A General Theory of Additive State Space Abstractions by Yang, - - PowerPoint PPT Presentation

A General Theory of Additive State Space Abstractions

by Yang, Culberson, Holte, Zahavi and Felner Jendrik Seipp

Artificial Intelligence Group University of Basel

November 14, 2013

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Introduction

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Example Pancake Puzzle

021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Abstractions

021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6

Coarser version of state space (e. g. PDB) Homomorphic mapping Preserve paths Underestimate goal-distances Goal-distance heuristic admissible

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Multiple abstractions

021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6 021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6

Max of estimates is admissible Sum is usually not admissible Costs counted multiple times

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Outline ⇒ Divide each operator’s cost among abstractions

1

All-or-nothing

2

Cost-splitting

3

Location-based costs

4

Results

5

Cost saturation

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

All-or-nothing

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

8-Puzzle – Maximum

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

8-Puzzle – Sum

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Pancake Puzzle

021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6

All operators change more than one object

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Pancake Puzzle

021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6

All operators change more than one object

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Cost-splitting

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Cost-splitting

ci(o) = bo

i × c(o)

bo

021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6 021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6

bl = 2, ba = 3 → ci(l) = 1/2, ci(a) = 1/3 h(021) = (1/3 + 1/2) + (1/2 + 1/3) = 5/3

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Cost-splitting

ci(o) = bo

i × c(o)

bo

021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6 021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6

bl = 2, ba = 3 → ci(l) = 1/2, ci(a) = 1/3 h(021) = (1/3 + 1/2) + (1/2 + 1/3) = 5/3

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Location-based costs

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Location-based costs

Assign each operator o a location loco ci(o) = c(o) if o changes loco to a distinguished value in abstraction i and 0 otherwise

021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6 021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6

loc(o) = left-most position. Move to middle state costs 1, everything else 0 h(021) = (1 + 0) + (1 + 0) = 2

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Location-based costs

Assign each operator o a location loco ci(o) = c(o) if o changes loco to a distinguished value in abstraction i and 0 otherwise

021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6 021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6

loc(o) = left-most position. Move to middle state costs 1, everything else 0 h(021) = (1 + 0) + (1 + 0) = 2

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Location-based costs

Assign each operator o a location loco ci(o) = c(o) if o changes loco to a distinguished value in abstraction i and 0 otherwise

021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6 021 201 102 120 210 012 l1 l2 l3 l4 l5 l6 a1 a2 a3 a4 a5 a6

loc(o) = left-most position. Move to middle state costs 1, everything else 0 h(021) = (1 + 0) + (1 + 0) = 2

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Results

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Results

cost loc TopSpin Puzzle

• X

Pancake Puzzle X

• Rubik’s Cube

X X

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Cost saturation

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Cost saturation

h = 3 h = 2 h = 1 h = 4 h = 0 1

• 3

2

• 2

2

• 7

7

• 6

1

• 5

2

• 7

1

• 3

5

• 1

5

• 1

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Cost saturation

h = 3 h = 2 h = 1 h = 4 h = 0 1

• 3

2

• 2

2

• 7

7 → 0

• 6

1

• 5

2

• 7

1

• 3

5 → 4

• 1

5 → 4

• 1

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Conclusion

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion

Conclusion

Cost partitioning → additive abstractions Usefulness varies between problems