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CPSC 420-500 Homework
SOLUTION
Total: 130 pts
Yoonsuck Choe
November 1, 2007
1 First-Order Logic
Important: In this section, assume that w, x, y, z are variables; A, B, C, D are constants; and f (·), g(·), h(·) are functions; and P (·), Q(·), R(·) are predicates.
1.1 Standard Forms
To do automatic theorem proving in first-order logic, you need to go through three steps to convert your initial first-order logic expression into a standard form. These are:
- Prenex normal form,
- Conjunctive normal form, and
- Skolemization.
Question 1 (12 pts): Convert to prenex normal form (4 points each):
- ∀x, ¬ (∃y, ¬P (x, y))
- ¬∀x (P (x) ∨ ¬(∃y, ¬Q(x, y)))
- ¬∀x (∃y, Q(x, y) → ¬P (x))
SOLUTION:
- ∀x, ¬ (∃y, ¬P (x, y)) ∀x, (∀y, ¬¬P (x, y)) ∀x, (∀y, P (x, y)) ∀x, ∀y, P (x, y)
- ¬∀x (P (x) ∨ ¬(∃y, ¬Q(x, y))) ∃x¬ (P (x) ∨ ¬(∃y, ¬Q(x, y))) ∃x (¬P (x) ∧ ¬¬(∃y, ¬Q(x, y))) ∃x (¬P (x) ∧ (∃y, ¬Q(x, y))) ∃x, ∃y, ¬P (x) ∧ ¬Q(x, y)
- ¬∀x (∃y, Q(x, y) → ¬P (x)) ∃x¬ (¬(∃y, Q(x, y)) ∨ ¬P (x)) ∃x (¬¬(∃y, Q(x, y)) ∧ ¬¬P (x)) ∃x (∃y, Q(x, y) ∧ P (x)) ∃x, ∃y, Q(x, y) ∧ P (x)
Question 2 (20 pts): Skolemize the expressions (4 points each):
- ∃xP (x)
- ∀x∃yP (x, y)
- ∃x, ∃y, ∀zP (x, y) ∧ Q(y, z)
- ∀x, ∃y, ∃zP (x, y) ∧ Q(y, z)
- ∀x, ∀y, ∃zP (x, y) ∧ Q(y, z)
SOLUTION:
1. P (A)
- P (x, f (x))
- P (A, B) ∧ Q(B, z)
- P (x, f (x)) ∧ Q(f (x), g(x)) The question was ambiguous, so you will get full points for this one if your answer was reasonable. The qeustion was supposed to be ∀x, ∃y, ∃zP (x, y) ∧ Q(y, z)
- P (x, y) ∧ Q(y, f (x, y))
Question 3 (9 pts): Convert the following into a standard form:
∀x, [¬P (x) → ¬(∃y, Q(x, y))]
SOLUTION: ∀x, [¬P (x) → ¬(∃y, Q(x, y))] ∀x, [¬¬P (x) ∨ ¬(∃y, Q(x, y))] ∀x, [P (x) ∨ (∀y, ¬Q(x, y))] ∀x, ∀y, [P (x) ∨ ¬Q(x, y)] P (x) ∨ ¬Q(x, y)
- ∀x, ∀y, (¬P (x) → (Q(x, y) ∨ R(y))) ∀x, ∀y, (¬¬P (x) ∨ (Q(x, y) ∨ R(y))) ∀x, ∀y, (P (x) ∨ Q(x, y) ∨ R(y)) P (x) ∨ Q(x, y) ∨ R(y)
- ∃x, ¬P (x) ¬P (A)
- ∀w, ∀z, (¬Q(w, z) ∨ R(w)) ¬Q(w, z) ∨ R(w)
- Negated conclusion: ¬R(v)
- Q(A, y) ∨ R(y) (1 and 2), σ = {x/A}
- Q(A, y) (4 and 5), σ = {v/y}
- ¬Q(w, z) (3 and 4), σ = {v/w}
- False (6 and 7), σ = {w/A, y/z}
2 Uncertainty and Probabilistic Reasoning
B T^ T^ F^ F^
E T F T F
P(A) . . .
.
P(B) .
P(E)
Alarm
Earthquake
JohnCalls MaryCalls
Burglary
A P(J) T^ F .90. A P(M) T^ F .70.
.
Figure 1: Belief Network. See problem 1.
Question 1 (10 pts): Given the Belief network as shown in figure 1, calculate the two joint probability values and answer the question. Note that in this section P (·) denotes the probability of the event. (5 points each):
- P (M aryCalls ∧ JohnCalls ∧ Alarm ∧ ¬Earthquake ∧ Burglary)
- P (¬M aryCalls ∧ ¬JohnCalls ∧ ¬Alarm ∧ ¬Earthquake ∧ Burglary)
SOLUTION:
- P (M aryCalls ∧ JohnCalls ∧ Alarm ∧ ¬Earthquake ∧ Burglary) = P (M |A)P (J|A)P (A|¬E, B)P (¬E)P (B) = 0. 7 × 0. 9 × 0. 94 × (1 − 0 .002) × 0. 001
- P (¬M aryCalls ∧ ¬JohnCalls ∧ ¬Alarm ∧ ¬Earthquake ∧ Burglary) = P (¬M |¬A)P (¬J|¬A)P (¬A|¬E, B)P (¬E)P (B) = (1 − 0 .01) × (1 − 0 .05) × (1 − 0 .94) × (1 − 0 .002) × 0. 001
Question 2 (5 pts): Why do belief networks give a much more compact representation of the joint proba- bility distribution, compared to a full joint probability table?
SOLUTION: Because the joint probability can be represented as a product of conditional probabilities where each node is conditioned on only a small number of other nodes (their parents) in the network due to the conditional independence structure of the domain, and thus much smaller probability tables for each node.
3 Learning
3.1 Decision Tree Learning
Consider the following set of examples where you are trying to make a decision whether to take a course or not.
Example# Usefulness in life Toughness Fun Decision (Take course?) 1 Stellar Okay Medium Yes 2 Not bad Very tough High Yes 3 Zero Light Medium No 4 Not bad Light Low No 5 Not bad Light High Yes 6 Not bad Very tough Low No 7 Zero Okay Medium No 8 Stellar Very tough Low Yes 9 Stellar Okay Medium Yes 10 Zero Light High No
Question 1 (12 pts): For each of the three attributes above, draw a decision tree rooted at that attribute with a single depth. See slide06, page 12, (a) and (b) which show some examples. (4 points each)
SOLUTION:
- Usefulness Stellar: (+: 1, 8, 9) (-: None) Not bad: (+: 2, 5) (-: 4, 6) Zero: (+: None) (-: 3, 7, 10)
- Toughness Okay: (+: 1, 9) (-: 7) Very tough: (+: 2, 8) (-: 6) Light: (+: 5) (-: 3, 4, 10)
- Fun Low: (+: 8) (-: 4, 6) Medium: (+: 1, 9) (-: 3, 7) High: (+: 2, 5) (-: 10)
