University of Phoenix MTH 221_WEEK 4 Assignment_ Q&A. Score 100%

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10.2 In Exercises 1-3 find the number of vertices, the number of edges, and the degree of each vertex in the given undirected graph. Identify all isolated and pendant vertices.

1.

2.

3.

Answer to 1-3: Vertices = 5, Edges = 13, degree(a) = 6, degree(b) = 6, degree(c) = 6, degree(d) = 5, degree(e) = 3, no isolated or pendant vertices, sum degrees = 26, 26/2 = 13 = edges 4.

Find the sum of the degrees of the vertices of each graph in Exercise 1-3 and verify that it equals twice the number of edges in the graph. a. Graph 1: sum of degrees sum degrees = 12, 12/2 = 6 = edges b. Graph 2: sum of degrees sum degrees = 26, 26/2 = 13 = edges c. Graph 3: sum of degrees sum degrees = 3 + 2 + 4 + 0 + 6 + 4 + 2 + 3 = 24, 24/2 = 12 = edges

10.5 In Exercises 1-8 determine whether the given graph has an Euler circuit. Construct such a circuit when one exists. If no Euler circuit exists, determine whether the graph has an Euler path and construct such a path if one exists.

1.

Since there are four vertices of odd degree (a, b, c, and e) and 4 > 2, this graph has neither an Euler circuit nor an Euler path.

2.

All the vertex degrees are even, so there is an Euler circuit. We can find one by trial and error, or by using and Algorithm. One such circuit is a, b, c, f, i, h, g, d, e, h, f, e, b, d, a.

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3.

Since there are two vertices of odd degree (a and d), this graph has no Euler circuit, but it does have an Euler path starting at a and ending at d. we can find such a path by inspection, or by using the splicing idea explained in this section. One such path is a, e, c, e, b, e, d, b, a, c, d.

4.

This graph has no Euler circuit, since the degree of vertex c (for one) is odd. There is an Euler path between the two vertices of odd degree. One such path is f, a, b, c, d, e, f, b, d, a, e, c.

11.1 45. Draw the first seven rooted Fibonacci tress.

T1

T2

T3

T4

T5

T6

T7

a.

We follow the recursive definition and produce the pictures shown above

for T 1 through T 7 .

46. How many vertices, leaves, and internal vertices does the rooted Fibonacci tree is a positive integer? What is its height?

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T n have, where n

a.

The number of vertices in the tree

T n satisfies the recurrence relation

v n =v n−1 +v n−2 +1 ( the “+1” is for the root), with v 1=v 2=1 . Thus the sequence begins

1,1,3,5,9,15,25,… . It is easy to prove by induction that v n =2 f n−1 , where

f n is the nth Fibonacci number. The number of leaves satisfies the recurrent relation

l n=l n−1 +l n−2 , with l 1=l 2=1 , so l n=f n . Since i n=v n −l n , we have i n=f n−1 . Finally, it is clear that the height of the tree T n is one more than the height of the tree

T n−1 for n ≥3 , with the height of T 2 being 0. Therefore, the height of

T n is n−2 for all n ≥2 (and of course the height of T 1 is 0).

11.2 1.

Build a binary search tree for the words banana, peach, apple, pear, coconut, mango, and papaya using alphabetical order. a. We first insert banana into the empty tree, giving us the tree with just a root, labeled banana.

Next we inset peach, which, being greater than banana in alphabetical order, become the right child of the root. We continue in this manner letting each new word find its place by coming down the tree, branching either right or left until it encounter a free positions. The final tree is as shown.

2.

Build a binary search tree for the words oenology, phrenology, campanology, ornithology, ichthyology, limnology, alchemy, and astrology using alphabetical order. a. We first insert oenology into the empty tree, giving us the tree with just a root, labeled oenology.

Next we inset phrenology, which, being greater than oenology in alphabetical order, become the right child of the root. We continue in this manner letting each new word find its place by coming down the tree, branching either right or left until it encounter a free positions. The final tree is as shown.

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3.

4.

8.

How many comparisons are needed to locate or to add each of these words in the search tree for Exercise 1, starting fresh each time? a. Pear i. To find pear, we compare it with the root (banana), then with the right child of the root (peach), and finally with the right child of that vertex (pear). Thus 3 comparisons are needed. b. Banana i. Only 1 comparison is needed, since the item being searched for is the root. c. Kumquat i. We fail to locate kumquat by comparing it successively to banana, peach, coconut, and mango. Once we determine that kumquat should be in the left subtree of mango, and find no vertices there, we know that kumquat is not in the tree. Thus 4 comparisons were used. d. Orange i. This one is similar to part (c), except that 5 comparisons are used. We compare orange successively to banana, peach, coconut, mango, and papaya.

How many comparison are needed to locate or to add each of the words in the search tree for Exercise 2, starting fresh each time? a. Palmistry i. We fail to locate palmistry by comparing it successively to oenology, phrenology, and ornithology. At this point it is known that the word is not in the tree, since ornithology has no right child. 3 comparison were used. b. Etymology i. We were also unable to locate etymology on the tree. There were also 3 comparisons used. c. Paleontology i. We were also unable to locate paleontology on the tree. There were also 3 comparisons used. d. Glaciology i. We were also unable to locate glaciology on the tree. There were also 3 comparisons used.

How many weighing’s of a balance scale are needed to find counterfeit coin among eight coins if the counterfeit coin is either heavier or lighter than the others? Describe an algorithm to find the counterfeit coin using this number of weighing’s.

11.3 In exercises 7-9 determine the order in which a preorder traversal visits the vertices of the given order rooted tree.

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7.

In preorder, the root comes first, then the left subtree in preorder. Thus the preorder is the vertices of the left subtree (the one rooted at

8.

b )in preorder, then c . Recursively, the preorder

in the subtree at

b is b , followed by d , followed by the vertices in the subtree rooted at e in

preorder, namely

e , f , g . putting this all together, we obtain the answer a , b , d , e , f , g , c .

See comments in the solution to Exercise 7 for the procedure. The only difference here is that some vertices have more than two children: after listing such a vertex, we list the vertices of its subtrees, in preorder, from left to right. The answer is

9.

a , followed by

a , b , d , e , i, j , m, n , o , c , f , g ,h , k , l , p .

See the comments in the solution to Exercise 7 for the procedure. The only difference here is that some vertices have more than two children: after listing such a vertex, we list the vertices of its subtree, in preorder, from left to right. The answer is

16. a) Represent the expression

a , b , e , k , l , m, f , g , n ,r , s , c , d , h , o , i, j , p , q .

( ( x+ 2 ) ↑ 3 )∗( y−( 3+ x ) ) −5

using a binary tree.

We build the tree from the top down while analyzing the expression by identifying the outer most operation at each stage. The outermost operation in this expression is the final subtraction.

Therefore the tree has – at its root, with two operands as the the subtrees at the root. The right operand is clearly 5, so the right child of the root is 5. The left operand is the result of multiplication, to the left subtree has * as its root. We continue recursively in this way until the entire tree is constructed.

Write this expression in: b) prefix notation – We can read off the answer from the picture we have just drawn by listing the vertices of the tree in preorder: First list the root, the left subtree in preorder, then the right subtree in preorder. Therefore the answer is

−¿ ↑+ x 23− y +3 x 5 .

c) postfix notation – We can read off the answer from the picture that we have just drawn simply by listing the vertices of the tree in postorder:

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x 2+m ↑ y 3 x ±∗5−¿ . expression is just the given expression, fully parenthesized:

d) infix notation – The infix

( ( ( ( x +2 ) ↑3 )∗( y −( 3+ x ) ) )−5) . This

corresponds to traversing the tree in inorder, putting in a left parenthesis whenever we go down to a left child and putting in a right parenthesis whenever we come up from a right child.

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