Assignment 4

1. Exercise 9.9 (12 marks)

We use a very simple ontology to make the examples easier:

a. Horse(x) Mammal(x)

Cow(x) Mammal(x)

Pig(x) Mammal(x)

b. Offspring(x, y) Horse(y) Horse(x)

c. Horse(Bluebeard)

d. Parent(Bluebeard,Charlie)

e. Offspring(x, y) Parent(y, x)

Parent(x, y) Offspring(y, x)

(Note we couldn’t do Offspring(x, y) Parent(y, x) because that is not in the form

expected by Generalized Modus Ponens.)

f. Mammal(x) Parent(G(x), x) (here G is a Skolem function).

1. Exercise 9.10  a, b (6mark)

This questions deals with the subject of looping in backward-chaining proofs. A loop is bound to occur whenever a subgoal arises that is a substitution instance of one of the goals on the stack. Not all loops can be caught this way, of course, otherwise we would have a way to solve the halting problem.

a. The proof tree is shown in Figure S9.1. The branch with Offspring(Bluebeard, y) and Parent(y,Bluebeard) repeats indefinitely, so the rest of the proof is never reached.

b. We get an infinite loop because of rule b, Offspring(x, y) Horse(y) Horse(x). The specific loop appearing in the figure arises because of the ordering of the clauses—it would be better to order Horse(Bluebeard) before the rule from b. However, a loop will occur no matter which way the rules are ordered if the theorem-prover is asked for all solutions.

1. Exercise 13.6 (8 marks)

a. This asks for the probability that Toothache is true.

P(toothache) = 0.108 + 0.012 + 0.016 + 0.064 = 0.2

b. This asks for the vector of probability values for the random variable Cavity. It has two values, which we list in the order <true, false>. First add up 0.108 + 0.012 + 0.072 + 0.008 = 0.2. Then we have P(Cavity) = <0.2, 0.8> .

c. This asks for the vector of probability values for Toothache, given that Cavity is true.

P(Toothache|cavity) = <(.108 + .012)/0.2, (0.072 + 0.008)/0.2> = <0.6, 0.4>

d. This asks for the vector of probability values for Cavity, given that either Toothache or Catch is true. First compute P(toothache catch) = 0.108+0.012+0.016+0.064+ 0.072 + 0.144 = 0.416. Then

P(Cavity|toothache catch)

=<(0.108 + 0.012 + 0.072)/0.416, (0.016 + 0.064 + 0.144)/0.416>

= <0.4615, 0.5384>

1. Exercise 13.8 (6 marks)

We are given the following information:

P(test|disease) = 0.99

P(test|disease) = 0.99

P(disease) = 0.0001

and the observation test. What the patient is concerned about is P(disease|test). Roughly speaking, the reason it is a good thing that the disease is rare is that P(disease|test) is proportional to P(disease), so a lower prior for disease will mean a lower value for P(disease|test). Roughly speaking, if 10,000 people take the test, we expect 1 to actually have the disease, and most likely test positive, while the rest do not have the disease, but 1% of them (about 100 people) will test positive anyway, so P(disease|test) will be about 1 in 100. More precisely, using the normalization equation from page 428:

P(disease|test)

= P(test|disease)P(disease) / (P(test|disease)P(disease)+P(test|¬disease)P(¬disease))

= 0.99×0.0001 / (0.99×0.0001+0.01×0.9999)

= .009804

The moral is that when the disease is much rarer than the test accuracy, a positive test result does not mean the disease is likely. A false positive reading remains much more likely. Here is an alternative exercise along the same lines: A doctor says that an infant who predominantly turns the head to the right while lying on the back will be right-handed, and one who turns to the left will be left-handed. Isabella predominantly turned her head to the left. Given that 90% of the population is right-handed, what is Isabella’s probability of being right-handed if the test is 90% accurate? If it is 80% accurate? The reasoning is the same, and the answer is 50% right-handed if the test is 90% accurate, 69% right-handed if the test is 80% accurate.

1. Exercise 14.1 a, b, c, d (10 marks)

Adding variables to an existing net can be done in two ways. Formally speaking, one should insert the variables into the variable ordering and rerun the network construction process from the point where the first new variable appears. Informally speaking, one never really builds a network by a strict ordering. Instead, one asks what variables are direct causes or influences on what other ones, and builds local parent/child graphs that way. It is usually easy to identify where in such a structure the new variable goes, but one must be very careful to check for possible induced dependencies downstream.

a. IcyWeather is not caused by any of the car-related variables, so needs no parents. It directly affects the battery and the starter motor. StarterMotor is an additional precondition for Starts. The new network is shown in Figure S14.1.

b. Reasonable probabilities may vary a lot depending on the kind of car and perhaps the personal experience of the assessor. The following values indicate the general order of magnitude and relative values that make sense:

A reasonable prior for IcyWeather might be 0.05 (perhaps depending on location and season).

P(Battery|IcyWeather) = 0.95, P(Battery|IcyWeather) = 0.997.

P(StarterMotor|IcyWeather) = 0.98, P(Battery|IcyWeather) = 0.999.

P(Radio|Battery) = 0.9999, P(Radio|Battery) = 0.05.

P(Ignition|Battery) = 0.998, P(Ignition|Battery) = 0.01.

P(Gas) = 0.995.

P(Starts|Ignition, StarterMotor,Gas) = 0.9999, other entries 0.0.

P(Moves|Starts) = 0.998.

c. With 8 Boolean variables, the joint has 28 1 = 255 independent entries.

d. Given the topology shown in Figure S14.1, the total number of independent CPT entries is 1+2+2+2+2+1+8+2= 20.

1. Exercise 14.3 (10marks)

1. Exercise 18.3 (4 marks)

1. Exercise 18.4 (5 marks)