Planning and Optimization D2. Abstractions: Additive Abstractions - PowerPoint PPT Presentation

Planning and Optimization D2. Abstractions: Additive Abstractions Gabriele R¨ oger and Thomas Keller Universit¨ at Basel October 29, 2018

Multiple Abstractions Additivity Outlook Summary Content of this Course Tasks Progression/ Regression Classical Complexity Heuristics Planning MDPs Uninformed Search Probabilistic Heuristic Search Monte-Carlo Methods

Multiple Abstractions Additivity Outlook Summary Content of this Course: Heuristics Abstractions Delete Relaxation in General Pattern Abstraction Databases Merge & Heuristics Landmarks Shrink Potential Heuristics Cost Partitioning

Multiple Abstractions Additivity Outlook Summary Multiple Abstractions

Multiple Abstractions Additivity Outlook Summary Multiple Abstractions One important practical question is how to come up with a suitable abstraction mapping α . Indeed, there is usually a huge number of possibilities, and it is important to pick good abstractions (i.e., ones that lead to informative heuristics). However, it is generally not necessary to commit to a single abstraction.

Multiple Abstractions Additivity Outlook Summary Combining Multiple Abstractions Maximizing several abstractions: Each abstraction mapping gives rise to an admissible heuristic. By computing the maximum of several admissible heuristics, we obtain another admissible heuristic which dominates the component heuristics. Thus, we can always compute several abstractions and maximize over the individual abstract goal distances. Adding several abstractions: In some cases, we can even compute the sum of individual estimates and still stay admissible. Summation often leads to much higher estimates than maximization, so it is important to understand under which conditions summation of heuristics is admissible.

Multiple Abstractions Additivity Outlook Summary Adding Several Abstractions: Example (1) ALR ARL LLR RRL ALL ARR LRR LLL RRR RLL BLL BRR LRL RLR BRL BLR h ∗ (LRR) = 4

Multiple Abstractions Additivity Outlook Summary Adding Several Abstractions: Example (2) ALR ALR ARL ARL LLR LLR RRL RRL ALL ARR ALL ARR LRR LLL LLL RRR RRR RLL RLL BLL BLL BRR BRR h α 2 (LRR) = 2 LRL LRL RLR RLR BRL BLR ALR ALR ARL ARL BRL BLR LLR LLR RRL RRL h α 1 (LRR) = 3 ALL ALL ARR ARR LLL RRR RLL LRR LLL RRR RLL BLL BLL BRR BRR LRL LRL RLR RLR BRL BLR BRL BLR

Multiple Abstractions Additivity Outlook Summary Adding Several Abstractions: Example (3) ALR ALR ARL ARL LLR LLR RRL RRL ALL ARR ALL ARR LRR LLL LLL RRR RRR RLL RLL BLL BLL BRR BRR h α 2 (LRR) = 1 LRL LRL RLR RLR BRL BLR ALR ALR ARL ARL BRL BLR LLR LLR RRL RRL h α 1 (LRR) = 3 ALL ALL ARR ARR LRR LLL RRR RLL LRR LLL RRR RLL BLL BLL BRR BRR LRL LRL RLR RLR BRL BLR BRL BLR

Multiple Abstractions Additivity Outlook Summary Additivity

Multiple Abstractions Additivity Outlook Summary Orthogonality of Abstractions Definition (Orthogonal) Let α 1 and α 2 be abstractions of transition system T . ℓ We say that α 1 and α 2 are orthogonal if for all transitions s − → t of T , we have α i ( s ) = α i ( t ) for at least one i ∈ { 1 , 2 } .

Multiple Abstractions Additivity Outlook Summary Affecting Transition Labels Definition (Affecting Transition Labels) Let T be a transition system, and let ℓ be one of its labels. ℓ We say that ℓ affects T if T has a transition s − → t with s � = t . Theorem (Affecting Labels vs. Orthogonality) Let α 1 and α 2 be abstractions of transition system T . If no label of T affects both T α 1 and T α 2 , then α 1 and α 2 are orthogonal. (Easy proof omitted.)

Multiple Abstractions Additivity Outlook Summary Orthogonality and Additivity Theorem (Additivity for Orthogonal Abstractions) Let h α 1 , . . . , h α n be abstraction heuristics of the same transition system such that α i and α j are orthogonal for all i � = j. Then � n i =1 h α i is a safe, goal-aware, admissible and consistent heuristic for Π .

Multiple Abstractions Additivity Outlook Summary Orthogonality and Additivity: Example (1) ALR ARL ALR ARL LLR LLR RRL RRL ALL ALL ARR ARR LLL RRR RLL LRR LLL RRR RLL BLL BLL BRR BRR h α 2 (LRR) = 2 LRL LRL RLR RLR ALR ARL BRL BLR ALR ARL BRL BLR LLR LLR RRL RRL h α 1 (LRR) = 3 ALL ALL ARR ARR LRR LLL LLL RRR RRR RLL RLL BLL BRR BLL BRR LRL LRL RLR RLR BRL BRL BLR BLR

Multiple Abstractions Additivity Outlook Summary Orthogonality and Additivity: Example (2) ALR ARL ALR ARL LLR LLR RRL RRL ALL ALL ARR ARR LLL RRR RLL LRR LLL RRR RLL BLL BLL BRR BRR h α 2 (LRR) = 1 LRL LRL RLR RLR ALR ARL BRL BLR ALR ARL BRL BLR LLR LLR RRL RRL h α 1 (LRR) = 3 ALL ALL ARR ARR LRR LRR LLL LLL RRR RRR RLL RLL BLL BRR BLL BRR LRL LRL RLR RLR BRL BRL BLR BLR

Multiple Abstractions Additivity Outlook Summary Orthogonality and Additivity: Proof (1) Proof. We prove goal-awareness and consistency; the other properties follow from these two. Let T = � S , L , c , T , s 0 , S ⋆ � be the concrete transition system. Let h = � n i =1 h α i . Goal-awareness: For goal states s ∈ S ⋆ , h ( s ) = � n i =1 h α i ( s ) = � n i =1 0 = 0 because all individual abstraction heuristics are goal-aware. . . .

Multiple Abstractions Additivity Outlook Summary Orthogonality and Additivity: Proof (1) Proof. We prove goal-awareness and consistency; the other properties follow from these two. Let T = � S , L , c , T , s 0 , S ⋆ � be the concrete transition system. Let h = � n i =1 h α i . Goal-awareness: For goal states s ∈ S ⋆ , h ( s ) = � n i =1 h α i ( s ) = � n i =1 0 = 0 because all individual abstraction heuristics are goal-aware. . . .

Multiple Abstractions Additivity Outlook Summary Orthogonality and Additivity: Proof (2) Proof (continued). o Consistency: Let s − → t ∈ T . We must prove h ( s ) ≤ c ( o ) + h ( t ). Because the abstractions are orthogonal, α i ( s ) � = α i ( t ) for at most one i ∈ { 1 , . . . , n } . Case 1: α i ( s ) = α i ( t ) for all i ∈ { 1 , . . . , n } . Then h ( s ) = � n i =1 h α i ( s ) = � n i =1 h ∗ T α i ( α i ( s )) = � n i =1 h ∗ T α i ( α i ( t )) = � n i =1 h α i ( t ) = h ( t ) ≤ c ( o ) + h ( t ) . . . .

Multiple Abstractions Additivity Outlook Summary Orthogonality and Additivity: Proof (2) Proof (continued). o Consistency: Let s − → t ∈ T . We must prove h ( s ) ≤ c ( o ) + h ( t ). Because the abstractions are orthogonal, α i ( s ) � = α i ( t ) for at most one i ∈ { 1 , . . . , n } . Case 1: α i ( s ) = α i ( t ) for all i ∈ { 1 , . . . , n } . Then h ( s ) = � n i =1 h α i ( s ) = � n i =1 h ∗ T α i ( α i ( s )) = � n i =1 h ∗ T α i ( α i ( t )) = � n i =1 h α i ( t ) = h ( t ) ≤ c ( o ) + h ( t ) . . . .

Multiple Abstractions Additivity Outlook Summary Orthogonality and Additivity: Proof (2) Proof (continued). o Consistency: Let s − → t ∈ T . We must prove h ( s ) ≤ c ( o ) + h ( t ). Because the abstractions are orthogonal, α i ( s ) � = α i ( t ) for at most one i ∈ { 1 , . . . , n } . Case 1: α i ( s ) = α i ( t ) for all i ∈ { 1 , . . . , n } . Then h ( s ) = � n i =1 h α i ( s ) = � n i =1 h ∗ T α i ( α i ( s )) = � n i =1 h ∗ T α i ( α i ( t )) = � n i =1 h α i ( t ) = h ( t ) ≤ c ( o ) + h ( t ) . . . .

Multiple Abstractions Additivity Outlook Summary Orthogonality and Additivity: Proof (2) Proof (continued). o Consistency: Let s − → t ∈ T . We must prove h ( s ) ≤ c ( o ) + h ( t ). Because the abstractions are orthogonal, α i ( s ) � = α i ( t ) for at most one i ∈ { 1 , . . . , n } . Case 1: α i ( s ) = α i ( t ) for all i ∈ { 1 , . . . , n } . Then h ( s ) = � n i =1 h α i ( s ) = � n i =1 h ∗ T α i ( α i ( s )) = � n i =1 h ∗ T α i ( α i ( t )) = � n i =1 h α i ( t ) = h ( t ) ≤ c ( o ) + h ( t ) . . . .

Multiple Abstractions Additivity Outlook Summary Orthogonality and Additivity: Proof (3) Proof (continued). Case 2: α i ( s ) � = α i ( t ) for exactly one i ∈ { 1 , . . . , n } . Let k ∈ { 1 , . . . , n } such that α k ( s ) � = α k ( t ). Then h ( s ) = � n i =1 h α i ( s ) i ∈{ 1 ,..., n }\{ k } h ∗ T α i ( α i ( s )) + h α k ( s ) = � i ∈{ 1 ,..., n }\{ k } h ∗ T α i ( α i ( t )) + c ( o ) + h α k ( t ) ≤ � = c ( o ) + � n i =1 h α i ( t ) = c ( o ) + h ( t ) , where the inequality holds because α i ( s ) = α i ( t ) for all i � = k and h α k is consistent.

Multiple Abstractions Additivity Outlook Summary Orthogonality and Additivity: Proof (3) Proof (continued). Case 2: α i ( s ) � = α i ( t ) for exactly one i ∈ { 1 , . . . , n } . Let k ∈ { 1 , . . . , n } such that α k ( s ) � = α k ( t ). Then h ( s ) = � n i =1 h α i ( s ) i ∈{ 1 ,..., n }\{ k } h ∗ T α i ( α i ( s )) + h α k ( s ) = � i ∈{ 1 ,..., n }\{ k } h ∗ T α i ( α i ( t )) + c ( o ) + h α k ( t ) ≤ � = c ( o ) + � n i =1 h α i ( t ) = c ( o ) + h ( t ) , where the inequality holds because α i ( s ) = α i ( t ) for all i � = k and h α k is consistent.

Multiple Abstractions Additivity Outlook Summary Outlook