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A*

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.

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Exercises: Artificial Intelligence A*
A* ALGORITHM A*
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
FIRST EXAMPLE ON A* A*
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>
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
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
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
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
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
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
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
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
PROBLEM A*
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
A* SEARCH A*
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
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
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
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
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
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
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
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
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
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
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
Exercises: Artificial Intelligence Iterated Deepening A*
IDA* ALGORITHM Iterated Deepening A*
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)
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
PROBLEM Iterated Deepening A*
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
IDA* SEARCH Iterated Deepening A*
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 = ¶¶¶¶
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!
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 = ¶¶¶¶
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
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
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 = ¶¶¶¶
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
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
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
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 = ¶¶¶¶
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 = ¶¶¶¶
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 = ¶¶¶¶
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
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
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
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
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 = ¶¶¶¶
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 = ¶¶¶¶
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 = ¶¶¶¶
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
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
Exercises: Artificial Intelligence Simplified Memory-bounded A*
SMA* ALGORITHM Simplified Memory-bounded A*
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
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
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
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
SMA* BY EXAMPLE Simplified Memory-bounded A*
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
SMA* by Example C S A GB 3 2 1 4 5 2 0 03 2 1 S3 3 0
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
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)
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
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
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
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
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
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
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
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
PROBLEM Simplified Memory-bounded A*
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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 ¶¶¶¶
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 (¶¶¶¶)
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
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
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 (¶¶¶¶)
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
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
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
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
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
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)
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) ¶¶¶¶
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
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
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) ¶¶¶¶
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
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
Exercises: Artificial Intelligence Monotonicity 1
PROBLEM Monotonicity 1
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)
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)
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)
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)
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)
Exercises: Artificial Intelligence Monotonicity 2
PROBLEM Monotonicity 2
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?
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.
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)
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)
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)
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)
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)
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)

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