

•

A function called getValue that accepts no parameters and returns an int value.

class Counter { private: int counter, limit; public: Counter(int, int); void increment(); void decrement(); int getValue(); };

Programming Quiz 9.9: Question 2:

(00000-10875)

Write the implementation (.cpp file) of the Counter class of the previous exercise.

The full specification of the class is: o

A data member counter of type int.

o

A data member named limit of type int.

o

A constructor that takes two int arguments and assigns the first one to counter and the second one to limit.

o

A function called increment that accepts no parameters and returns no value. If the data member counter is less than limit, increment just adds one to the instance variable counter.

o

A function called decrement that accepts no parameters and returns no value.

If counter is greater than 0, decrement subtracts one from the counter.

o

A function called getValue that accepts no parameters. It returns the value of the instance variable counter.

Don’t include the header file in this exercise, because REVEL assumes that the header and implementation are placed in the same file.

Counter::Counter(int theCounter, int theLimit) { counter =Manual theCounter; Solutions for Introduction to C++ Programming and Data

Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang limit = theLimit; } void Counter::increment() { if (counter < limit) counter++; } void Counter::decrement() { if (counter > 0) counter--; } int Counter::getValue() { return counter; }

Programming Quiz 9.11: Question 1: (00000-10882) Write the header file (.h file) of a class Acc2 containing: •

A data member named sum of type int.

•

A constructor accepting no parameters.

•

A function named getSum that accepts no parameters and returns an int.

class Acc2 { private: int sum; public: Acc2(); int getSum(); };

Programming Quiz 9.11: Question 2: (00000-10883) Write the implementation (.cpp file) of the Acc2 class of the previous exercise.

The full specification of the class is: •

A data member named sum of type int.

•

A constructor that accepts no parameters. The constructor initializes the data member sum to 0.

•

A function named getSum that accepts no parameters and returns an int. getSum returns the value of sum.

Don’t include the header file in this exercise, because REVEL assumes that the header and

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang implementation are placed in the same file.

Acc2::Acc2()

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { sum = 0; } int Acc2::getSum() { return sum; }

Chapter 9: Programming Project 1:

(345021-00039)

(The Rectangle class) Design a class named Rectangle to represent a rectangle. The class contains: • Two double data fields named width and height that specify the width and height of the rectangle.

• A no-arg constructor that creates a rectangle with width 1 and height 1.

• A constructor that creates a rectangle with the specified width and height.

• The accessor and mutator functions for all the data fields.

• A function named getArea()that returns the area of this rectangle.

• A function named getPerimeter() that returns the perimeter.

Implement the class. Write a test program that creates two Rectangle objects. Assign width 4 and height 40 to the first object and width 3.5 and height 35.9 to the second. Display the width, height, area, and perimeters of the first object and then the second object.

Use the code from https://liangcpp.pearsoncmg.com/test/Exercise09_01.txt to complete your program.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise09_01 using namespace std; class Rectangle { public: Rectangle(); Rectangle(double newWidth, double newHeight); double getWidth(); double getHeight(); double getArea(); double getPerimeter(); private: double width, height; }; Rectangle::Rectangle() { width = height = 1; } // Constructor Rectangle::Rectangle(double newWidth,to double newHeight) Solutions Manual for Introduction C++ Programming and Data { Structures 5th edition by Daniel Liang width = newWidth;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang height = newHeight; } double Rectangle::getWidth() { return width; } double Rectangle::getHeight() { return height; } double Rectangle::getArea() { return width * height; } double Rectangle::getPerimeter() { return 2 * (width + height); } int main() { Rectangle myRectangle(4, 40); cout << "The area of a rectangle with width " << myRectangle.getWidth() << " and height " << myRectangle.getHeight() << " is " << myRectangle.getArea() << " and perimeter is " << myRectangle.getPerimeter() << endl; Rectangle yourRectangle(3.5, 35.9); cout << "The area of a rectangle with width " << yourRectangle.getWidth() << " and height " << yourRectangle.getHeight() << " is " << yourRectangle.getArea() << " and perimeter is " << yourRectangle.getPerimeter() << endl; return 0; }

Chapter 9: Programming Project 2: (345021-00040) (The EvenNumber class) Define the EvenNumber class for representing an even number. The class contains: • A data field value of the int type that represents the integer value stored in the object.

• A no-arg constructor that creates an EvenNumber object for the value 0.

• A constructor that constructs an EvenNumber object with the specified value.

• A function named getValue() to return an int value for this object.

• A function named getNext() to return an EvenNumber object that represents the next even number after the current even number in this object.

• A function named getPrevious() to return an EvenNumber object that represents the previous even number before the current even number in this object.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Implement the class. Write a test program that creates an EvenNumber object for value 16 and invokes

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang the getNext() and getPrevious() functions to obtain and displays these numbers.

Use the code from https://liangcpp.pearsoncmg.com/test/Exercise09_11.txt to complete your program.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; class EvenNumber { public: EvenNumber(); EvenNumber(int newValue); int getValue(); EvenNumber getNext(); EvenNumber getPrevious(); private: int value; }; // No-arg constructor EvenNumber::EvenNumber() { value = 0; } // Constructor EvenNumber::EvenNumber(int newValue) { value = newValue; } int EvenNumber::getValue() { return value; } EvenNumber EvenNumber::getNext() { return EvenNumber(value + 2); } EvenNumber EvenNumber::getPrevious() { return EvenNumber(value - 2); } int main() { EvenNumber t(16); cout << t.getValue() << endl; cout << t.getNext().getValue() << endl; cout << t.getPrevious().getValue() << endl; return 0; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Chapter 10

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 10.2: Question 1:

(00000-10721)

Assume that an int variable pos, as well as the strings line and clause, have already been declared. Write a sequence of statements that finds the first comma in line, and assigns to clause the portion of line up to, but not including the comma. pos = line.find(","); clause = line.substr(0, pos); Programming Quiz 10.2: Question 2:

(00000-11115)

Assume that s is a string. Write an expression whose value is true if and only if the value of s would come between "mortgage" and "mortuary" in the dictionary. "mortgage" <= s && s <= "mortuary"

Programming Quiz 10.2: Question 3:

(00000-10865)

Assume that sentence is a variable of type string that has been assigned a value.

Assume furthermore that this value is a string consisting of words separated by single space characters with a period at the end.

For example: "This is a possible value of sentence."

Assume that there is another variable declared, secondWord, also of type string.

Write the statements needed so that the second word of the value of sentence is assigned to secondWord.

So, if the value of sentence were "Broccoli is delicious." your code would assign the value "is" to secondWord. int k = sentence.find(" "); string temp = sentence.substr(k + 1); secondWord = temp.substr(0, temp.find(" "));

Programming Quiz 10.4: Question 1:

(00000-11230)

Assume a class Window with a constructor that accepts two int arguments: width and height (in that order).

Declare an array named winarr, consisting of 3 Window objects, where the first element is an 80x20 window, the second a 10x10 window, and the third a 133x40 window.

Window winarr[] = {Window(80, 20), Window(10, 10), Window(133, 40)};

Programming Quiz 10.4: Question 2:

(00000-11231)

Assume a class Window with accessor function getWidth that accepts no parameters and returns an int. Assume further an array of 3 Window elements named winarr, has been declared and initialized. Write a sequence of statements that prints out the width of the widest window in the array. int maxIndex = 0; for Introduction to C++ Programming and Data Solutions Manual for (int i = 1; edition i < 3; i++) Structures 5th

by Daniel Liang if (winarr[i].getWidth() > winarr[maxIndex].getWidth())

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang maxIndex = i; cout << winarr[maxIndex].getWidth() << endl;

Programming Quiz 10.5: Question 1:

(00000-10878)

Write the header file (.h file) of a class Counter containing: •

A data member counter of type int.

•

A data member named counterID of type int.

•

A static int data member named nCounters.

•

A constructor that takes an int argument.

•

A function called increment that accepts no parameters and returns no value.

•

A function called decrement that accepts no parameters and returns no value.

•

A function called getValue that accepts no parameters and returns an int.

•

A function named getCounterID that accepts no parameters and returns an int.

class Counter { private: int counter, counterID; static int nCounters; public: Counter(int); void increment(); void decrement(); int getValue(); int getCounterID(); }; Programming Quiz 10.5: Question 2:

(00000-10879)

Write the implementation (.cpp file) of the Counter class of the previous exercise.

The full specification of the class is: o

A data member counter of type int.

o

A data member named counterID of type int.

o

A static int data member named nCounters which is initialized to 0.

o

A constructor that takes an int argument and assigns its value to counter. It also adds one to the static variable nCounters and assigns the (new) value of nCounters to counterID.

o

A function called increment that accepts no parameters and returns no value. increment adds one to the instance variable counter.

o

A function called decrement that accepts no parameters and returns no value. decrement subtracts one from the counter.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang o

A function called getValue that accepts no parameters and returns an int. It returns the value of the instance variable counter.

o

A function named getCounterID that accepts no parameters and returns an int. getCounterID returns the value of the data member counterID.

Note: Don't use the #include "Counter.h", because REVEL assumes that your Counter header is in the same file with the implementation. You need to initialize static data member nCounters as int Counter::nCounters = 0; This line has to be put as the first line in your code.

int Counter::nCounters = 0; Counter::Counter(int a) { counter = a; nCounters++; counterID = nCounters; } void Counter::increment() { counter++; } void Counter::decrement() { counter--; } int Counter::getValue() { return counter ; } int Counter::getCounterID() { return counterID ; }

Programming Quiz 10.5: Question 3: (00000-10876) Write the header file (.h file) of a class Counter containing: •

A data member counter of type int.

•

A data member named limit of type int.

•

A static int data member named nCounters.

•

A constructor that takes two int arguments.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang •

A function called increment that accepts no parameters and returns no value.

•

A function called decrement that accepts no parameters and returns no value.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang •

A function called getValue that accepts no parameters and returns an int.

•

A static function named getNCounters that accepts no parameters and returns an int.

class Counter { private: int counter, limit; static int nCounters; public: Counter(int, int); void increment(); void decrement(); int getValue(); static int getNCounters(); }; Programming Quiz 10.5: Question 4: (00000-10877) Write the implementation (.cpp file) of the Counter class of the previous exercise.

The full specification of the class is: o

A data member counter of type int.

o

A data member named limit of type int.

o

A static int data member named nCounters that is initialized to 0.

o

A constructor that takes two int arguments and assigns the first one to counter and the second one to limit. It also adds one to the static variable nCounters

o

A function called increment that accepts no parameters and returns no value. If the data member counter is less than limit, increment just adds one to the instance variable counter.

o

A function called decrement that accepts no parameters and returns no value.

If counter is greater than 0, decrement subtracts one from the counter.

o

A function called getValue that accepts no parameters. It returns the value of the instance variable counter.

o

A static function named getNCounters that accepts no parameters and return an int. getNCounters returns the value of the static variable nCounters.

Note: Don't use the #include "Counter.h", because REVEL assumes that your Counter header is in the same file with the implementation. You need to initialize static data member nCounters as int Counter::nCounters = 0; This line has to be put as the first line in your code.

int Counter::nCounters = 0;

Solutions Manual for Introduction to C++ Programming and Data Counter::Counter(int c, int l) Structures 5th edition

by Daniel Liang {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang counter = c; limit = l; nCounters++; } void Counter::increment() { if (counter < limit) counter++; } void Counter::decrement() { if (counter > 0) counter--; } int Counter::getValue() { return counter; } int Counter::getNCounters() { return nCounters; }

Chapter 10: Programming Project 1: (345021-00041) (Anagrams) Write a function that checks whether two words are anagrams. Two words are anagrams if they contain the same letters in any order. For example, “silent” and “listen” are anagrams.

The header of the function is as follows: bool isAnagram(const string& s1, const string& s2) Write a test program that prompts the user to enter two strings and checks whether they are anagrams.

Sample Run 1 Enter a string s1: silent Enter a string s2: listen silent and listen are anagrams Sample Run 2 Enter a string s1: split Enter a string s2: lisp split and lisp are not anagrams For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang #include <iostream> #include <string> using namespace std; string sort(string& s) { for (int i = s.length() - 1; i >= 1; i--) { // Find the maximum in the list[0..i] char currentMax = s[0]; int currentMaxIndex = 0; for (int j = 1; j <= i; j++) { if (currentMax < s[j]) { currentMax = s[j]; currentMaxIndex = j; } } // Swap list[i] with list[currentMaxIndex] if necessary; if (currentMaxIndex != i) { s[currentMaxIndex] = s[i]; s[i] = currentMax; } } return s; } bool isAnagram(const string& s1, const string& s2) { if (s1.length() != s2.length()) return false; string newS1 = string(s1); string newS2 = string(s2); sort(newS1); sort(newS2); return newS1 == newS2; } int main() { // Prompt the user to enter two strings cout << "Enter a string s1: "; string s1; getline(cin, s1); // Prompt the user to enter two strings cout << "Enter a string s2: "; string s2; getline(cin, s2);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << s1 << " and " << s2 << " are " << (isAnagram(s1, s2) ? "anagrams" : "not anagrams") << endl;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return 0; }

Chapter 10: Programming Project 2:

(345021-00042)

(Geometry: n-sided regular polygon) The description of this project is given in Programming Exercise 10.13 in the Chapter 10 Programming Exercise from the Book section. For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <cmath> using namespace std; class RegularPolygon { private: int n; double side; double x; double y; public: RegularPolygon() { n = 3; side = 1; x = 0; y = 0; } RegularPolygon(int n, double side) { this->n = n; this->side = side; x = 0; y = 0; } RegularPolygon(int n, double side, double x, double y) { this->n = n; this->side = side; this->x = x; this->y = y; } int getN() const { return n; } void setN(int n) { Solutions Manual for Introduction to C++ Programming and Data this->n = n;edition by Daniel Liang Structures 5th }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang double getSide() const { return side; } void setSide(double side) { this->side = side; } double getX() const { return x; } void setX(double x) { this->x = x; } double getY() const { return y; } void setY(double y) { this->y = y; } double getPerimeter() const { return n * side; } double getArea() const { return n * side * side / tan(3.14159 / n) / 4; } }; int main() { RegularPolygon polygon1; RegularPolygon polygon2(6, 4); RegularPolygon polygon3(10, 4, 5.6, 7.8); cout << "Polygon 1 perimeter: " << polygon1.getPerimeter() << endl; cout << "Polygon 1 area: " << polygon1.getArea() << endl; cout << "Polygon 2 perimeter: " << polygon2.getPerimeter() << endl; cout << "Polygon 2 area: " << polygon2.getArea() << endl; cout << "Polygon 3 perimeter: " << polygon3.getPerimeter() << endl; cout << "Polygon 3 area: " << polygon3.getArea() << endl; return 0;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Chapter 11 Programming Quiz 11.2: Question 1:

(00000-10693)

Assume that an int variable diff has already been declared.

Assume further a variable diffPointer of type "pointer to int" has also already been declared.

Write a statement that assigns the address of diff to diffPointer. diffPointer = &diff;

Programming Quiz 11.2: Question 2:

(00000-10697)

Assume that strikeCounter has already been declared to be a "pointer to int".

Assume further that strikeCounter has been initialized -- its value is the address of some int variable.

Write a statement that adds 22 to the value of the variable that strikeCounter is pointing to. *strikeCounter += 22;

Programming Quiz 11.2: Question 3:

(00000-10698)

Assume that ip1, ip2, and ip3 have already been declared to be of type "pointer to int".

Assume further that each of these pointer variables have been initialized -- each points to some int variable.

Write a statement that computes the sum of the variables that ip1 and ip2 point to, and assigns that value (the sum) to the variable that ip3 points to. *ip3 = *ip1 + *ip2;

Programming Quiz 11.2: Question 4:

(00000-10789)

The variables xp and yp have both been declared as pointers to int, and have been assigned values.

Write the code to exchange the two int values pointed by xp and yp. (so that after the swap xp still points at the same location, but it now contains the int value originally contained in the location pointed to by yp; and vice versa-- in other words, in this exercise you are swapping the ints, not the pointers).

Declare any necessary variables.

int t = *xp; *xp = *yp; *yp = t;

Programming Quiz 11.5: Question 1:

(00000-10699)

Assume that ip has been declared to be a pointer to int and that result has been declared to be an

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang array of 100 elements.

Assume further that ip has been initialized to point to an element in the first half of the array.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write an expression whose value is the element in the array after the element that ip points to. *(ip + 1)

Programming Quiz 11.5: Question 2:

(00000-10702)

Assume that ip has been declared to be a pointer to int and that result has been declared to be an array of 100 elements.

Assume further that ip has been initialized to point to an element in the first half of the array.

Write an expression whose value is the sum of the element that ip points to plus the next two elements. *ip + *(ip + 1) + *(ip + 2)

Programming Quiz 11.5: Question 3:

(00000-10705)

Assume that ip has been declared to be a pointer to int and that enrollment has been declared to be an array of 20 elements.

Write a statement that makes ip point to the first element in the array. ip = &enrollment[0];

Programming Quiz 11.5: Question 4:

(00000-10706)

Assume that ip has been declared to be a pointer to int and that enrollment has been declared to be an array of 20 elements.

Write a statement that makes ip point to the last element in the array. ip = &enrollment[19];

Programming Quiz 11.6: Question 1:

(00000-10672)

Write the definition of a function zeroIt, which is used to zero out a variable.

The function is used as follows: int x = 5; zeroIt(&x); /* x is now equal to 0 */ void zeroIt(int* p) { *p = 0; }

Programming Quiz 11.6: Question 2:

(00000-10673)

Write the definition of a function doubleIt, which doubles the value of its argument but returns nothing so that it can be used as follows: int x = 5; doubleIt(&x); /* x is now equal to 10 */ void doubleIt(int* p) { Solutions Manual for Introduction to C++ Programming and Data *p = 2 * *p;

Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang }

Programming Quiz 11.6: Question 3:

(00000-10674)

Write the definition of a function tripleIt, which triples its argument but returns nothing so that it can be used as follows: int x = 5; tripleIt(&x); /* x is now equal to 15 */ void tripleIt(int* p) { *p = 3 * *p; }

Programming Quiz 11.6: Question 4:

(00000-10675)

Write the definition of a function divide that takes four arguments and returns no value. The first two arguments are of type int. The last two arguments arguments are pointers to int and are set by the function to the quotient and remainder of dividing the first argument by the second argument. The function does not return a value.

The function can be used as follows: int numerator = 42, denominator = 5, quotient, remainder; divide(numerator, denominator, &quotient, &remainder); /* quotient is now 8 and remainder is now 2 */ void divide(int n, int d, int* q, int* r) { *q = n / d; *r = n % d; }

Programming Quiz 11.9: Question 1: (00000-10782) Given the variable ip, already declared as a pointer to an int, write the code to dynamically allocate memory for a single int value, assign the resulting pointer to ip, and initialize the int value to 27. ip = new int(27);

Programming Quiz 11.9: Question 2:

(00000-10784)

Declare a variable, bp, as a pointer to bool, dynamically allocate memory for a single bool value, assign the resulting pointer to bp and initialize the value to true. bool* bp = new bool(true);

Programming Quiz 11.9: Question 3:

(00000-10786)

Declare variable dp to refer to an array of double. Assuming the int variable n has been assigned a value, allocate an array of n double values and assign the resulting pointer to dp.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang double* dp = new double[n];

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 11.9: Question 4:

(00000-10791)

The variable cp_arr has been declared as an array of 26 pointers to char.

Allocate 26 character values, initialized to the letters 'A' through 'Z' and assign their pointers to the elements of cp_arr (in that order). for (char c = 'A'; c <= 'Z'; c++) cp_arr[c - 'A'] = new char(c);

Programming Quiz 11.10: Question 1:

(00000-10783)

The class Date has a single constructor that accepts the int values: a month- (1 for January through 12 for December), a day of the month (1-31), and a year (in that order).

Given the Date pointer variable datep, dynamically allocate a Date object with the initial value of March 12, 2006, and assign the resulting pointer to datep. datep = new Date(3, 12, 2006);

Programming Quiz 11.12: Question 1:

(00000-11228)

Assume the existence of a class named Window with functions named close and freeResources, both of which accept no parameters and return no value. Write a destructor for the class that invokes close followed by freeResources.

Note: Don’t use the Window:: qualification in your code, because REVEL assumes that your code will be inserted directly in the Window class definition as an inline implementation. ~Window() { close(); freeResources(); }

Programming Quiz 11.15: Question 1:

(00000-11232)

Assume the existence of a class GraphicProgram with a string member, executableName, and data member, windowPtr, of type Window* (where Window is also a class type).

Assume further that the Window class has a function called clone, that takes no parameters and returns a pointer to a new copy of the Window.

Write a copy constructor for the GraphicProgram class that uses the clone function to make a proper copy of the Window member (rather than simple member assignment).

The executableName member may be copied using simple member assignment.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Note: Don’t use the GraphicProgram:: qualification in your code, because REVEL assumes that your code will be inserted directly in the GraphicProgram class definition as an inline implementation.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang GraphicProgram(const GraphicProgram& pgm) { executableName = pgm.executableName; windowPtr = pgm.windowPtr->clone(); } Chapter 11: Programming Project 1:

(345021-00043)

(Geometry: find the bounding rectangle) A bounding rectangle is the minimum rectangle that encloses a set of points in a two-dimensional plane, as shown in Figure 11.9d.

Write a function that returns a bounding rectangle for a set of points in a two-dimensional plane, as follows: const int SIZE = 2; Rectangle2D getRectangle(const double points[][SIZE]); Write another function that returns a pointer to the bounding rectangle as follows: Rectangle2D* getRectanglePointer(const double points[][SIZE]); The Rectangle2D class is defined in Programming Exercise 11.9.

Write a test program that prompts the user to enter five points and displays the bounding rectangle’s center, width, and height.

Sample Run Enter five points: 1.0 2.5 3 4 5 6 7 8 9 10 The bounding rectangle’s center (5.0, 6.25), width 8.0, height 7.5 The bounding rectangle’s center (5.0, 6.25), width 8.0, height 7.5

Use the code from https://liangcpp.pearsoncmg.com/test/Exercise11_11.txt to complete your program. For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #ifndef RECTANGLE2D_H #define RECTANGLE2D_H class Rectangle2D { public: Rectangle2D(); Rectangle2D(double x, double y, double width, double height); double getX() const; void setX(double x); double getY() const; void setY(double y); double getWidth() const; Solutions Manual for Introduction to C++ Programming and Data

Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang void setWidth(double width); double getHeight() const; void setHeight(double height); double getPerimeter() const; double getArea() const; bool contains(double x, double y) const; bool contains(const Rectangle2D& r) const; bool overlaps(const Rectangle2D& r) const; static double distance(double x1, double y1, double x2, double y2); private: double x, y; // Center of the rectangle double width, height; }; #endif #include <cmath> #include <algorithm> Rectangle2D::Rectangle2D(double x, double y, double width, double height) { this->x = x; this->y = y; this->width = width; this->height = height; } double Rectangle2D::getX() const { return x; } void Rectangle2D::setX(double x) { this->x = x; } double Rectangle2D::getY() const { return y; } void Rectangle2D::setY(double y) { this->y = y; } double Rectangle2D::getWidth() const { return width; } void Rectangle2D::setWidth(double width) { this->width = width;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } double Rectangle2D::getHeight() const

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { return height; } void Rectangle2D::setHeight(double height) { this->height = height; } double Rectangle2D::getPerimeter() const { return 2 * (width + height); } double Rectangle2D::getArea() const { return width * height; } bool Rectangle2D::contains(double x, double y) const { return abs(x - this->x) <= width / 2 && abs(y - this->y) <= height / 2; } bool Rectangle2D::contains(const Rectangle2D & r) const { return contains(r.x - r.width / 2, r.y + r.height / 2) && contains(r.x r.width / 2, r.y - r.height / 2) && contains(r.x + r.width / 2, r.y + r.height / 2) && contains(r.x + r.width / 2, r.y - r.height / 2); } bool Rectangle2D::overlaps(const Rectangle2D & r) const { return abs(r.x - x) <= (r.width + width) / 2 && abs(r.y - y) <= (r.height + height) / 2; } double Rectangle2D::distance(double x1, double y1, double x2, double y2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); } // MyRectangle2D getRectangle(const double points[][SIZE], int numberOfPoints); const int SIZE = 2; double minX(const double points[][SIZE], int numberOfPoints) { double minX = points[0][0]; for (int i = 0; i < numberOfPoints; i++) if (minX > points[i][0]) minX = points[i][0]; return minX;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } double maxX(const double points[][SIZE], int numberOfPoints)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { double maxX = points[0][0]; for (int i = 0; i < numberOfPoints; i++) if (maxX < points[i][0]) maxX = points[i][0]; return maxX; } double minY(const double points[][SIZE], int numberOfPoints) { double minY = points[0][1]; for (int i = 0; i < numberOfPoints; i++) if (minY > points[i][1]) minY = points[i][1]; return minY; } double maxY(const double points[][SIZE], int numberOfPoints) { double maxY = points[0][1]; for (int i = 0; i < numberOfPoints; i++) if (maxY < points[i][1]) maxY = points[i][1]; return maxY; } Rectangle2D* getRectanglePointer(const double points[][SIZE], int numberOfPoints) { double minimalX = minX(points, numberOfPoints); double minimalY = minY(points, numberOfPoints); double maximalX = maxX(points, numberOfPoints); double maximalY = maxY(points, numberOfPoints); return new Rectangle2D((minimalX + maximalX) / 2, (minimalY + maximalY) / 2, maximalX - minimalX, maximalY - minimalY); } Rectangle2D getRectangle(const double points[][SIZE], int numberOfPoints) { double minimalX = minX(points, numberOfPoints); double minimalY = minY(points, numberOfPoints); double maximalX = maxX(points, numberOfPoints); double maximalY = maxY(points, numberOfPoints); return Rectangle2D((minimalX + maximalX) / 2, (minimalY + maximalY) / 2, maximalX - minimalX, maximalY - minimalY); }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang #include <iostream> using namespace std;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang int main() { double points[5][2]; cout << "Enter five points: "; for (int i = 0; i < 5; i++) { cin >> points[i][0] >> points[i][1]; } Rectangle2D boundingRectangle = getRectangle(points, 5); cout << "The bounding rectangle's center (" << boundingRectangle.getX() << ", " << boundingRectangle.getY() << "), width " << boundingRectangle.getWidth() << ", height " << boundingRectangle.getHeight() << endl; Rectangle2D* boundingRectangleP = getRectanglePointer(points, 5); cout << "The bounding rectangle's center (" << boundingRectangleP->getX() << ", " << boundingRectangleP->getY() << "), width " << boundingRectangleP->getWidth() << ", height " << boundingRectangleP->getHeight() << endl; return 0; }

Chapter 12

Programming Quiz 12.6: Question 1: (00000-10809) Declare a vector named scores of twenty-five elements of type int. vector<int> scores(25);

Programming Quiz 12.6: Question 2: (00000-10815) Given a vector a, declared to contain 34 elements, write an expression that refers to the last element of the vector. a[33]

Programming Quiz 12.6: Question 3: (00000-10818) Given that a vector of int named a has been declared, assign 3 to its first element. a[0] = 3;

Programming Quiz 12.6: Question 4: (00000-10816)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Assume that the vector arr has been declared. In addition, assume that VECTOR_SIZE has been defined to be an int that equals the number of elements in arr.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write a statement that assigns the next to last element of the vector to the variable x (x has already been declared). x = arr[VECTOR_SIZE - 2]; Programming Quiz 12.6: Question 5: (00000-10825) Given that a vector of int named a has been declared with 12 elements and that the int variable k holds a value between 0 and 6.

Assign 9 to the element just after a[k]. a[k+1] = 9;

Programming Quiz 12.7: Question 1: (00000-10989) Assume v is a vector that has been declared and initialized.

Write an expression whose value is the number of values that have been stored in v. v.size() Programming Quiz 12.7: Question 2: (00000-10992) Assume v is a vector that has been declared and initialized. Write an expression whose value is true if there are any values stored in v. !v.empty()

Programming Quiz 12.7: Question 3: (00000-10993) Assume v is a vector that has been declared and initialized.

Write a statement that removes all the values stored in v. v.clear();

Programming Quiz 12.7: Question 4: (00000-10994) Assume v is a vector of int that has been declared and initialized.

Write a statement that adds the value 42 to the vector. v.push_back(42);

Chapter 12: Programming Project 1: (345021-00044) (Evaluate expression)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Modify LiveExample 12.12 EvaluateExpression.cpp to add operators ^ for exponent and % for modulus.

For example, 3 ^ 2 is 9 and 3 % 2 is 1.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang The ^ operator has the highest precedence and the % operator has the same precedence as the * and / operators. For simplicity, assume the expression has at most one ^ operator.

Sample Run Enter an expression: (5 * 2 ^ 3 + 2 * 3 % 2) * 4 (5 * 2 ^ 3 + 2 * 3 % 2) * 4 = 160 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html.

#include <vector> #include <iostream> #include <cmath> using namespace std; #ifndef STACK_H #define STACK_H template<typename T> class Stack { public: Stack(); bool empty() const; T peek() const; void push(T value); T pop(); int getSize() const; private: vector<T> elements; }; template<typename T> Stack<T>::Stack() { } template<typename T> bool Stack<T>::empty() const { return elements.size() == 0; } template<typename T> T Stack<T>::peek() const { return elements[elements.size() - 1]; } template<typename T> void Stack<T>::push(T value) { elements.push_back(value); Solutions Manual for Introduction to C++ Programming and Data }

Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename T> T Stack<T>::pop() { T temp = elements[elements.size() - 1]; elements.pop_back(); return temp; } template<typename T> int Stack<T>::getSize() const { return elements.size(); } #endif #include <string> #include <cctype> using namespace std; vector<string> split(const string& expression); int evaluateExpression(const string& expression); void processAnOperator( Stack<int>& operandStack, Stack<char>& operatorStack); int main() { string expression; cout << "Enter an expression: "; getline(cin, expression); cout << expression << " = " << evaluateExpression(expression) << endl; return 0; } vector<string> split(const string& expression) { vector<string> v; // A vector to store split items as strings string numberString; // A numeric string for (unsigned int i = 0; i < expression.length(); i++) { if (isdigit(expression[i])) numberString.append(1, expression[i]); // Append a digit else { if (numberString.size() > 0) { v.push_back(numberString); // Store the numeric string numberString.erase(); // Empty the numeric string }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang if (!isspace(expression[i])) {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang string s; s.append(1, expression[i]); v.push_back(s); // Store an operator and parenthesis } } } if (numberString.size() > 0) v.push_back(numberString); return v; } int evaluateExpression(const string& expression) { // Create operandStack to store operands Stack<int> operandStack; // Create operatorStack to store operators Stack<char> operatorStack; // Extract operands and operators vector<string> tokens = split(expression); // Phase 1: Scan tokens for (unsigned int i = 0; i < tokens.size(); i++) { if (tokens[i][0] == '+' || tokens[i][0] == '-') { while (!operatorStack.empty() && (operatorStack.peek() == '+' || operatorStack.peek() == '-' || operatorStack.peek() == '*' || operatorStack.peek() == '/' || operatorStack .peek() == '^' || operatorStack.peek() == '%')) { processAnOperator(operandStack, operatorStack); } operatorStack.push(tokens[i][0]); } else if (tokens[i][0] == '*' || tokens[i][0] == '/' || tokens[i][0] == '%') { while (!operatorStack.empty() && (operatorStack.peek() == '*' || operatorStack.peek() == '/' || operatorStack.peek() == '%' || operatorStack.peek() == '^')) { processAnOperator(operandStack, operatorStack); } operatorStack.push(tokens[i][0]); } else if (tokens[i][0] == '^') { while (!operatorStack.empty() && operatorStack.peek() == '^') { processAnOperator(operandStack, operatorStack); }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang operatorStack.push(tokens[i][0]); } else if (tokens[i][0] == '(')

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { operatorStack.push('('); // Push '(' to stack } else if (tokens[i][0] == ')') { while (operatorStack.peek() != '(') { processAnOperator(operandStack, operatorStack); } operatorStack.pop(); // Pop the '(' symbol from the stack } else { operandStack.push(atoi(tokens[i].c_str())); } } while (!operatorStack.empty()) { processAnOperator(operandStack, operatorStack); } return operandStack.pop(); } void processAnOperator( Stack<int>& operandStack, Stack<char>& operatorStack) { char op = operatorStack.pop(); int op1 = operandStack.pop(); int op2 = operandStack.pop(); if (op == '+') operandStack.push(op2 + op1); else if (op == '-') operandStack.push(op2 - op1); else if (op == '*') operandStack.push(op2 * op1); else if (op == '/') operandStack.push(op2 / op1); else if (op == '%') operandStack.push(op2 % op1); else if (op == '^') operandStack.push(static_cast<int>(round(pow(op2, op1)))); }

Chapter 12: Programming Project 2: (345021-00045) (Pattern recognition: consecutive four equal numbers) Rewrite the isConsecutiveFour function in Programming Exercise 8.21 using a vector as follows: bool isConsecutiveFour(const vector<vector<int>>& values)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write a test program like the one in Programming Exercise 8.21.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Sample Run 1 Enter row size: 6 Enter column size: 7 0 1 0 3 1 6 1 0 1 6 8 6 0 1 5 6 2 1 8 2 9 6 5 6 1 1 9 1 1 3 6 1 4 0 7 3 3 3 3 4 0 7 The array has consecutive fours Sample Run 2 Enter row size: 6 Enter column size: 7 0 1 0 3 1 6 1 0 1 6 8 6 0 1 5 6 2 1 8 2 9 6 5 6 1 1 9 1 1 3 6 1 4 0 7 5 5 5 3 4 0 7 The array does not have consecutive fours For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise12_23 #include <vector> #include <cmath> #include <algorithm> // for using max and min using namespace std; bool isConsecutiveFour(const vector<int> values) { for (int i = 0; i < values.size() - 3; i++) { bool isEqual = true; for (int j = i; j < i + 3; j++) { if (values[j] != values[j + 1]) { isEqual = false; break; } } if (isEqual) return true; } return false; } bool isConsecutiveFour(const vector<vector<int>>& values) { int numberOfRows = values.size(); int numberOfColumns = values[0].size();

Solutions for Introduction to C++ Programming and Data // Check Manual rows Structures edition

by Daniel Liang for (int i5th = 0; i < numberOfRows; i++)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { if (isConsecutiveFour(values[i])) return true; } // Check columns for (int j = 0; j < numberOfColumns; j++) { vector<int> column(numberOfRows); // Get a column into an array for (int i = 0; i < numberOfRows; i++) column[i] = values[i][j]; if (isConsecutiveFour(column)) return true; } // Check major diagonal (lower part) for (int i = 0; i < numberOfRows - 3; i++) { int numberOfElementsInDiagonal = min(numberOfRows - i, numberOfColumns); vector<int> diagonal(numberOfElementsInDiagonal); for (int k = 0; k < numberOfElementsInDiagonal; k++) diagonal[k] = values[k + i][k]; if (isConsecutiveFour(diagonal)) return true; } // Check major diagonal (upper part) for (int j = 1; j < numberOfColumns - 3; j++) { int numberOfElementsInDiagonal = min(numberOfColumns - j, numberOfRows); vector<int> diagonal(numberOfElementsInDiagonal); for (int k = 0; k < numberOfElementsInDiagonal; k++) diagonal[k] = values[k][k + j]; if (isConsecutiveFour(diagonal)) return true; } // Check sub-diagonal (left part) for (int j = 3; j < numberOfColumns; j++) { int numberOfElementsInDiagonal = min(j + 1, numberOfRows); vector<int> diagonal(numberOfElementsInDiagonal); for (int k = 0; k < numberOfElementsInDiagonal; k++) diagonal[k] = values[k][j - k]; if (isConsecutiveFour(diagonal)) return true; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Check sub-diagonal (right part) for (int i = 1; i < numberOfRows - 3; i++)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { int numberOfElementsInDiagonal = min(numberOfRows - i, numberOfColumns); vector<int> diagonal(numberOfElementsInDiagonal); for (int k = 0; k < numberOfElementsInDiagonal; k++) diagonal[k] = values[k + i][numberOfColumns - k - 1]; if (isConsecutiveFour(diagonal)) return true; } return false; } int main() { cout << "Enter the number of rows: "; int numberOfRows; cin >> numberOfRows; cout << "Enter the number of columns: "; int numberOfColumns; cin >> numberOfColumns; cout << "Enter the array values: " << endl; vector<vector<int>> board(numberOfRows); for (int i = 0; i < board.size(); i++) { board[i] = vector<int>(numberOfColumns); for (int j = 0; j < numberOfColumns; j++) cin >> board[i][j]; } cout << (isConsecutiveFour(board) ?

"The array has consecutive fours" : "The array does not have consecutive fours")

<< endl;

return 0; }

Chapter 12: Programming Project 3: (345021-00046) (Intersection) Write a function that returns the intersection of two vectors using the following header: template<typename T> vector<T> intersect(const vector<T>& v1, const vector<T>& v2) The intersection of two vectors contains the common elements that appear in both vectors. The elements in the result appear in the same order as in v1.

For example, the intersection of two vectors {2, 3, 1, 5} and {3, 4, 5} is {3, 5}.

Write a test program that prompts the user to enter two vectors, each with five strings, and displays their intersection.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Sample Run Enter five strings for vector1:

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Atlanta Dallas Chicago Boston Denver Enter five strings for vector2: Dallas Tampa Miami Boston Richmond The common strings are Dallas Boston For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html.

#include <iostream> // Exercise12_31 #include <string> #include <vector> using namespace std; template<typename T> int indexOf(const vector<T> v, T name) { for (int i = 0; i < v.size(); i++) if (v[i] == name) return i; return -1; } template<typename T> vector<T> intersection(const vector<T>& v1, const vector<T>& v2) { vector<T> v; for (int i = 0; i < v1.size(); i++) if (indexOf(v2, v1[i]) >= 0) v.push_back(v1[i]); return v; } int main() { cout << "Enter five strings for vector1: "; string s; vector<string> v1; for (int i = 0; i < 5; i++) { cin >> s; v1.push_back(s); } cout << "Enter five strings for vector2: "; vector<string> v2; for (int i = 0; i < 5; i++) { cin >> s; v2.push_back(s); } vector<string> v = intersection(v1, v2); cout << "The common strings are ";

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang for (int i = 0; i < v.size(); i++) cout << v[i] << " "; cout << endl;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return 0; } Chapter 13 Programming Quiz 13.2: Question 1: (00000-11075) Given a bool variable isReadable, write some statements that assign true to isReadable if the file "topsecret" exists and can be read by the program and assigns false to isReadable otherwise. ifstream infile; infile.open("topsecret"); isReadable = !infile.fail(); Programming Quiz 13.2: Question 2: (00000-11076) Given an int variable x, write some statements that attempt to open a file named "table20" and read a value into x; if that turns out not to be possible, your code should then read a value from console input into x. ifstream infile; infile.open("table20"); if (infile.fail()) cin >> x; else infile >> x;

Programming Quiz 13.2: Question 3: (00000-11125) Given the availability of an ofstream object named output and a string variable name tweet, write the statements to open a file named "mytweet", display the prompt tweet: and then read an entire line into tweet and then write it out to the file mytweet. output.open("mytweet"); cout << "tweet:"; getline(cin, tweet); output << tweet << endl; output.close(); Programming Quiz 13.2: Question 4:

(00000-11126)

Given the availability of a file named numbers, write the statements to read an int from standard input and then read in that many values from numbers and display their total. ifstream infile; infile.open("numbers"); int total = 0; int n, x; cin >> n; for (int i = 0; i < n; i++) { infile >> x; total += x; } Solutions Manual for Introduction to C++ Programming and Data cout << total << endl;

Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 13.4: Question 1: (00000-11032) A variable c of type char has been declared. Write the code to read in the next character from standard input and store it in c, regardless of whether it is a whitespace character. cin.get(c);

Programming Quiz 13.4: Question 2: (00000-11033) Declare a string named line and write a statement that reads in the next line of console input into this variable. string line; getline(cin, line); Chapter 13: Programming Project 1: (345021-00054) (Process scores in a text file) Suppose that a text file Exercise13_3.txt contains an unspecified number of scores. Write a program that reads the scores from the file and displays their total and average. Scores are separated by blanks.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <fstream> using namespace std; int main() { ifstream input; // Open a file input.open("Exercise13_3.txt"); if (input.fail()) { cout << "File does not exist" << endl; cout << "Exit program" << endl; return 0; } double number = 0; double total = 0; int count = 0; while (!input.eof()) // Continue if not end of file { input >> number; total += number; count++; } input.close();

Solutions Manual for Introduction to C++ Programming and Data cout << "Total is " << total << endl; Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "Average is " << total / count << endl; return 0; } Chapter 13: Programming Project 2: (345021-00055) (Create a text file) Write a program to create a file named Exercise13_1.txt if it does not exist. If it does exist, append new data to it. Write 100 integers 0 to 99 into the file using text I/O. Integers are separated by a space.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <fstream> using namespace std; int main() { ofstream output; // Create a file output.open("Exercise13_1.txt", ios::out | ios::app); for (int i = 0; i < 100; i++) { output << i << " "; } output.close(); cout << "Done" << endl; return 0; }

Chapter 14 Programming Quiz 14.8: Question 1: (00000-11220) Define and implement the following Window class: - int data members, width and height.

- a constructor that accepts two int parameters (width followed by height) and uses them to initialize the data members.

- a friend function, areSameSize, that accepts two Window objects and returns a bool indicating if they are of the same size. Two Window objects are of the same size if their widths and heights match. class Window { public: Window(int width, int height): width(width), height(height) {} friend bool areSameSize(const Window& w1, const Window& w2) { Solutions for to C++ Programming return Manual w1.width == Introduction w2.width && w1.height == w2.height;and Data Structures 5th edition by Daniel Liang }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang private: int width, height; };

Programming Quiz 14.9: Question 1: (00000-11226) Assume the existence of a Window class with int data members width and height.

Overload the << operator for the Window class-- i.e., write a nonmember ostream-returning function that accepts a reference to an ostream object and a constant reference to a Window object and sends the following to the ostream: 'a (width x height) window' (without the quotes and with width and height replaced by the actual width and height of the window.

Thus for example, if the window had width=80 and height=20, << would send 'a (80 x 20) window' to the ostream.

Don't forget to have the function return the proper value as well.

Note: The Window class is already in place. Declare and implement a friend << operator function inside the Window class. friend ostream& operator<<(ostream& os, const Window& window) { os << "a (" << window.width << " x " << window.height << ") window"; return os; } Programming Quiz 14.9: Question 2: (00000-11227) Assume the existence of a Window class with int data members width and height.

Overload the >> operator for the Window class-- i.e., write a nonmember istream-returning function that accepts a reference to an istream object and a reference to a Window object and reads the next two values from the istream into the width and height members respectively.

Note: The Window class is already in place. Declare and implement a friend << operator function inside the Window class. friend istream& operator>>(istream& is, Window& window) { is >> window.width >> window.height; return is; } Chapter 14: Programming Project 1: (345021-00168) (Math: The Complex class) The description of this project is given in Programming Exercise 14.7 in the Chapter 14 Programming

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Exercise from the Book section.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Use the code from https://liangcpp.pearsoncmg.com/test/Exercise14_07_5e.txt to complete your program.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <sstream> #include <cmath> #include <string> #include <iostream> using namespace std; class Complex { public: Complex(); Complex(double a, double b); Complex(double a); double getRealPart() const; double getImaginaryPart() const; Complex add(const Complex& secondComplex) const; Complex subtract(const Complex& secondComplex) const; Complex multiply(const Complex& secondComplex) const; Complex divide(const Complex& secondComplex) const; double abs() const; string toString() const; Complex& operator+=(Complex& secondComplex); Complex& operator-=(Complex& secondComplex); Complex& operator*=(Complex& secondComplex); Complex& operator/=(Complex& secondComplex); double& operator[](const int&

index);

Complex& operator++(); // Prefix ++ Complex& operator--(); // Prefix -Complex operator++(int dummy); // Postfix ++ Complex operator--(int dummy); // Postfix -Complex operator+(); // Unary + Complex operator-(); // Unary friend ostream& operator<<(ostream& stream, const Complex& complex); friend istream& operator >> (istream& stream, Complex& complex); private: double a; double b; }; // Non-member operator functions Complex operator+(const Complex& c1, const Complex& c2); Complex operator-(const Complex& c1, const Complex& c2); Complex operator*(const Complex& c1, const Complex& c2); Complex operator/(const Complex& c1, const Complex& c2);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang bool operator<(const Complex& c1, const Complex& c2); bool operator<=(const Complex& c1, const Complex& c2);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang bool operator==(const Complex& c1, const Complex& c2); bool operator!=(const Complex& c1, const Complex& c2); bool operator>(const Complex& c1, const Complex& c2); bool operator>=(const Complex& c1, const Complex& c2); // Nonmember + function Complex operator+(const Complex& c1, const Complex& c2) { return c1.add(c2); } Complex operator-(const Complex& c1, const Complex& c2) { return c1.subtract(c2); } Complex operator*(const Complex& c1, const Complex& c2) { return c1.multiply(c2); } Complex operator/(const Complex& c1, const Complex& c2) { return c1.divide(c2); } bool operator<(const Complex& c1, const Complex& c2) { return c1.abs() < c2.abs(); } bool operator<=(const Complex& c1, const Complex& c2) { return c1.abs() <= c2.abs(); } bool operator==(const Complex& c1, const Complex& c2) { return c1.abs() == c2.abs(); } bool operator!=(const Complex& c1, const Complex& c2) { return c1.abs() != c2.abs(); } bool operator>(const Complex& c1, const Complex& c2) { return c1.abs() > c2.abs(); } bool operator>=(const Complex& c1, const Complex& c2) { return c1.abs() >= c2.abs(); }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Complex::Complex() { a = 0;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang b = 0; } Complex::Complex(double a, double b) { this->a = a; this->b = b; } Complex::Complex(double a) { this->a = a; this->b = 0; } double Complex::getRealPart() const { return a; } double Complex::getImaginaryPart() const { return b; } Complex Complex::add(const Complex& secondComplex) const { double newA = a + secondComplex.getRealPart(); double newB = b + secondComplex.getImaginaryPart(); return Complex(newA, newB); } Complex Complex::subtract(const Complex& secondComplex) const { double newA = a - secondComplex.getRealPart(); double newB = b - secondComplex.getImaginaryPart(); return Complex(newA, newB); } Complex Complex::multiply(const Complex& secondComplex) const { double newA = a * secondComplex.getRealPart() - b * secondComplex.getImaginaryPart(); double newB = b * secondComplex.getRealPart() + a * secondComplex.getImaginaryPart(); return Complex(newA, newB); } Complex Complex::divide(const Complex& secondComplex) const { double newA = (a * secondComplex.getRealPart() + b * secondComplex.getImaginaryPart()) / (pow(secondComplex.getRealPart(), 2.0) + pow(secondComplex.getImaginaryPart(), 2.0)); double newB = (b * secondComplex.getRealPart() - a * secondComplex.getImaginaryPart()) / (pow(secondComplex.getRealPart(), 2.0) + pow(secondComplex.getImaginaryPart(), 2.0));

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return Complex(newA, newB); }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang double Complex::abs() const { return sqrt(a * a + b * b); } /* string Complex::toString() const { stringstream ss; ss << a; if (b != 0) ss << " + " << b << "i"; return ss.str(); } */ string Complex::toString() const { string result = to_string(a); // to_string is a new function in C++ 11 if (b != 0) result += " + " + to_string(b) + "i"; return result; } Complex& Complex::operator+=(Complex& secondComplex) { *this = this->add(secondComplex); return *this; } Complex& Complex::operator-=(Complex& secondComplex) { *this = this->subtract(secondComplex); return *this; } Complex& Complex::operator*=(Complex& secondComplex) { *this = this->multiply(secondComplex); return *this; } Complex& Complex::operator/=(Complex& secondComplex) { *this = this->divide(secondComplex); return *this; } double& Complex::operator[](const int& index) { if (index == 0) return a; else if (index == 1) return b; else { cout << "subscript error" << endl; exit(0);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Complex& Complex::operator++() // Prefix ++ { a += 1; return *this; } Complex& Complex::operator--() // Prefix -{ a -= 1; return *this; } Complex Complex::operator++(int dummy) // Postfix ++ { Complex temp(a, b); // Complex temp(*this) a += 1; return temp; } Complex Complex::operator--(int dummy) // Postfix -{ Complex temp(a, b); a -= 1; return temp; } Complex Complex::operator+() // Unary + { return *this; } Complex Complex::operator-() // Unary { a *= -1; b *= -1; return *this; } ostream& operator<<(ostream& out, const Complex& complex) { if (complex.b == 0) out << complex.a; else { if (complex.a == 0) { if (complex.b == 1) out << complex.a << "i"; else if (complex.b == -1) out << complex.a << "-i"; else out << complex.b << "i"; } else { if (complex.b == 1) out << complex.a << " + i"; else if (complex.b == -1)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang out << complex.a << " - i"; else if (complex.b < 0) out << complex.a << " - " << -complex.b << "i";

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang else out << complex.a << " + " << complex.b << "i"; } } return out; } istream& operator >> (istream& in, Complex& complex) { cout << "Enter a and b for (a + bi): "; in >> complex.a; in >> complex.b; return in; } int main() { Complex number1; cout << "Enter the first complex number: "; cin >> number1; Complex number2; cout << "Enter the second complex number: "; cin >> number2; cout << "(" << number1 << ")" << " + " << "(" << number2 << ") = " << (number1 + number2) << endl; cout << "(" << number1 << ")" << " - " << "(" << number2 << ") = " << (number1 - number2) << endl; cout << "(" << number1 << ")" << " * " << "(" << number2 << ") = " << (number1 * number2) << endl; cout << "(" << number1 << ")" << " / " << "(" << number2 << ") = " << (number1 / number2) << endl; cout << "|" << number1 << "|" << " = " << number1.abs() << endl; cout << "|" << number2 << "|" << " = " << number2.abs() << endl; number1[0] = 3.4; cout << number1++ << endl; cout << ++number2 << endl; cout << (3 + number2) << endl; cout << (number2 += number1) << endl; cout << (number2 *= number1) << endl; number1 = number2; cout << number1 << endl; cout << number2 << endl; cout << (number1 < number2) << endl; cout << (number1 <= number2) << endl; cout << (number1 == number2) << endl; cout << (number1 != number2) << endl; cout << (number1 > number2) << endl; cout << (number1 >= number2) << endl; return 0;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } Chapter 14: Programming Project 2: (345021-00169)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

(Algebra: vertex form equations) The description of this project is given in Programming Exercise 14.11 in the Chapter 14 Programming Exercise from the Book section.

Use the code from https://liangcpp.pearsoncmg.com/test/Exercise14_11.txt to complete your program.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #ifndef RATIONALWITHOPERATORS_H #define RATIONALWITHOPERATORS_H #include <string> #include <iostream> using namespace std; class Rational { public: Rational(); Rational(int numerator, int denominator); int getNumerator() const; int getDenominator() const; Rational add(const Rational& secondRational) const; Rational subtract(const Rational& secondRational) const; Rational multiply(const Rational& secondRational) const; Rational divide(const Rational& secondRational) const; int compareTo(const Rational& secondRational) const; bool equals(const Rational& secondRational) const; int intValue() const; double doubleValue() const; string toString() const; Rational(int numerator); // Suitable for type conversion // Define function operators for augmented operators Rational& operator+=(const Rational& secondRational); Rational& operator-=(const Rational& secondRational); Rational& operator*=(const Rational& secondRational); Rational& operator/=(const Rational& secondRational); // Define function operator [] int& operator[](int index); // Define function operators for prefix ++ and -Rational& operator++(); Rational& operator--(); // Define function operators for postfix ++ and -Rational operator++(int dummy); Rational operator--(int dummy); // Define function operators for unary + and Rational operator+(); Rational operator-();

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Define the << and >> operators

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang friend ostream& operator<<(ostream&, const Rational&); friend istream& operator>>(istream&, Rational&); private: int numerator; int denominator; static int gcd(int n, int d); }; // Define nonmember function operators for relational operators bool operator<(const Rational& r1, const Rational& r2); bool operator<=(const Rational& r1, const Rational& r2); bool operator>(const Rational& r1, const Rational& r2); bool operator>=(const Rational& r1, const Rational& r2); bool operator==(const Rational& r1, const Rational& r2); bool operator!=(const Rational& r1, const Rational& r2); // Define nonmember function operators for arithmetic operators Rational operator+(const Rational& r1, const Rational& r2); Rational operator-(const Rational& r1, const Rational& r2); Rational operator*(const Rational& r1, const Rational& r2); Rational operator/(const Rational& r1, const Rational& r2); #endif #include <cstdlib> // For the abs function Rational::Rational() { numerator = 0; denominator = 1; } Rational::Rational(int numerator, int denominator) { int factor = gcd(numerator, denominator); this->numerator = (denominator > 0 ? 1 : -1) * numerator / factor; this->denominator = abs(denominator) / factor; } int Rational::getNumerator() const { return numerator; } int Rational::getDenominator() const { return denominator; } // Find GCD of two numbers int Rational::gcd(int n, int d) { int n1 = abs(n); int n2 = abs(d); int gcd = 1;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang for (int k = 1; k <= n1 && k <= n2; k++) {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang if (n1 % k == 0 && n2 % k == 0) gcd = k; } return gcd; } Rational Rational::add(const Rational& secondRational) const { int n = numerator * secondRational.getDenominator() + denominator * secondRational.getNumerator(); int d = denominator * secondRational.getDenominator(); return Rational(n, d); } Rational Rational::subtract(const Rational& secondRational) const { int n = numerator * secondRational.getDenominator() - denominator * secondRational.getNumerator(); int d = denominator * secondRational.getDenominator(); return Rational(n, d); } Rational Rational::multiply(const Rational& secondRational) const { int n = numerator * secondRational.getNumerator(); int d = denominator * secondRational.getDenominator(); return Rational(n, d); } Rational Rational::divide(const Rational& secondRational) const { int n = numerator * secondRational.getDenominator(); int d = denominator * secondRational.numerator; return Rational(n, d); } int Rational::compareTo(const Rational& secondRational) const { Rational temp = subtract(secondRational); if (temp.getNumerator() < 0) return -1; else if (temp.getNumerator() == 0) return 0; else return 1; } bool Rational::equals(const Rational& secondRational) const { if (compareTo(secondRational) == 0) return true; else return false; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang int Rational::intValue() const { return getNumerator() / getDenominator();

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } double Rational::doubleValue() const { return 1.0 * getNumerator() / getDenominator(); } string Rational::toString() const { if (denominator == 1) return to_string(numerator); // See Ch7 for to_string else return to_string(numerator) + "/" + to_string(denominator); } Rational::Rational(int numerator) // Suitable for type conversion { this->numerator = numerator; this->denominator = 1; } // Define function operators for augmented operators Rational& Rational::operator+=(const Rational& secondRational) { *this = add(secondRational); return *this; } Rational& Rational::operator-=(const Rational& secondRational) { *this = subtract(secondRational); return *this; } Rational& Rational::operator*=(const Rational& secondRational) { *this = multiply(secondRational); return *this; } Rational& Rational::operator/=(const Rational& secondRational) { *this = divide(secondRational); return *this; } // Define function operator [] int& Rational::operator[](int index) { if (index == 0) return numerator; else return denominator; } // Define function operators for prefix ++ and --

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Rational& Rational::operator++() { numerator += denominator;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return *this; } Rational& Rational::operator--() { numerator -= denominator; return *this; } // Define function operators for postfix ++ and -Rational Rational::operator++(int dummy) { Rational temp(numerator, denominator); numerator += denominator; return temp; } Rational Rational::operator--(int dummy) { Rational temp(numerator, denominator); numerator -= denominator; return temp; } // Define function operators for unary + and Rational Rational::operator+() { return *this; } Rational Rational::operator-() { return Rational(-numerator, denominator); } // Define the output and input operator ostream& operator<<(ostream& out, const Rational& rational) { if (rational.denominator == 1) out << rational.numerator; else out << rational.numerator << "/" << rational.denominator; return out; } istream& operator>>(istream& in, Rational& rational) { cout << "Enter numerator: "; in >> rational.numerator; cout << "Enter denominator: "; in >> rational.denominator; return in; } // Define function operators for relational operators

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang bool operator<(const Rational& r1, const Rational& r2) { return (r1.compareTo(r2) < 0);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } bool operator<=(const Rational& r1, const Rational& r2) { return (r1.compareTo(r2) <= 0); } bool operator>(const Rational& r1, const Rational& r2) { return (r1.compareTo(r2) > 0); } bool operator>=(const Rational& r1, const Rational& r2) { return (r1.compareTo(r2) >= 0); } bool operator==(const Rational& r1, const Rational& r2) { return (r1.compareTo(r2) == 0); } bool operator!=(const Rational& r1, const Rational& r2) { return (r1.compareTo(r2) != 0); } // Define non-member function operators for arithmetic operators Rational operator+(const Rational& r1, const Rational& r2) { return r1.add(r2); } Rational operator-(const Rational& r1, const Rational& r2) { return r1.subtract(r2); } Rational operator*(const Rational& r1, const Rational& r2) { return r1.multiply(r2); } Rational operator/(const Rational& r1, const Rational& r2) { return r1.divide(r2); } int main() { cout << "Enter a, b, c: "; int a, b, c; cin >> a >> b >> c; Rational h(-b, 2 * a); Rational k(4 * a * c - b * b, 4 * a);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "h is " << h << " k is " << k << endl;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return 0; } Chapter 15 Programming Quiz 15.2: Question 1: (00000-11206) Assume the existence of a BankAccount class.

Define a derived class SavingsAccount that contains two instance variables: the first a double, named interestRate, and the second an int named interestType. The value of the type variable can be 1 for simple interest and 2 for compound interest.

There is also a constructor that accepts two parameters: a double that is used to initialize the interestRate variable, and a string that you may assume will contain either "Simple", or "Compound", and which should be used to initialize the type variable appropriately.

There should also be a pair of functions getInterestRate and getInterestType that return the values of the corresponding data members (as double and int respectively).

Note: You need to define the SavingsAccount class and also implement the SavingsAccount’s constructor and the getInterestRate and getInterestType functions. class SavingsAccount : public BankAccount { public: SavingsAccount(double interestRate, string interestTypeStr); double getInterestRate(); int getInterestType(); private: double interestRate; int interestType; }; SavingsAccount::SavingsAccount(double interestRate, string interestTypeStr): interestRate(interestRate) { if (interestTypeStr == "Simple") interestType = 1; else interestType = 2; } double SavingsAccount::getInterestRate() { return interestRate; } int SavingsAccount::getInterestType() { return interestType; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Programming Quiz 15.2: Question 2: (00000-11207) Assume the existence of a Window class with a function getWidth that returns the width of the window.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Define a derived class WindowWithBorder that contains a single additional integer instance variable named borderWidth and a constructor that accepts an integer parameter used to initialize the instance variable.

There is also a function getUseableWidth that returns the width of the window minus the width of the border.

Note: You need to define the WindowWithBorder class and also implement the WindowWithBorder’s constructor and the getUseableWidth function. class WindowWithBorder: public Window { public: WindowWithBorder(int borderWidth); int getUseableWidth(); private: int borderWidth; }; WindowWithBorder::WindowWithBorder(int borderWidth) { this->borderWidth = borderWidth; } int WindowWithBorder::getUseableWidth() { return getWidth() - borderWidth; } Programming Quiz 15.2: Question 3: (00000-11208) Assume the existence of a Phone class.

Define a derived class CameraPhone that contains two data members: an int named imageSize, representing the size in megabytes of each picture, and an int named memorySize, representing the number of megabytes in the camera's memory.

There is a constructor that accepts two int parameters corresponding to the above two data members and which are used to initialize the respective data members. There is also a function named numPictures that returns (as an int) the number of pictures the camera's memory can hold.

Note: You need to define the CameraPhone class and also implement the CameraPhone’s constructor and the numPictures function. class CameraPhone: public Phone { public: CameraPhone(int imageSize, int memorySize); int numPictures(); private: int imageSize; int memorySize; };

Solutions Manual for Introduction to C++ Programming and Data CameraPhone::CameraPhone(int imageSize, int memorySize): Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang imageSize(imageSize), memorySize(memorySize) { } int CameraPhone::numPictures() { return memorySize / imageSize; } Chapter 15: Programming Project 1: (345021-00048) (Extend MyPoint) In Programming Exercise 9.4, the MyPoint class was created to model a point in a two-dimensional plane. The MyPoint class has the properties x and y that represent x- and y-coordinates, two getter functions for x and y, and the function for returning the distance between two points.

Define a class named ThreeDPoint to model a point in a three-dimensional space.

Let ThreeDPoint be derived from MyPoint with the following additional features: - A data field named z that represents the z-coordinate.

- A no-arg constructor that constructs a point with coordinates (0, 0, 0).

- A constructor that constructs a point with three specified coordinates.

- A constant getter function that returns the z value.

- A constant distance(const MyPoint&) function to return the distance between this point and the other point in the three-dimensional space.

Implement the class. Write a test program that creates two points (0, 0, 0) and (10, 30, 25.5) and displays the distance between them.

Use the code from https://liangcpp.pearsoncmg.com/test/Exercise15_03.txt to complete your program.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <cmath> using namespace std; class MyPoint { private: double x; double y; public: MyPoint() { x = y = 0; } MyPoint(double x, double y) { this->x = x; this->y = y; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition MyPoint&

by Daniel double distance(const p2)Liang const

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { return sqrt((x - p2.x) * (x - p2.x) + (y - p2.y) * (y - p2.y)); } double getX() const { return x; } double getY() const { return y; } }; class ThreeDPoint : public MyPoint { private: double z; public: ThreeDPoint(): MyPoint(0, 0), z(0) { } ThreeDPoint(double x, double y, double z): MyPoint(x, y) { this->z = z; } double getZ() const { return z; } double distance(const ThreeDPoint& p2) const { return sqrt((getX() - p2.getX()) * (getX() - p2.getX()) + (getY() - p2.getY()) * (getY() - p2.getY()) + (getZ() - p2.getZ()) * (getZ() - p2.getZ())); } }; int main() { cout << "distance is " << ThreeDPoint().distance(ThreeDPoint(10, 30, 25.5)) << endl; return 0; }

Chapter 16 Programming Quiz 16.2: Question 1: (00000-11217) Assume the existence of a class RangeException with a constructor that accepts minimum,

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang maximum and violating integer values (in that order).

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write a function, void verify(int min, int max), which reads in integers from the console input and compares them against its two parameters. Use the following template to read the input from the console repeatedly until the end. void verify(int min, int max) { int value; // (cin >> value) reads an input into value and // it returns false when the input ends. // See the last paragraph in Section 13.2.4. while (cin >> value) { // Verify value } }

As long as the numbers are between min and max (inclusively), the function continues to read in values.

If an input value is encountered that is less than min or greater than max, the function throws a RangeException with the min and max values, and the violating (i.e. out of range) input. void verify(int min, int max) { int value; while (cin >> value) { if (value < min || value > max) throw RangeException(min, max, value); } } Programming Quiz 16.2: Question 2:

(00000-11214)

Write a function double average(int arr[], int n) that returns the average of the even elements of the array. If the array has no even elements, throw 0. double average(int arr[], int n) { int total = 0, count = 0; for (int i = 0; i < n; i++) if (arr[i] % 2 == 0) { total += arr[i]; count++; } if (count == 0) throw count; return static_cast<double>(total) / count; } Chapter 16: Programming Project 1: (345021-00056) (invalid_argument )

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang LiveExample 6.18 gives the hex2Dec(const string& hexString) function that returns a decimal number from a hex string. Implement the hex2Dec function to throw an invalid_argument exception if the string is not a hex string.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write a test program that prompts the user to enter a hex number as a string and display the number in decimal. If the function throws an exception, display "Not a hex number". You need to write the code to catch the exception in the main function.

Use the code from https://liangcpp.pearsoncmg.com/test/Exercise16_01.txt to complete your program.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include<iostream> #include<stdexcept> #include<string> #include<cctype> using namespace std; int hex2Dec(const string& hexString); int convertHexToDec(char ch); int hex2Dec(const string& hexString) { int value = 0; for (unsigned int i = 0; i < hexString.size(); i++) value = value * 16 + convertHexToDec(toupper(hexString[i])); return value; } int convertHexToDec(char ch) { if (ch >= 'A' && ch <= 'F') return 10 + (ch - 'A'); else if (ch <= '9' && ch >= '0') return ch - '0'; else throw invalid_argument("Not a hex number"); } int main() { string hexString; try { cout << "Enter a hex number: "; cin >> hexString; cout << hex2Dec(hexString) << endl; } catch(invalid_argument& ex) { cout << ex.what() << endl; } return 0; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Chapter 16: Programming Project 2: (345021-00057) (HexFormatException)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Implement the hex2Dec function in Programming Exercise 16.1 to throw a HexFormatException if the string is not a hex string.

Define a custom exception class named HexFormatException. Write a test program that prompts the user to enter a hex number as a string and displays the number in decimal. If the function throws an exception, display "Not a hex number". You need to write the code to catch the exception in the main function.

Use the code from https://liangcpp.pearsoncmg.com/test/Exercise16_05.txt to complete your program.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <string> #include <stdexcept> #include <typeinfo> using namespace std; class HexFormatException : public runtime_error { public: HexFormatException(char ch) : runtime_error("Not a hex number") { this->ch = ch; } char getCh() { return ch; } private: char ch; }; int convertHexToDec(char ch) { if (ch == 'A') { return 10; } else if (ch == 'B') { return 11; } else if (ch == 'C') { return 12; } else if (ch == 'D') { return 13; } else if (ch == 'E') Solutions Manual for Introduction to C++ Programming and Data { Structures 5th edition by Daniel Liang return 14; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang else if (ch == 'F') { return 15; } else if (ch >= '0' && ch <= '9') return ch - '0'; else throw HexFormatException(ch); } int hex2Dec(const string& hexString) { int value = convertHexToDec(hexString[0]); for (int i = 1; i < hexString.size(); i++) { value = value * 16 + convertHexToDec(hexString[i]); } return value; } int main() { string hexString; try { cout << "Enter a hex number: "; cin >> hexString; cout << hex2Dec(hexString) << endl; } catch (runtime_error& ex) { cout << ex.what() << endl; } return 0; } Chapter 17 Programming Quiz 17.2: Question 1: (00000-10724) Write a function called printStars. The function receives an int parameter. If the parameter is positive, the function prints (to console output) the given number of asterisks; otherwise the function does nothing. The function does not return a value.

Thus, if the printStars(8) is called, ******** (8 asterisks) will be printed.

The function must not use a loop of any kind (for, while, do-while) to accomplish its job. Instead, it gets the job done by examining its parameter, returning if the parameter's value is not positive.

If the parameter is positive, it •

prints a single asterisk (and no other characters)

•

then (recursively) calls itself to print the remaining asterisks.

void printStars(int n) Introduction to C++ Programming and Data Solutions Manual for { Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang if (n > 0) { cout << '*'; printStars(n - 1); } } Programming Quiz 17.2: Question 2: (00000-10762) The sum of the numbers from 1 to n can be performed recursively as follows: •

The sum from 1 to 1 is 1.

•

The sum from 1 to n is n more than the sum from 1 to n-1.

Write an int function named sum that accepts an int parameter, n, and recursively calculates and returns the sum of the numbers from 1 to n. int sum(int n) { if (n == 1) return 1; else return n + sum(n - 1); } Programming Quiz 17.2: Question 3: (00000-10763) Given non-negative integers x and n, x taken to the nth power can be defined as: •

x to the 0th power is 1

•

x to the nth power can be obtained by multiplying x to the n-1th power with x

Write a long valued function named power that accepts two int parameters x and n (in that order) and recursively calculates and returns the value of x taken to the nth power. long power(int x, int n) { if (n == 0) return 1; else return x * power(x, n-1); } Programming Quiz 17.2: Question 4: (00000-10766) Write a recursive, bool-returned function, containsVowel, that accepts a string and returns true if the string contains a vowel. A character is a vowel if it is A/a, E/e, I/I, O/o, U/u.

A string contains a vowel if: •

The first character of the string is a vowel, or

•

The rest of the string (beyond the first character) contains a vowel

bool containsVowel(string s) { Solutions for false; Introduction to C++ Programming and Data if (s == Manual "") return Structures 5th edition by Daniel Liang if (string("aeiouAEIOU").find(s[0]) != string::npos)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return true; return containsVowel(s.substr(1)); } Programming Quiz 17.2: Question 5: (00000-10767) Write a recursive, string-returned function, replace, that accepts a string and returns a new string consisting of the original string with each blank replaced with an asterisk (*) Replacing the blanks in a string involves: •

Nothing if the string is empty

•

Otherwise: o

If the first character is not a blank, simply concatenate it with the result of replacing the rest of the string

o

If the first character is a blank, concatenate an * with the result of replacing the rest of the string

string replace(string s) { if (s == "") return ""; if (s[0] == ' ') return '*' + replace(s.substr(1)); return s[0] + replace(s.substr(1)); } Programming Quiz 17.3: Question 1: (00000-10837) Assume the availability of a function named printStars that can be passed a non-negative integer n and print a line of asterisks.

Write a function named printTriangle that receives a non-negative integer n and prints a triangle of asterisks as follows: first a line of n asterisks, followed by a line of n-1 asterisks, and then a line of n-2 asterisks, and so on. You need to separate lines using cout << endl or cout << "\n".

For example, if the function received 5 it would print: * * * * * * * * * * * * * * * The function must not use a loop of any kind (for, while, do-while) to accomplish its job.

The function should invoke printStars to accomplish the task of printing a single line. void printTriangle(int n) { if (n > 0) { printStars(n); Solutions Manual cout << "\n"; for Introduction to C++ Programming and Data Structures 5th edition printTriangle(n - 1);by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } } Programming Quiz 17.3: Question 2: (00000-10835) The nth harmonic number is defined non-recursively as: 1 +1/2 + 1/3 + 1/4 + ... + 1/n.

Come up with a recursive definition and use it to guide you to write a function definition for a doublereturned function named harmonic that accepts an int parameter n and recursively calculates and returns the nth harmonic number. double harmonic(int n) { return n == 1 ? 1.0 : 1.0 / n + harmonic(n - 1); } Programming Quiz 17.3: Question 3: (00000-10836) The "odd/even factorial" of a positive integer n is represented as n!! and is defined non-recursively as: (n)(n-2)(n-4)...(4)(2) if n is even and is (n)(n-2)(n-4)...(5)(3)(1) if n is odd.

For example 7!! equals 7*5*3*1 or 105 and 6!! equals 6*4*2 or 48.

Come up with a recursive definition for n!! and use it to guide you to write a function definition for a function called oddevenfact that recursively calculates the odd/even factorial value of its single int parameter.

The value returned by oddevenfact is a long. long oddevenfact(int n) { return n < 3 ? n : n * oddevenfact(n - 2); } Programming Quiz 17.3: Question 4: (00000-10765) Write a recursive int-returned function len that accepts a string and returns the number of characters in the string.

The length of a string is: •

0 if the string is the empty string ("").

•

1 more than the length of the rest of the string beyond the first character.

int len(string s) { if (s == "") return 0; return 1 + len(s.substr(1)); } Programming Quiz 17.3: Question 5: (00000-10771) The sum of the elements of an int valued array recursively calculated as follows:

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang •

The sum of an array of size 0 is 0;

•

Otherwise, the sum is the value of the first element added to the sum of the rest of the array.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write an int valued function named sum that accepts an int array, and the number of elements in the array and returns the sum of the elements of the array. int sum(int* a, int n) { if (n == 0) return 0; return a[0] + sum(a + 1, n - 1); } Chapter 17: Proramming Project 1:

(345021-00049)

(Compute greatest common divisor using recursion) The gcd(m, n) can also be defined recursively as follows: If m % n is 0, gcd (m, n) is n. Otherwise, gcd(m, n) is gcd(n, m % n).

Write a recursive function to find the GCD. Write a test program that prompts the user to enter two integers and displays their GCD. For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise17_03 using namespace std; int gcd(int m, int n) { if (m % n == 0) return n; else return gcd(n, m % n); } int main() { // Enter the first number cout << "Enter the first number: "; int m; cin >> m; // Enter the first number cout << "Enter the second number: "; int n; cin >> n; cout << "The GCD of " << m << " and " << n << " is " << gcd(m, n) << endl; return 0; }

Chapter 17: Proramming Project 2:

(345021-00050)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang (Sum the digits in an integer using recursion)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Write a recursive function that computes the sum of the digits in an integer. Use the following function header: int sumDigits(int n) For example, sumDigits(234) returns 2 + 3 + 4 = 9.

Write a test program that prompts the user to enter an integer and displays its sum.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise17_11 using namespace std; int sumDigits(long n) { int result = 0; if (n != 0) { result = sumDigits(n / 10) + (int)(n % 10); } return result; } int main() { cout << "Ente an integer: "; int n; cin >> n; cout << "The sum of digits in " << n << " is " << sumDigits(n) << endl; return 0; } Chapter 17: Proramming Project 3:

(345021-00051)

(Occurrences of a specified character in a string) Rewrite Programming Exercise 17.10 using a helper function to pass the substring high index to the function.

You need to define the following two functions. The second one is a recursive helper function. int count(const string& s, char a) int count(const string& s, char a, int high) Write a test program that prompts the user to enter a string and a character, and displays the number of occurrences of the character in the string.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> Solutions Manual for Introduction to C++ Programming and Data #include <string> Structures 5th edition by Daniel Liang using namespace std;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang int count(const string& s, char ch, int high) { if (high >= 0) return (s[high] == ch ? 1 : 0) + count(s, ch, high - 1); else return 0; } int count(const string& s, char ch) { return count(s, ch, s.size() - 1); } int main() { cout << "Enter a string: "; string s; getline(cin, s); cout << "Enter a character: "; char ch; cin >> ch; int times = count(s, ch); cout << ch << " appears " << times << (times > 1 ? " times" : " time") << " in " << s << endl; return 0; } Chapter 17: Proramming Project 4:

(345021-00052)

(Decimal to binary) Write a recursive function that converts a decimal number into a binary number as a string.

The function header is: string decimalToBinary(int value) Write a test program that prompts the user to enter a decimal number and displays its binary equivalent.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise17_21 #include <string> #include <sstream> using namespace std; string decimalToBinary(int value) { if (value == 0) return ""; else { Solutions for Introduction C++ Programming return Manual decimalToBinary(value / 2)to + (value % 2 == 0 ? and "0" :Data "1"); Structures 5th edition by Daniel Liang }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } int main() { cout << "Enter a decimal integer: "; int decimal; cin >> decimal; cout << decimal << " is binary " << decimalToBinary(decimal) << endl; return 0; }

Chapter 17: Proramming Project 5:

(345021-00053)

(Binary to decimal) Write a recursive function that parses a binary number as a string into a decimal integer.

The function header is: int binaryToDecimal(const string& binaryString) Write a test program that prompts the user to enter a binary string and displays its decimal equivalent.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <string> using namespace std; int binaryToDecimal(const string& binaryString, int low, int high) { if (high < low) return 0; else return binaryToDecimal(binaryString, low, high - 1) * 2 + (binaryString[high] - '0'); } int binaryToDecimal(const string& binaryString) { return binaryToDecimal(binaryString, 0, binaryString.length() - 1); } int main() { cout << "Enter a binary number: "; string binary; cin >> binary; cout << binary << " is decimal " << binaryToDecimal(binary) << endl; return 0; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Chapter 18 Chapter 18: Proramming Project 1: (345021-00058)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang (Maximum consecutive increasingly ordered substring) Write a program that prompts the user to enter a string and displays the maximum consecutive increasingly ordered substring.

Sample Run 1 Enter a string: abcabcdgabxy The maximum consecutive increasing ordered substring is abcdg Sample Run 2 Enter a string: abcabcdgabmnsxy The maximum consecutive increasing ordered substring is abmnsxy For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <string> #include <vector> using namespace std; string maxConsecutiveSortedSubstring(const string& s) { vector<int> maxConsecutiveLength(s.length()); for (int i = 0; i < s.length(); i++) { maxConsecutiveLength[i] = 0; } int current = 0; for (int i = 1; i < s.length(); i++) { if (s.at(i) <= s.at(i - 1)) // next character is smaller { current = i; } else { // s[current], s[current+1], ..., s[i-1] is a mazimal increase sequenece for (int j = i - 1; j >= current; j--) maxConsecutiveLength[j]++; } } // maxConsecutiveLength[i] denotes the length of a max sequeuece starting at index i int currentMaxLength = maxConsecutiveLength[0]; int index = 0; for (int i = 0; i < s.length(); i++) { if (maxConsecutiveLength[i] > currentMaxLength) { currentMaxLength = maxConsecutiveLength[i]; Solutions Manual for Introduction to C++ Programming and Data index = i; Structures 5th edition by Daniel Liang }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } return s.substr(index, currentMaxLength + 1); } int main() { cout << "Enter a string: "; string s; getline(cin, s); cout << "The maximum consecutive increasing ordered substring is " << maxConsecutiveSortedSubstring(s) << endl; return 0; } Chapter 18: Proramming Project 2:

(345021-00059)

(Pattern matching) Write a program that prompts the user to enter two strings and tests whether the second string is a substring in the first string. Suppose the characters in the second string are distinct. Your algorithm needs to be in O(n) time.

Sample Run Enter a string s1: Welcome to C++ Enter a string s2: come matched at index 3 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <string> using namespace std; /** The worst-case complexity is O(n), where n is s.length() @param s String @param pattern String @return int **/ int match(const string s, const string pattern) { int k = 0; for (int i = 0; i < s.length(); i++) { if (k == pattern.length()) return i - pattern.length(); else { if (s.at(i) == pattern.at(k)) k++; else k = 0; Solutions Manual for Introduction to C++ Programming and Data } Structures 5th edition by Daniel Liang }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang if (k == pattern.length()) return s.length() - pattern.length(); else return -1; } int main() { cout << "Enter a string s1: "; string s1; getline(cin, s1); cout << "Enter a string s2: "; string s2; getline(cin, s2); int index = match(s1, s2); if (index > 0) { cout << "Matched at index " << index << endl; } else { cout << "No match" << endl; } return 0; } Chapter 18: Proramming Project 3: (345021-00060) (Same number subsequence) Write an O(n) time program that prompts the user to enter a sequence of integers ending with 0 and finds longest subsequence with the same number.

Sample Run Enter a series of numbers ending with 0: 2 4 4 8 8 8 8 2 4 4 0 The longest same number sequence starts at index 3 with 4 values of 8 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> using namespace std; int main() { cout << "Enter a series of numbers ending with 0: "; int longestSequenceCount = 0; int longestSequenceValue = 0; int longestSequenceIndex = 0; int currentLongestSequenceCount = 1;

Solutions Manual for Introduction to C++ Programming and Data int currentLongestSequenceValue = 0; Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang int value; int index = 0; int previous = 0; do { cin >> value; if (value == previous) currentLongestSequenceCount++; else if (currentLongestSequenceCount > longestSequenceCount) { longestSequenceCount = currentLongestSequenceCount; longestSequenceValue = currentLongestSequenceValue; longestSequenceIndex = index - currentLongestSequenceCount; currentLongestSequenceCount = 1; } currentLongestSequenceValue = value; previous = value; index++; } while (value != 0); cout << "The longest same number sequence starts at index " << longestSequenceIndex << " with " << longestSequenceCount << " values of " << longestSequenceValue << endl; return 0; } Chapter 19 Chapter 19: Proramming Project 1: (345021-00061) (Generic quick sort) Write a generic function for quick sort using following header: template<typename T> void quickSort(T list[], int arraySize); Write a test program that prompts the user to read 10 integers, applies the quick sort, and displays the sorted numbers.

Sample Run Enter 10 integers: 2 4 4 8 8 8 8 2 4 4 0 The sorted array is 0 2 2 4 4 4 4 8 8 8 8 For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise19_03 using namespace std; // Function prototypes template<typename T> Introduction to C++ Programming and Data Solutions Manual for void quickSort(T list[], intDaniel arraySize); Structures 5th edition by Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename T> void quickSort(T list[], int first, int last); template<typename T> int partition(T list[], int first, int last); template<typename T> void quickSort(T list[], int arraySize) { quickSort(list, 0, arraySize - 1); } template<typename T> void quickSort(T list[], int first, int last) { if (last > first) { int pivotIndex = partition(list, first, last); quickSort(list, first, pivotIndex - 1); quickSort(list, pivotIndex + 1, last); } } /* Partition the array list[first..last] */ template<typename T> int partition(T list[], int first, int last) { T pivot = list[first]; // Choose the first element as the pivot int low = first + 1; // Index for forward search int high = last; // Index for backward search while (high > low) { // Search forward from left while (low <= high && list[low] <= pivot) low++; // Search backward from right while (low <= high && list[high] > pivot) high--; // Swap two elements in the list if (high > low) { T temp = list[high]; list[high] = list[low]; list[low] = temp; } } while (high > first && list[high] >= pivot) high--; // Swap pivot with list[high] if (pivot > list[high]) {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang list[first] = list[high]; list[high] = pivot; return high;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } else { return first; } } template<typename T> void printArray(T list[], int arraySize) { for (int i = 0; i < arraySize; i++) { cout << list[i] << " "; } cout << endl; } int main() { int list1[10]; cout << "Enter 10 integers: "; for (unsigned i = 0; i < 10; i++) cin >> list1[i]; quickSort(list1, 10); printArray(list1, 10); return 0; }

Chapter 20 Chapter 20: Proramming Project 1:

(345021-00062)

(Add set-like operations in LinkedList) Add and implement the following functions in LinkedList: // Add the elements in otherList to this list. void addAll(const LinkedList<T>& otherList) // Remove all the elements in otherList from this list void removeAll(const LinkedList<T>& otherList) // Retain the elements in this list if they are also in otherList void retainAll(const LinkedList<T>& otherList) Add three function operators: +, -, and ^ for set union, difference, and intersection. Overload the = operator to perform a deep copy of a list. Add the [] operator for accessing/modifying an element.

Use the code from https://liangcpp.pearsoncmg.com/test/Exercise20_07.txt to complete your program.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang #ifndef LINKEDLIST_H #define LINKEDLIST_H #include <stdexcept> using std::logic_error; using std::out_of_range; using std::iterator; using std::forward_iterator_tag; template<typename T> struct Node { T element; // Element contained in the node Node<T>* next; // Pointer to the next node Node(T element) : Node(element, nullptr) { } // Constructor Node(T element, Node<T>* next) : element(element), next(next) { } // Constructor }; template<typename T> class Iterator : public iterator<forward_iterator_tag, T> { Node<T>* current; public: Iterator(Node<T>* p) : current(p) { } Iterator operator++() { current = current->next; return *this; } T& operator*() const { return current->element; } bool operator==(const Iterator<T>& iterator) const { return current == iterator.current; } bool operator!=(const Iterator<T>& iterator) const { return current != iterator.current; } }; template<typename T> class LinkedList { Node<T>* head = nullptr, * tail = nullptr; int size = 0; public:

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang LinkedList() = default; LinkedList(const LinkedList<T>& list);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang virtual ~LinkedList(); void addFirst(T element); void addLast(T element); T getFirst() const; T getLast() const; T removeFirst(); T removeLast(); void add(T element); void add(int index, T element); void clear(); bool contains(T element) const; T get(int index) const; int indexOf(T element) const; bool isEmpty() const; int lastIndexOf(T element) const; void remove(T element); int getSize() const; T removeAt(int index); T set(int index, T element); Iterator<T> begin() const; Iterator<T> end() const; void addAll(const LinkedList<T>&); void removeAll(const LinkedList<T>&); void retainAll(const LinkedList<T>&); LinkedList<T>& operator=(const LinkedList<T>&); LinkedList<T> operator+(const LinkedList<T>&) const; LinkedList<T> operator-(const LinkedList<T>&) const; LinkedList<T> operator^(const LinkedList<T>&) const; T& operator[](int) const; }; template<typename T> LinkedList<T>::LinkedList(const LinkedList<T>& list) { for (auto iter = list.begin(); iter != list.end(); ++iter) add(*iter); } template<typename T> LinkedList<T>::~LinkedList() { clear(); } template<typename T> void LinkedList<T>::addFirst(T element) { head = new Node<T>(element, head); if (isEmpty()) tail = head; size++; } template<typename T>

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang void LinkedList<T>::addLast(T element) { if (isEmpty())

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang head = tail = new Node<T>(element); else tail = tail->next = new Node<T>(element); ++size; } template<typename T> T LinkedList<T>::getFirst() const { if (isEmpty()) throw logic_error("No elements in the list"); return head->element; } template<typename T> T LinkedList<T>::getLast() const { if (isEmpty()) throw logic_error("No elements in the list"); return tail->element; } template<typename T> void LinkedList<T>::add(T element) { addLast(element); } template<typename T> void LinkedList<T>::add(int index, T element) { if (index < 0 || index > size) throw out_of_range("Insertion index out of range"); else if (index == size) { tail = tail->next = new Node<T>(element); } else { Node<T>** indir = &head; while (--index >= 0) indir = &(*indir)->next; // indir is &head iff index was 0, and &beforeNode.next otherwise. *indir = new Node<T>(element, *indir); } size++; } template<typename T> void LinkedList<T>::clear() { tail = nullptr; while (head != nullptr)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { Node<T>* temp = head; head = head->next;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang delete temp; } size = 0; } template<typename T> T LinkedList<T>::get(int index) const { if (index < 0 || index >= size) throw out_of_range("Index out of range"); auto iter = begin(); while (--index >= 0) ++iter; return *iter; } template<typename T> int LinkedList<T>::indexOf(T element) const { int idx = 0; for (auto iter = begin(); iter != end(); ++iter, ++idx) { if (*iter == element) return idx; } return -1; } template<typename T> bool LinkedList<T>::isEmpty() const { return size == 0; } template<typename T> int LinkedList<T>::getSize() const { return size; } template<typename T> T LinkedList<T>::removeFirst() { if (isEmpty()) throw logic_error("No elements in the list"); Node<T>* temp = head; head = head->next; if (size == 1) tail = nullptr; --size; T element = move(temp->element); delete temp; return element; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename T> T LinkedList<T>::removeLast()

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { if (isEmpty()) throw logic_error("No elements in the list"); Node<T>* temp = tail; if (size == 1) head = tail = nullptr; else { Node<T>* before = head; while (before->next != tail) before = before->next; tail = before; tail->next = nullptr; } --size; T element = move(temp->element); delete temp; return element; } template<typename T> T LinkedList<T>::removeAt(int index) { if (index == size - 1) return removeLast(); else if (index < 0 || index >= size) throw out_of_range("Index out of range"); else { Node<T>** indir = &head; while (--index >= 0) indir = &(*indir)->next; // indir is &head iff index was 0, and &beforeNode.next otherwise.

Node<T>* temp = *indir; *indir = &(*indir)->next; --size; T element = move(temp->element); delete temp; return element; } } template<typename T> Iterator<T> LinkedList<T>::begin() const { return Iterator<T>(head); }; template<typename T> Iterator<T> LinkedList<T>::end() const { return Iterator<T>(nullptr); };

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // The functions remove(T element), lastIndexOf(T element), // contains(T element), and set(int index, T element) are // left as an exercise

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename T> bool LinkedList<T>::contains(T element) const { for (auto iter = begin(); iter != end(); ++iter) { if (*iter == element) return true; } return false; } template<typename T> int LinkedList<T>::lastIndexOf(T element) const { int idx = 0, lastIdx = -1; for (auto iter = begin(); iter != end(); ++idx, ++iter) { if (*iter == element) lastIdx = idx; } return lastIdx; } template<typename T> void LinkedList<T>::remove(T element) { if (isEmpty()) ; else if (head->element == element) removeFirst(); else if (tail->element == element) removeLast(); else { Node<T>* indir = &head; do { indir = &(*indir)->next; if (*indir == tail) return; // Element not found. } while ((*indir)->element != element); // indir is &beforeNode.next.

Node<T>* temp = *indir; *indir = &(*indir)->next; --size; delete temp; } } template<typename T> T LinkedList<T>::set(int index, T element) { return *this[index] = element; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename T> void LinkedList<T>::addAll(const LinkedList<T>& other)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { for (auto iter = other.begin(); iter != other.end(); ++iter) add(*iter); } template<typename T> void LinkedList<T>::removeAll(const LinkedList<T>& other) { Node<T>* prev = nullptr; Node<T>** indir = &head; while (*indir != nullptr) { if (other.contains((*indir)->element)) { Node<T>* temp = *indir; if (tail == *indir) tail = prev; *indir = (*indir)->next; --size; delete temp; } else { prev = *indir; indir = &(*indir)->next; } } } template<typename T> void LinkedList<T>::retainAll(const LinkedList<T>& other) { // Identical to removeAll except condition inverted.

Node<T>* prev = nullptr; Node<T>** indir = &head; while (*indir != nullptr) { if (!other.contains((*indir)->element)) { Node<T>* temp = *indir; if (tail == *indir) tail = prev; *indir = (*indir)->next; --size; delete temp; } else { prev = *indir; indir = &(*indir)->next; } } } template<typename T> LinkedList<T>& LinkedList<T>::operator=(const LinkedList<T>& other)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { clear(); addAll(other);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return *this; } template<typename T> LinkedList<T> LinkedList<T>::operator+(const LinkedList<T>& other) const { LinkedList<T> uni(*this); uni.addAll(other); return uni; } template<typename T> LinkedList<T> LinkedList<T>::operator-(const LinkedList<T>& other) const { // If *this has no duplicates, the result will have no duplicates (setlike).

LinkedList<T> diff(*this); diff.removeAll(other); return diff; } template<typename T> LinkedList<T> LinkedList<T>::operator^(const LinkedList<T>& other) const { // If *this has no duplicates, the result will have no duplicates (setlike).

LinkedList<T> intersect; for (auto iter = other.begin(); iter != other.end(); ++iter) if (contains(*iter)) intersect.add(*iter); return intersect; } template<typename T> T& LinkedList<T>::operator[](int index) const { if (index < 0 || index >= size) throw out_of_range("Index out of range"); auto iter = begin(); while (--index >= 0) ++iter; return *iter; } #endif // Contents of exercise20.7.cpp /** * Class: CSCI2490 C++ Programming * Instructor: Y. Daniel Liang */ #include <iostream> #include <string> //#include "LinkedList.h"

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang using namespace std; template<typename T>

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang void printList(const LinkedList<T>& list) { Iterator<T> current = list.begin(); while (current != list.end()) { cout << *current << " "; ++current; } cout << endl; } int main() { // Create a list for strings LinkedList<string> list; cout << "Enter five lines of strings: "; for (unsigned i = 0; i < 5; i++) { string s; getline(cin, s); list.add(s); } /* list.add("Beijing"); list.add("Tokyo"); list.add("New York"); list.add("London"); list.add("Paris");*/ // Create a list for strings LinkedList<string> list2; cout << "Enter five lines of strings: "; for (unsigned i = 0; i < 5; i++) { string s; getline(cin, s); list2.add(s); } /* list2.add("Beijing"); list2.add("Shanghai"); list2.add("Paris"); list2.add("Berlin"); list2.add("Rome");*/ LinkedList<string> list1(list); cout << "list1: "; printList(list1); cout << "list2: "; printList(list2); cout << "After list1.addAll(list2), list1 is "; printList(list1);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang list1 = list; cout << "list1: "; printList(list1);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "list2: "; printList(list2); list1.removeAll(list2); cout << "After list1.removeAll(list2), list1 is "; printList(list1); list1 = list; cout << "list1: "; printList(list1); cout << "list2: "; printList(list2); list1.retainAll(list2); cout << "After list1.retainAll(list2), list1 is "; printList(list1); /* list1 = list; cout << "list1: "; printList(list1); cout << "list2: "; printList(list2); list1 = list1 + list2; cout << "After list1 = list1 + list2, list1 is "; printList(list1); list1 = list; cout << "list1: "; printList(list1); cout << "list2: "; printList(list2); list1 = list1 - list2; cout << "After list1 = list1 - list2, list1 is "; printList(list1); list1 = list; cout << "list1: "; printList(list1); cout << "list2: "; list1 = list1 ^ list2; cout << "After list1 = list1 ^ list2, list1 is "; printList(list1); list1 = list; cout << list1[0] << endl; list1[0] = list[2]; cout << list1[0] << endl; */ return 0; } Chapter 20 Chapter 20: Proramming Project 2: (345021-00155) (Implement ArrayList)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang The book introduces linked lists and gave an implementation for linked lists. Implement ArrayList for performing list operations. The elements in an ArrayList are stored in an array. The array is dynamically

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang created. If the actual size of the elements in an ArrayList exceeds the array capacity, create a new array that is twice as large as the current array and copy the contents of the current array into this new array. The UML diagram for the ArrayList class is shown in Figure 20.19.

Use the code from https://liangcpp.pearsoncmg.com/test/Exercise20_11.txt to complete your code.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #ifndef ArrayList_H #define ArrayList_H #include <stdexcept> #include <iostream> #include <string> using namespace std; template<typename T> class Iterator : public std::iterator<std::forward_iterator_tag, T> { public: Iterator(int index, T* list) { this->list = list; cursor = index; } Iterator operator++() // Prefix ++s { cursor++; return *this; } Iterator operator++(int dummy) // Postfix ++ { Iterator temp(*this); cursor++; return temp; } T& operator*() { return list[cursor]; } bool operator==(const Iterator<T>& iterator) { return list[cursor] == list[iterator.cursor]; } bool operator!=(const Iterator<T>& iterator) { return list[cursor] != list[iterator.cursor]; } private: int cursor; T* list; Manual for Introduction to C++ Programming and Data Solutions }; Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename T> class ArrayList { public: ArrayList(); ArrayList(ArrayList<T>& arr_list); ~ArrayList(); void add(T element); void add(int index, T element); void clear(); bool contains(T element) const; T get(int index) const; int indexOf(T element) const; bool isEmpty() const; int lastIndexOf(T element) const; void remove(T element); int getSize() const; T removeAt(int index); T set(int index, T element); T& operator[](int index); Iterator<T> begin() const { return Iterator<T>(0, list); } Iterator<T> end() const { return Iterator<T>(size, list); } private: T* list; int capacity; int size; void ensureCapacity(); }; template<typename T> ArrayList<T>::ArrayList() : size(0), capacity(4) { list = new T[capacity]; } template<typename T> ArrayList<T>::ArrayList(ArrayList<T>& arr_list) { list = new T[arr_list.capacity]; size = arr_list.size; capacity = arr_list.capacity; for (int i = 0; i < size; i++) { list[i] = arr_list.list[i]; } } template<typename T>

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang ArrayList<T>::~ArrayList() { delete[] list;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } template<typename T> void ArrayList<T>::add(T element) { ensureCapacity(); list[size++] = element; } template<typename T> void ArrayList<T>::add(int index, T element) { ensureCapacity(); if (index >= size) { add(element); } else { for (int i = size - 1; i >= index; i--) list[i + 1] = list[i]; list[index] = element; size++; } } template<typename T> void ArrayList<T>::ensureCapacity() { if (size >= capacity) { T* old = list; capacity = 2 * size + 1; list = new T[size * 2]; for (int i = 0; i < size; i++) list[i] = old[i]; delete[] old; } } template<typename T> void ArrayList<T>::clear() { delete[] list; list = new T[capacity]; size = 0; } template<typename T> bool ArrayList<T>::contains(T element) const { for (int i = 0; i < list.size; list++) { if (list[i].equals(element)) return true;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } return false; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename T> T ArrayList<T>::get(int index) const { if (index < 0 || index >= size) throw runtime_error("Index out of range"); else return list[index]; } template<typename T> int ArrayList<T>::indexOf(T element) const { for (int i = 0; i < size; i++) { if (list[i].equals(element)) return i; } return -1; } template<typename T> bool ArrayList<T>::isEmpty() const { return size <= 0; } template<typename T> int ArrayList<T>::lastIndexOf(T element) const { int index = -1; for (int i = 0; i < size; i++) { if (list[i].equals(element)) index = i; } return index; } template<typename T> void ArrayList<T>::remove(T element) { int index = -1; for (int i = 0; i < size; i++) { if (list[i] == element) { index = i; break; } } if (index > -1) { for (int i = index; i < size; i++) list[i] = list[i + 1]; size--; } }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename T> int ArrayList<T>::getSize() const {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return size; } template<typename T> T ArrayList<T>::removeAt(int index) { if (index < 0 || index >= size) throw runtime_error("Index out of range"); else { T temp = list[index]; for (int i = index; i < size; i++) list[i] = list[i + 1]; size--; return temp; } } template<typename T> T ArrayList<T>::set(int index, T element) { if (index < 0 || index >= size) throw runtime_error("Index out of range"); else { T temp = list[index]; list[index] = element; return temp; } } template<typename T> T& ArrayList<T>::operator[](int index) { if (index < 0 || index >= size) throw runtime_error("Index out of range"); else return list[index]; } #endif void printList(const ArrayList<string>& list) { for (int i = 0; i < list.getSize(); i++) { cout << list.get(i) << " "; } cout << endl; } int main() { // Create a list for strings ArrayList<string> list;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Add elements to the list list.add("America"); // Add it to the list

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "(1) "; printList(list); list.add(0, "Canada"); // Add it to the beginning of the list cout << "(2) "; printList(list); list.add("Russia"); // Add it to the end of the list cout << "(3) "; printList(list); list.add("France"); // Add it to the end of the list cout << "(4) "; printList(list); list.add(2, "Germany"); // Add it to the list at index 2 cout << "(5) "; printList(list); list.add(5, "Norway"); // Add it to the list at index 5 cout << "(6) "; printList(list); cout << "Enter a new string: "; string s; getline(cin, s); list.add(0, s); // Same as list.addFirst(s) cout << "(7) "; printList(list); // Remove elements from the list int index; cout << "Enter an index: "; cin >> index; list.removeAt(index); // Same as list.remove("Netherlands ") in this case cout << "(8) "; printList(list); list.removeAt(2); // Remove the element at index 2 cout << "(9) "; printList(list); list.removeAt(list.getSize() - 1); // Remove the last element cout << "(10) "; printList(list); cout << list[2] << endl; list[2] = "Spain"; // Traverse a list using iterators for (string& s : list) { cout << s << " "; } auto p = list.begin();

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << endl << "The first element: " << *(p++) << endl; cout << "The third element: " << *(++p) << endl;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return 0; }

Chapter 21 Chapter 21: Proramming Project 1:

(345021-00156)

(Height of BST) Add an instance function in the BST class to return the height of the tree. // Return the height of this binary tree. Height of an empty tree is -1. // The height of a non-empty tree is the length of the path from the // root to its furthest leaf. int height() Use the code from https://liangcpp.pearsoncmg.com/test/Exercise21_03.txt to complete your program.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <string> #include <algorithm> using namespace std; #ifndef BST_H #define BST_H template<typename T> class TreeNode { public: T element; // Element contained in the node TreeNode<T>* left; // Pointer to the left child TreeNode<T>* right; // Pointer to the right child TreeNode() // No-arg constructor { left = NULL; next = NULL; } TreeNode(T element) // Constructor { this->element = element; left = NULL; right = NULL; } }; template<typename T> class BST { public: BST(); Solutions Manual for Introduction to C++ Programming and Data BST(T elements[], int arraySize); Structures 5th edition bool insert(T element);by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang void inorder(); void preorder(); void postorder(); int getSize(); bool search(T element); void breadthFirstTraversal(); int height(); private: TreeNode<T>* root; int size; void inorder(TreeNode<T>* root); void postorder(TreeNode<T>* root); void preorder(TreeNode<T>* root); bool search(T element, TreeNode<T>* root); int height(TreeNode<T>* root); }; template<typename T> BST<T>::BST() { root = NULL; size = 0; } template<typename T> BST<T>::BST(T elements[], int arraySize) { root = NULL; size = 0; for (int i = 0; i < arraySize; i++) { insert(elements[i]); } } /* Insert element into the binary tree * Return true if the element is inserted successfully * Return false if the element is already in the list */ template<typename T> bool BST<T>::insert(T element) { if (root == NULL) root = new TreeNode<T>(element); // Create a new root else { // Locate the parent node TreeNode<T>* parent = NULL; TreeNode<T>* current = root; while (current != NULL) if (element < current->element) { parent = current; current = current->left; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang else if (element > current->element) { parent = current;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang current = current->right; } else return false; // Duplicate node not inserted // Create the new node and attach it to the parent node if (element < parent->element) parent->left = new TreeNode<T>(element); else parent->right = new TreeNode<T>(element); } size++; return true; // Element inserted } /* Inorder traversal */ template<typename T> void BST<T>::inorder() { inorder(root); } /* Inorder traversal from a subtree */ template<typename T> void BST<T>::inorder(TreeNode<T>* root) { if (root == NULL) return; inorder(root->left); cout << root->element << " "; inorder(root->right); } /* Postorder traversal */ template<typename T> void BST<T>::postorder() { postorder(root); } /** Inorder traversal from a subtree */ template<typename T> void BST<T>::postorder(TreeNode<T>* root) { if (root == NULL) return; postorder(root->left); postorder(root->right); cout << root->element << " "; } /* Preorder traversal */

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename T> void BST<T>::preorder() {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang preorder(root); } /* Preorder traversal from a subtree */ template<typename T> void BST<T>::preorder(TreeNode<T>* root) { if (root == NULL) return; cout << root->element << " "; preorder(root->left); preorder(root->right); } /* Get the number of nodes in the tree */ template<typename T> int BST<T>::getSize() { return size; } template<typename T> int BST<T>::height() { return height(root); } template<typename T> int BST<T>::height(TreeNode<T>* root) { if (root == NULL) return -1; else return 1 + max(height(root->right), height(root->left)); } #endif int main() { BST<string> tree; cout << "The height of an empty tree is " << tree.height() << endl; cout << "Enter six strings: "; for (int i = 0; i < 6; i++) { string s; cin >> s; tree.insert(s); } cout << "\nThe height of the tree is " << tree.height() << endl; tree.insert("Green"); cout << "The height of the tree is " << tree.height() << endl; tree.preorder(); // Traverses the tree

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang string names[] = { "Tom", "George", "Jean", "Jane", "Kevin", "Peter", "Susan",

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang "Jen", "Kim", "Michael", "Michelle" }; BST<string> tree1(names, 11); cout << "\nThe preorder traversal of tree1 is "; tree1.preorder(); cout << "\nThe height of tree1 is " << tree1.height() << endl; int numbers[] = { 50, 45, 35, 48, 59, 51, 58 }; BST<int> tree2(numbers, 7); cout << "The preorder traversal for tree2 is "; tree2.preorder(); cout << "\nThe height of tree2 is " << tree2.height() << endl; return 0; }

Chapter 21 Chapter 21: Proramming Project 2:

(345021-00157)

(Find the nonleaves) Add an instance function in the BST class to return the number of the nonleaves as follows: // Returns the number of non-leaf nodes int getNumberofNonLeaves() Use the code from https://liangcpp.pearsoncmg.com/test/Exercise21_09.txt to complete your program.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <string> using namespace std; #ifndef BST_H #define BST_H #include <vector> #include <stdexcept> using namespace std; template<typename T> class TreeNode { public: T element; // Element contained in the node TreeNode<T>* left; // Pointer to the left child TreeNode<T>* right; // Pointer to the right child TreeNode(const T& e) // Constructor { this->element = e; left = nullptr; right = nullptr; }

Solutions Manual for Introduction to C++ Programming and Data }; Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template <typename T> class Iterator : public std::iterator<std::forward_iterator_tag, T> { public: Iterator(TreeNode<T>* p) { if (p == nullptr) current = -1; // The end else { // Get all the elements in inorder treeToVector(p); current = 0; } } Iterator operator++() { current++; if (current == v.size()) current = -1; // The end return *this; } T& operator*() { return v[current]; } bool operator==(const Iterator<T>& iterator) const { return current == iterator.current; } bool operator!=(const Iterator<T>& iterator) const { return current != iterator.current; } private: int current; vector<T> v; void treeToVector(const TreeNode<T>* p) { if (p != nullptr) { treeToVector(p->left); v.push_back(p->element); treeToVector(p->right); } } }; template <typename T> class BST {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang public: BST(); BST(const T elements[], int arraySize);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang BST(const BST<T>& tree); ~BST(); bool search(const T& e) const; virtual bool insert(const T& e); virtual bool remove(const T& e); void inorder() const; void preorder() const; void postorder() const; int getSize() const; void clear(); vector<TreeNode<T>*>* path(const T& e) const; int getNumberOfNonLeaves(); Iterator<T> begin() const { return Iterator<T>(root); }; Iterator<T> end() const { return Iterator<T>(nullptr); }; protected: TreeNode<T>* root; int size; virtual TreeNode<T>* createNewNode(const T& e); private: void inorder(const TreeNode<T>* root) const; void postorder(const TreeNode<T>* root) const; void preorder(const TreeNode<T>* root) const; void copy(const TreeNode<T>* root); void clear(const TreeNode<T>* root); int getNumberOfNonLeaves(TreeNode<T>* root); }; template <typename T> BST<T>::BST() { root = nullptr; size = 0; } template <typename T> BST<T>::BST(const T elements[], int arraySize) { root = nullptr; size = 0; for (int i = 0; i < arraySize; i++) { insert(elements[i]); } }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang /* Copy constructor */ template <typename T> BST<T>::BST(const BST<T>& tree)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { root = nullptr; size = 0; copy(tree.root); // Recursively copy nodes to this tree } /* Copies the element from the specified tree to this tree */ template <typename T> void BST<T>::copy(const TreeNode<T>* root) { if (root != nullptr) { insert(root->element); copy(root->left); copy(root->right); } } /* Destructor */ template <typename T> BST<T>::~BST() { clear(); } /* Return true if the element is in the tree */ template <typename T> bool BST<T>::search(const T& e) const { TreeNode<T>* current = root; // Start from the root while (current != nullptr) if (e < current->element) { current = current->left; // Go left } else if (e > current->element) { current = current->right; // Go right } else // Element e matches current.element return true; // Element e is found return false; // Element e is not in the tree } template <typename T> TreeNode<T>* BST<T>::createNewNode(const T& e) { return new TreeNode<T>(e); } // Insert element e into the binary tree // Return true if the element is inserted successfully // Return false if the element is already in the list template <typename T>

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang bool BST<T>::insert(const T& e) { if (root == nullptr)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang root = createNewNode(e); // Create a new root else { // Locate the parent node TreeNode<T>* parent = nullptr; TreeNode<T>* current = root; while (current != nullptr) if (e < current->element) { parent = current; current = current->left; } else if (e > current->element) { parent = current; current = current->right; } else return false; // Duplicate node not inserted // Create the new node and attach it to the parent node if (e < parent->element) parent->left = createNewNode(e); else parent->right = createNewNode(e); } size++; return true; // Element inserted } /* Inorder traversal */ template <typename T> void BST<T>::inorder() const { inorder(root); } /* Inorder traversal from a subtree */ template <typename T> void BST<T>::inorder(const TreeNode<T>* root) const { if (root == nullptr) return; inorder(root->left); cout << root->element << " "; inorder(root->right); } /* Postorder traversal */ template <typename T> void BST<T>::postorder() const { postorder(root); } /** Inorder traversal from a subtree */

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template <typename T> void BST<T>::postorder(const TreeNode<T>* root) const {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang if (root == nullptr) return; postorder(root->left); postorder(root->right); cout << root->element << " "; } /* Preorder traversal */ template <typename T> void BST<T>::preorder() const { preorder(root); } /* Preorder traversal from a subtree */ template <typename T> void BST<T>::preorder(const TreeNode<T>* root) const { if (root == nullptr) return; cout << root->element << " "; preorder(root->left); preorder(root->right); } /* Get the number of nodes in the tree */ template <typename T> int BST<T>::getSize() const { return size; } /* Remove all nodes from the tree */ template <typename T> void BST<T>::clear() { // Left as exercise } /* Return a path from the root leading to the specified element */ template <typename T> vector<TreeNode<T>*>* BST<T>::path(const T& e) const { vector<TreeNode<T>*>* v = new vector<TreeNode<T>*>(); TreeNode<T>* current = root; while (current != nullptr) { v->push_back(current); if (e < current->element) current = current->left; else if (e > current->element) current = current->right; else break; } return v;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } /* Delete an element e from the binary tree.

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang * Return true if the element is deleted successfully * Return false if the element is not in the tree */ template <typename T> bool BST<T>::remove(const T& e) { // Locate the node to be deleted and also locate its parent node TreeNode<T>* parent = nullptr; TreeNode<T>* current = root; while (current != nullptr) { if (e < current->element) { parent = current; current = current->left; } else if (e > current->element) { parent = current; current = current->right; } else break; // Element e is in the tree pointed by current } if (current == nullptr) return false; // Element e is not in the tree // Case 1: current has no left children if (current->left == nullptr) { // Connect the parent with the right child of the current node if (parent == nullptr) { root = current->right; } else { if (e < parent->element) parent->left = current->right; else parent->right = current->right; } delete current; // Delete current } else { // Case 2: The current node has a left child // Locate the rightmost node in the left subtree of // the current node and also its parent TreeNode<T>* parentOfRightMost = current; TreeNode<T>* rightMost = current->left; while (rightMost->right != nullptr) { parentOfRightMost = rightMost;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang rightMost = rightMost->right; // Keep going to the right }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Replace the element in current by the element in rightMost current->element = rightMost->element; // Eliminate rightmost node if (parentOfRightMost->right == rightMost) parentOfRightMost->right = rightMost->left; else // Special case: parentOfRightMost->right == current parentOfRightMost->left = rightMost->left; delete rightMost; // Delete rightMost } size--; return true; // Element inserted } template<typename T> int BST<T>::getNumberOfNonLeaves() { return getNumberOfNonLeaves(root); } template<typename T> int BST<T>::getNumberOfNonLeaves(TreeNode<T>* root) { if (root == NULL) return 0; else if (root->right == NULL && root->left == NULL) return 0; else return 1 + getNumberOfNonLeaves(root->right) + getNumberOfNonLeaves(root->left); } #endif int main() { BST<string> tree1; cout << "Enter six strings: "; for (int i = 0; i < 6; i++) { string s; cin >> s; tree1.insert(s); } tree1.insert("Michael"); tree1.insert("George"); tree1.insert("Tom"); cout << "Inorder (sorted): "; tree1.inorder(); cout << "\nPostorder: "; tree1.postorder();

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "\nPreorder: "; tree1.preorder();

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "\nThe number of nodes is " << tree1.getSize(); int numbers[] = { -8, -9, 2, 4, 3, 1, 8, 5, 6, 7 }; BST < int > tree2(numbers, 10); cout << "\nPreorder (sorted): "; tree2.preorder(); cout << "\nNumber of non-leaves nodes in tree1: " << tree1.getNumberOfNonLeaves() << endl; cout << "Number of non-leaves nodes in tree2: " << tree2.getNumberOfNonLeaves() << endl; return 0; }

Chapter 22 Chapter 22: Proramming Project 1: (345021-00158) (Maximum and minimum) Implement the following functions that find the maximum and minimum elements in a first-class container: template<typename ElementType, typename ContainerType> ElementType maxElement(const ContainerType& container) template<typename ElementType, typename ContainerType> ElementType minElement(const ContainerType& container) Use the code from https://liangcpp.pearsoncmg.com/test/Exercise22_01.txt to complete your program.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <vector> #include <string> using namespace std; template<typename ElementType, typename ContainerType> ElementType maxElement(const ContainerType& container) { ElementType max = container[0]; for (int i = 0; i < container.size(); i++) if (max < container[i]) max = container[i]; return max; }

Solutions Manual for Introduction to ContainerType> C++ Programming and Data template<typename ElementType, typename Structures edition by Daniel Liang container) ElementType 5th minElement(const ContainerType&

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { ElementType min = container[0]; for (int i = 0; i < container.size(); i++) if (min > container[i]) min = container[i]; return min; } int main() { vector<string> stringVector{ "red", "green", "blue", "yellow", "orange", "purple", "gray", "white" }; cout << "The maximum string is " << (maxElement<string, vector<string>>(stringVector)) << endl; vector<int> intVector(8); cout << "Enter eight integers: "; for (unsigned i = 0; i < 8; i++) cin >> intVector[i]; cout << "The maximum int is " << (maxElement<int, vector<int> >(intVector)) << endl; cout << "The minimum int is " << (minElement<int, vector<int> >(intVector)) << endl; return 0; }

Chapter 23 Chapter 23: Proramming Project 1: (345021-00159) (generate and generate_n functions) Implement the generate and generate_n functions. template<typename ForwardIterator, typename function> void generate(ForwardIterator beg, ForwardIterator end, function gen) template<typename ForwardIterator, typename size, typename function> void generate_n(ForwardIterator beg, size n, function gen) Use the code from https://liangcpp.pearsoncmg.com/test/Exercise23_09.txt to complete your program.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise23_09 #include <list> #include <iterator> using namespace std; template <typename ForwardIterator, typename function> void generate(ForwardIterator beg, ForwardIterator end, function gen); template <typename ForwardIterator, typename function> Solutions Manual for Introduction to C++ Programming and Data void generate(ForwardIterator beg, ForwardIterator end, function gen) Structures 5th edition by Daniel Liang {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang for (ForwardIterator p = beg; p != end; p++) *p = gen(); } template <typename ForwardIterator, typename size, typename function> void generate_n(ForwardIterator beg, size n, function gen); template <typename ForwardIterator, typename size, typename function> void generate_n(ForwardIterator beg, size n, function gen) { ForwardIterator p = beg; for (int i = 0; i < n; i++) { *p = gen(); p++; } } int nextNum() { static int n = 20; return n++; } int main() { int values[6] = {0}; cout << "Enter six integers: "; for (unsigned i = 0; i < 6; i++) cin >> values[i]; list<int> intList(values, values + 6); ostream_iterator<int> output(cout, " "); cout << "Initial contents, values: "; copy(values, values + 6, output); cout << "\nInitial contents, intList: "; copy(intList.begin(), intList.end(), output); generate(values + 2, values + 4, nextNum); generate_n(intList.begin(), 2, nextNum); cout << "\nAfter the generate function, values: "; copy(values, values + 6, output); cout << "\nAfter the generate_n function, intList: "; copy(intList.begin(), intList.end(), output); return 0; }

Chapter 23 Chapter 23: Proramming Project 2:

(345021-00160)

(Implement replace and replace_if functions)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Implement the replace and replace_if functions. template<typename ForwardIterator, typename T>

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang void replace(ForwardIterator beg, ForwardIterator end, const T& oldValue, const T& newValue) template<typename ForwardIterator, typename boolFunction, typename T> void replace_if(ForwardIterator beg, ForwardIterator end, boolFunction f, const T& newValue) Use the code from https://liangcpp.pearsoncmg.com/test/Exercise23_11.txt to complete your code.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <vector> #include <iterator> using namespace std; template <typename ForwardIterator, typename T> void replace(ForwardIterator beg, ForwardIterator end, const T &oldValue, const T &newValue); template <typename ForwardIterator, typename T> void replace(ForwardIterator beg, ForwardIterator end, const T &oldValue, const T &newValue) { for (ForwardIterator p = beg; p != end; p++) { if (*p == oldValue) { *p = newValue; } } } template <typename ForwardIterator, typename boolFunction, typename T> void replace_if(ForwardIterator beg, ForwardIterator end, boolFunction f, const T &newValue); template <typename ForwardIterator, typename boolFunction, typename T> void replace_if(ForwardIterator beg, ForwardIterator end, boolFunction f, const T &newValue) { for (ForwardIterator p = beg; p != end; p++) if (f(*p)) { *p = newValue; } } bool greaterThan4(double value) { return value > 4; } int main() { double values[6] = {0}; Solutions Manual for Introduction to C++ Programming and Data double values1[6] = {0}; Structures 5th edition

by Daniel cout << "Enter six integers: "; Liang for (unsigned i = 0; i < 6; i++)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { cin >> values[i]; values1[i] = values[i]; } ostream_iterator<double> output(cout, " "); cout << "Initial contents, values: "; copy(values, values + 6, output); replace(values, values + 6, values[3], 51.5); cout << "\nAfter the replace function, values: "; copy(values, values + 6, output); cout << "Initial contents, values1: "; copy(values1, values1 + 6, output); replace_if(values1, values1 + 6, greaterThan4, 555); cout << "\nAfter the replace function, values: "; copy(values1, values1 + 6, output); return 0; }

Chapter 24 Chapter 24: Proramming Project 1:

(345021-00161)

(Implement getKeys) Implement the getKeys function defined in LiveExample 24.1.

Use the code from https://liangcpp.pearsoncmg.com/test/Exercise24_09.txt to complete your code.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #ifndef MYMAP_H #define MYMAP_H #include <vector> #include <string> #include <sstream> #include <stdexcept> using namespace std; int DEFAULT_INITIAL_CAPACITY = 4; float DEFAULT_MAX_LOAD_FACTOR = 0.75f; unsigned MAXIMUM_CAPACITY = 1 << 30; template<typename K, typename V> class Entry // Represent an entry with key and value { public: Entry(const K& key, const V& value) { this->key = key; this->value = value; Solutions Manual for Introduction to C++ Programming and Data }

Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang string toString() const { stringstream ss; ss << "[" << key << ", " << value << "]"; return ss.str(); } K key; V value; }; template<typename K, typename V> class MyMap { public: MyMap(); MyMap(int initialCapacity); MyMap(int initialCapacity, float loadFactorThreshold); V put(const K& key, const V& value); V& get(const K& key) const; int getSize() const; bool isEmpty() const; vector<Entry<K, V>> getEntries() const; vector<K> getKeys() const; vector<V> getValues() const; string toString() const; bool containsKey(const K& key) const; bool containsValue(const V& value) const; void remove(const K& key); void clear(); private: int size; float loadFactorThreshold; int capacity; // Hash table is an array with each cell as a vector vector<Entry<K, V>>* table; int hash(int hashCode) const; unsigned hashCode(const K& key) const; int supplementalHash(int h) const; int trimToPowerOf2(int initialCapacity); void rehash(); void removeEntries(); }; template<typename K, typename V> MyMap<K, V>::MyMap() { capacity = DEFAULT_INITIAL_CAPACITY; table = new vector<Entry<K, V>>[capacity]; loadFactorThreshold = DEFAULT_MAX_LOAD_FACTOR; size = 0; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename K, typename V> MyMap<K, V>::MyMap(int initialCapacity)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { capacity = initialCapacity; table = new vector<Entry<K, V>>[capacity]; loadFactorThreshold = DEFAULT_MAX_LOAD_FACTOR; size = 0; } template<typename K, typename V> MyMap<K, V>::MyMap(int initialCapacity, float loadFactorThreshold) { if (initialCapacity > MAXIMUM_CAPACITY) capacity = MAXIMUM_CAPACITY; else capacity = trimToPowerOf2(initialCapacity); this->loadFactorThreshold = loadFactorThreshold; table = new vector<Entry<K, V>>[capacity]; size = 0; } template<typename K, typename V> V MyMap<K, V>::put(const K& key, const V& value) { if (containsKey(key)) { // The key is already in the map int bucketIndex = hash(hashCode(key)); for (Entry<K, V>& entry : table[bucketIndex]) { if (entry.key == key) { V oldValue = entry.value; // Replace old value with new value entry.value = value; // Return the old value for the key return oldValue; } } } // Check load factor if (size >= capacity * loadFactorThreshold) { if (capacity == MAXIMUM_CAPACITY) throw runtime_error("Exceeding maximum capacity"); rehash(); } int bucketIndex = hash(hashCode(key)); // Add a new entry (key, value) to hashTable[index] table[bucketIndex].push_back(Entry<K, V>(key, value)); size++; // Increase size return value;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } template<typename K, typename V>

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang V& MyMap<K, V>::get(const K& key) const { int bucketIndex = hash(hashCode(key)); for (Entry<K, V>& entry : table[bucketIndex]) if (entry.key == key) return entry.value; throw runtime_error("Key not found"); } template<typename K, typename V> bool MyMap<K, V>::isEmpty() const { return size == 0; } template<typename K, typename V> vector<Entry<K, V>> MyMap<K, V>::getEntries() const { vector<Entry<K, V>> v; for (int i = 0; i < capacity; i++) { for (Entry<K, V>& entry : table[i]) v.push_back(entry); } return v; } template<typename K, typename V> bool MyMap<K, V>::containsKey(const K& key) const { try { get(key); return true; } catch (runtime_error& ex) { return false; // Key not found } } template<typename K, typename V> bool MyMap<K, V>::containsValue(const V& value) const { for (int i = 0; i < capacity; i++) { for (Entry<K, V> entry : table[i]) if (entry.value == value) return true; } return false;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } template<typename K, typename V>

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang void MyMap<K, V>::remove(const K& key) { int bucketIndex = hash(hashCode(key)); // Remove the first entry that matches the key from a bucket if (table[bucketIndex].size() > 0) { for (auto p = table[bucketIndex].begin(); p != table[bucketIndex].end(); p++) if (p->key == key) { table[bucketIndex].erase(p); size--; // Decrease size break; // Remove just one entry that matches the key } } } template<typename K, typename V> void MyMap<K, V>::clear() { size = 0; removeEntries(); } template<typename K, typename V> void MyMap<K, V>::removeEntries() { for (int i = 0; i < capacity; i++) { table[i].clear(); } } template<typename K, typename V> vector<K> MyMap<K, V>::getKeys() const { vector<K> v; for (int i = 0; i < capacity; i++) { for (Entry<K, V>& e : table[i]) v.push_back(e.key); } return v; } template<typename K, typename V> vector<V> MyMap<K, V>::getValues() const { // Left as exercise } template<typename K, typename V> string MyMap<K, V>::toString() const

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { stringstream ss; ss << "[";

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang for (int i = 0; i < capacity; i++) { for (Entry<K, V>& entry : table[i]) ss << entry.toString(); } ss << "]"; return ss.str(); } template<typename K, typename V> unsigned MyMap<K, V>::hashCode(const K& key) const { return std::hash<K>()(key); } template<typename K, typename V> int MyMap<K, V>::hash(int hashCode) const { return supplementalHash(hashCode) & (capacity - 1); } template<typename K, typename V> int MyMap<K, V>::supplementalHash(int h) const { h ^= (h >> 20) ^ (h >> 12); return h ^ (h >> 7) ^ (h >> 4); } template<typename K, typename V> int MyMap<K, V>::trimToPowerOf2(int initialCapacity) { int capacity = 1; while (capacity < initialCapacity) { capacity <<= 1; } return capacity; } template<typename K, typename V> void MyMap<K, V>::rehash() { vector<Entry<K, V>> set = getEntries(); // Get entries capacity <<= 1; // Double capacity delete[] table; // Delete old hash table table = new vector<Entry<K, V>>[capacity]; // Create a new hash table size = 0; // Reset size to 0 for (Entry<K, V>& entry : set) put(entry.key, entry.value); // Store to new table } template<typename K, typename V> int MyMap<K, V>::getSize() const

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { return size; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang #endif #include <iostream> #include <string> using namespace std; int main() { // Create a map and add five name and age to the map MyMap<string, int> map; string name; int age; for (unsigned i = 0; i < 5; i++) { cout << "Enter name and age: "; cin >> name >> age; map.put(name, age); } vector<string> v = map.getKeys(); cout << "The keys are: " << endl; for (string& s : v) cout << s << " "; cout << endl; return 0; } Chapter 25 Chapter 25: Proramming Project 1:

(345021-00162)

(Test AVL tree) Define a new class named MyBST that extends the BST class with the following function: // Returns true if the tree is an AVL tree bool isAVLTree() Use the code from https://liangcpp.pearsoncmg.com/test/Exercise25_01Extra.txt to complete your code.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <algorithm> // For max and min in VS #include <cstdlib> // For max and min in GNU C++ using namespace std; #ifndef BST_H #define BST_H #include <vector> #include <stdexcept> template<typename T>

Solutions Manual for Introduction to C++ Programming and Data class TreeNode Structures 5th edition by Daniel Liang {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang public: T element; // Element contained in the node TreeNode<T>* left; // Pointer to the left child TreeNode<T>* right; // Pointer to the right child TreeNode(const T& e) // Constructor { this->element = e; left = nullptr; right = nullptr; } }; template <typename T> class Iterator: public std::iterator<std::forward_iterator_tag, T> { public: Iterator(TreeNode<T>* p) { if (p == nullptr) current = -1; // The end else { // Get all the elements in inorder treeToVector(p); current = 0; } } Iterator operator++() { current++; if (current == v.size()) current = -1; // The end return *this; } T &operator*() { return v[current]; } bool operator==(const Iterator<T>& iterator) const { return current == iterator.current; } bool operator!=(const Iterator<T>& iterator) const { return current != iterator.current; } private: int current; vector<T> v; void treeToVector(const TreeNode<T>* p)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { if (p != nullptr) {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang treeToVector(p->left); v.push_back(p->element); treeToVector(p->right); } } }; template <typename T> class BST { public: BST(); BST(const T elements[], int arraySize); BST(const BST<T>& tree); ~BST(); bool search(const T& e) const; virtual bool insert(const T& e); virtual bool remove(const T& e); void inorder() const; void preorder() const; void postorder() const; int getSize() const; void clear(); vector<TreeNode<T>*>* path(const T& e) const; Iterator<T> begin() const { return Iterator<T>(root); }; Iterator<T> end() const { return Iterator<T>(nullptr); }; //protected: TreeNode<T>* root; int size; virtual TreeNode<T>* createNewNode(const T& e); private: void inorder(const TreeNode<T>* root) const; void postorder(const TreeNode<T>* root) const; void preorder(const TreeNode<T>* root) const; void copy(const TreeNode<T>* root); void clear(const TreeNode<T>* root); }; template <typename T> BST<T>::BST() { root = nullptr; size = 0; } template <typename T>

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang BST<T>::BST(const T elements[], int arraySize) { root = nullptr;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang size = 0; for (int i = 0; i < arraySize; i++) { insert(elements[i]); } } /* Copy constructor */ template <typename T> BST<T>::BST(const BST<T>& tree) { root = nullptr; size = 0; copy(tree.root); // Recursively copy nodes to this tree } /* Copies the element from the specified tree to this tree */ template <typename T> void BST<T>::copy(const TreeNode<T>* root) { if (root != nullptr) { insert(root->element); copy(root->left); copy(root->right); } } /* Destructor */ template <typename T> BST<T>::~BST() { clear(); } /* Return true if the element is in the tree */ template <typename T> bool BST<T>::search(const T& e) const { TreeNode<T>* current = root; // Start from the root while (current != nullptr) if (e < current->element) { current = current->left; // Go left } else if (e > current->element) { current = current->right; // Go right } else // Element e matches current.element return true; // Element e is found return false; // Element e is not in the tree }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template <typename T> TreeNode<T>* BST<T>::createNewNode(const T& e)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { return new TreeNode<T>(e); } // Insert element e into the binary tree // Return true if the element is inserted successfully // Return false if the element is already in the list template <typename T> bool BST<T>::insert(const T& e) { if (root == nullptr) root = createNewNode(e); // Create a new root else { // Locate the parent node TreeNode<T>* parent = nullptr; TreeNode<T>* current = root; while (current != nullptr) if (e < current->element) { parent = current; current = current->left; } else if (e > current->element) { parent = current; current = current->right; } else return false; // Duplicate node not inserted // Create the new node and attach it to the parent node if (e < parent->element) parent->left = createNewNode(e); else parent->right = createNewNode(e); } size++; return true; // Element inserted } /* Inorder traversal */ template <typename T> void BST<T>::inorder() const { inorder(root); } /* Inorder traversal from a subtree */ template <typename T> void BST<T>::inorder(const TreeNode<T>* root) const { if (root == nullptr) return; inorder(root->left); cout << root->element << " ";

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang inorder(root->right); }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang /* Postorder traversal */ template <typename T> void BST<T>::postorder() const { postorder(root); } /** Inorder traversal from a subtree */ template <typename T> void BST<T>::postorder(const TreeNode<T>* root) const { if (root == nullptr) return; postorder(root->left); postorder(root->right); cout << root->element << " "; } /* Preorder traversal */ template <typename T> void BST<T>::preorder() const { preorder(root); } /* Preorder traversal from a subtree */ template <typename T> void BST<T>::preorder(const TreeNode<T>* root) const { if (root == nullptr) return; cout << root->element << " "; preorder(root->left); preorder(root->right); } /* Get the number of nodes in the tree */ template <typename T> int BST<T>::getSize() const { return size; } /* Remove all nodes from the tree */ template <typename T> void BST<T>::clear() { // Left as exercise } /* Return a path from the root leading to the specified element */ template <typename T> vector<TreeNode<T>*>* BST<T>::path(const T& e) const { vector<TreeNode<T>*>* v = new vector<TreeNode<T>*>(); TreeNode<T>* current = root; while (current != nullptr)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { v->push_back(current); if (e < current->element)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang current = current->left; else if (e > current->element) current = current->right; else break; } return v; } /* Delete an element e from the binary tree. * Return true if the element is deleted successfully * Return false if the element is not in the tree */ template <typename T> bool BST<T>::remove(const T& e) { // Locate the node to be deleted and also locate its parent node TreeNode<T>* parent = nullptr; TreeNode<T>* current = root; while (current != nullptr) { if (e < current->element) { parent = current; current = current->left; } else if (e > current->element) { parent = current; current = current->right; } else break; // Element e is in the tree pointed by current } if (current == nullptr) return false; // Element e is not in the tree // Case 1: current has no left children if (current->left == nullptr) { // Connect the parent with the right child of the current node if (parent == nullptr) { root = current->right; } else { if (e < parent->element) parent->left = current->right; else parent->right = current->right; } delete current; // Delete current }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang else { // Case 2: The current node has a left child

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Locate the rightmost node in the left subtree of // the current node and also its parent TreeNode<T>* parentOfRightMost = current; TreeNode<T>* rightMost = current->left; while (rightMost->right != nullptr) { parentOfRightMost = rightMost; rightMost = rightMost->right; // Keep going to the right } // Replace the element in current by the element in rightMost current->element = rightMost->element; // Eliminate rightmost node if (parentOfRightMost->right == rightMost) parentOfRightMost->right = rightMost->left; else // Special case: parentOfRightMost->right == current parentOfRightMost->left = rightMost->left; delete rightMost; // Delete rightMost } size--; return true; // Element inserted } #endif #include <iostream> using namespace std; template<typename T> class AVLTreeNode : public TreeNode<T> { public: int height; // height of the node AVLTreeNode(const T& e) : TreeNode<T>(e) // Constructor { height = 0; } }; template <typename T> class MyBST : public BST<T> { private: // Returns the height of this binary tree, i.e., the // number of the nodes in the longest path of the root to a leaf int height() { return height(this->root); }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang int height(TreeNode<T>* root) { if (root == nullptr)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return -1; else return 1 + max(height(root->left), height(root->right)); } public: // Returns true if the tree is an AVL tree bool isAVLTree() { return isAVLTree(this->root); } // Returns true if the tree is an AVL tree at the specified root bool isAVLTree(TreeNode<T>* root) { if (root == nullptr) return true; int height1 = height(root->right); int height2 = height(root->left); return isAVLTree(root->left) && isAVLTree(root->right) && abs(height1 - height2) <= 1; } }; int main() { MyBST<int> tree; cout << "How many integers to be inserted into the tree? "; int number; cin >> number; cout << "Enter " << number << " integers: "; int count = 0; while (count++ < number) { int e; cin >> e; tree.insert(e); } cout << "Is this tree an AVL tree? " << (tree.isAVLTree() ? "true": "false") << endl; return 0; } Chapter 26 Chapter 26: Proramming Project 1:

(345021-00163)

(Test whether a graph is connected) The description of this exercise is Chapter 26: Programming Project 1 from https://liangcpp.pearsoncmg.com/test/Extra.pdf.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #ifndef GRAPH_H

Solutions Manual for Introduction to C++ Programming and Data #define GRAPH_H Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang #include <vector> #include <queue> // For implementing BFS #include <stdexcept> #include <sstream> // For converting a number to a string #include <algorithm> #include <iostream> using namespace std; #ifndef TREE_H #define TREE_H class Tree { public: // Construct an empty tree Tree() { } // Construct a tree with root, parent, and searchOrder Tree(int root, const vector<int>& parent, const vector<int>& searchOrders) { this->root = root; this->parent = parent; this->searchOrders = searchOrders; } // Return the root of the tree int getRoot() const { return root; } // Return the parent of vertex v int getParent(int v) const { return parent[v]; } // Return search order vector<int> getSearchOrders() const { return searchOrders; } // Return number of vertices found int getNumberOfVerticesFound() const { return searchOrders.size(); } // Return the path of vertices from v to the root in a vector vector<int> getPath(int v) const { vector<int> path;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang do {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang path.push_back(v); v = parent[v]; } while (v != -1); return path; } // Print the whole tree void printTree() const { cout << "Root is: " << root << endl; cout << "Edges: "; for (unsigned i = 0; i < searchOrders.size(); i++) { if (parent[searchOrders[i]] != -1) { // Display an edge cout << "(" << parent[searchOrders[i]] << ", " << searchOrders[i] << ") "; } } cout << endl; } private: int root; // The root of the tree vector<int> parent; // Store the parent of each vertex vector<int> searchOrders; // Store the search order }; #endif #ifndef EDGE_H #define EDGE_H class Edge { public: int u; int v; Edge(int u, int v) { this->u = u; this->v = v; } }; #endif template<typename V> class Graph { public: // Construct an empty graph Graph(); // Construct a graph from vertices in a vector and

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // edges in 2-D array Graph(const vector<V>& vertices, const int edges[][2], int numberOfEdges);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Construct a graph with vertices 0, 1, ..., n-1 and // edges in 2-D array Graph(int numberOfVertices, const int edges[][2], int numberOfEdges); // Construct a graph from vertices and edges objects Graph(const vector<V>& vertices, const vector<Edge>& edges); // Construct a graph with vertices 0, 1, ..., n-1 and // edges in a vector Graph(int numberOfVertices, const vector<Edge>& edges); // Return the number of vertices in the graph int getSize() const; // Return the degree for a specified vertex int getDegree(int v) const; // Return the vertex for the specified index V getVertex(int index) const; // Return the index for the specified vertex int getIndex(V v) const; // Return the vertices in the graph vector<V> getVertices() const; // Return the neighbors of vertex v vector<int> getNeighbors(int v) const; // Print the edges void printEdges() const; // Print the adjacency matrix void printAdjacencyMatrix() const; // Clear the graph void clear(); // Adds a vertex to the graph virtual bool addVertex(const V& v); // Adds an edge from u to v to the graph bool addEdge(int u, int v); // Obtain a depth-first search tree // To be discussed in Section 24.6 Tree dfs(int v) const; // Starting bfs search from vertex v // To be discussed in Section 24.7 Tree bfs(int v) const; protected: vector<V> vertices; // Store vertices vector<vector<Edge*>> neighbors; // Adjacency edge lists bool createEdge(Edge* e); // Add an edge

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang private: // Create adjacency lists for each vertex from an edge array

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang void createAdjacencyLists(int numberOfVertices, const int edges[][2], int numberOfEdges); // Create adjacency lists for each vertex from an Edge vector void createAdjacencyLists(int numberOfVertices, const vector<Edge>& edges); // Recursive function for DFS search void dfs(int v, vector<int>& parent, vector<int>& searchOrders, vector<bool>& isVisited) const; }; template<typename V> Graph<V>::Graph() { } template<typename V> Graph<V>::Graph(const vector<V>& vertices, const int edges[][2], int numberOfEdges) { for (unsigned i = 0; i < vertices.size(); i++) addVertex(vertices[i]); createAdjacencyLists(vertices.size(), edges, numberOfEdges); } template<typename V> Graph<V>::Graph(int numberOfVertices, const int edges[][2], int numberOfEdges) { for (int i = 0; i < numberOfVertices; i++) addVertex(i); // vertices is {0, 1, 2, ..., n-1} createAdjacencyLists(numberOfVertices, edges, numberOfEdges); } template<typename V> Graph<V>::Graph(const vector<V>& vertices, const vector<Edge>& edges) { for (unsigned i = 0; i < vertices.size(); i++) addVertex(vertices[i]); createAdjacencyLists(vertices.size(), edges); } template<typename V> Graph<V>::Graph(int numberOfVertices, const vector<Edge>& edges) { for (int i = 0; i < numberOfVertices; i++) addVertex(i); // vertices is {0, 1, 2, ..., n-1} createAdjacencyLists(numberOfVertices, edges); } template<typename V>

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang void Graph<V>::createAdjacencyLists(int numberOfVertices, const int edges[][2], int numberOfEdges) {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang for (int i = 0; i < numberOfEdges; i++) { int u = edges[i][0]; int v = edges[i][1]; addEdge(u, v); } } template<typename V> void Graph<V>::createAdjacencyLists(int numberOfVertices, const vector<Edge>& edges) { for (unsigned i = 0; i < edges.size(); i++) { int u = edges[i].u; int v = edges[i].v; addEdge(u, v); } } template<typename V> int Graph<V>::getSize() const { return vertices.size(); } template<typename V> int Graph<V>::getDegree(int v) const { return neighbors[v].size(); } template<typename V> V Graph<V>::getVertex(int index) const { return vertices[index]; } template<typename V> int Graph<V>::getIndex(V v) const { for (unsigned i = 0; i < vertices.size(); i++) { if (vertices[i] == v) return i; } return -1; // If vertex is not in the graph } template<typename V> vector<V> Graph<V>::getVertices() const { return vertices; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename V> vector<int> Graph<V>::getNeighbors(int u) const {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang vector<int> result; for (Edge* e: neighbors[u]) result.push_back(e->v); return result; } template<typename V> void Graph<V>::printEdges() const { for (unsigned u = 0; u < neighbors.size(); u++) { cout << "Vertex " << getVertex(u) << "(" << u << "): "; for (Edge* e: neighbors[u]) { cout << "(" << getVertex(e->u) << ", " << getVertex(e->v) << ") "; } cout << endl; } } template<typename V> void Graph<V>::printAdjacencyMatrix() const { // Use vector for 2-D array vector<vector<int>> adjacencyMatrix(getSize()); // Initialize 2-D array for adjacency matrix for (int i = 0; i < getSize(); i++) { adjacencyMatrix[i] = vector<int>(getSize()); } for (unsigned i = 0; i < neighbors.size(); i++) { for (Edge* e: neighbors[i]) { adjacencyMatrix[i][e->v] = 1; } } for (unsigned i = 0; i < adjacencyMatrix.size(); i++) { for (unsigned j = 0; j < adjacencyMatrix[i].size(); j++) { cout << adjacencyMatrix[i][j] << " "; } cout << endl; } } template<typename V> void Graph<V>::clear() { vertices.clear(); for (int i = 0; i < getSize(); i++)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang for (Edge* e: neighbors[i]) delete e; neighbors.clear();

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } template<typename V> bool Graph<V>::addVertex(const V& v) { if (find(vertices.begin(), vertices.end(), v) == vertices.end()) { vertices.push_back(v); neighbors.push_back(vector<Edge*>(0)); return true; } else { return false; } } template<typename V> bool Graph<V>::createEdge(Edge* e) { if (e->u < 0 || e->u > getSize() - 1) { stringstream ss; ss << e->u; throw invalid_argument("No such edge: " + ss.str()); } if (e->v < 0 || e->v > getSize() - 1) { stringstream ss; ss << e->v; throw invalid_argument("No such edge: " + ss.str()); } vector<int> listOfNeighbors = getNeighbors(e->u); if (find(listOfNeighbors.begin(), listOfNeighbors.end(), e->v) == listOfNeighbors.end()) { neighbors[e->u].push_back(e); return true; } else { return false; } } template<typename V> bool Graph<V>::addEdge(int u, int v) { return createEdge(new Edge(u, v)); } template<typename V> Tree Graph<V>::dfs(int v) const {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang vector<int> searchOrders; vector<int> parent(vertices.size()); for (unsigned i = 0; i < vertices.size(); i++)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang parent[i] = -1; // Initialize parent[i] to -1 // Mark visited vertices vector<bool> isVisited(vertices.size()); for (unsigned i = 0; i < vertices.size(); i++) { isVisited[i] = false; } // Recursively search dfs(v, parent, searchOrders, isVisited); // Return a search tree return Tree(v, parent, searchOrders); } template<typename V> void Graph<V>::dfs(int v, vector<int>& parent, vector<int>& searchOrders, vector<bool>& isVisited) const { // Store the visited vertex searchOrders.push_back(v); isVisited[v] = true; // Vertex v visited for (Edge* e: neighbors[v]) { // w is a neighbor of v. Note that e->u is v and e->v. int w = e->v; if (!isVisited[w]) { parent[w] = v; // The parent of vertex i is v dfs(w, parent, searchOrders, isVisited); // Recursive search } } } template<typename V> Tree Graph<V>::bfs(int v) const { vector<int> searchOrders; vector<int> parent(vertices.size()); for (int i = 0; i < getSize(); i++) parent[i] = -1; // Initialize parent[i] to -1 queue<int> queue; // Stores vertices to be visited vector<bool> isVisited(getSize()); queue.push(v); // Enqueue v isVisited[v] = true; // Mark it visited while (!queue.empty()) { int u = queue.front(); // Get from the front of the queue queue.pop(); // remove the front element searchOrders.push_back(u); // u searched for (Edge* e: neighbors[u]) { // w is a neighbor of v. Note that e->u is v and e->v. int w = e->v;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang if (!isVisited[w]) { queue.push(w); // Enqueue w

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang parent[w] = u; // The parent of w is u isVisited[w] = true; // Mark it visited } } } return Tree(v, parent, searchOrders); } #endif #include <sstream> #include <vector> #include <string> using namespace std; int main() { // Read the number of vertices cout << "Enter the number of vertices: "; int numberOfVertices; cin >> numberOfVertices; cout << "The number of vertices is " << numberOfVertices << endl; vector<Edge> list; cout << "Enter the edges in " << numberOfVertices << " lines: "; for (int k = 0; k <= numberOfVertices; k++) { string s; getline(cin, s); stringstream ss(s); int startingVertex; ss >> startingVertex; while (!ss.eof()) { int adjacentVertex; ss >> adjacentVertex; list.push_back(Edge(startingVertex, adjacentVertex)); } } Graph<int> graph(numberOfVertices, list); graph.printEdges(); Tree tree = graph.dfs(0); if (tree.getNumberOfVerticesFound() == numberOfVertices) cout << "The graph is connected" << endl; else cout << "The graph is not connected" << endl; return 0; }

Chapter 26: Programming Project 2: (345021-00164)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang (Find paths)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang The description of this exercise is Chapter 26: Programming Project 2 from https://liangcpp.pearsoncmg.com/test/Extra.pdf.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer https://liangcpp.pearsoncmg.com/faq.html. #ifndef GRAPH_H #define GRAPH_H #include <vector> #include <queue> // For implementing BFS #include <stdexcept> #include <sstream> // For converting a number to a string #include <iostream> #include <algorithm> using namespace std; #ifndef TREE_H #define TREE_H class Tree { public: // Construct an empty tree Tree() { } // Construct a tree with root, parent, and searchOrder Tree(int root, const vector<int>& parent, const vector<int>& searchOrders) { this->root = root; this->parent = parent; this->searchOrders = searchOrders; } // Return the root of the tree int getRoot() const { return root; } // Return the parent of vertex v int getParent(int v) const { return parent[v]; } // Return search order vector<int> getSearchOrders() const { return searchOrders; } // Return number of vertices found

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang int getNumberOfVerticesFound() const { return searchOrders.size();

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } // Return the path of vertices from v to the root in a vector vector<int> getPath(int v) const { vector<int> path; do { path.push_back(v); v = parent[v]; } while (v != -1); return path; } // Print the whole tree void printTree() const { cout << "Root is: " << root << endl; cout << "Edges: "; for (unsigned i = 0; i < searchOrders.size(); i++) { if (parent[searchOrders[i]] != -1) { // Display an edge cout << "(" << parent[searchOrders[i]] << ", " << searchOrders[i] << ") "; } } cout << endl; } private: int root; // The root of the tree vector<int> parent; // Store the parent of each vertex vector<int> searchOrders; // Store the search order }; #endif #ifndef EDGE_H #define EDGE_H class Edge { public: int u; int v; Edge(int u, int v) { this->u = u; this->v = v; } }; #endif

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename V> class Graph

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { public: // Construct an empty graph Graph(); // Construct a graph from vertices in a vector and // edges in 2-D array Graph(const vector<V>& vertices, const int edges[][2], int numberOfEdges); // Construct a graph with vertices 0, 1, ..., n-1 and // edges in 2-D array Graph(int numberOfVertices, const int edges[][2], int numberOfEdges); // Construct a graph from vertices and edges objects Graph(const vector<V>& vertices, const vector<Edge>& edges); // Construct a graph with vertices 0, 1, ..., n-1 and // edges in a vector Graph(int numberOfVertices, const vector<Edge>& edges); // Return the number of vertices in the graph int getSize() const; // Return the degree for a specified vertex int getDegree(int v) const; // Return the vertex for the specified index V getVertex(int index) const; // Return the index for the specified vertex int getIndex(V v) const; // Return the vertices in the graph vector<V> getVertices() const; // Return the neighbors of vertex v vector<int> getNeighbors(int v) const; // Print the edges void printEdges() const; // Print the adjacency matrix void printAdjacencyMatrix() const; // Clear the graph void clear(); // Adds a vertex to the graph virtual bool addVertex(const V& v); // Adds an edge from u to v to the graph bool addEdge(int u, int v); // Obtain a depth-first search tree // To be discussed in Section 24.6 Tree dfs(int v) const;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Starting bfs search from vertex v // To be discussed in Section 24.7

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Tree bfs(int v) const; // Return a shortest path from u to v vector<int> getPath(int u, int v) const; protected: vector<V> vertices; // Store vertices vector<vector<Edge*>> neighbors; // Adjacency edge lists bool createEdge(Edge* e); // Add an edge private: // Create adjacency lists for each vertex from an edge array void createAdjacencyLists(int numberOfVertices, const int edges[][2], int numberOfEdges); // Create adjacency lists for each vertex from an Edge vector void createAdjacencyLists(int numberOfVertices, const vector<Edge>& edges); // Recursive function for DFS search void dfs(int v, vector<int>& parent, vector<int>& searchOrders, vector<bool>& isVisited) const; }; template<typename V> Graph<V>::Graph() { } template<typename V> Graph<V>::Graph(const vector<V>& vertices, const int edges[][2], int numberOfEdges) { for (unsigned i = 0; i < vertices.size(); i++) addVertex(vertices[i]); createAdjacencyLists(vertices.size(), edges, numberOfEdges); } template<typename V> Graph<V>::Graph(int numberOfVertices, const int edges[][2], int numberOfEdges) { for (int i = 0; i < numberOfVertices; i++) addVertex(i); // vertices is {0, 1, 2, ..., n-1} createAdjacencyLists(numberOfVertices, edges, numberOfEdges); } template<typename V> Graph<V>::Graph(const vector<V>& vertices, const vector<Edge>& edges) { for (unsigned i = 0; i < vertices.size(); i++) addVertex(vertices[i]); createAdjacencyLists(vertices.size(), edges);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } template<typename V>

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang Graph<V>::Graph(int numberOfVertices, const vector<Edge>& edges) { for (int i = 0; i < numberOfVertices; i++) addVertex(i); // vertices is {0, 1, 2, ..., n-1} createAdjacencyLists(numberOfVertices, edges); } template<typename V> void Graph<V>::createAdjacencyLists(int numberOfVertices, const int edges[][2], int numberOfEdges) { for (int i = 0; i < numberOfEdges; i++) { int u = edges[i][0]; int v = edges[i][1]; addEdge(u, v); } } template<typename V> void Graph<V>::createAdjacencyLists(int numberOfVertices, const vector<Edge>& edges) { for (unsigned i = 0; i < edges.size(); i++) { int u = edges[i].u; int v = edges[i].v; addEdge(u, v); } } template<typename V> int Graph<V>::getSize() const { return vertices.size(); } template<typename V> int Graph<V>::getDegree(int v) const { return neighbors[v].size(); } template<typename V> V Graph<V>::getVertex(int index) const { return vertices[index]; } template<typename V> int Graph<V>::getIndex(V v) const { for (unsigned i = 0; i < vertices.size(); i++) { if (vertices[i] == v)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return i; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return -1; // If vertex is not in the graph } template<typename V> vector<V> Graph<V>::getVertices() const { return vertices; } template<typename V> vector<int> Graph<V>::getNeighbors(int u) const { vector<int> result; for (Edge* e : neighbors[u]) result.push_back(e->v); return result; } template<typename V> void Graph<V>::printEdges() const { for (unsigned u = 0; u < neighbors.size(); u++) { cout << "Vertex " << getVertex(u) << "(" << u << "): "; for (Edge* e : neighbors[u]) { cout << "(" << getVertex(e->u) << ", " << getVertex(e->v) << ") "; } cout << endl; } } template<typename V> void Graph<V>::printAdjacencyMatrix() const { // Use vector for 2-D array vector<vector<int>> adjacencyMatrix(getSize()); // Initialize 2-D array for adjacency matrix for (int i = 0; i < getSize(); i++) { adjacencyMatrix[i] = vector<int>(getSize()); } for (unsigned i = 0; i < neighbors.size(); i++) { for (Edge* e : neighbors[i]) { adjacencyMatrix[i][e->v] = 1; } } for (unsigned i = 0; i < adjacencyMatrix.size(); i++) { for (unsigned j = 0; j < adjacencyMatrix[i].size(); j++) {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << adjacencyMatrix[i][j] << " "; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << endl; } } template<typename V> void Graph<V>::clear() { vertices.clear(); for (int i = 0; i < getSize(); i++) for (Edge* e : neighbors[i]) delete e; neighbors.clear(); } template<typename V> bool Graph<V>::addVertex(const V& v) { if (find(vertices.begin(), vertices.end(), v) == vertices.end()) { vertices.push_back(v); neighbors.push_back(vector<Edge*>(0)); return true; } else { return false; } } template<typename V> bool Graph<V>::createEdge(Edge* e) { if (e->u < 0 || e->u > getSize() - 1) { stringstream ss; ss << e->u; throw invalid_argument("No such edge: " + ss.str()); } if (e->v < 0 || e->v > getSize() - 1) { stringstream ss; ss << e->v; throw invalid_argument("No such edge: " + ss.str()); } vector<int> listOfNeighbors = getNeighbors(e->u); if (find(listOfNeighbors.begin(), listOfNeighbors.end(), e->v) == listOfNeighbors.end()) { neighbors[e->u].push_back(e); return true; } else { return false;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename V> bool Graph<V>::addEdge(int u, int v) { return createEdge(new Edge(u, v)); } template<typename V> Tree Graph<V>::dfs(int v) const { vector<int> searchOrders; vector<int> parent(vertices.size()); for (unsigned i = 0; i < vertices.size(); i++) parent[i] = -1; // Initialize parent[i] to -1 // Mark visited vertices vector<bool> isVisited(vertices.size()); for (unsigned i = 0; i < vertices.size(); i++) { isVisited[i] = false; } // Recursively search dfs(v, parent, searchOrders, isVisited); // Return a search tree return Tree(v, parent, searchOrders); } template<typename V> void Graph<V>::dfs(int v, vector<int>& parent, vector<int>& searchOrders, vector<bool>& isVisited) const { // Store the visited vertex searchOrders.push_back(v); isVisited[v] = true; // Vertex v visited for (Edge* e : neighbors[v]) { // w is a neighbor of v. Note that e->u is v and e->v. int w = e->v; if (!isVisited[w]) { parent[w] = v; // The parent of vertex i is v dfs(w, parent, searchOrders, isVisited); // Recursive search } } } template<typename V> Tree Graph<V>::bfs(int v) const { vector<int> searchOrders; vector<int> parent(vertices.size()); for (int i = 0; i < getSize(); i++) parent[i] = -1; // Initialize parent[i] to -1 queue<int> queue; // Stores vertices to be visited

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang vector<bool> isVisited(getSize()); queue.push(v); // Enqueue v isVisited[v] = true; // Mark it visited

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang while (!queue.empty()) { int u = queue.front(); // Get from the front of the queue queue.pop(); // remove the front element searchOrders.push_back(u); // u searched for (Edge* e : neighbors[u]) { // w is a neighbor of v. Note that e->u is v and e->v. int w = e->v; if (!isVisited[w]) { queue.push(w); // Enqueue w parent[w] = u; // The parent of w is u isVisited[w] = true; // Mark it visited } } } return Tree(v, parent, searchOrders); } template<typename V> vector<int> Graph<V>::getPath(int u, int v) const { Tree tree = this->bfs(u); vector<int> list; int current = v; while (tree.getParent(current) != -1) { list.insert(list.begin(), current); current = tree.getParent(current); } if (current == u) { list.insert(list.begin(), current); return list; } else return vector<int>(0); } #endif #include <vector> #include <string> using namespace std; int main() { // Vertices for graph in Figure 26.1 vector<string> vertices = { "Seattle", "San Francisco", "Los Angeles", "Denver", "Kansas City", "Chicago", "Boston", "New York", "Atlanta", "Miami", "Dallas", "Houston" }; // Edge array for graph in Figure 26.1

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang int edges[][2] = { {0, 1}, {0, 3}, {0, 5}, {1, 0}, {1, 2}, {1, 3},

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang {2, 1}, {2, 3}, {2, 4}, {2, 10}, {3, 0}, {3, 1}, {3, 2}, {3, 4}, {3, 5}, {4, 2}, {4, 3}, {4, 5}, {4, 7}, {4, 8}, {4, 10}, {5, 0}, {5, 3}, {5, 4}, {5, 6}, {5, 7}, {6, 5}, {6, 7}, {7, 4}, {7, 5}, {7, 6}, {7, 8}, {8, 4}, {8, 7}, {8, 9}, {8, 10}, {8, 11}, {9, 8}, {9, 11}, {10, 2}, {10, 4}, {10, 8}, {10, 11}, {11, 8}, {11, 9}, {11, 10} }; const int NUMBER_OF_EDGES = 46; // 46 edges in Figure 26.1 Graph<string> graph(vertices, edges, NUMBER_OF_EDGES); cout << "Enter a starting city: "; string startingCity; getline(cin, startingCity); cout << "Enter an ending city: "; string endingCity; getline(cin, endingCity); vector<int> list = graph.getPath(graph.getIndex(startingCity), graph.getIndex(endingCity)); cout << "The path is "; for (int i : list) { cout << graph.getVertex(i) << " "; } return 0; }

Chapter 27 Chapter 27: Proramming Project 1: (345021-00165) (Find a minimum spanning tree) The description of this exercise is Chapter 27: Programming Project 1 from https://liangcpp.pearsoncmg.com/test/Extra.pdf.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> #include <vector> #include <cmath> #include <algorithm> using namespace std; #ifndef TREE_H #define TREE_H class Tree

Solutions Manual for Introduction to C++ Programming and Data { Structures 5th edition by Daniel Liang public:

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Construct an empty tree Tree() { } // Construct a tree with root, parent, and searchOrder Tree(int root, const vector<int>& parent, const vector<int>& searchOrders) { this->root = root; this->parent = parent; this->searchOrders = searchOrders; } // Return the root of the tree int getRoot() const { return root; } // Return the parent of vertex v int getParent(int v) const { return parent[v]; } // Return search order vector<int> getSearchOrders() const { return searchOrders; } // Return number of vertices found int getNumberOfVerticesFound() const { return searchOrders.size(); } // Return the path of vertices from v to the root in a vector vector<int> getPath(int v) const { vector<int> path; do { path.push_back(v); v = parent[v]; } while (v != -1); return path; } // Print the whole tree void printTree() const { cout << "Root is: " << root << endl;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "Edges: "; for (unsigned i = 0; i < searchOrders.size(); i++) {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang if (parent[searchOrders[i]] != -1) { // Display an edge cout << "(" << parent[searchOrders[i]] << ", " << searchOrders[i] << ") "; } } cout << endl; } private: int root; // The root of the tree vector<int> parent; // Store the parent of each vertex vector<int> searchOrders; // Store the search order }; #endif #ifndef EDGE_H #define EDGE_H class Edge { public: int u; int v; Edge(int u, int v) { this->u = u; this->v = v; } }; #endif #ifndef GRAPH_H #define GRAPH_H #include <vector> #include <queue> // For implementing BFS #include <stdexcept> #include <sstream> // For converting a number to a string template<typename V> class Graph { public: // Construct an empty graph Graph(); // Construct a graph from vertices in a vector and // edges in 2-D array Graph(const vector<V>& vertices, const int edges[][2], int numberOfEdges); // Construct a graph with vertices 0, 1, ..., n-1 and // edges in 2-D array Graph(int numberOfVertices, const int edges[][2], int numberOfEdges);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Construct a graph from vertices and edges objects Graph(const vector<V>& vertices, const vector<Edge>& edges);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Construct a graph with vertices 0, 1, ..., n-1 and // edges in a vector Graph(int numberOfVertices, const vector<Edge>& edges); // Return the number of vertices in the graph int getSize() const; // Return the degree for a specified vertex int getDegree(int v) const; // Return the vertex for the specified index V getVertex(int index) const; // Return the index for the specified vertex int getIndex(V v) const; // Return the vertices in the graph vector<V> getVertices() const; // Return the neighbors of vertex v vector<int> getNeighbors(int v) const; // Print the edges void printEdges() const; // Print the adjacency matrix void printAdjacencyMatrix() const; // Clear the graph void clear(); // Adds a vertex to the graph virtual bool addVertex(const V& v); // Adds an edge from u to v to the graph bool addEdge(int u, int v); // Obtain a depth-first search tree // To be discussed in Section 24.6 Tree dfs(int v) const; // Starting bfs search from vertex v // To be discussed in Section 24.7 Tree bfs(int v) const; protected: vector<V> vertices; // Store vertices vector<vector<Edge*>> neighbors; // Adjacency edge lists bool createEdge(Edge* e); // Add an edge private: // Create adjacency lists for each vertex from an edge array void createAdjacencyLists(int numberOfVertices, const int edges[][2], int numberOfEdges);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Create adjacency lists for each vertex from an Edge vector void createAdjacencyLists(int numberOfVertices, const vector<Edge>& edges);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Recursive function for DFS search void dfs(int v, vector<int>& parent, vector<int>& searchOrders, vector<bool>& isVisited) const; }; template<typename V> Graph<V>::Graph() { } template<typename V> Graph<V>::Graph(const vector<V>& vertices, const int edges[][2], int numberOfEdges) { for (unsigned i = 0; i < vertices.size(); i++) addVertex(vertices[i]); createAdjacencyLists(vertices.size(), edges, numberOfEdges); } template<typename V> Graph<V>::Graph(int numberOfVertices, const int edges[][2], int numberOfEdges) { for (int i = 0; i < numberOfVertices; i++) addVertex(i); // vertices is {0, 1, 2, ..., n-1} createAdjacencyLists(numberOfVertices, edges, numberOfEdges); } template<typename V> Graph<V>::Graph(const vector<V>& vertices, const vector<Edge>& edges) { for (unsigned i = 0; i < vertices.size(); i++) addVertex(vertices[i]); createAdjacencyLists(vertices.size(), edges); } template<typename V> Graph<V>::Graph(int numberOfVertices, const vector<Edge>& edges) { for (int i = 0; i < numberOfVertices; i++) addVertex(i); // vertices is {0, 1, 2, ..., n-1} createAdjacencyLists(numberOfVertices, edges); } template<typename V> void Graph<V>::createAdjacencyLists(int numberOfVertices, const int edges[][2], int numberOfEdges) { for (int i = 0; i < numberOfEdges; i++) { int u = edges[i][0];

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang int v = edges[i][1]; addEdge(u, v); }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } template<typename V> void Graph<V>::createAdjacencyLists(int numberOfVertices, const vector<Edge>& edges) { for (unsigned i = 0; i < edges.size(); i++) { int u = edges[i].u; int v = edges[i].v; addEdge(u, v); } } template<typename V> int Graph<V>::getSize() const { return vertices.size(); } template<typename V> int Graph<V>::getDegree(int v) const { return neighbors[v].size(); } template<typename V> V Graph<V>::getVertex(int index) const { return vertices[index]; } template<typename V> int Graph<V>::getIndex(V v) const { for (unsigned i = 0; i < vertices.size(); i++) { if (vertices[i] == v) return i; } return -1; // If vertex is not in the graph } template<typename V> vector<V> Graph<V>::getVertices() const { return vertices; } template<typename V> vector<int> Graph<V>::getNeighbors(int u) const { vector<int> result; for (Edge* e : neighbors[u]) result.push_back(e->v);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return result; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename V> void Graph<V>::printEdges() const { for (unsigned u = 0; u < neighbors.size(); u++) { cout << "Vertex " << getVertex(u) << "(" << u << "): "; for (Edge* e : neighbors[u]) { cout << "(" << getVertex(e->u) << ", " << getVertex(e->v) << ") "; } cout << endl; } } template<typename V> void Graph<V>::printAdjacencyMatrix() const { // Use vector for 2-D array vector<vector<int>> adjacencyMatrix(getSize()); // Initialize 2-D array for adjacency matrix for (int i = 0; i < getSize(); i++) { adjacencyMatrix[i] = vector<int>(getSize()); } for (unsigned i = 0; i < neighbors.size(); i++) { for (Edge* e : neighbors[i]) { adjacencyMatrix[i][e->v] = 1; } } for (unsigned i = 0; i < adjacencyMatrix.size(); i++) { for (unsigned j = 0; j < adjacencyMatrix[i].size(); j++) { cout << adjacencyMatrix[i][j] << " "; } cout << endl; } } template<typename V> void Graph<V>::clear() { vertices.clear(); for (int i = 0; i < getSize(); i++) for (Edge* e : neighbors[i]) delete e; neighbors.clear(); } template<typename V>

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang bool Graph<V>::addVertex(const V& v) { if (find(vertices.begin(), vertices.end(), v) == vertices.end())

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { vertices.push_back(v); neighbors.push_back(vector<Edge*>(0)); return true; } else { return false; } } template<typename V> bool Graph<V>::createEdge(Edge* e) { if (e->u < 0 || e->u > getSize() - 1) { stringstream ss; ss << e->u; throw invalid_argument("No such edge: " + ss.str()); } if (e->v < 0 || e->v > getSize() - 1) { stringstream ss; ss << e->v; throw invalid_argument("No such edge: " + ss.str()); } vector<int> listOfNeighbors = getNeighbors(e->u); if (find(listOfNeighbors.begin(), listOfNeighbors.end(), e->v) == listOfNeighbors.end()) { neighbors[e->u].push_back(e); return true; } else { return false; } } template<typename V> bool Graph<V>::addEdge(int u, int v) { return createEdge(new Edge(u, v)); } template<typename V> Tree Graph<V>::dfs(int v) const { vector<int> searchOrders; vector<int> parent(vertices.size()); for (unsigned i = 0; i < vertices.size(); i++) parent[i] = -1; // Initialize parent[i] to -1 // Mark visited vertices

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang vector<bool> isVisited(vertices.size()); for (unsigned i = 0; i < vertices.size(); i++) {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang isVisited[i] = false; } // Recursively search dfs(v, parent, searchOrders, isVisited); // Return a search tree return Tree(v, parent, searchOrders); } template<typename V> void Graph<V>::dfs(int v, vector<int>& parent, vector<int>& searchOrders, vector<bool>& isVisited) const { // Store the visited vertex searchOrders.push_back(v); isVisited[v] = true; // Vertex v visited for (Edge* e : neighbors[v]) { // w is a neighbor of v. Note that e->u is v and e->v. int w = e->v; if (!isVisited[w]) { parent[w] = v; // The parent of vertex i is v dfs(w, parent, searchOrders, isVisited); // Recursive search } } } template<typename V> Tree Graph<V>::bfs(int v) const { vector<int> searchOrders; vector<int> parent(vertices.size()); for (int i = 0; i < getSize(); i++) parent[i] = -1; // Initialize parent[i] to -1 queue<int> queue; // Stores vertices to be visited vector<bool> isVisited(getSize()); queue.push(v); // Enqueue v isVisited[v] = true; // Mark it visited while (!queue.empty()) { int u = queue.front(); // Get from the front of the queue queue.pop(); // remove the front element searchOrders.push_back(u); // u searched for (Edge* e : neighbors[u]) { // w is a neighbor of v. Note that e->u is v and e->v. int w = e->v; if (!isVisited[w]) { queue.push(w); // Enqueue w parent[w] = u; // The parent of w is u isVisited[w] = true; // Mark it visited }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang return Tree(v, parent, searchOrders); } #endif #ifndef WEIGHTEDEDGE_H #define WEIGHTEDEDGE_H class WeightedEdge : public Edge { public: double weight; // The weight on edge (u, v) // Create a weighted edge on (u, v) WeightedEdge(int u, int v, double weight) : Edge(u, v) { this->weight = weight; } // Compare edges based on weights bool operator<(const WeightedEdge& edge) const { return (*this).weight < edge.weight; } bool operator<=(const WeightedEdge& edge) const { return (*this).weight <= edge.weight; } bool operator>(const WeightedEdge& edge) const { return (*this).weight > edge.weight; } bool operator>=(const WeightedEdge& edge) const { return (*this).weight >= edge.weight; } bool operator==(const WeightedEdge& edge) const { return (*this).weight == edge.weight; } bool operator!=(const WeightedEdge& edge) const { return (*this).weight != edge.weight; } }; #endif #ifndef SHORTESTPATHTREE_H #define SHORTESTPATHTREE_H class ShortestPathTree : public Tree

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { public: // Create an empty ShortestPathTree

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang ShortestPathTree() { } // Construct a tree with root, parent, searchOrders, // and cost ShortestPathTree(int root, const vector<int>& parent, const vector<int>& searchOrders, const vector<double>& cost) : Tree(root, parent, searchOrders) { this->cost = cost; } // Return the cost for the path from the source to vertex v. double getCost(int v) const { return cost[v]; } private: vector<double> cost; }; #endif #ifndef MST_H #define MST_H class MST : public Tree { public: // Create an empty MST MST() { } // Construct a tree with root, parent, searchOrders, // and total weight MST(int root, const vector<int>& parent, const vector<int>& searchOrders, double totalWeight) : Tree(root, parent, searchOrders) { this->totalWeight = totalWeight; } // Return the total weight of the tree double getTotalWeight() { return totalWeight; } private: double totalWeight; }; #endif #ifndef WEIGHTEDGRAPH_H #define WEIGHTEDGRAPH_H

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename V> class WeightedGraph : public Graph<V>

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { public: // Construct an empty graph WeightedGraph(); // Construct a graph from vertices and edges objects WeightedGraph(const vector<V>& vertices, const vector<WeightedEdge>& edges); // Construct a graph with vertices 0, 1, ..., n-1 and // edges in a vector WeightedGraph(int numberOfVertices, const vector<WeightedEdge>& edges); // Print all edges in the weighted tree void printWeightedEdges(); // Add a weighted edge bool addEdge(int u, int v, double w); // Get a minimum spanning tree rooted at vertex 0 MST getMinimumSpanningTree(); // Get a minimum spanning tree rooted at a specified vertex MST getMinimumSpanningTree(int startingVertex); // Find single-source shortest paths ShortestPathTree getShortestPath(int sourceVertex); }; template<typename V> WeightedGraph<V>::WeightedGraph() { } template<typename V> WeightedGraph<V>::WeightedGraph(const vector<V>& vertices, const vector<WeightedEdge>& edges) { // Add vertices to the graph for (unsigned i = 0; i < vertices.size(); i++) { addVertex(vertices[i]); } // Add edges to the graph for (unsigned i = 0; i < edges.size(); i++) { addEdge(edges[i].u, edges[i].v, edges[i].weight); } } template<typename V> WeightedGraph<V>::WeightedGraph(int numberOfVertices, const vector<WeightedEdge>& edges) { // Add vertices to the graph

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang for (int i = 0; i < numberOfVertices; i++) this->addVertex(i); // vertices is {0, 1, 2, ..., n-1}

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Add edges to the graph for (unsigned i = 0; i < edges.size(); i++) { addEdge(edges[i].u, edges[i].v, edges[i].weight); } } template<typename V> void WeightedGraph<V>::printWeightedEdges() { for (int i = 0; i < this->getSize(); i++) { // Display all edges adjacent to vertex with index i cout << "Vertex " << this->getVertex(i) << "(" << i << "): "; // Display all weighted edges for (Edge* e : this->neighbors[i]) { cout << "(" << e->u << ", " << e->v << ", " << static_cast<WeightedEdge*>(e)->weight << ") "; } cout << endl; } } template<typename V> bool WeightedGraph<V>::addEdge(int u, int v, double w) { return this->createEdge(new WeightedEdge(u, v, w)); } template<typename V> MST WeightedGraph<V>::getMinimumSpanningTree() { return getMinimumSpanningTree(0); } template<typename V> MST WeightedGraph<V>::getMinimumSpanningTree(int startingVertex) { vector<int> T; // T contains the vertices in the tree vector<int> parent(this->getSize()); // Parent of a verte parent[startingVertex] = -1; // startingVertex is the root double totalWeight = 0; // Total weight of the tree thus far // cost[v] stores the cost by adding v to the tree vector<double> cost(this->getSize()); for (unsigned i = 0; i < cost.size(); i++) { cost[i] = INFINITY; // Initial cost set to infinity } cost[startingVertex] = 0; // Cost of source is 0 // Expand T while (T.size() < this->getSize())

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { // Find smallest cost v in V - T int u = -1; // Vertex to be determined

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang double currentMinCost = INFINITY; for (int i = 0; i < this->getSize(); i++) { if (find(T.begin(), T.end(), i) == T.end() && cost[i] < currentMinCost) { currentMinCost = cost[i]; u = i; } } T.push_back(u); // Add a new vertex to T totalWeight += cost[u]; // Add cost[u] to the tree // Adjust cost[v] for v that is adjacent to u and v in V - T for (Edge* e : this->neighbors[u]) { if (find(T.begin(), T.end(), e->v) == T.end() && cost[e->v] > static_cast<WeightedEdge*>(e)->weight) { cost[e->v] = static_cast<WeightedEdge*>(e)->weight; parent[e->v] = u; } } } // End of while return MST(startingVertex, parent, T, totalWeight); } template<typename V> ShortestPathTree WeightedGraph<V>::getShortestPath(int sourceVertex) { // T stores the vertices whose path found so far vector<int> T; // parent[v] stores the previous vertex of v in the path vector<int> parent(this->getSize()); parent[sourceVertex] = -1; // The parent of source is set to -1 // cost[v] stores the cost of the path from v to the source vector<double> cost(this->getSize()); for (unsigned i = 0; i < cost.size(); i++) { cost[i] = INFINITY; // Initial cost set to infinity } cost[sourceVertex] = 0; // Cost of source is 0 // Expand T while (T.size() < this->getSize()) { // Find smallest cost v in V - T int u = -1; // Vertex to be determined double currentMinCost = INFINITY; for (int i = 0; i < this->getSize(); i++) { if (find(T.begin(), T.end(), i) == T.end()

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang && cost[i] < currentMinCost) { currentMinCost = cost[i];

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang u = i; } } if (u == -1) break; T.push_back(u); // Add a new vertex to T // Adjust cost[v] for v that is adjacent to u and v in V - T for (Edge* e : this->neighbors[u]) { if (find(T.begin(), T.end(), e->v) == T.end() && cost[e->v] > cost[u] + static_cast<WeightedEdge*>(e)->weight) { cost[e->v] = cost[u] + static_cast<WeightedEdge*>(e)->weight; parent[e->v] = u; } } } // End of while // Create a ShortestPathTree return ShortestPathTree(sourceVertex, parent, T, cost); } #endif #include <iostream> #include <string> #include <regex> // See LiveExample H.2 // Split a string into substrings using the specified delimiter vector<string> split(const string& text, const string& delim) { regex delimeter(delim); // Create a regex for separating words sregex_token_iterator tokenIterator(text.begin(), text.end(), delimeter, -1); sregex_token_iterator end; vector<string> result; while (tokenIterator != end) { string token = *tokenIterator; // Get a token if (token.size() > 0) result.push_back(token); // Save token in the vector tokenIterator++; // Iterate to next token } return result; } int main() { string line;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Read the number of vertices cout << "Enter the number of vertices: "; getline(cin, line, '\n'); // Cannot use line-input after token-input

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang int numberOfVertices = atoi(line.c_str()); cout << "The number of vertices is " << numberOfVertices << endl; vector<WeightedEdge> list; cout << "Enter the triplets in one line: "; getline(cin, line, '\n'); vector<string> triplets = split(line, "[\|]"); // Get the triplets for (string triplet : triplets) { vector<string> tokens = split(triplet, "[,]"); int u = atoi(tokens[0].c_str()); int v = atoi(tokens[1].c_str()); double w = atof(tokens[2].c_str()); list.push_back(WeightedEdge(u, v, w)); list.push_back(WeightedEdge(v, u, w)); } WeightedGraph<int> graph(numberOfVertices, list); graph.printWeightedEdges(); MST tree = graph.getMinimumSpanningTree(); cout << "Total weight in MST is " << tree.getTotalWeight() << endl; tree.printTree(); return 0; }

Chapter 27 Chapter 27: Proramming Project 2: (345021-00166) (Find shortest paths) The description of this exercise is Chapter 27: Programming Project 2 from https://liangcpp.pearsoncmg.com/test/Extra.pdf.

For a hint on this program, please see https://liangcpp.pearsoncmg.com/cpprevel2e.html. If you get a logic or runtime error, please refer to https://liangcpp.pearsoncmg.com/faq.html. #include <iostream> // Exercise27_02Extra #include <vector> #include <cmath> #include <algorithm> using namespace std; #ifndef TREE_H #define TREE_H class Tree { public: // Construct an empty tree Tree() Solutions Manual for Introduction to C++ Programming and Data { Structures 5th edition by Daniel Liang }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Construct a tree with root, parent, and searchOrder Tree(int root, const vector<int>& parent, const vector<int>& searchOrders) { this->root = root; this->parent = parent; this->searchOrders = searchOrders; } // Return the root of the tree int getRoot() const { return root; } // Return the parent of vertex v int getParent(int v) const { return parent[v]; } // Return search order vector<int> getSearchOrders() const { return searchOrders; } // Return number of vertices found int getNumberOfVerticesFound() const { return searchOrders.size(); } // Return the path of vertices from v to the root in a vector vector<int> getPath(int v) const { vector<int> path; do { path.push_back(v); v = parent[v]; } while (v != -1); return path; } // Print the whole tree void printTree() const { cout << "Root is: " << root << endl; cout << "Edges: "; for (unsigned i = 0; i < searchOrders.size(); i++) { if (parent[searchOrders[i]] != -1)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { // Display an edge cout << "(" << parent[searchOrders[i]] << ", " <<

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang searchOrders[i] << ") "; } } cout << endl; } private: int root; // The root of the tree vector<int> parent; // Store the parent of each vertex vector<int> searchOrders; // Store the search order }; #endif #ifndef EDGE_H #define EDGE_H class Edge { public: int u; int v; Edge(int u, int v) { this->u = u; this->v = v; } }; #endif #ifndef GRAPH_H #define GRAPH_H #include <vector> #include <queue> // For implementing BFS #include <stdexcept> #include <sstream> // For converting a number to a string template<typename V> class Graph { public: // Construct an empty graph Graph(); // Construct a graph from vertices in a vector and // edges in 2-D array Graph(const vector<V>& vertices, const int edges[][2], int numberOfEdges); // Construct a graph with vertices 0, 1, ..., n-1 and // edges in 2-D array Graph(int numberOfVertices, const int edges[][2], int numberOfEdges); // Construct a graph from vertices and edges objects Graph(const vector<V>& vertices, const vector<Edge>& edges);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Construct a graph with vertices 0, 1, ..., n-1 and // edges in a vector Graph(int numberOfVertices, const vector<Edge>& edges);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Return the number of vertices in the graph int getSize() const; // Return the degree for a specified vertex int getDegree(int v) const; // Return the vertex for the specified index V getVertex(int index) const; // Return the index for the specified vertex int getIndex(V v) const; // Return the vertices in the graph vector<V> getVertices() const; // Return the neighbors of vertex v vector<int> getNeighbors(int v) const; // Print the edges void printEdges() const; // Print the adjacency matrix void printAdjacencyMatrix() const; // Clear the graph void clear(); // Adds a vertex to the graph virtual bool addVertex(const V& v); // Adds an edge from u to v to the graph bool addEdge(int u, int v); // Obtain a depth-first search tree // To be discussed in Section 24.6 Tree dfs(int v) const; // Starting bfs search from vertex v // To be discussed in Section 24.7 Tree bfs(int v) const; protected: vector<V> vertices; // Store vertices vector<vector<Edge*>> neighbors; // Adjacency edge lists bool createEdge(Edge* e); // Add an edge private: // Create adjacency lists for each vertex from an edge array void createAdjacencyLists(int numberOfVertices, const int edges[][2], int numberOfEdges); // Create adjacency lists for each vertex from an Edge vector void createAdjacencyLists(int numberOfVertices, const vector<Edge>& edges);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Recursive function for DFS search void dfs(int v, vector<int>& parent, vector<int>& searchOrders, vector<bool>& isVisited) const;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang }; template<typename V> Graph<V>::Graph() { } template<typename V> Graph<V>::Graph(const vector<V>& vertices, const int edges[][2], int numberOfEdges) { for (unsigned i = 0; i < vertices.size(); i++) addVertex(vertices[i]); createAdjacencyLists(vertices.size(), edges, numberOfEdges); } template<typename V> Graph<V>::Graph(int numberOfVertices, const int edges[][2], int numberOfEdges) { for (int i = 0; i < numberOfVertices; i++) addVertex(i); // vertices is {0, 1, 2, ..., n-1} createAdjacencyLists(numberOfVertices, edges, numberOfEdges); } template<typename V> Graph<V>::Graph(const vector<V>& vertices, const vector<Edge>& edges) { for (unsigned i = 0; i < vertices.size(); i++) addVertex(vertices[i]); createAdjacencyLists(vertices.size(), edges); } template<typename V> Graph<V>::Graph(int numberOfVertices, const vector<Edge>& edges) { for (int i = 0; i < numberOfVertices; i++) addVertex(i); // vertices is {0, 1, 2, ..., n-1} createAdjacencyLists(numberOfVertices, edges); } template<typename V> void Graph<V>::createAdjacencyLists(int numberOfVertices, const int edges[][2], int numberOfEdges) { for (int i = 0; i < numberOfEdges; i++) { int u = edges[i][0]; int v = edges[i][1]; addEdge(u, v); } }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang template<typename V> void Graph<V>::createAdjacencyLists(int numberOfVertices,

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang const vector<Edge>& edges) { for (unsigned i = 0; i < edges.size(); i++) { int u = edges[i].u; int v = edges[i].v; addEdge(u, v); } } template<typename V> int Graph<V>::getSize() const { return vertices.size(); } template<typename V> int Graph<V>::getDegree(int v) const { return neighbors[v].size(); } template<typename V> V Graph<V>::getVertex(int index) const { return vertices[index]; } template<typename V> int Graph<V>::getIndex(V v) const { for (unsigned i = 0; i < vertices.size(); i++) { if (vertices[i] == v) return i; } return -1; // If vertex is not in the graph } template<typename V> vector<V> Graph<V>::getVertices() const { return vertices; } template<typename V> vector<int> Graph<V>::getNeighbors(int u) const { vector<int> result; for (Edge* e : neighbors[u]) result.push_back(e->v); return result; } template<typename V>

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang void Graph<V>::printEdges() const { for (unsigned u = 0; u < neighbors.size(); u++)

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { cout << "Vertex " << getVertex(u) << "(" << u << "): "; for (Edge* e : neighbors[u]) { cout << "(" << getVertex(e->u) << ", " << getVertex(e->v) << ") "; } cout << endl; } } template<typename V> void Graph<V>::printAdjacencyMatrix() const { // Use vector for 2-D array vector<vector<int>> adjacencyMatrix(getSize()); // Initialize 2-D array for adjacency matrix for (int i = 0; i < getSize(); i++) { adjacencyMatrix[i] = vector<int>(getSize()); } for (unsigned i = 0; i < neighbors.size(); i++) { for (Edge* e : neighbors[i]) { adjacencyMatrix[i][e->v] = 1; } } for (unsigned i = 0; i < adjacencyMatrix.size(); i++) { for (unsigned j = 0; j < adjacencyMatrix[i].size(); j++) { cout << adjacencyMatrix[i][j] << " "; } cout << endl; } } template<typename V> void Graph<V>::clear() { vertices.clear(); for (int i = 0; i < getSize(); i++) for (Edge* e : neighbors[i]) delete e; neighbors.clear(); } template<typename V> bool Graph<V>::addVertex(const V& v) { if (find(vertices.begin(), vertices.end(), v) == vertices.end()) {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang vertices.push_back(v); neighbors.push_back(vector<Edge*>(0)); return true;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } else { return false; } } template<typename V> bool Graph<V>::createEdge(Edge* e) { if (e->u < 0 || e->u > getSize() - 1) { stringstream ss; ss << e->u; throw invalid_argument("No such edge: " + ss.str()); } if (e->v < 0 || e->v > getSize() - 1) { stringstream ss; ss << e->v; throw invalid_argument("No such edge: " + ss.str()); } vector<int> listOfNeighbors = getNeighbors(e->u); if (find(listOfNeighbors.begin(), listOfNeighbors.end(), e->v) == listOfNeighbors.end()) { neighbors[e->u].push_back(e); return true; } else { return false; } } template<typename V> bool Graph<V>::addEdge(int u, int v) { return createEdge(new Edge(u, v)); } template<typename V> Tree Graph<V>::dfs(int v) const { vector<int> searchOrders; vector<int> parent(vertices.size()); for (unsigned i = 0; i < vertices.size(); i++) parent[i] = -1; // Initialize parent[i] to -1 // Mark visited vertices vector<bool> isVisited(vertices.size()); for (unsigned i = 0; i < vertices.size(); i++) { isVisited[i] = false;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } // Recursively search

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang dfs(v, parent, searchOrders, isVisited); // Return a search tree return Tree(v, parent, searchOrders); } template<typename V> void Graph<V>::dfs(int v, vector<int>& parent, vector<int>& searchOrders, vector<bool>& isVisited) const { // Store the visited vertex searchOrders.push_back(v); isVisited[v] = true; // Vertex v visited for (Edge* e : neighbors[v]) { // w is a neighbor of v. Note that e->u is v and e->v. int w = e->v; if (!isVisited[w]) { parent[w] = v; // The parent of vertex i is v dfs(w, parent, searchOrders, isVisited); // Recursive search } } } template<typename V> Tree Graph<V>::bfs(int v) const { vector<int> searchOrders; vector<int> parent(vertices.size()); for (int i = 0; i < getSize(); i++) parent[i] = -1; // Initialize parent[i] to -1 queue<int> queue; // Stores vertices to be visited vector<bool> isVisited(getSize()); queue.push(v); // Enqueue v isVisited[v] = true; // Mark it visited while (!queue.empty()) { int u = queue.front(); // Get from the front of the queue queue.pop(); // remove the front element searchOrders.push_back(u); // u searched for (Edge* e : neighbors[u]) { // w is a neighbor of v. Note that e->u is v and e->v. int w = e->v; if (!isVisited[w]) { queue.push(w); // Enqueue w parent[w] = u; // The parent of w is u isVisited[w] = true; // Mark it visited } } } return Tree(v, parent, searchOrders);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } #endif

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang #ifndef WEIGHTEDEDGE_H #define WEIGHTEDEDGE_H class WeightedEdge : public Edge { public: double weight; // The weight on edge (u, v) // Create a weighted edge on (u, v) WeightedEdge(int u, int v, double weight) : Edge(u, v) { this->weight = weight; } // Compare edges based on weights bool operator<(const WeightedEdge& edge) const { return (*this).weight < edge.weight; } bool operator<=(const WeightedEdge& edge) const { return (*this).weight <= edge.weight; } bool operator>(const WeightedEdge& edge) const { return (*this).weight > edge.weight; } bool operator>=(const WeightedEdge& edge) const { return (*this).weight >= edge.weight; } bool operator==(const WeightedEdge& edge) const { return (*this).weight == edge.weight; } bool operator!=(const WeightedEdge& edge) const { return (*this).weight != edge.weight; } }; #endif #ifndef SHORTESTPATHTREE_H #define SHORTESTPATHTREE_H class ShortestPathTree : public Tree { public: // Create an empty ShortestPathTree ShortestPathTree()

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang { }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Construct a tree with root, parent, searchOrders, // and cost ShortestPathTree(int root, const vector<int>& parent, const vector<int>& searchOrders, const vector<double>& cost) : Tree(root, parent, searchOrders) { this->cost = cost; } // Return the cost for the path from the source to vertex v. double getCost(int v) const { return cost[v]; } private: vector<double> cost; }; #endif #ifndef MST_H #define MST_H class MST : public Tree { public: // Create an empty MST MST() { } // Construct a tree with root, parent, searchOrders, // and total weight MST(int root, const vector<int>& parent, const vector<int>& searchOrders, double totalWeight) : Tree(root, parent, searchOrders) { this->totalWeight = totalWeight; } // Return the total weight of the tree double getTotalWeight() { return totalWeight; } private: double totalWeight; }; #endif #ifndef WEIGHTEDGRAPH_H #define WEIGHTEDGRAPH_H template<typename V> class WeightedGraph : public Graph<V> {

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang public: // Construct an empty graph WeightedGraph();

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang // Construct a graph from vertices and edges objects WeightedGraph(const vector<V>& vertices, const vector<WeightedEdge>& edges); // Construct a graph with vertices 0, 1, ..., n-1 and // edges in a vector WeightedGraph(int numberOfVertices, const vector<WeightedEdge>& edges); // Print all edges in the weighted tree void printWeightedEdges(); // Add a weighted edge bool addEdge(int u, int v, double w); // Get a minimum spanning tree rooted at vertex 0 MST getMinimumSpanningTree(); // Get a minimum spanning tree rooted at a specified vertex MST getMinimumSpanningTree(int startingVertex); // Find single-source shortest paths ShortestPathTree getShortestPath(int sourceVertex); }; template<typename V> WeightedGraph<V>::WeightedGraph() { } template<typename V> WeightedGraph<V>::WeightedGraph(const vector<V>& vertices, const vector<WeightedEdge>& edges) { // Add vertices to the graph for (unsigned i = 0; i < vertices.size(); i++) { addVertex(vertices[i]); } // Add edges to the graph for (unsigned i = 0; i < edges.size(); i++) { addEdge(edges[i].u, edges[i].v, edges[i].weight); } } template<typename V> WeightedGraph<V>::WeightedGraph(int numberOfVertices, const vector<WeightedEdge>& edges) { // Add vertices to the graph for (int i = 0; i < numberOfVertices; i++) this->addVertex(i); // vertices is {0, 1, 2, ..., n-1} // Add edges to the graph

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang for (unsigned i = 0; i < edges.size(); i++) { addEdge(edges[i].u, edges[i].v, edges[i].weight);

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } } template<typename V> void WeightedGraph<V>::printWeightedEdges() { for (int i = 0; i < this->getSize(); i++) { // Display all edges adjacent to vertex with index i cout << "Vertex " << this->getVertex(i) << "(" << i << "): "; // Display all weighted edges for (Edge* e : this->neighbors[i]) { cout << "(" << e->u << ", " << e->v << ", " << static_cast<WeightedEdge*>(e)->weight << ") "; } cout << endl; } } template<typename V> bool WeightedGraph<V>::addEdge(int u, int v, double w) { return this->createEdge(new WeightedEdge(u, v, w)); } template<typename V> MST WeightedGraph<V>::getMinimumSpanningTree() { return getMinimumSpanningTree(0); } template<typename V> MST WeightedGraph<V>::getMinimumSpanningTree(int startingVertex) { vector<int> T; // T contains the vertices in the tree vector<int> parent(this->getSize()); // Parent of a verte parent[startingVertex] = -1; // startingVertex is the root double totalWeight = 0; // Total weight of the tree thus far // cost[v] stores the cost by adding v to the tree vector<double> cost(this->getSize()); for (unsigned i = 0; i < cost.size(); i++) { cost[i] = INFINITY; // Initial cost set to infinity } cost[startingVertex] = 0; // Cost of source is 0 // Expand T while (T.size() < this->getSize()) { // Find smallest cost v in V - T int u = -1; // Vertex to be determined double currentMinCost = INFINITY;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang for (int i = 0; i < this->getSize(); i++) { if (find(T.begin(), T.end(), i) == T.end()

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang && cost[i] < currentMinCost) { currentMinCost = cost[i]; u = i; } } T.push_back(u); // Add a new vertex to T totalWeight += cost[u]; // Add cost[u] to the tree // Adjust cost[v] for v that is adjacent to u and v in V - T for (Edge* e : this->neighbors[u]) { if (find(T.begin(), T.end(), e->v) == T.end() && cost[e->v] > static_cast<WeightedEdge*>(e)->weight) { cost[e->v] = static_cast<WeightedEdge*>(e)->weight; parent[e->v] = u; } } } // End of while return MST(startingVertex, parent, T, totalWeight); } template<typename V> ShortestPathTree WeightedGraph<V>::getShortestPath(int sourceVertex) { // T stores the vertices whose path found so far vector<int> T; // parent[v] stores the previous vertex of v in the path vector<int> parent(this->getSize()); parent[sourceVertex] = -1; // The parent of source is set to -1 // cost[v] stores the cost of the path from v to the source vector<double> cost(this->getSize()); for (unsigned i = 0; i < cost.size(); i++) { cost[i] = INFINITY; // Initial cost set to infinity } cost[sourceVertex] = 0; // Cost of source is 0 // Expand T while (T.size() < this->getSize()) { // Find smallest cost v in V - T int u = -1; // Vertex to be determined double currentMinCost = INFINITY; for (int i = 0; i < this->getSize(); i++) { if (find(T.begin(), T.end(), i) == T.end() && cost[i] < currentMinCost) { currentMinCost = cost[i]; u = i;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang } }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang if (u == -1) break; T.push_back(u); // Add a new vertex to T // Adjust cost[v] for v that is adjacent to u and v in V - T for (Edge* e : this->neighbors[u]) { if (find(T.begin(), T.end(), e->v) == T.end() && cost[e->v] > cost[u] + static_cast<WeightedEdge*>(e)->weight) { cost[e->v] = cost[u] + static_cast<WeightedEdge*>(e)->weight; parent[e->v] = u; } } } // End of while // Create a ShortestPathTree return ShortestPathTree(sourceVertex, parent, T, cost); } #endif #include <iostream> #include <string> #include <regex> // See LiveExample H.2 // Split a string into substrings using the specified delimiter vector<string> split(const string& text, const string& delim) { regex delimeter(delim); // Create a regex for separating words sregex_token_iterator tokenIterator(text.begin(), text.end(), delimeter, -1); sregex_token_iterator end; vector<string> result; while (tokenIterator != end) { string token = *tokenIterator; // Get a token if (token.size() > 0) result.push_back(token); // Save token in the vector tokenIterator++; // Iterate to next token } return result; } int main() { string line; // Read the number of vertices cout << "Enter the number of vertices: "; getline(cin, line, '\n'); // Cannot use line-input after token-input int numberOfVertices = atoi(line.c_str());

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "The number of vertices is " << numberOfVertices << endl; vector<WeightedEdge> list;

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang cout << "Enter the triplets in one line: "; getline(cin, line, '\n'); vector<string> triplets = split(line, "[\|]"); // Get the triplets for (string triplet : triplets) { vector<string> tokens = split(triplet, "[,]"); int u = atoi(tokens[0].c_str()); int v = atoi(tokens[1].c_str()); double w = atof(tokens[2].c_str()); list.push_back(WeightedEdge(u, v, w)); list.push_back(WeightedEdge(v, u, w)); } WeightedGraph<int> graph(numberOfVertices, list); // Read two vertices cout << "Enter two vertices (integer indexes): "; int v1; int v2; cin >> v1 >> v2; graph.printWeightedEdges(); ShortestPathTree tree = graph.getShortestPath(v1); vector<int> path = tree.getPath(v2); cout << "A path from " << v1 << " to " << v2 << ": "; for (int i = path.size() - 1; i >= 0; i--) cout << path[i] << " "; cout << endl; return 0; }

Solutions Manual for Introduction to C++ Programming and Data Structures 5th edition by Daniel Liang