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Question 1

1 / 1 pts

Is the following statement true or fa~se?

A transition system is a directed graph whose vatices correspond to the states of the wortd and whose edges are labelled by actions.

"--~-----,;

'

Correcll ..

~

TnJe

1 / 1 pts

Question 2 Is the fo llowing statement true or false?

The actron ..hofd" in the foHowfng CJfrigo program is non--0et:erministic_

%

n: p n.dulum Ip % sort!. aoo abJ ct d boo! n(~fJ 1) • hold T), •a MO 1

,

t(T).

(T).

%by do

cha -0. m,.1

+

eO m,.1

In

False

pos.ltfnn

QuesUon 3

m

Considef' the simple transition s~rstem in .he foll owing image.. ~-'ihf b ans«Ner optrnn represents k nitial state is exog~nous"'1

-

-

p-

p-

t

-

f I pts

Giveri this Bayes·an nerwork. how many probabiHty parameters do we need to specify for this network?

Question 5

Consider the fol lo•Nin,g MltJN it,·ith ti.vo object constants ail ice alld l:t;JD.

\'Jtlich answer option js th.~ unoormaHzed prob.abHity measure {i.e.., \'ieigh.t} of tl7e mterpretation {in Ltence(alioe, bob.I} ? f~:, y ranges over botn object

r:onstant.sJ ?

HJNT:

Vle are oonsidEri111g tllir weight unde-r Mall'kov Log ic Network

8ll'.pf8)

Question 6 In ALC. •Nflich ca egocy does ...Student(John,■ ll,ewng to?

Coii::~pt c11:se1t1a11

111fJ\S

1/lpts

Question 7

[n reg,ar.d t o syrna f.eJationship, which 0 .1: t he folloi-\ling statem-ems js

co rect?

Cln

c, ov ,

3

0

C lP

w

f.Jr ~~ :-o~r. :I ~:;~7 A. G.v ..::.SP r.m:9 n:.

...,.

I

1

1

--,

-l

q-

--

31

-

i ii,

&a.mmethe forlowing probabil" y distnbution tanle.

0

1/Jtiat fs the probability tr.at t he- \'teaiher ·s stmny given t hat the temperature is hot?

P(Wea her= sunny ITempera1ure =hot} = ?

0.6 0

C _5

Question 10

,

1/lpts

Ql!lestion 11

Wnat v~iU be the stab[~ madef · · p is true at f me O and a(O) rs true in th.f follm-,ing program,

p(t, , ) :- a{O).

{p , O); p( , O)} 1. { (0}}. {p(

1)} :- p{

0).

{p( • 1)} ;. p( • 0).

:- no 1{p(t. 1): p(, 1H1"

n_

r-

Question 12

Consider the follO'-mng clingo program \'thh 5 nJl es. bird( X} :- fcSH:!entBird(X). blrd(X} :- mig.ratorJBird(X).

-- residentBird(X}, m.igratoryBtrd{X)

residemB1rdao). mi graroryBiraQo).

What are the stab~ models?

: r .:i .:

Un8cti :s iai:-le

1 / 1 pts

1 /1pts

Question lJ Which rufe is not alJowed in dingo giveo the fol lm-vmg ruft(1

bool.ean(true; alss),

E:

Q.iu:ecstian 14 Which mateme-nt of concept satisfi ability rs 0011-eci:?

L

JI

} ~

.:

2

1 / 1 rpts

Question 1 5 Ir. - ~ D,-_,r~ Ii.: l'J~I:: •·~rn.' icJ- af ~ ·::1 k r, •·11:1 iJF"b ~ p J ~itb11 f ~ ,:"I !Jk: T "s; ~ r.ci 'nerr~I fl r~r:'?

'

{ r ght

)

~

·gh

• •-

2,1

{le

~

{ le

1 } =·

!i'. ; ~p ~~iti:5 ·

~~

.1 old{T), -

t

•

- -0--r -1-

- -~

•

Quiz 5oore: 15 u: o 1 5

Question 1

1 / 1 pts

Is the following statement true or fa~se?

A transition system is a directed graph whose vatices correspond to the states of the wortd and whose edges are labelled by actions.

"--~-----,;

'

Correcll ..

~

TnJe

1 / 1 pts

Question 2 Is the fo llowing statement true or false?

The actron ..hofd" in the foHowfng CJfrigo program is non--0et:erministic_

%

n: p n.dulum Ip % sort!. aoo abJ ct d boo! n(~fJ 1) • hold T), •a MO 1

,

t(T).

(T).

%by do

cha -0. m,.1

+

eO m,.1

In

False

pos.ltfnn

QuesUon 3

m

Considef' the simple transition s~rstem in .he foll owing image.. ~-'ihf b ans«Ner optrnn represents k nitial state is exog~nous"'1

-

-

p-

p-

t

-

f I pts

Giveri this Bayes·an nerwork. how many probabiHty parameters do we need to specify for this network?

Question 5

Consider the fol lo•Nin,g MltJN it,·ith ti.vo object constants ail ice alld l:t;JD.

\'Jtlich answer option js th.~ unoormaHzed prob.abHity measure {i.e.., \'ieigh.t} of tl7e mterpretation {in Ltence(alioe, bob.I} ? f~:, y ranges over botn object

r:onstant.sJ ?

HJNT:

Vle are oonsidEri111g tllir weight unde-r Mall'kov Log ic Network

8ll'.pf8)

Question 6 In ALC. •Nflich ca egocy does ...Student(John,■ ll,ewng to?

Coii::~pt c11:se1t1a11

111fJ\S

1/lpts

Question 7

[n reg,ar.d t o syrna f.eJationship, which 0 .1: t he folloi-\ling statem-ems js

co rect?

Cln

c, ov ,

3

0

C lP

w

f.Jr ~~ :-o~r. :I ~:;~7 A. G.v ..::.SP r.m:9 n:.

...,.

I

1

1

--,

-l

q-

--

31

-

i ii,

&a.mmethe forlowing probabil" y distnbution tanle.

0

1/Jtiat fs the probability tr.at t he- \'teaiher ·s stmny given t hat the temperature is hot?

P(Wea her= sunny ITempera1ure =hot} = ?

0.6 0

C _5

Question 10

,

1/lpts

Ql!lestion 11

Wnat v~iU be the stab[~ madef · · p is true at f me O and a(O) rs true in th.f follm-,ing program,

p(t, , ) :- a{O).

{p , O); p( , O)} 1. { (0}}. {p(

1)} :- p{

0).

{p( • 1)} ;. p( • 0).

:- no 1{p(t. 1): p(, 1H1"

n_

r-

Question 12

Consider the follO'-mng clingo program \'thh 5 nJl es. bird( X} :- fcSH:!entBird(X). blrd(X} :- mig.ratorJBird(X).

-- residentBird(X}, m.igratoryBtrd{X)

residemB1rdao). mi graroryBiraQo).

What are the stab~ models?

: r .:i .:

Un8cti :s iai:-le

1 / 1 pts

1 /1pts

Question lJ Which rufe is not alJowed in dingo giveo the fol lm-vmg ruft(1

bool.ean(true; alss),

E:

Q.iu:ecstian 14 Which mateme-nt of concept satisfi ability rs 0011-eci:?

L

JI

} ~

.:

2

1 / 1 rpts

Question 1 5 Ir. - ~ D,-_,r~ Ii.: l'J~I:: •·~rn.' icJ- af ~ ·::1 k r, •·11:1 iJF"b ~ p J ~itb11 f ~ ,:"I !Jk: T "s; ~ r.ci 'nerr~I fl r~r:'?

'

{ r ght

)

~

·gh

• •-

2,1

{le

~

{ le

1 } =·

!i'. ; ~p ~~iti:5 ·

~~

.1 old{T), -

t

•

- -0--r -1-

- -~

•

Quiz 5oore: 15 u: o 1 5

Question 1

1 / 1 pts

Is the following statement true or fa~se?

A transition system is a directed graph whose vatices correspond to the states of the wortd and whose edges are labelled by actions.

"--~-----,;

'

Correcll ..

~

TnJe

1 / 1 pts

Question 2 Is the fo llowing statement true or false?

The actron ..hofd" in the foHowfng CJfrigo program is non--0et:erministic_

%

n: p n.dulum Ip % sort!. aoo abJ ct d boo! n(~fJ 1) • hold T), •a MO 1

,

t(T).

(T).

%by do

cha -0. m,.1

+

eO m,.1

In

False

pos.ltfnn

QuesUon 3

m

Considef' the simple transition s~rstem in .he foll owing image.. ~-'ihf b ans«Ner optrnn represents k nitial state is exog~nous"'1

-

-

p-

p-

t

-

f I pts

Giveri this Bayes·an nerwork. how many probabiHty parameters do we need to specify for this network?

Question 5

Consider the fol lo•Nin,g MltJN it,·ith ti.vo object constants ail ice alld l:t;JD.

\'Jtlich answer option js th.~ unoormaHzed prob.abHity measure {i.e.., \'ieigh.t} of tl7e mterpretation {in Ltence(alioe, bob.I} ? f~:, y ranges over botn object

r:onstant.sJ ?

HJNT:

Vle are oonsidEri111g tllir weight unde-r Mall'kov Log ic Network

8ll'.pf8)

Question 6 In ALC. •Nflich ca egocy does ...Student(John,■ ll,ewng to?

Coii::~pt c11:se1t1a11

111fJ\S

1/lpts

Question 7

[n reg,ar.d t o syrna f.eJationship, which 0 .1: t he folloi-\ling statem-ems js

co rect?

Cln

c, ov ,

3

0

C lP

w

f.Jr ~~ :-o~r. :I ~:;~7 A. G.v ..::.SP r.m:9 n:.

...,.

I

1

1

--,

-l

q-

--

31

-

i ii,

&a.mmethe forlowing probabil" y distnbution tanle.

0

1/Jtiat fs the probability tr.at t he- \'teaiher ·s stmny given t hat the temperature is hot?

P(Wea her= sunny ITempera1ure =hot} = ?

0.6 0

C _5

Question 10

,

1/lpts

Ql!lestion 11

Wnat v~iU be the stab[~ madef · · p is true at f me O and a(O) rs true in th.f follm-,ing program,

p(t, , ) :- a{O).

{p , O); p( , O)} 1. { (0}}. {p(

1)} :- p{

0).

{p( • 1)} ;. p( • 0).

:- no 1{p(t. 1): p(, 1H1"

n_

r-

Question 12

Consider the follO'-mng clingo program \'thh 5 nJl es. bird( X} :- fcSH:!entBird(X). blrd(X} :- mig.ratorJBird(X).

-- residentBird(X}, m.igratoryBtrd{X)

residemB1rdao). mi graroryBiraQo).

What are the stab~ models?

: r .:i .:

Un8cti :s iai:-le

1 / 1 pts

1 /1pts

Question lJ Which rufe is not alJowed in dingo giveo the fol lm-vmg ruft(1

bool.ean(true; alss),

E:

Q.iu:ecstian 14 Which mateme-nt of concept satisfi ability rs 0011-eci:?

L

JI

} ~

.:

2

1 / 1 rpts

Question 1 5 Ir. - ~ D,-_,r~ Ii.: l'J~I:: •·~rn.' icJ- af ~ ·::1 k r, •·11:1 iJF"b ~ p J ~itb11 f ~ ,:"I !Jk: T "s; ~ r.ci 'nerr~I fl r~r:'?

'

{ r ght

)

~

·gh

• •-

2,1

{le

~

{ le

1 } =·

!i'. ; ~p ~~iti:5 ·

~~

.1 old{T), -

t

•

- -0--r -1-

- -~

•

Quiz 5oore: 15 u: o 1 5

FinalExam_4/28/21: CSE 579: Knowledge Representation (2021 Spring)

Page 1 of 14

2021Spring-T-CSE579-34823 FinalExam_4/28/21

T Account Dashboard Courses Groups Calendar 2 unread Inbox messages.2 History Commons 10 release Help notes.10 ASU Home My ASU Colleges & Schools Map & Locations Contact Us

FinalExam_4/28/21

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FinalExam_4/28/21: CSE 579: Knowledge Representation (2021 Spring)

Page 2 of 14

Due Apr 28 at 7:05pm Points 15 Questions 15 Available Apr 28 at 6:10pm - Apr 28 at 7:05pm about 1 hour Time Limit 55 Minutes

Instructions 

Your camera must be ON until the end of the exam. Please turn off your phone and show your ID (student ID or passport or Driving license) at the beginning of the exam. OFF camera means that you are absent, and your grade is zero.

This quiz was locked Apr 28 at 7:05pm.

Attempt History LATEST

Attempt

Time

Score

Attempt 1

38 minutes

15 out of 15

Score for this quiz: 15 out of 15 Submitted Apr 28 at 6:49pm This attempt took 38 minutes.

Question 1 1 / 1 pts

Is the following statement true or false?

A transition system is a directed graph whose vertices correspond to the states of the world and whose edges are labelled by actions.

True

Correct!

False

0

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FinalExam_4/28/21: CSE 579: Knowledge Representation (2021 Spring)

Page 3 of 14

Question 2 1 / 1 pts Is the following statement true or false?

The action “hold” in the following Clingo program is non-deterministic.

% File: pendulum.Ip % sorts and object declaration boolean (t;f).

% effects of hold rlght(T +1) :- hold(T), right(T). left(T+1) :- hold(T), left(T). % by default, pendulum changes the position {left.(T+1)} :-l"ight(T), T=O .. m-1. {~ight(T+1)} :- left(T), T=O .. m-1. % fluents are exogenous initially 1{right(O);left(O)}1. % uniqueness and existence of values for fluents :- not 1{right(T);left(T)}1, T=1 .. m. % exogenous action {hold(T)} :-T=O.. m-1.

True

0 Correct!

False

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FinalExam_4/28/21: CSE 579: Knowledge Representation (2021 Spring)

Page 4 of 14



Consider the simple transition system in the following image.

Which answer option represents “initial state is exogenous”?

I -

 

 1{p(t,0); p(f,0)}1

Correct!

{a(0)}

0 {p(t,1)} :- p(t,0)

0 p(t,1) :- a(0)

0

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FinalExam_4/28/21: CSE 579: Knowledge Representation (2021 Spring)

Page 5 of 14

1 / 1 pts



Given this Bayesian network, how many probability parameters do we need to spec for this network?

WetGrass

 

 5

0 9

Correct!

7

0 8

0

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FinalExam_4/28/21: CSE 579: Knowledge Representation (2021 Spring)

Page 6 of 14

Question 5 1 / 1 pts

Consider the following MLN with two object constants alice and bob.

Which answer option is the unnormalized probability measure (i.e., weight) of the interpretation {influence(alice, bob)} ? (x, y ranges over both object constants) ?

2 : smoke(y) f- smoke(x) /\ influence(x, y) 1 : smok,e(alice)

HINT: We are considering the weight under Markov Logic Network.

exp(4)

0 exp(8)

Correct!

exp(0)

0 exp(2)

0

In ALC, which category does “Student(John)” belong to?

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FinalExam_4/28/21: CSE 579: Knowledge Representation (2021 Spring)

Page 7 of 14

Concept inclusion

0 Tbox

0 Role assertion

0 Concept assertion

Correct!

Question 7 1 / 1 pts

In regard to syntax relationship, which of the following statements is correct?



1

(>=2 RC) in descrip tion logics ca n be, written as R m in 2 i

'l,

V R. C in description logics can be written as Ra ll C in 0\1\

.,

- R.C in description logics ca n be written as R some C in

c,{

C1 r-



•

-, C2 can be w ri tte n as C1 SubClassOf C2 in OWL.  

3

Correct!

2

0 1

0 4

0

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FinalExam_4/28/21: CSE 579: Knowledge Representation (2021 Spring)

Page 8 of 14

Question 8 1 / 1 pts

The completion of an ASP program Π is defined as the union of Π and

for each ground atom A. Given ASP program Π:

p

-.q

q

-.·r

What is the formula after the completion?

1

p r

---iq

p B -,q q B ,r

r +--1p B -,q q+-1-,r

,r

v\

p B -,q q+-1-,p

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FinalExam_4/28/21: CSE 579: Knowledge Representation (2021 Spring)

Page 9 of 14

4

0 3

Correct!

1

0 2

0



Examine the following probability distribution table.

Weather

Temperature

µ(w)

sunny

hot

0.10

sunny

mild

0.20

sunny

cold

0.10

cloudy

hot

0.05

cloudy

mild

0.35

cloudy

cold

0.20

What is the probability that the weather is sunny given that the temperature is hot?

P(Weather = sunny | Temperature = hot) = ?

 



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FinalExam_4/28/21: CSE 579: Knowledge Representation (2021 Spring)

Page 10 of 14

0.25

0 0.20

0 0.10

0 0.67

Correct!

Question 10 1 / 1 pts



Which of the following options is the transformation of 7rx(Man(a) A J hasCh ilc

f

Man(a) A J y (hasChild(x, y) A female(y)).

(},-

Man(a) A =l x (hasChi ld(x, y) A fema le(y)).

:}

V a Man(a) A J y (hasChi ld(x, y) A fema le(y)) .

• .i,;

V x Man(x) A J y (hasChi ld(x, y) A fema le(y)).



 

1

Correct!

2

0 3

0 4

0

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FinalExam_4/28/21: CSE 579: Knowledge Representation (2021 Spring)

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Question 11 1 / 1 pts

What will be the stable model if p is true at time 0 and a(0) is true in the following program?



p(t, 1) :- a(O).

1{p(t, 0); p(f, 0)} 1.

{a(O)}.

{ p(t, 1)} :- p(t, 0).

{p(f,. 1)} :- p(f, 0).. :- not 1 {o(t. 1 ): o<f. 1n1. - - - - - -



 

{p(t, 0), a(0), p(t, 1), p(f, 1)}

0 {p(t, 0), a(0), p(t, 1)}

Correct!

{p(t, 0), a(0), p(f, 1)}

0 {p(t, 1)}

0

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FinalExam_4/28/21: CSE 579: Knowledge Representation (2021 Spring)

Page 12 of 14

Consider the following clingo program with 5 rules. bird(X) :- residentBird(X). bird(X) :- migratoryBird(X). :- residentBird(X), migratoryBird(X). residentBird(jo). migratoryBird(jo).

What are the stable models?

{ residentBird(jo), migratoryBird(jo)}

0 Unsatisfiable

Correct!

{ bird(jo)}

0 {bird(jo), residentBird(jo), migratoryBird(jo)}

0

Question 13 1 / 1 pts

Which rule is not allowed in clingo given the following rule? boolean(true; false).

1{hasBananas(BB, 0): boolean(BB)}1.

0 :-not 1{hasBananas(BB, T): boolean(BB)}1, T=1..m

0 Correct!

1{hasBananas(BB, 0)}1.

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FinalExam_4/28/21: CSE 579: Knowledge Representation (2021 Spring)

Page 13 of 14

:- not 1{hasBananas(BB, T)}1, T=1..m.

0

Question 14 1 / 1 pts



Which statement of concept satisfiability is correct? ,

A concept C is satisfiable with respect to a knowledge base Keven if K itself is not satisfiable.

2_3

'1'

A concept C is satisfiable with respect to a knowledge base K if there exists a model I of K such Pa rentOfOnlyMa leChildren _ Person n 'if hasChild.Male.

Consider the knowledge base K below.

Male ~ -. Female The concept



Male n , Female



 4

0 2

Correct!

1

0 3

0

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FinalExam_4/28/21: CSE 579: Knowledge Representation (2021 Spring)

Page 14 of 14



In the Pendulum problem, which of the following options represents that the positio of pendulum is a non-inertial fluent?

{right(T +1 )} :- right(T), T=O .. im -1. right(T +1) :- hold(T), right(T}. {I eft(T+ 1)} ::- right(T), T=O .. m -1 .. {I eft(T+ 1)} ::- Ieft(T), T=O .. m-1 .  

 2

0 4

0 1

0 Correct!

3

Quiz Score: 15 out of 15

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Module 5 Practice Quiz Due No due date Points 10 Available after Oct 16 at 12am

Questions 10 Time Limit None

Allowed Attempts Unlimited

Take the Quiz Again

Attempt History LATEST

Attempt

Time

Score

Attempt 1

less than 1 minute

0 out of 10

Submitted Dec 6 at 10:22pm Unanswered

0 / 1 pts

Question 1

Consider the transition system in the following image, which option contains all the fluent(s)?

a= f

p

a= f

a= t = f 1-----------.;;,.j p=t

a= t

both p=t and a=t orrect Answer

0

p

p, a, t, and f a

Unanswered

0 / 1 pts

Question 2

Which options are the stable models that correspond to the following transition system? Choose all that apply.

a= f

p= f

a= f

a= t

p=t

a= t orrect Answer

[) {p(f,0), p(t,1), a(0)} {p(f,0), p(t,1)}

orrect Answer orrect Answer

Unanswered

[) {p(f,0), p(f,1)} [) {p(t,0), p(t,1), a(0)}

Question 3

0 / 1 pts

Consider the same transition system in question 2, which option, along with the following constraint, represents the commonsense law of inertia?

:- T=1..m, not 1{p(t, T); p(f, T)}1.

{a(0)}. p(t,1) :- a(0).

1{p(t,0); p(f,0)}1. orrect Answer

Unanswered

0

{p(t,1)} :- p(t,0).

0 / 1 pts

Question 4

Choose the transition system described by the following ASP program.

Note that: p denotes a fluent symbol, and a denotes an action symbol.

(Hint: the 4th rule is different from that in the slides.) boolean(t;f). p(t,T+1) :- a(T), T=0..m-1.

1{p(B,0):boolean(B)}1.

1{p(B,T):boolean(B)}1 :- T=1..m. {a(T)} :- T=0..m-1. {p(B,T+1)} :- p(B,T), T=0..m-1.

a= f

a=f

p=f

a=t

a=t

p=t

a=f

p =f

a=f

a= t

p=t

a= t orrect Answer

0

a=f

a=f

a= t p =f

p=t

a= f a= t a= t

a= t

p =f

a= f

p=t

a= f

Unanswered

Question 5

0 / 1 pts

Consider the following clingo program for “Monkey and Bananas” problem.

%% sort and object declaration boolean(t;f). object(monkey;bananas;box). location(l1;l2;l3). %% state constraints loc(bananas,L,T) :- hasBananas(t,T), loc(monkey,L,T). loc(monkey,L,T) :- onBox(t,T), loc(box,L,T). %% effect and preconditions of actions % walk loc(monkey,L,T+1) :- walk(L,T). :- walk(L,T), loc(monkey,L,T). :- walk(L,T), onBox(t,T). % pushBox loc(box,L,T+1) :- pushBox(L,T). loc(monkey,L,T+1) :- pushBox(L,T). :- pushBox(L,T), loc(monkey,L,T). :- pushBox(L,T), onBox(t,T). :- pushBox(L,T), loc(monkey,L1,T), loc(box,L2,T), L1 != L2. % climbOn onBox(t,T+1) :- climbOn(T). :- climbOn(T), onBox(t,T). :- climbOn(T), loc(monkey,L1,T), loc(box,L2,T), L1 != L2. % climbOff onBox(f,T+1) :- climbOff(T). :- climbOff(T), onBox(f,T). % graspBananas hasBananas(t,T+1) :- graspBananas(T). :- graspBananas(T), hasBananas(t,T). :- graspBananas(T), onBox(f,T). :- graspBananas(T), loc(monkey,L1,T), loc(bananas,L2,T), L1 != L2. % disallow concurrent actions :- walk(L,T), pushBox(L,T). :- walk(L,T), climbOn(T).

:- pushBox(L,T), climbOn(T). :- climbOff(T), graspBananas(T). %% domain independent axioms % fluents are initially exogenous 1{hasBananas(BB,0):boolean(BB)}1.

1{onBox(BB,0):boolean(BB)}1.

1{loc(O,LL,0):location(LL)}1 :- object(O). % uniqueness and existence of fluent values :- not 1{loc(O,LL,T)}1, object(O), T = 1..m. :- not 1{onBox(BB,T)}1, T = 1..m. :- not 1{hasBananas(BB,T)}1, T = 1..m. % actions are exogenous {walk(L,T)} :- location(L), T = 0..m-1. {pushBox(L,T)} :- location(L), T = 0..m-1. {climbOn(T)} :- T = 0..m-1. {climbOff(T)} :- T = 0..m-1. {graspBananas(T)} :- T = 0..m-1. % commonsense law of inertia {hasBananas(B,T+1)} :- hasBananas(B,T), T=0..m-1. {onBox(B,T+1)} :- onBox(B,T), T=0..m-1. {loc(O,L,T+1)} :- loc(O,L,T), T=0..m-1.

Which option is a set of rules that makes the following rule redundant? :- climbOn(T), climbOff(T).

onBox(t,T+1) :- climbOn(T). onBox(f,T+1) :- climbOff(T).

1{onBox(BB,0):boolean(BB)}1.

onBox(t,T+1) :- climbOn(T). :- climbOn(T), onBox(t,T). :- climbOn(T), loc(monkey,L1,T), loc(box,L2,T), L1 != L2. onBox(f,T+1) :- climbOff(T). :- climbOff(T), onBox(f,T).

onBox(t,T+1) :- climbOn(T). :- climbOn(T), onBox(t,T). onBox(f,T+1) :- climbOff(T). :- climbOff(T), onBox(f,T). orrect Answer

onBox(t,T+1) :- climbOn(T). onBox(f,T+1) :- climbOff(T).

0

Unanswered

:- not 1{onBox(BB,T)}1, T = 1..m.

Question 6

0 / 1 pts

Consider the following clingo program for “Blocks World” problem.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % sort and object declaration %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % every block is a location location(B) :- block(B). % the table is a location location(table). %%%%%%%%%%%%%%%%%%%%%%%%%% % state description %%%%%%%%%%%%%%%%%%%%%%%%%% % two blocks can't be on the same block at the same time :- 2{on(BB,B,T)}, block(B), T = 0..m.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % effect and preconditions of action %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % effect of moving a block on(B,L,T+1) :- move(B,L,T).

% concurrent actions are limited by num of grippers :- not {move(BB,LL,T)} grippers, T = 0..m-1. % a block can be moved only when it is clear :- move(B,L,T), on(B1,B,T). % a block can't be moved onto a block that is being moved also :- move(B,B1,T), move(B1,L,T).

%%%%%%%%%%%%%%%%%%%%%%%%%%%% % domain independent axioms %%%%%%%%%%%%%%%%%%%%%%%%%%%% % fluents are initially exogenous 1{on(B,LL,0):location(LL)}1 :- block(B). % uniqueness and existence of value constraints :- not 1{on(B,LL,T)}1, block(B), T=1..m. % actions are exogenous {move(B,L,T)} :- block(B), location(L), T = 0..m-1. % commonsense law of inertia {on(B,L,T+1)} :- on(B,L,T), T < m.

Suppose we have two grippers G1 and G2. According to the ASP formalization, can G1 move B onto C at the same time G2 moves C onto the table?

Yes orrect Answer

Unanswered

0

No

Question 7

0 / 1 pts

Consider the following clingo program for “Blocks World” problem.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % sort and object declaration %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % every block is a location location(B) :- block(B). % the table is a location location(table). %%%%%%%%%%%%%%%%%%%%%%%%%% % state description %%%%%%%%%%%%%%%%%%%%%%%%%% % two blocks can't be on the same block at the same time :- 2{on(BB,B,T)}, block(B), T = 0..m.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % effect and preconditions of action %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % effect of moving a block on(B,L,T+1) :- move(B,L,T). % concurrent actions are limited by num of grippers :- not {move(BB,LL,T)} grippers, T = 0..m-1. % a block can be moved only when it is clear :- move(B,L,T), on(B1,B,T). % a block can't be moved onto a block that is being moved also :- move(B,B1,T), move(B1,L,T).

%%%%%%%%%%%%%%%%%%%%%%%%%%%% % domain independent axioms %%%%%%%%%%%%%%%%%%%%%%%%%%%% % fluents are initially exogenous 1{on(B,LL,0):location(LL)}1 :- block(B). % uniqueness and existence of value constraints :- not 1{on(B,LL,T)}1, block(B), T=1..m. % actions are exogenous {move(B,L,T)} :- block(B), location(L), T = 0..m-1. % commonsense law of inertia {on(B,L,T+1)} :- on(B,L,T), T < m.

Suppose we have two grippers G1 and G2. According to the ASP formalization, can G1 move B onto D at the same time G2 moves C onto the table? orrect Answer

0

Yes

No

Unanswered

Question 8

0 / 1 pts

Consider the following clingo program for “Blocks World” problem.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % sort and object declaration %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % every block is a location location(B) :- block(B). % the table is a location location(table). %%%%%%%%%%%%%%%%%%%%%%%%%% % state description %%%%%%%%%%%%%%%%%%%%%%%%%% % two blocks can't be on the same block at the same time :- 2{on(BB,B,T)}, block(B), T = 0..m.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % effect and preconditions of action %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % effect of moving a block on(B,L,T+1) :- move(B,L,T). % concurrent actions are limited by num of grippers :- not {move(BB,LL,T)} grippers, T = 0..m-1.

% a block can be moved only when it is clear :- move(B,L,T), on(B1,B,T). % a block can't be moved onto a block that is being moved also :- move(B,B1,T), move(B1,L,T).

%%%%%%%%%%%%%%%%%%%%%%%%%%%% % domain independent axioms %%%%%%%%%%%%%%%%%%%%%%%%%%%% % fluents are initially exogenous 1{on(B,LL,0):location(LL)}1 :- block(B). % uniqueness and existence of value constraints :- not 1{on(B,LL,T)}1, block(B), T=1..m. % actions are exogenous {move(B,L,T)} :- block(B), location(L), T = 0..m-1. % commonsense law of inertia {on(B,L,T+1)} :- on(B,L,T), T < m.

Suppose there are 2 boxes A and B where A is on the table and B is on A. If you want to use clingo to generate a plan to move A onto B, which option is the set of rules you need to add?

on(a, table, 0; b, a, 0; a, b, m; b, table, m).

:- not on(a, table, 0; b, a, 0). on(a, b, m; b, table, m). orrect Answer

:- not on(a, table, 0). :- not on(b, a, 0). :- not on(a, b, m).

0

:- not on(b, table, m).

:- not on(a, table, 0; b, a, 0; a, b, 1; b, table, 1).

Unanswered

Question 9

0 / 1 pts

Consider the following clingo program discussed in the module “Expressive Possibilities”.

boolean(t;f). % sorts and object declarations object(jack;car). location(home;work). % effect and precondition of go loc(jack,L,T+1) :- go(L,T). {loc(car,L,T+1)} :- go(L,T), loc(car,L1,T), loc(jack,L1,T), T=0..m-1. :- go(L,T), loc(jack,L,T). % fluents are exogenous initially 1{loc(O,LL,0):location(LL)}1 :- object(O). % uniqueness and existence of fluent values :- not 1{loc(O,LL,T)}1, object(O), T=1..m. % actions are exogenous {go(L,T)} :- location(L), T=0..m-1. % fluents are inertial {loc(O,L,T+1)} :- loc(O,L,T), T=0..m-1.

Which option correctly explains why the effect of action “go” is nondeterministic?

The rule below is nondeterministic since the location of O may or may not be L even if it was at L. {loc(O,L,T+1)} :- loc(O,L,T), T=0..m-1.

The rule below represents that the action “go” is exogenous, which means it is not determined. {go(L,T)} :- location(L), T=0..m-1.

The rule below defines a precondition of action “go” that jack cannot go to the place that he is currently at. :- go(L,T), loc(jack,L,T). orrect Answer

0 The rule below uses a choice rule in the head to represent that the location of the car may or may not change to the location that jack goes to. {loc(car,L,T+1)} :- go(L,T), loc(car,L1,T), loc(jack,L1,T), T=0..m-1.

Unanswered

Question 10

0 / 1 pts

Consider the following clingo program for “pendulum” problem.

% sorts and object declaration boolean(t;f). % effect of hold right(T+1) :- hold(T), right(T). left(T+1) :- hold(T), left(T). % by default, pendulum changes the position {left(T+1)} :- right(T), T=0..m-1. {right(T+1)} :- left(T), T=0..m-1. % fluents are exogenous initially 1{right(0);left(0)}1. % uniqueness and existence of values for fluents :- not 1{right(T);left(T)}1, T=1..m. % exogenous action {hold(T)} :- T=0..m-1.

Suppose we observe that the pendulum is on the right at time 0. We also observe that it is also on the right at time 2. When m equals to 2, how many stable models of the above program will agree with these observations?

4 orrect Answer

0 2 1 0

Module 6 Practice Quiz Due No due date Points 10 Available after Oct 23 at 12am

Questions 10 Time Limit None

Allowed Attempts Unlimited

Take the Quiz Again

Attempt History LATEST

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Attempt 1

less than 1 minute

0 out of 10

Submitted Dec 6 at 10:25pm I

Unanswered

Question 1

0 / 1 pts

Consider the following MLN program with two object constants alice and bob. Compute the unnormalized probability measure (i.e., weight) of the interpretation I = {smoke(alice), smoke(bob)} under MLN. (x, y range over both object constants.)

2: smoke(y) +- smoke(x) I\ influence(x, y) 1 : smoke( alice)

a : influence( alice, bob)

orrect Answer

0

exp(9)

exp(1) exp(3) 0

I

Unanswered

Question 2

0 / 1 pts

Consider the following LPMLN program with two object constants alice and bob. Compute the unnormalized probability measure (i.e., weight) of the interpretation I = {smoke(alice), smoke(bob)} under LPMLN. (x, y range over both object constants.)

2: smoke(y) +- smoke(x) I\ influence(x, y) 1 : smoke( alice) a : influence( alice, bob)

exp(𝛂+1) exp(3) orrect Answer

0

0

exp(𝛂+9)

Unanswered

Question 3

0 / 1 pts

Consider the following LPMLN program. Apply the completion method to convert the LPMLN program into an equivalent Markov Logic Network. Which option is the correct MLN program?

a:p

1: p +- ,q orrect Answer

a:p

(

1: p +-- ,q a: p-+ TV ,q a: q -+..l a:p

1: p +-- ,q a: p-+ ,q a: q -+..l

a:p

1 : p +-- ,q a:p-+T a: p-+ ,q a: q -t..l a:p

1: p +-- ,q a: p-+ TV ,q

I

Unanswered

Question 4

0 / 1 pts

Which option is the most probable stable model of the following LPMLN program?

1 :p a.:pf-q Q :_l_f- ,q

∅ {p, q} {q} orrect Answer

0

{p}

I Unanswered

Question 5

0 / 1 pts

Which option is the weight of the most probable stable model of the following LPMLN program?

l:q -2: pf- q -3 :_l_f- r

exp(1) orrect Answer

0 exp(-2) 1 -2

Unanswered

Question 6

0 / 1 pts

Consider the following LPMLN program. Translate it into an ASP program containing weak constraints. Which option is the correct translated ASP program?

1 :p -2 : p t---- q 3 :_it- p

unsat(1) :- not p. p :- not unsat(1). :~ unsat(1). [-1]

unsat(2) :- q, not p. p :- q, not unsat(2). :~ unsat(2). [2]

unsat(3) :- p. :- p, not unsat(3). :~ unsat(3). [-3]

unsat(1) :- not p. :~ unsat(1). [1]

unsat(2) :- q, not p. :~ unsat(2). [-2]

unsat(3) :- p. :~ unsat(3). [3] orrect Answer

unsat(1) :- not p. p :- not unsat(1). :~ unsat(1). [1]

unsat(2) :- q, not p. p :- q, not unsat(2). :~ unsat(2). [-2]

unsat(3) :- p. :- p, not unsat(3).

0

:~ unsat(3). [3]

l:p -2: p +-- q 3 :..l+-- p a:p ➔ TVq

a : q --+..l

Unanswered

Question 7

0 / 1 pts

Consider the following LPMLN program. Which option is the most probable stable model of the program?

1 :p -2 : p +---- q 3 :..it- p {p, q} {q} orrect Answer

0

∅

{p}

Unanswered

Question 8

0 / 1 pts

Which option about probability is impossible no matter what random variables A and B are?

P(A, B) < P(B) P(A | B) < P(B | A) orrect Answer

0

P(A | B) < P(A, B)

P(A, B) = P(A | B)

Unanswered

Question 9

0 / 1 pts

Consider the following full joint distribution for Boolean variables A, B, and C. Which option is closest to the value of P(A = 1 | B=1, C = 0)?

A

B

C

P(A,B,C)

0

0

0

0.2

0

0

0

0.1

0

0

0.15 0.05

0

0

0

0.05 0.25

0

0.05 0.15

orrect Answer

0

0.25

0.5 2.5 0.05

I

Unanswered

0 / 1 pts

Question 10

Consider the same full joint distribution for Boolean variables A, B, and C as in question 9. Are A and B independent of each other?

A

B

C

P(A,B,C)

0

0

0

0.2

0

0

0

0.1

0

0

0.15 0.05

0

0

0

0.05 0.25

0

0.05 0.15

orrect Answer

0

True

False

Module 7 Practice Quiz Due No due date Points 10 Available after Nov 4 at 11:59pm

Questions 10 Time Limit None

Allowed Attempts Unlimited

Take the Quiz Again

Attempt History LATEST

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Question 1

0 / 1 pts

Consider the following TBox.

Ac :3R.B A C\/R.C Which option is the First-Order formula that is translated from this TBox?

\lx(A(x) ➔ :3y(R(x, y) I\ B(y))) I\ \lx(A(x) ➔ \ly(R(x, y) I\ C(y))) orrect Answer

r,.

\lx(A(x) ➔ :3y(R(x, y) I\ B(y))) I\ \lx(A(x) ➔ \ly(R(x, y) ➔ C(y)))

:3x(A(x) ➔ :3y(R(x,y) ➔ B(y))) I\ \lx(A(x) ➔ \ly(R(x,y) ➔ C(y)))

:3x(A(x) ➔ :3y(R(x, y) I\ B(y))) I\ \t'x(A(x) ➔ \t'y(R(x, y) ➔ C(y)))

I

Unanswered

Question 2

0 / 1 pts

Which option is entailed by the following TBox?

AC :3R.B A C\:/R.C orrect Answer

(

A c :3R. (B n C)

A c 'v'R. (B n C)

I

Unanswered

Question 3 Consider the following ALC role assertions. teaches(Joe, KR) teaches(Joe, AI) ≤0 teaches(Joe)

Which option is correct? orrect Answer

0 It is an ABox and is not satisfiable.

It is a TBox and is not satisfiable.

It is an ABox and is satisfiable.

It is a TBox and is satisfiable.

0 / 1 pts

I

Unanswered

Question 4

0 / 1 pts

Which kind of reasoning in ontologies should you use in answering the previous question (question 3)?

Answering concept queries Taxonomies orrect Answer

0 Knowledge Base Satisfiability Concept Satisfiability

I

Unanswered

Question 5

0 / 1 pts

Which option is a suitable ALC concept in description logic for the class of humans “happy pet owner”, using the following concept names and role name? concept names: Person, Happy, Animal role name: own

Happy ⊓ Person ⊓ owns.Animal Person ⊓ Happy ⊓ Animal ⊓ owns(Person, Animal) orrect Answer

0 Person ⊓ Happy ⊓ ∃owns.Animal Happy ⊓ Person ⊓ owns(Person, Animal)

Unanswered

Question 6

0 / 1 pts

Which option about the use of ontologies in industrial applications is correct?

The axims in WordNet are written in First-Order Logic.

The nodes in a knowledge graph captures entities, attributes, and relationships.

orrect Answer

0 DBpedia automatically evolves as Wikipedia changes.

The axioms in CYC are automatically generated from web sources.

I

Unanswered

Question 7

0 / 1 pts

Which option is correct about RDF?

User defines class hierarchies in RDF to disallow nonsensical statements.

Resources can be thought of as objects, thus is disjoint from properties.

Every statement has an Internationalized Resource Identifier. orrect Answer

0 We can use reification to turn a sequence of information into statements in RDF.

Unanswered

Question 8

0 / 1 pts

Using the following concept names and role names, which option can correctly capture the knowledge “cars are exactly those vehicles that have wheels and are powered by an engine”? concept names: Vehicle, Car, Wheel, Engine

role names: hasPart, poweredBy

Car ⊒ Vehicle ⊓ ∀hasPart.Wheel ⊓ ∀poweredBy.Engine orrect Answer

0 Car ≡ Vehicle ⊓ ∃hasPart.Wheel ⊓ ∃poweredBy.Engine Car ⬄ Vehicle ⊓ hasPart.Wheel ⊓ poweredBy.Engine Car ⊑ Vehicle ⊓ ∃hasPart.Wheel ⊓ ∀poweredBy.Engine

I

Unanswered

Question 9

0 / 1 pts

Using the following concept names and role names, which option can correctly capture the knowledge “bicycles have exactly two wheels”? concept names: Bicycle, Wheel role names: hasPart orrect Answer

0 Bicycle ⊑ (=2 hasPart.Wheel) Bicycle ⊑ ∃hasPart.Wheel ⊓ ∃hasPart.Wheel Bicycle ≡ (=2 hasPart.Wheel) Bicycle ≡ ∀hasPart.Wheel ⊓ (=2 hasPart.Wheel)

Unanswered

Question 10 Which of the following statements is correct?

A ⊓ ∃r.(B ⊔ C) is subsumed by A ⊓ ∃r.B A ⊓ ∃r.B is subsumed by A ⊓ ∀r.B

0 / 1 pts

A ⊓ ∀r.B is subsumed by A ⊓ ∃r.B orrect Answer

0 A ⊓ ∃r.A ⊓ ∀r.B is subsumed by A ⊓ ∃r.B

Quiz #5 Due Oct 30 at 11:59pm Points 10 Available until Oct 30 at 11:59pm

Questions 10

Time Limit 300 Minutes

This quiz was locked Oct 30 at 11:59pm.

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Score for this quiz: 10 out of 10 Submitted Oct 30 at 7:36pm This attempt took 22 minutes.

1 / 1 pts

Question 1

Consider the following image, which options do not represents "actions are exogenous”?

a= f

p= f

a= f

a= t

p=t

a= t Correct!

p(t,1) :- a(0).

Correct!

{p(t,1)} :- p(t,0).

Correct!

1{p(t,0); p(f,0)}1.

{a(0)}.

Question 2

1 / 1 pts

Which option is not correct about action description languages?

They are used to describe the preconditions and effects of actions.

Correct!

The “commonsense law of inertia” describes the effect of actions.

Correct!

The fluents in an action language have values, and the values are either t or f.

Correct!

They always map to answer set programs.

Question 3

1 / 1 pts

Choose the transition system described by the following ASP program.

Note that p denotes a fluent symbol, and a denotes an action symbol.

(Hint: the 2nd rule is different from that in the slides.)

boolean(t;f). p(f, T+1) :- a(T), T=0..m-1.

1{p(B,0): boolean(B)}1. :- T=1..m, not 1{p(B, T): boolean(B)}1. {a(T)} :- T=0..m-1. {p(B, T+1)} :- p(B, T), T=0..m-1.

a=f

p=f

a=f

a= t

p=t

a=t Correct!

a= f

a=f

p=f

a= t

p=t

a= t

a=f

a=f a= f p =f

p=t

a= t a= t

a= t

a=t

p=f

a= f

p=t

a=f

Question 4

1 / 1 pts

Consider the same clingo program below that you have seen in question 3. How many stable models are there when m equals to 1?

boolean(t;f). p(f, T+1) :- a(T), T=0..m-1.

1{p(B,0): boolean(B)}1. :- T=1..m, not 1{p(B, T): boolean(B)}1. {a(T)} :- T=0..m-1. {p(B, T+1)} :- p(B, T), T=0..m-1.

8 6 2 Correct!

4

I Question 5

1 / 1 pts

Consider the following clingo program for “Monkey and Bananas” problem.

%% sort and object declaration boolean(t;f). object(monkey;bananas;box). location(l1;l2;l3). %% state constraints loc(bananas,L,T) :- hasBananas(t,T), loc(monkey,L,T). loc(monkey,L,T) :- onBox(t,T), loc(box,L,T). %% effect and preconditions of actions % walk loc(monkey,L,T+1) :- walk(L,T). :- walk(L,T), loc(monkey,L,T). :- walk(L,T), onBox(t,T). % pushBox loc(box,L,T+1) :- pushBox(L,T). loc(monkey,L,T+1) :- pushBox(L,T). :- pushBox(L,T), loc(monkey,L,T). :- pushBox(L,T), onBox(t,T). :- pushBox(L,T), loc(monkey,L1,T), loc(box,L2,T), L1 != L2. % climbOn onBox(t,T+1) :- climbOn(T). :- climbOn(T), onBox(t,T). :- climbOn(T), loc(monkey,L1,T), loc(box,L2,T), L1 != L2. % climbOff onBox(f,T+1) :- climbOff(T). :- climbOff(T), onBox(f,T). % graspBananas hasBananas(t,T+1) :- graspBananas(T). :- graspBananas(T), hasBananas(t,T). :- graspBananas(T), onBox(f,T). :- graspBananas(T), loc(monkey,L1,T), loc(bananas,L2,T), L1 != L2. % disallow concurrent actions :- walk(L,T), pushBox(L,T). :- walk(L,T), climbOn(T).

:- pushBox(L,T), climbOn(T). :- climbOff(T), graspBananas(T). %% domain independent axioms % fluents are initially exogenous 1{hasBananas(BB,0):boolean(BB)}1.

1{onBox(BB,0):boolean(BB)}1.

1{loc(O,LL,0):location(LL)}1 :- object(O). % uniqueness and existence of fluent values :- not 1{loc(O,LL,T)}1, object(O), T = 1..m. :- not 1{onBox(BB,T)}1, T = 1..m. :- not 1{hasBananas(BB,T)}1, T = 1..m. % actions are exogenous {walk(L,T)} :- location(L), T = 0..m-1. {pushBox(L,T)} :- location(L), T = 0..m-1. {climbOn(T)} :- T = 0..m-1. {climbOff(T)} :- T = 0..m-1. {graspBananas(T)} :- T = 0..m-1. % commonsense law of inertia {hasBananas(B,T+1)} :- hasBananas(B,T), T=0..m-1. {onBox(B,T+1)} :- onBox(B,T), T=0..m-1. {loc(O,L,T+1)} :- loc(O,L,T), T=0..m-1.

Which option is a set of rules that makes the following rule redundant? :- walk(L,T), pushBox(L1,T), L!=L1.

:- walk(L,T), onBox(t,T). :- pushBox(L,T), onBox(t,T).

loc(monkey,L,T+1) :- walk(L,T). :- walk(L,T), loc(monkey,L,T). loc(monkey,L,T+1) :- pushBox(L,T). :- pushBox(L,T), loc(monkey,L,T).

Correct!

loc(monkey,L,T+1) :- walk(L,T). loc(monkey,L,T+1) :- pushBox(L,T). :- not 1{loc(O,LL,T)}1, object(O), T = 1..m.

:- walk(L,T), loc(monkey,L,T). :- pushBox(L,T), loc(monkey,L,T). :- not 1{loc(O,LL,T)}1, object(O), T = 1..m.

Question 6

1 / 1 pts

Consider the clingo program for “Monkey and Bananas” problem in question 5. Suppose monkey and bananas are in location L1, and the box is in location L2. If you want to use clingo to generate a plan for the monkey to get the bananas, which option is the set of rules you need to add?

Correct!

:- not loc(monkey, l1, 0; bananas, l1, 0; box, l2, 0). :- not hasBananas(f, 0; t, m).

:- not loc(monkey, l1, 0; bananas, l1, 0; box, l2, 0). hasBananas(f, 0). hasBananas(t, m).

loc(monkey, l1, 0). loc(bananas, l1, 0). loc(box, l2, 0). hasBananas(f, 0; t, m).

:- not loc(monkey, l1, 0). :- not loc(bananas, l1, 0). :- not loc(box, l2, 0). :- not hasBananas(f, 0). :- not hasBananas(t, 1).

Question 7

1 / 1 pts

Consider the following clingo program for “Blocks World” problem.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % sort and object declaration %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % every block is a location location(B) :- block(B). % the table is a location location(table). %%%%%%%%%%%%%%%%%%%%%%%%%% % state description %%%%%%%%%%%%%%%%%%%%%%%%%% % two blocks can't be on the same block at the same time :- 2{on(BB,B,T)}, block(B), T = 0..m.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % effect and preconditions of action %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % effect of moving a block on(B,L,T+1) :- move(B,L,T). % concurrent actions are limited by num of grippers :- not {move(BB,LL,T)} grippers, T = 0..m-1. % a block can be moved only when it is clear :- move(B,L,T), on(B1,B,T). % a block can't be moved onto a block that is being moved also :- move(B,B1,T), move(B1,L,T).

%%%%%%%%%%%%%%%%%%%%%%%%%%%% % domain independent axioms %%%%%%%%%%%%%%%%%%%%%%%%%%%% % fluents are initially exogenous 1{on(B,LL,0):location(LL)}1 :- block(B).

% uniqueness and existence of value constraints :- not 1{on(B,LL,T)}1, block(B), T=1..m. % actions are exogenous {move(B,L,T)} :- block(B), location(L), T = 0..m-1. % commonsense law of inertia {on(B,L,T+1)} :- on(B,L,T), T < m.

Suppose we have two grippers G1 and G2. According to the ASP formalization, when A, B are on the table and C is on B, can G1 move C onto the table at the same time G2 moves A onto B?

Correct!

Yes

No

Question 8

1 / 1 pts

Consider the following clingo program discussed in the module "Expressive Possibilities".

boolean(t;f). % sorts and object declarations object(jack;car). location(home;work). % effect and precondition of go loc(jack,L,T+1) :- go(L,T). {loc(car,L,T+1)} :- go(L,T), loc(car,L1,T), loc(jack,L1,T), T=0..m-1. :- go(L,T), loc(jack,L,T). % fluents are exogenous initially 1{loc(O,LL,0):location(LL)}1 :- object(O). % uniqueness and existence of fluent values :- not 1{loc(O,LL,T)}1, object(O), T=1..m.

% actions are exogenous {go(L,T)} :- location(L), T=0..m-1. % fluents are inertial {loc(O,L,T+1)} :- loc(O,L,T), T=0..m-1.

Suppose Jack and his car are at home at time 0 and he goes to work at time 0. How many stable models are there that following this condition when m equals to 1?

0 Correct!

2

1 4

Question 9 Consider the following clingo program for “lifting” problem. % sort and object declarations boolean(t;f). end(leftEnd;rightEnd). height(low;high). % state condition onTable(f,T) :- level(leftEnd,H,T), level(rightEnd,H1,T), H!=H1. % effect and precondition of lift level(E,high,T+1) :- lift(E,T). :- lift(E,T), level(E,high,T). % fluents are exogenous initially 1{level(E,HH,0): height(HH)}1 :- end(E).

1{onTable(BB,0): boolean(BB)}1. % uniqueness and existence of fluent values :- not 1{level(E,HH,T)}1, end(E), T=1..m. :- not 1{onTable(BB,T)}1, T=1..m.

1 / 1 pts

% actions are exogenous {lift(E,T)} :- end(E), T=0..m-1. % fluents are inertial {level(E,H,T+1)} :- level(E,H,T), T=0..m-1. {onTable(B,T+1)} :- onTable(B,T), T=0..m-1.

When m equals to 1, how many stable models of this program satisfy the following scenario? % scenario onTable(t,0). level(E,low,0) :- end(E). :- not onTable(t,m). :- level(E,low,m).

0 2 4 Correct!

1

Question 10

1 / 1 pts

Considering “monkey grasp bananas” problem in the following image, which option is a fluent?

- ----

. •tr::~ ..; . -;-.::.;-.::..~: ~-, r•••~

u

•,

!..... :-- - ~J ,,..,-

u.

Monkey grasps the bananas.

Monkey moves the box.

Correct!

The location of the box.

The jump of the monkey.

Quiz Score: 10 out of 10

Quiz #6 Due Nov 13 at 11:59pm Points 10 Available until Nov 13 at 11:59pm

Questions 10

Time Limit 300 Minutes

This quiz was locked Nov 13 at 11:59pm.

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25 minutes

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Score for this quiz: 10 out of 10 Submitted Nov 13 at 7:53pm This attempt took 25 minutes.

1 / 1 pts

Question 1 Examine the following probability distribution table.

Weather

Temperature

µ(w)

sunny

hot

0.10

sunny

mild

0.20

sunny

cold

0.10

cloudy

hot

0.05

cloudy

mild

0.35

cloudy

cold

0.20

What is the probability that the weather is cloudy P(Weather = cloudy) = ?

Correct!

0.6

0.4

0.3 0.25

Since all of the events are mutually exclusive, to find out the probability that the weather is cloudy, we simply add up all rows that have cloudy weather, which are rows 4, 5, and 6. As a result, P( Weather = cloudy) = 0.05+0.35+0.20 = 0.6

1 / 1 pts

Question 2 Examine the following probability distribution table.

Weather

Temperature

µ(w)

sunny

hot

0.10

sunny

mild

0.20

sunny

cold

0.10

cloudy

hot

0.05

cloudy

mild

0.35

cloudy

cold

0.20

Given the following probabilities, compute the value of P (Temperature = mild | Weather = cloudy) by applying the Bayes rule.

P (Weather = cloudy)= 0.6 P (Weather = cloudy | Temperature = mild) = 0.64 P (Temperature = mild) = 0.55

0.7 0.3 0.38

Correct!

0.59

By applying the Bayes rule P (Temperature = mild | Weather = cloudy) = [P (Weather = cloudy | Temperature = mild) X P (Temperature = mild) ] /P (Weather = cloudy) = (0.64 X 0.55)/ 0.6 = ~0.59

1 / 1 pts

Question 3 Is the following statement true or false?

Given the following table,

Correct!

and

are marginally independent.

C1

C2

P (C1, C2)

head

head

0.25

head

tail

0.25

tail

head

0.25

tail

tail

0.25

True

False

To prove whether C1 and C2 are marginally independent, one of many ways is to show P(C1 = head) = P(C1=head|C2=head) We know P(C1 = head) = 0.25 + 0.25 = 0.5 By applying the Bayes rule P(C1=head|C2=head) = 0.25/0.5 = 0.5 As a result, we can conclude C1​and C2​marginally independent

Question 4 Given the following Bayesian network, how many probability parameters do we need to specify for this network?

Correct!

9

6

1 / 1 pts

4 16

We need one probability parameter to specify A. For B and C, since both of them are dependent on A, we need to use two probability parameters to specify each of them, so there are 4 of them in total.

Finally, D is dependent on B and C, we need 4 probability parameters. As a result, 4+4+1= 9.

Question 5 Consider the Herbrand universe {a,b} . Select all the correct grounding for the following formula.

\:/ x Smokes (x) ---+ Cancer (x) Correct!

Smokes (b) -----+ Cancer (b) Smokes (a) -----+ Cancer (b) Smokes (b) -----+ Cancer (a)

Correct!

Smokes (a) -----+ Cancer (a)

If we groud the given formula, we will have

Smokes (a) ---+ Cancer (a) Smokes (b) ---+ Cancer (b)

1 / 1 pts

Question 6

1 / 1 pts

Markov logic combines first-order logic with Markov networks.

Correct!

True

False

Question 7

1 / 1 pts

Given the Herbrand universe {a, b} and the following ungrounded formula:

\;/ x Smokes (x) ---+ Cancer (x) the Herbrand interpretation w = smokes(b) is a Herbrand model of the above formula.

True Correct!

@ False

The correct grounding for the given formula is

Smokes (a) ---+ Cancer (a) Smokes (b) ---+ Cancer (b) In order to satisfy the second grounded formula, if smokes(b) is true, cancer(b) has to be true as well.

Question 8

1 / 1 pts

Which of the following choices is not a tight program?

p +--- q, r q +--- r,s s +--- ,p

p +--- ,q q +--- ,r Correct!

p +--- q q +--- p p +--- ,r

p +--- ,q q +--- ,p

pf- q qf-p pf- ,r The above program has a loop, so it cannot be a tight program.

In the rest of the given choices, we don't have any loops.

Question 9 ASP can be viewed as a special case of MLN.

True Correct!

False

1 / 1 pts

ASP can be viewed as a special case of LPMLN.

Question 10

1 / 1 pts

Which option about probability is possible no matter what random variables A and B are? (select all that are correct) Correct!

Correct!

Correct!

P(A, B) = P(A | B)

P(A | B) < P(B | A)

P(A, B) < P(B)

P(A | B) < P(A, B)

P(A|B) = P(A,B) / P(B) If P(B)=1, then the equation in this option is true. ======================= P(A|B) = P(A,B) / P(B) P(B|A) = P(A,B) / P(A) If P(B) > P(A), then the equation in this option is true. ===================== P(A,B) = P(B) * P(A|B) If P(A|B) < 1, then the equation in this option is true.

However, P(A|B) = P(A,B) / P(B) is impossible Since 0 < P(B) < 1, P(A|B) cannot be less than P(A,B).

Quiz Score: 10 out of 10

Quiz #7 Due Nov 27 at 11:59pm Points 10 Questions 10 Available Nov 3 at 12:59am - Nov 27 at 11:59pm Time Limit None

This quiz was locked Nov 27 at 11:59pm.

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Score for this quiz: 10 out of 10 Submitted Nov 27 at 10:35pm This attempt took 75 minutes.

Question 1 What are Internationalized Resource Identifiers in RDF?

None of the above ~ Object Names ~ Subject ~ Predicate

Refer Lecture 7-2 : General Introduction for explanation.

1 / 1 pts

Question 2

1 / 1 pts

Which of the following are not true about knowledge graph?

Knowledge in graph form ~ Captures entities and attributes but not relationships ~ Node is not an entity

Typed edges capture a relationship between entities.

Refer Lecture 7-3 for explanation.

Question 3 Consider the following ALC concepts. hasChild(Joe, Ahn) hasChild(Joe, Eva) hasChild(Joe, Mary) ≤2 hasChild(Joe)

Which option is correct?

It is an ABox and is satisfiable.

It is an ABox and is not satisfiable.

It is a TBox and is satisfiable.

It is a TBox and is not satisfiable.

1 / 1 pts

Refer Lecture 7-4 : ALC Semantics for explanation.

Question 4

1 / 1 pts

Which kind of reasoning in ontologies should you use if you are asked “whether Barack’s last name is Obama” with some knowledge about Barack?

Answering concept queries @ Instance checking

Taxonomies Concept satisfiability

Refer section 7-4 for explanation.

Question 5

1 / 1 pts

Which option is a suitable ALC concept in description logic for the class of humans “pet owner who only owns cat”, using the following concept names and role name? concept names: Person, Cat, Animal role name: owns

Person ⊓ Cat ⊓ owns(Person, Cat) Person ⊓ ∃owns.Animal ⊓ ∀owns .Cat

Person ⊓ ∀owns .Cat Person ⊓ owns(Person, Cat)

Refer section 7-4 for explanation.

Question 6

1 / 1 pts

Direct semantics are defined for the functional style syntax including malformed syntax.

True @ False

Refer Lecture 7-4 : ALC Semantics for explanation.

Question 7 Which option is correct about RDFS?

The class hierarchy is a binary tree.

“Only humans can teach a class” can be guaranteed by range restriction.

“is taught by” is a subclass of “involves”

1 / 1 pts

Instances of a class inherit the properties of that class.

Refer Lecture 7-2 for explanation.

Question 8

1 / 1 pts

Using the following concept names and role names, which option can correctly capture the knowledge “drivers of cars are adults”? concept names: Person, Car, Adult role names: controls

∀controls.Car ⊑ Adult Adult ⊑ Person Person ⊓ controls.Car ⊓ Adult @ Person ⊓ ∃controls.Car ⊑ Adult

Adult ≡ Person ⊓ ∀controls.Car

Refer Lecture 7-5 for explanation.

Question 9

1 / 1 pts

Using the following concept names and role names, which option can correctly capture the knowledge “cars have between three and four wheels”? concept names: Car, Wheel

role names: hasPart

Car ≡ (≥3 hasPart.Wheel) Car ≡ (≤4 hasPart.Wheel) Car ⊓ ∀hasPart.Wheel ⊑ Wheel ≥ 3 ⊓ Wheel ≤ 4 Car ⊑ (3 ≤ hasPart.Wheel ≤ 4) Car ⊑ (≥3 hasPart.Wheel) ⊓ (≤4 hasPart.Wheel)

Refer Lecture 7-5 for explanation.

1 / 1 pts

Question 10 Match the following DL terminologies with corresponding OWL terminologies. concept

role

constant

theory

I class I object property r

V

V

individual

I ontology

V

Quiz Score: 10 out of 10