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 a 3 201 102 l 1 l 5 a 4 l 2 l 6 021 012 a 1 a 5 l 3 a 2 a 6 120 210 l 4

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion Abstractions a 3 201 102 l 1 l 5 a 4 l 2 l 6 021 012 a 1 a 5 l 3 a 2 a 6 120 210 l 4 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 a 3 a 3 201 102 201 102 l 1 l 5 l 1 l 5 a 4 a 4 l 2 l 6 l 2 l 6 021 012 021 012 a 1 a 5 a 1 a 5 l 3 l 3 a 2 a 6 a 2 a 6 120 210 120 210 l 4 l 4 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 All-or-nothing 1 Cost-splitting 2 Location-based costs 3 Results 4 Cost saturation 5

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 a 3 201 102 l 1 l 5 a 4 l 2 l 6 021 012 a 1 a 5 l 3 a 2 a 6 120 210 l 4 All operators change more than one object

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion Pancake Puzzle a 3 201 102 l 1 l 5 a 4 l 2 l 6 021 012 a 1 a 5 l 3 a 2 a 6 120 210 l 4 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 c i ( o ) = b o i × c ( o ) b o a 3 a 3 201 102 201 102 l 1 l 5 l 1 l 5 a 4 a 4 l 2 l 6 l 2 l 6 021 012 021 012 a 1 a 5 a 1 a 5 l 3 l 3 a 2 a 6 a 2 a 6 120 210 120 210 l 4 l 4 b l = 2, b a = 3 → c i ( l ) = 1 / 2 , c i ( 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 c i ( o ) = b o i × c ( o ) b o a 3 a 3 201 102 201 102 l 1 l 5 l 1 l 5 a 4 a 4 l 2 l 6 l 2 l 6 021 012 021 012 a 1 a 5 a 1 a 5 l 3 l 3 a 2 a 6 a 2 a 6 120 210 120 210 l 4 l 4 b l = 2, b a = 3 → c i ( l ) = 1 / 2 , c i ( 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 loc o c i ( o ) = c ( o ) if o changes loc o to a distinguished value in abstraction i and 0 otherwise a 3 a 3 201 102 201 102 l 1 l 5 l 1 l 5 a 4 a 4 l 2 l 6 l 2 l 6 021 012 021 012 a 1 l 3 a 5 a 1 l 3 a 5 a 2 a 6 a 2 a 6 120 210 120 210 l 4 l 4 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 loc o c i ( o ) = c ( o ) if o changes loc o to a distinguished value in abstraction i and 0 otherwise a 3 a 3 201 102 201 102 l 1 l 5 l 1 l 5 a 4 a 4 l 2 l 6 l 2 l 6 021 012 021 012 a 1 l 3 a 5 a 1 l 3 a 5 a 2 a 6 a 2 a 6 120 210 120 210 l 4 l 4 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 loc o c i ( o ) = c ( o ) if o changes loc o to a distinguished value in abstraction i and 0 otherwise a 3 a 3 201 102 201 102 l 1 l 5 l 1 l 5 a 4 a 4 l 2 l 6 l 2 l 6 021 012 021 012 a 1 l 3 a 5 a 1 l 3 a 5 a 2 a 6 a 2 a 6 120 210 120 210 l 4 l 4 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 = 1 1 1 2 o 3 o 5 1 o 3 o 2 2 5 h = 3 h = 2 h = 0 o 7 o 1 7 5 o 7 o 6 2 o 1 h = 4

Introduction All-or-nothing Cost-splitting Location-based costs Results Cost saturation Conclusion Cost saturation h = 1 1 1 2 o 3 o 5 1 o 3 o 2 2 5 → 4 h = 3 h = 2 h = 0 o 7 o 1 7 → 0 5 → 4 o 7 o 6 2 o 1 h = 4

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