Final Exam 361
Model Answer
Solution Q1.a
Depth limit: 0
2
Depth limit 1
4
2
2
1
4
1
2
1
3
4
1
2
3
1
1
3
1
4
2
3
4
3
1
3
3
4
1
3
2
2
2
1
3
Depth limit 2
4
4
4
2
1
4
2
4
3
2
4
2
3
2
1
1
3
4
4
2
3
1
4
3
1
2
4
3
1
1
2
3
4
1
3
Depth limit 3
4
4
1
4
1
2
2
3
2
1
4
3
4
1
3
2
3
2
2
2
3
4
1
3
1
4
3
2
1
1
4
2
3
4
3
1
2
3
4
Solution Q1.b
• How to start:
start with the initial solution (2, 4, 1, 3)
• How to improve:
improve using Local transformation (swap
operator). Select the next state as the best
solution in the neighborhood using Manhattan
distance.
• When to stop:
Halt when the next solution is not better than
current one.
Solution Q2
Cutoff = 610,
Use DFS
Bremen
120 + 720
Hannover
Dusseldorf
320 + 540
Cutoff = 665,
Use DFS
Bremen
120 + 720
Frankfurt
365 + 380
0 + 610
Hamburg
Leipzig
155 + 720 255 + 410
Shweren
270 + 680
Hannover 0 + 610
Dusseldorf
320 + 540
Frankfurt
365 + 380
Berlin
255 + 185 + 590
Hamburg
Leipzig
155 + 720 255 + 410
Dresden
255 + 140 + 400
Munchen
Solution Q3. BT + FC
5
4
6
2
3
7
8
1
10
9
13
11
12
Solution Q3. BT + MCV
5
4
6
2
3
7
8
1
10
9
13
11
12
1, 8, 11, 3, 5, 10, 6, 2, 4, 7, 9, 12, 13
Solution Question 4.a
• Prove 1: “there is no pit in (2,2)”
This sentence can easily be deduced using inference rules
• Equivalence rules (bi-conditional)
• Modus Tollens
• And-Elimination
and the following sentences:
– (B1,2 ⇔ (P1,1  P2,2  P1,3))
– ¬ B1,2
Solution Question 4.a
Prove 2: “there is a wumpus in (1,3)”
This sentence can easily be deduced using inference rules:
•
•
•
•
Equivalence rules (bi-conditional)
Modus Tollens, Modus Ponens
And-Elimination
Resolution
and the following sentences:
– (S1,2 ⇔ (W1,1  W2,2  W1,3))
– (S2,1 ⇔ (W1,1  W2,2  W3,1))
– ¬ W1,1, ¬ S2,1, S1,2
Solution Question 5.a
• Working Memory: A, B, D, G, I, K, L
Cycle 1: R7  B
Cycle 2: R1, R7  D
Cycle 3: R1, R4, R7  G
Cycle 4: R1, R4, R6, R7  I
Cycle 5: R1, R4, R6, R7, R8, R9  K
Cycle 6: R1, R4, R6, R7, R8, R9  L
Cycle 7: R1, R4, R6, R7, R8, R9  STOP no applicable rule
Solution Question 5.b
• Working Memory: A
To prove K, we have to prove first G and I (rule 8)
To prove G, prove B and D (rule 4)
B can be proved using (rule 7)
To prove D, prove first A and B (rule 1): as A is present in WM and B
already proved, therefore D is true so G is true
To prove I, prove G and not(H) first (rule 6): G already proved and as H is
not present in WM then not(H) is true, so I is true.
G and I are true, therefore K is true