Planning and Optimization October 16, 2019 C2. Delete Relaxation: - PowerPoint PPT Presentation

Planning and Optimization October 16, 2019 — C2. Delete Relaxation: Properties of Relaxed Planning Tasks C2.1 The Domination Lemma Planning and Optimization C2. Delete Relaxation: Properties of Relaxed Planning Tasks C2.2 The Relaxation Lemma Malte Helmert and Thomas Keller C2.3 Further Properties Universit¨ at Basel C2.4 Greedy Algorithm October 16, 2019 C2.5 Summary M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 1 / 28 M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 2 / 28 Content of this Course Content of this Course: Heuristics Delete Relaxation Relaxed Tasks Foundations Heuristics Relaxed Logic Abstraction Task Graphs Classical Heuristics Relaxation Heuristics Constraints Landmarks Planning Network Constraints Explicit MDPs Flows Probabilistic Factored MDPs Potential Heuristics M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 3 / 28 M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 4 / 28

C2. Delete Relaxation: Properties of Relaxed Planning Tasks The Domination Lemma C2. Delete Relaxation: Properties of Relaxed Planning Tasks The Domination Lemma On-Set and Dominating States Definition (On-Set) The on-set of a valuation s is the set of propositional variables C2.1 The Domination Lemma that are true in s , i.e., on ( s ) = s − 1 ( { T } ). � for states of propositional planning tasks: states can be viewed as sets of (true) state variables Definition (Dominate) A valuation s ′ dominates a valuation s if on ( s ) ⊆ on ( s ′ ). � all state variables true in s are also true in s ′ M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 5 / 28 M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 6 / 28 C2. Delete Relaxation: Properties of Relaxed Planning Tasks The Domination Lemma C2. Delete Relaxation: Properties of Relaxed Planning Tasks The Domination Lemma Domination Lemma (1) Domination Lemma (2) Proof (continued). Lemma (Domination) ◮ Base case χ = v ∈ V : if s | = v , then v ∈ on ( s ). With on ( s ) ⊆ on ( s ′ ), we get v ∈ on ( s ′ ) and hence s ′ | Let s and s ′ be valuations of a set of propositional variables V , = v . ◮ Inductive case χ = χ 1 ∧ χ 2 : by induction hypothesis, our and let χ be a propositional formula over V claim holds for the proper subformulas χ 1 and χ 2 of χ . which does not contain negation symbols. = χ and s ′ dominates s, then s ′ | If s | = χ . s | = χ = ⇒ s | = χ 1 ∧ χ 2 = ⇒ s | = χ 1 and s | = χ 2 Proof. I.H. (twice) s ′ | = χ 1 and s ′ | = ⇒ = χ 2 Proof by induction over the structure of χ . s ′ | ◮ Base case χ = ⊤ : then s ′ | = ⇒ = χ 1 ∧ χ 2 = ⊤ . s ′ | = ⇒ = χ. ◮ Base case χ = ⊥ : then s �| = ⊥ . . . . ◮ Inductive case χ = χ 1 ∨ χ 2 : analogous M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 7 / 28 M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 8 / 28

C2. Delete Relaxation: Properties of Relaxed Planning Tasks The Relaxation Lemma C2. Delete Relaxation: Properties of Relaxed Planning Tasks The Relaxation Lemma Add Sets and Delete Sets Definition (Add Set and Delete Set for an Effect) Consider a propositional planning task with state variables V . Let e be an effect over V , and let s be a state over V . C2.2 The Relaxation Lemma The add set of e in s , written addset ( e , s ), and the delete set of e in s , written delset ( e , s ), are defined as the following sets of state variables: addset ( e , s ) = { v ∈ V | s | = effcond ( v , e ) } delset ( e , s ) = { v ∈ V | s | = effcond ( ¬ v , e ) } Note: For all states s and operators o applicable in s , we have on ( s � o � ) = ( on ( s ) \ delset ( eff ( o ) , s )) ∪ addset ( eff ( o ) , s ). M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 9 / 28 M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 10 / 28 C2. Delete Relaxation: Properties of Relaxed Planning Tasks The Relaxation Lemma C2. Delete Relaxation: Properties of Relaxed Planning Tasks The Relaxation Lemma Relaxation Lemma Proof of Relaxation Lemma (1) For this and the following chapters on delete relaxation, we assume implicitly that we are working with propositional planning tasks in positive normal form. Proof. Let V be the set of state variables. Lemma (Relaxation) Let s be a state, and let s ′ be a state that dominates s. Part 1: Because o is applicable in s , we have s | = pre ( o ). Because pre ( o ) is negation-free and s ′ dominates s , 1 If o is an operator applicable in s, we get s ′ | = pre ( o ) from the domination lemma. then o + is applicable in s ′ and s ′ � o + � dominates s � o � . Because pre ( o + ) = pre ( o ), this shows that o + is applicable in s ′ . 2 If π is an operator sequence applicable in s, . . . then π + is applicable in s ′ and s ′ � π + � dominates s � π � . 3 If additionally π leads to a goal state from state s, then π + leads to a goal state from state s ′ . M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 11 / 28 M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 12 / 28

C2. Delete Relaxation: Properties of Relaxed Planning Tasks The Relaxation Lemma C2. Delete Relaxation: Properties of Relaxed Planning Tasks The Relaxation Lemma Proof of Relaxation Lemma (2) Proof of Relaxation Lemma (3) Proof (continued). Proof (continued). To prove that s ′ � o + � dominates s � o � , We then get: we first compare the relevant add sets: on ( s � o � ) = ( on ( s ) \ delset ( eff ( o ) , s )) ∪ addset ( eff ( o ) , s ) addset ( eff ( o ) , s ) = { v ∈ V | s | = effcond ( v , eff ( o )) } ⊆ on ( s ) ∪ addset ( eff ( o ) , s ) = effcond ( v , eff ( o + )) } = { v ∈ V | s | (1) ⊆ on ( s ′ ) ∪ addset ( eff ( o + ) , s ′ ) ⊆ { v ∈ V | s ′ | = effcond ( v , eff ( o + )) } (2) = on ( s ′ � o + � ) , = addset ( eff ( o + ) , s ′ ) , and thus s ′ � o + � dominates s � o � . where (1) uses effcond ( v , eff ( o )) ≡ effcond ( v , eff ( o + )) and (2) uses the dominance lemma (note that effect conditions This concludes the proof of Part 1. . . . are negation-free for operators in positive normal form). . . . M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 13 / 28 M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 14 / 28 C2. Delete Relaxation: Properties of Relaxed Planning Tasks The Relaxation Lemma C2. Delete Relaxation: Properties of Relaxed Planning Tasks The Relaxation Lemma Proof of Relaxation Lemma (4) Proof of Relaxation Lemma (5) Proof (continued). Part 2: by induction over n = | π | Proof (continued). Base case: π = �� Part 3: Let γ be the goal formula. The empty plan is trivially applicable in s ′ , and From Part 2, we obtain that t ′ = s ′ � π + � dominates t = s � π � . s ′ � �� + � = s ′ dominates s � �� � = s by prerequisite. By prerequisite, t is a goal state and hence t | = γ . Inductive case: π = � o 1 , . . . , o n +1 � By the induction hypothesis, � o + 1 , . . . , o + Because the task is in positive normal form, γ is negation-free, n � is applicable in s ′ , and hence t ′ | and t ′ = s ′ � � o + 1 , . . . , o + = γ because of the domination lemma. n � � dominates t = s � � o 1 , . . . , o n � � . Also, o n +1 is applicable in t . Therefore, t ′ is a goal state. n +1 is applicable in t ′ and s ′ � π + � = t ′ � o + Using Part 1, o + n +1 � dominates s � π � = t � o n +1 � . This concludes the proof of Part 2. . . . M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 15 / 28 M. Helmert, T. Keller (Universit¨ at Basel) Planning and Optimization October 16, 2019 16 / 28