The document describes the A* and Iterated Deepening A* (IDA*) search algorithms. It provides examples of applying each algorithm to search problems. For A*, it shows the steps of expanding nodes from a queue and sorting by cost. For IDA*, it iteratively increases a depth limit and re-runs the search until the goal is found.

1. Exercises: Artificial Intelligence
A*

2. A* ALGORITHM
A*

3. A* Algorithm
• Input:
– QUEUE: Path only containing root
• Algorithm:
– WHILE (QUEUE not empty && first path not reach goal) DO
• Remove first path from QUEUE
Create paths to all children• Create paths to all children
• Reject paths with loops
• Add paths and sort QUEUE (by f = cost + heuristic)
• IF QUEUE contains paths: P, Q
AND P ends in node Ni && Q contains node Ni
AND cost_P ≥ cost_Q
THEN remove P
– IF goal reached THEN success ELSE failure

4. FIRST EXAMPLE ON A*
A*

5. A* algorithm by Example
S7
7
0
f = accumulated path cost + heuristic
QUEUE = path containing root
CS
A GB10
7 5
9
0
6
9
1 1
12
5
QUEUE: <S>

6. A* algorithm by Example
S
f = accumulated path cost + heuristic
A10
11
1
B9
10
1
Remove first path, Create paths to all
children, Reject loops and Add paths.
Sort QUEUE by f
CS
A GB10
7 5
9
0
6
9
1 1
12
5
QUEUE: <SB,SA>
A10 B9

7. A* algorithm by Example
S
f = accumulated path cost + heuristic
A10
11
1
B
Remove first path, Create paths to all
children, Reject loops and Add paths.
Sort QUEUE by f
CS
A GB10
7 5
9
0
6
9
1 1
12
5
QUEUE: <SA,SBC,SBG,SBA>
A10 B
C5
12
7
G0
13
13
A10
10
20

8. A* algorithm by Example
S
f = accumulated path cost + heuristic
A10
11
1
B
IF P terminating in I with cost_P &&
Q containing I with cost_Q AND
cost_P ≥ cost_Q THEN remove P
CS
A GB10
7 5
9
0
6
9
1 1
12
5
QUEUE: <SA,SBC,SBG,SBA>
A10 B
C5
12
7
G0
13
13
A10
10
20

9. A* algorithm by Example
S
f = accumulated path cost + heuristic
A B
Remove first path, Create paths to all
children, Reject loops and Add paths.
Sort QUEUE by f
CS
A GB10
7 5
9
0
6
9
1 1
12
5
QUEUE: <SBC,SBG,SAB>
A B
C5
12
7
G0
13
13
B9
10
19

10. A* algorithm by Example
S
f = accumulated path cost + heuristic
A B
IF P terminating in I with cost_P &&
Q containing I with cost_Q AND
cost_P ≥ cost_Q THEN remove P
CS
A GB10
7 5
9
0
6
9
1 1
12
5
QUEUE: <SBC,SBG,SAB>
A B
C5
12
7
G0
13
13
B9
10
19

11. A* algorithm by Example
S
f = accumulated path cost + heuristic
B
Remove first path, Create paths to all
children, Reject loops and Add paths.
Sort QUEUE by f
CS
A GB10
7 5
9
0
6
9
1 1
12
5
QUEUE: <SBCG,SBG>
B
C G0
13
13
G0
12
12

12. A* algorithm by Example
S
f = accumulated path cost + heuristic
B
IF P terminating in I with cost_P &&
Q containing I with cost_Q AND
cost_P ≥ cost_Q THEN remove P
CS
A GB10
7 5
9
0
6
9
1 1
12
5
QUEUE: <SBCG,SBG>
B
C G0
13
13
G0
12
12

13. A* algorithm by Example
S
f = accumulated path cost + heuristic
B
SUCCESS
CS
A GB10
7 5
9
0
6
9
1 1
12
5
QUEUE: <SBCG>
B
C
G0
12
12

14. PROBLEM
A*

15. Problem
• Perform the A* Algorithm on the following
figure. Explicitly write down the queue at each
step.
A
10
6
C
S
A
GB
0
6
D
E
F
1
4
2
13
4
17
10
5
6
6
6
7
4
6
3

16. A* SEARCH
A*

17. A* Search
S
QUEUE:
S
17
17
0
C
S
A
GB
0
6
D
E
F
1
4
2
13
10
4
17
10
5
6
6
6
7
4
6
3
S

18. A* Search
S
C
S
A
GB
0
6
D
E
F
1
4
2
13
10
4
17
10
5
6
6
6
7
4
6
3
A10
16
6
B13
18
5
C4
14
10
QUEUE:
SCSC
SA
SB

19. A* Search
S
C
S
A
GB
0
6
D
E
F
1
4
2
13
10
4
17
10
5
6
6
6
7
4
6
3
A10
16
6
B13
18
5
C QUEUE:
SA
D2
18
16
SA
SCD
SB

20. A* Search
S
C
S
A
GB
0
6
D
E
F
1
4
2
13
10
4
17
10
5
6
6
6
7
4
6
3
A B13
18
5
C QUEUE:
SAE
D2
18
16
E4
16
12
SAE
SCD
SB

21. A* Search
S
C
S
A
GB
0
6
D
E
F
1
4
2
13
10
4
17
10
5
6
6
6
7
4
6
3
A B13
18
5
C QUEUE:
SAEF
D2
18
16
E
F1
17
16
B13
31
18
SAEF
SCD
SB
SAEB

22. A* Search
S
C
S
A
GB
0
6
D
E
F
1
4
2
13
10
4
17
10
5
6
6
6
7
4
6
3
A B13
18
5
C QUEUE:
SCD
D2
18
16
E
F
G0
19
19
D2
24
22
SCD
SB
SAEFG
SAEFD

23. A* Search
S
C
S
A
GB
0
6
D
E
F
1
4
2
13
10
4
17
10
5
6
6
6
7
4
6
3
A B13
18
5
C QUEUE:
SB
DE
F
G0
19
19
F1
23
22
B13
36
23
SB
SAEFG
SCDF
SCDB

24. A* Search
S
C
S
A
GB
0
6
D
E
F
1
4
2
13
10
4
17
10
5
6
6
6
7
4
6
3
A B QUEUE:
SBD
E
F
G0
19
19
E4
15
11
D2
14
12
SBD
SBE
SAEFG

25. A* Search
S
C
S
A
GB
0
6
D
E
F
1
4
2
13
10
4
17
10
5
6
6
6
7
4
6
3
A B QUEUE:
SBE
E
F
G0
19
19
E4
15
11
D
F1
19
18
C4
22
18
SBE
SBDF
SAEFG
SBDC

26. A* Search
S
C
S
A
GB
0
6
D
E
F
1
4
2
13
10
4
17
10
5
6
6
6
7
4
6
3
A B QUEUE:
SBEF
E
F
G0
19
19
ED
F1
19
18
C4
22
18
F1
16
15
A10
27
17
SBEF
SAEFG
SBDF
SBDC
SBEA

27. A* Search
S
C
S
A
GB
0
6
D
E
F
1
4
2
13
10
4
17
10
5
6
6
6
7
4
6
3
A B QUEUE:
SBEFG
E
F
G0
19
19
ED
C4
22
18
FA10
27
17
G0
18
18
D 23
2
21
SBEFG
SAEFG
SBDC
SBEFD
SBEA

28. Exercises: Artificial Intelligence
Iterated Deepening A*

29. IDA* ALGORITHM
Iterated Deepening A*

30. IDA* Algorithm
• f-bound ôôôô f(S)
• Algorithm:
– WHILE (goal is not reached) DO
• f-bound ôôôô f-limitted_search(f-bound)• f-bound ôôôô f-limitted_search(f-bound)
– Perform f-limited search with f-bound
(See next slide)

31. f-limitted Search Algorithm
• Input:
– QUEUE ô Path only containing root
– f-bound ô Natural number
– f-new ô ¶
• Algorithm:• Algorithm:
– WHILE (QUEUE not empty && goal not reached) DO
• Remove first path from QUEUE
• Create paths to children
• Reject paths with loops
• Add paths with f(path) ≤ f-bound to front of QUEUE (depth-first)
• f-new ô minimum( {f-new} » {f(P) | P is rejected path} )
– IF goal reached THEN success ELSE report f-new

32. PROBLEM
Iterated Deepening A*

33. Problem
• Perform the IDA* Algorithm on the following
figure.
C S A109
G B D
5 1
4
8
5
S A B C D G
heuristic 0 0 4 3 0 0

34. IDA* SEARCH
Iterated Deepening A*

35. IDA* Search
S0
0
0
C S A
G B D
5
109
1
4
8
5
f-bound = 0
f-new = ¶¶¶¶
0 0
040
3
f-new = ¶¶¶¶

36. IDA* Search
S
A0
10
10
B4
12
8
C3
12
9
C S A
G B D
5
109
1
4
8
5
0 0
040
3
f-bound = 0
f-new = 10
A0 B4 C3
f-new = 10
Children are explored
depth-first!

37. IDA* Search
S0
0
0
C S A
G B D
5
109
1
4
8
5
0 0
040
3
f-bound = 10
f-new = ¶¶¶¶f-new = ¶¶¶¶

38. IDA* Search
S
A0
10
10
B4
12
8
C3
12
9
C S A
G B D
5
109
1
4
8
5
0 0
040
3
f-bound = 10
f-new = 12
A0 B4 C3
f-new = 12

39. IDA* Search
S
A B4
12
8
C3
12
9
C S A
G B D
5
109
1
4
8
5
0 0
040
3
f-bound = 10
f-new = 11
A B4 C3
D0
11
11
f-new = 11

40. IDA* Search
S0
0
0
C S A
G B D
5
109
1
4
8
5
0 0
040
3
f-bound = 11
f-new = ¶¶¶¶f-new = ¶¶¶¶

41. IDA* Search
S
A0
10
10
B4
12
8
C3
12
9
C S A
G B D
5
109
1
4
8
5
0 0
040
3
f-bound = 11
f-new = 12
A0 B4 C3
f-new = 12

42. IDA* Search
S
A B4
12
8
C3
12
9
C S A
G B D
5
109
1
4
8
5
0 0
040
3
f-bound = 11
f-new = 12
A B4 C3
D0
11
11
f-new = 12

43. IDA* Search
S
A B4
12
8
C3
12
9
C S A
G B D
5
109
1
4
8
5
0 0
040
3
f-bound = 11
f-new = 12
A B4 C3
D
B4
19
15
f-new = 12

44. IDA* Search
S0
0
0
C S A
G B D
5
109
1
4
8
5
0 0
040
3
f-bound = 12
f-new = ¶¶¶¶f-new = ¶¶¶¶

45. IDA* Search
S
A0
10
10
B4
12
8
C3
12
9
C S A
G B D
109
1
4
8
5
0 0
040
3
f-bound = 12
f-new = ¶¶¶¶
A0 B4 C3
f-new = ¶¶¶¶

46. IDA* Search
S
A B4
12
8
C3
12
9
C S A
G B D
109
1
4
8
5
0 0
040
3
f-bound = 12
f-new = ¶¶¶¶
A B4 C3
D0
11
11
f-new = ¶¶¶¶

47. IDA* Search
S
A B4
12
8
C3
12
9
C S A
G B D
109
1
4
8
5
0 0
040
3
f-bound = 12
f-new = 19
A B4 C3
D
B4
19
15
f-new = 19

48. IDA* Search
S
A B C3
12
9
C S A
G B D
109
1
4
8
5
0 0
040
3
f-bound = 12
f-new = 13
A B C3
D
B4
19
15
D G0
12
12
0
13
13
f-new = 13

49. IDA* Search
S
A B C3
12
9
C S A
G B D
109
1
4
8
5
0 0
040
3
f-bound = 12
f-new = 13
A B C3
D
B4
19
15
D G0
12
12
0
13
13
A0
13
13
f-new = 13

50. IDA* Search
S
A B C3
12
9
C S A
G B D
109
1
4
8
5
0 0
040
3
f-bound = 12
f-new = 13
A B C3
D
B4
19
15
D G0
12
12
0
13
13
A0
13
13
G0
14
14
f-new = 13

51. IDA* Search
S0
0
0
C S A
G B D
109
1
4
8
5
0 0
040
3
f-bound = 13
f-new = ¶¶¶¶f-new = ¶¶¶¶

52. IDA* Search
S
A0
10
10
B4
12
8
C3
12
9
C S A
G B D
109
1
4
8
5
0 0
040
3
f-bound = 13
f-new = ¶¶¶¶
A0 B4 C3
f-new = ¶¶¶¶

53. IDA* Search
S
A B4
12
8
C3
12
9
C S A
G B D
109
1
4
8
5
0 0
040
3
f-bound = 13
f-new = ¶¶¶¶
A B4 C3
D0
11
11
f-new = ¶¶¶¶

54. IDA* Search
S
A B4
12
8
C3
12
9
C S A
G B D
109
1
4
8
5
0 0
040
3
f-bound = 13
f-new = 19
A B4 C3
D
B4
19
15
f-new = 19

55. IDA* Search
S
A B C3
12
9
C S A
G B D
109
1
4
8
5
0 0
040
3
f-bound = 13
f-new = 19
A B C3
D
B4
19
15
D G0
12
12
0
13
13
f-new = 19

56. Exercises: Artificial Intelligence
Simplified Memory-bounded A*

57. SMA* ALGORITHM
Simplified Memory-bounded A*

58. SMA* Algorithm
• Optimizes A* to work within reduced memory
• Key Idea:
– IF memory full for extra node (C)
– Remove highest f-value leaf (A)– Remove highest f-value leaf (A)
– Remember best-forgotten child in
each parent node (15 in S)
S
A 15 B 13
C 18
13
(15)
E.g. Memory of 3 nodes only

59. SMA* Algorithm
• Generate Children 1 by 1
– Expanding: add 1 child at the time to QUEUE
– Avoids memory overflow
– Allows monitoring if nodes need deletion– Allows monitoring if nodes need deletion
S
A 15 B 13
13
First add A later B

60. SMA* Algorithm
• Too long paths: Give up
– Extending path cannot fit in memory
• give up (C)
– Set f-value node (C) to ¶¶¶¶
S 13
• Remembers:
path cannot be found here B 13
C ¶¶¶¶
D
18
E.g. Memory of 3 nodes only

61. SMA* Algorithm
• Adjust f-values
– IF all children Mi of node N have been explored
– AND "i: f(S…Mi) > f(S…N)
– THEN reset (through N ï through children)– THEN reset (through N ï through children)
• f(S…N) = min{f(S…Mi)|Mi child of N}
Better estimate for f(S)
S
A 15 B 24
1513

62. SMA* BY EXAMPLE
Simplified Memory-bounded A*

63. SMA* by Example
• Perform SMA* (memory: 3 nodes) on the
following figure.
A
24
C
S GB3
15
S A B C G
heuristic 3 0 2 1 0
2

64. SMA* by Example
C
S
A
GB
3
2
1
4
5
2
0
03 2
1
S3 3
0

65. SMA* by Example
C
S
A
GB
3
2
1
4
5
2
0
03 2
1
S3 3
0
A0 4
4
Generate children
(One by one)
A0 4

66. SMA* by Example
C
S
A
GB
3
2
1
4
5
2
0
03 2
1
S3 3
0
A0 4
4
B2 5
3
A0 4 B2 5
Generate children
(One by one)

67. SMA* by Example
C
S
A
GB
3
2
1
4
5
2
0
03 2
1
S3 3
0
A0 4
4
B2 5
3
C1 6
5
(5)
A0 4 B2 5 C1 6
Generate children
(One by one)
Memory full

68. SMA* by Example
C
S
A
GB
3
2
1
4
5
2
0
03 2
1
S3
0
A0 4
4
C1 6
5
(5)
43
A0 4 C1 6
All children are
explored
Adjust f-values

69. SMA* by Example
C
S
A
GB
3
2
1
4
5
2
0
03 2
1
S3
0
A0 4
4
C1 6
5
(5)
4
A0 4 C1 6
G0 6
6
Generate children
(One by one)
Memory full

70. SMA* by Example
C
S
A
GB
3
2
1
4
5
2
0
03 2
1
S3
0
A0 4
4
(5)
4
6A0 4
G0 6
6
6
All children are
explored
Adjust f-values

71. SMA* by Example
C
S
A
GB
3
2
1
4
5
2
0
03 2
1
S3
0
A0 6
4
(5)
54
A0 6
G0 6
6
All children are
explored (update)
Adjust f-values

72. SMA* by Example
C
S
A
GB
3
2
1
4
5
2
0
03 2
1
S3
0
A0
4
6
5
B2 5
3(6)
A0
G0 6
6
6 B2 5
Generate children
(One by one)
Memory full

73. SMA* by Example
C
S
A
GB
3
2
1
4
5
2
0
03 2
1
S3
0
A0
4
(6)
6
5
B2 5
3(6)
C1 6
5
A0 6 B2 5 C1 6
Generate children
(One by one)
Memory full

74. SMA* by Example
C
S
A
GB
3
2
1
4
5
2
0
03 2
1
S3
0 (6)
5
B2 5
3
C1 6
5
B2 5 C1 6
G0 5
5
Generate children
(One by one)
Memory full

75. PROBLEM
Simplified Memory-bounded A*

76. Problem
• Perform SMA* (memory: 4 nodes) on the
following figure.
C 32
A E
10 10 9 2
D
S FG
18
S A B C D E F H G
heuristic 12 5 5 5 2 2 1 1 0
B H
8 16 3 1

77. Problem
S12 12
0
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1

78. Problem
S12 12
0
A5 15
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
A5 15

79. Problem
S12 12
0
A5 15
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 13
8
A5 15 B5 13

80. Problem
S12 12
0
A5 15
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 13
8
13
A5 15 B5 13

81. Problem
S12 13
0
A5 15
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 13
8
A5 15 B5 13
D2 18
16

82. Problem
S12 13
0
A5 15
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 13
8 (18)
A5 15 B5 13
D2 18
16
G0 24
24

83. Problem
S12 13
0
A5 15
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 13
8 (18)
18A5 15 B5 13
G0 24
24
18

84. Problem
S12 13
0
A5 15
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 18
8 (18)
15
A5 15 B5 18
G0 24
24

85. Problem
S12 15
0
A5 15
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 18
8 (18)
A5 15 B5 18
G0 24
24
C5 17
12

86. Problem
S12 15
0
A5 15
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 18
8 (18)
(18)
A5 15 B5 18
G0 20
20
C5 17
12

87. Problem
S12 15
0
A5 15
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
(18)
17A5 15
G0 20
20
C5 17
12
17

88. Problem
S12 15
0
A5 17
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
(18)
17
A5 17
G0 20
20
C5 17
12

89. Problem
S12 17
0
A5 17
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
(18)
(20)
A5 17
G0 20
20
C5 17
12
E2 17
15

90. Problem
S12 17
0
A5 17
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
(18)
(20)
A5 17
C5 17
12
E2 17
15
¶¶¶¶

91. Problem
S12 17
0
A5 17
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
(18)
(20)
A5 17
C5 17
12
E2
15
¶¶¶¶ G0 21
21
(¶¶¶¶)

92. Problem
S12 17
0
A5 17
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
(18)
(20)
A5 17
C5 17
12
G0 21
21
(¶¶¶¶)
21

93. Problem
S12 17
0
A5 17
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
(18)
(20)
20A5 17
C5 21
12
G0 21
21
(¶¶¶¶)
20

94. Problem
S12 17
0
A5 20
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
(18)
(20)
18
A5 20
C5 21
12
G0 21
21
(¶¶¶¶)

95. Problem
S12 18
0
A5 20
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1(20)
B5 13
8
A5 20
C5 21
12
G0 21
21
(21)
B5 13

96. Problem
S12 18
0
A5 20
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1(20)
B5 13
8
A5 20
C5 21
12
D2 18
16(21)
B5 13

97. Problem
S12 18
0
A5 20
10
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1(20)
B5 13
8
(20)
A5 20
D2 18
16
B5 13
G0 24
24

98. Problem
S12 18
0
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 13
8
(20)
18
D2 18
16
B5 13
G0 24
24
18

99. Problem
S12 18
0
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 18
8
(20)
(24)
D2 18
16
B5 18
G0 24
24
G0 19
19

100. Problem
S12 18
0
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 18
8
(20)
(24)
D2 18
16
B5 18
H1 18
17
G0 19
19
(19)

101. Problem
S12 18
0
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 18
8
(20)
(24)
D2 18
16
B5 18
H1 18
17
(19)
¶¶¶¶

102. Problem
S12 18
0
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 18
8
(20)
(24)
D2 18
16
B5 18
H1
17
(19)
¶¶¶¶
19

103. Problem
S12 18
0
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 18
8
(20)
(24)
19
D2 19
16
B5 18
H1
17
(19)
¶¶¶¶
19

104. Problem
S12 18
0
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 19
8
(20)
(24)
19
D2 19
16
B5 19
H1
17
(19)
¶¶¶¶

105. Problem
S12 19
0
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 19
8
(20)
(24)
D2 19
16
B5 19
H1
17
¶¶¶¶ G0
19
19

106. Problem
S12 19
0
D
S
C
FG
3
1
2
8
B
A E
H
10
8
10 9 2
16 3 1
12
5 5 2
0 1
25 1
B5 19
8
(20)
(24)
D2 19
16
B5 19
G0
19
19

107. Exercises: Artificial Intelligence
Monotonicity 1

108. PROBLEM
Monotonicity 1

109. Problem
• Prove that:
– IF a heuristic function h satisfies the monotonicity
restriction
• h(x) ≤ cost(x…y) + h(y)• h(x) ≤ cost(x…y) + h(y)
– THEN f is monotonously non-decreasing
• f(s…x) ≤ f(s…x…y)

110. Monotonicity 1
• Given:
– h satisfies the monotonicity restriction
• Proof:
f(S…A) = cost(S…A) + h(A)f(S…A) = cost(S…A) + h(A)

111. Monotonicity 1
• Given:
– h satisfies the monotonicity restriction
• Proof:
f(S…A) = cost(S…A) + h(A)f(S…A) = cost(S…A) + h(A)
≤ cost(S…A) + cost(A…B) + h(B)

112. Monotonicity 1
• Given:
– h satisfies the monotonicity restriction
• Proof:
f(S…A) = cost(S…A) + h(A)f(S…A) = cost(S…A) + h(A)
≤ cost(S…A) + cost(A…B) + h(B)
≤ cost(S…A…B) + h(B)

113. Monotonicity 1
• Given:
– h satisfies the monotonicity restriction
• Proof:
f(S…A) = cost(S…A) + h(A)f(S…A) = cost(S…A) + h(A)
≤ cost(S…A) + cost(A…B) + h(B)
≤ cost(S…A…B) + h(B)
≤ f(S…A…B)

114. Exercises: Artificial Intelligence
Monotonicity 2

115. PROBLEM
Monotonicity 2

116. Problem
• Prove or refute:
– IF f is monotonously non-decreasing
• f(s…x) ≤ f(s…xy)
– THEN h is an admissable heuristic– THEN h is an admissable heuristic
• h is an underestimate of the remaining path to the goal
with the smallest cost
• Can an extra constraint on h change this?

117. Monotonicity 2
• Given:
– f is mononously non-decreasing
• Proof (Counter-example):
S7 7
0
D6 8
2
B6 7
1
G5
3
8
f is monotonously non-decreasing,
yet h is not an admissable heuristic.

118. Monotonicity 2
• Given:
– f is mononously non-decreasing
– Extra constraint: h(G) = 0
• Proof:• Proof:
f(S…A) ≤ f(S…AB) ≤ … ≤ f(S…AB…G)

119. Monotonicity 2
• Given:
– f is mononously non-decreasing
– Extra constraint: h(G) = 0
• Proof:• Proof:
f(S…A) ≤ f(S…AB) ≤ … ≤ f(S…AB…G) ñ
f(S…A) ≤ f(S…G)

120. Monotonicity 2
• Given:
– f is mononously non-decreasing
– Extra constraint: h(G) = 0
• Proof:• Proof:
f(S…A) ≤ f(S…AB) ≤ … ≤ f(S…AB…G) ñ
f(S…A) ≤ f(S…G) ñ
cost(S…A) + h(A) ≤ cost(S…G) + h(G)

121. Monotonicity 2
• Given:
– f is mononously non-decreasing
– Extra constraint: h(G) = 0
• Proof:• Proof:
f(S…A) ≤ f(S…AB) ≤ … ≤ f(S…AB…G) ñ
f(S…A) ≤ f(S…G) ñ
cost(S…A) + h(A) ≤ cost(S…G) + h(G) ñ
cost(S…A) + h(A) ≤ cost(S…A) + cost(A…G) + h(G)

122. Monotonicity 2
• Given:
– f is mononously non-decreasing
– Extra constraint: h(G) = 0
• Proof:• Proof:
f(S…A) ≤ f(S…AB) ≤ … ≤ f(S…AB…G) ñ
f(S…A) ≤ f(S…G) ñ
cost(S…A) + h(A) ≤ cost(S…G) + h(G) ñ
cost(S…A) + h(A) ≤ cost(S…A) + cost(A…G) + h(G) ñ
h(A) ≤ cost(A…G) + h(G)

123. Monotonicity 2
• Given:
– f is mononously non-decreasing
– Extra constraint: h(G) = 0
• Proof:• Proof:
f(S…A) ≤ f(S…AB) ≤ … ≤ f(S…AB…G) ñ
f(S…A) ≤ f(S…G) ñ
cost(S…A) + h(A) ≤ cost(S…G) + h(G) ñ
cost(S…A) + h(A) ≤ cost(S…A) + cost(A…G) + h(G) ñ
h(A) ≤ cost(A…G) + h(G) ñ
h(A) ≤ cost(A…G)