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Chapter 7 Propositional Satisfiability Techniques

Lecture slides for Automated Planning: Theory and Practice. Chapter 7 Propositional Satisfiability Techniques. Dana S. Nau CMSC 722, AI Planning University of Maryland, Fall 2004. Motivation. Propositional satisfiability: given a boolean formula

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Chapter 7 Propositional Satisfiability Techniques

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  1. Lecture slides for Automated Planning: Theory and Practice Chapter 7Propositional Satisfiability Techniques Dana S. Nau CMSC 722, AI Planning University of Maryland, Fall 2004

  2. Motivation • Propositional satisfiability: given a boolean formula • e.g., (P  Q)  (Q  R  S)  (R P), does there exist a model • i.e., an assignment of truth values to the propositions that makes the formula true? • This was the very first problem shown to be NP-complete • Lots of research on algorithms for solving it • Algorithms are known for solving all but a small subset in average-case polynomial time • Therefore, • Try translating classical planning problems into satisfiability problems, and solving them that way

  3. Outline • Encoding planning problems as satisfiability problems • Extracting plans from truth values • Satisfiability algorithms • Davis-Putnam • Local search • GSAT • Combining satisfiability with planning graphs • BlackBox

  4. Overall Approach • Bounded planning problem (P,n): • P is a planning problem; n is a positive integer • Find a solution for P of length n • Do iterative deepening like we did with Graphplan: • for n = 0, 1, 2, …, • encode (P,n) as a satisfiability problem  • if  is satisfiable, then • From the set of truth values that satisfies , a solution plan can be constructed, so return it and exit • We’ll use propositional logic • but we’ll write propositions that look like ground atoms • e.g., at(r1,loc1) is the proposition that’s true iff r1 is at loc1

  5. Fluents • If π = a0, a1, …, an–1 is a solution for (P,n), then it generates the following states: s0, s1 = (s0,a0), s2 = (s1,a1), …, sn = (sn–1, an–1) • Fluents: propositions used to describe what’s true in each si • at(r1,loc1,i) is a fluent that’s true iff at(r1,loc1) is in si • We’ll use lito denote the fluent for literal l in state si • e.g., if l = at(r1,loc1) then li = at(r1,loc1,i) • ai is a fluent saying that a is the i’th step of π • e.g., if a = move(r1,loc2,loc1) then ai = move(r1,loc2,loc1,i)

  6. Encoding Planning Problems • Encode (P,n) as a formula  such that π = a0, a1, …, an–1 is a solution for (P,n) if and only if  can be satisfied in a way that makes the fluents a0, …, an–1 true • Let • A = {all actions in the planning domain} • S = {all states in the planning domain} • L = {all literals in the language} •  is the conjunct of many other formulas …

  7. Formulas in  • Formula describing the initial state: /\{l0 | ls0} /\{l0 | lL – s0} • Formula describing the goal: /\{ln | lg+} /\{ln | lg–} • For every action a in A, formulas describing what changes a would make if it were the i’th step of the plan: • ai /\{pi | p Precond(a)} /\{ei+1 | e Effects(a)} • Complete exclusion axiom: • For all actions a and b, formulas saying they can’t occur at the same time ai bi • this guarantees there can be only one action at a time • Is this enough?

  8. Frame Axioms • Frame axioms: • Formulas describing what doesn’t changebetween steps i and i+1 • Several ways to write these • One way: explanatory frame axioms • One axiom for every literal l • Says that if l changes between si and si+1, then the action at step i must be responsible: (lili+1V{ai |l in effects+(a)}) (lili+1V{ai |l in effects–(a)})

  9. Example • Planning domain: • one robot r1 • two adjacent locations l1, l2 • one operator (move the robot) • Encode (P,n) where n = 1 • Initial state: {at(r1,l1)} Encoding: at(r1,l1,0) at(r1,l2,0) • Goal: {at(r1,l2)} Encoding: at(r1,l2,1) at(r1,l1,1) • Operator: see next slide

  10. Example (continued) • Operator: move(r,l,l’) precond: at(r,l) effects: at(r,l’), at(r,l) Encoding: move(r1,l1,l2,0)  at(r1,l1,0)  at(r1,l2,1) at(r1,l1,1) move(r1,l2,l1,0)  at(r1,l2,0)  at(r1,l1,1) at(r1,l2,1) move(r1,l1,l1,0)  at(r1,l1,0)  at(r1,l1,1) at(r1,l1,1) move(r1,l2,l2,0)  at(r1,l2,0)  at(r1,l2,1) at(r1,l2,1) move(l1,r1,l2,0)  … move(l2,l1,r1,0)  … move(l1,l2,r1,0)  … move(l2,l1,r1,0)  … • How to avoid generating the last four actions? • Assign data types to the constant symbolslike we did for state-variable representation contradictions(easy to detect) nonsensical

  11. Example (continued) • Locations: l1, l2 • Robots: r1 • Operator: move(r : robot, l : location, l’ : location) precond: at(r,l) effects: at(r,l’), at(r,l) Encoding: move(r1,l1,l2,0)  at(r1,l1,0)  at(r1,l2,1) at(r1,l1,1) move(r1,l2,l1,0)  at(r1,l2,0)  at(r1,l1,1) at(r1,l2,1)

  12. Example (continued) • Complete-exclusion axiom: move(r1,l1,l2,0) move(r1,l2,l1,0) • Explanatory frame axioms: at(r1,l1,0)  at(r1,l1,1)  move(r1,l2,l1,0) at(r1,l2,0)  at(r1,l2,1)  move(r1,l1,l2,0) at(r1,l1,0) at(r1,l1,1)  move(r1,l1,l2,0) at(r1,l2,0) at(r1,l2,1)  move(r1,l2,l1,0)

  13. Extracting a Plan • Suppose we find an assignment of truth values that satisfies . • This means P has a solution of length n • For i=1,…,n, there will be exactly one action a such that ai = true • This is the i’th action of the plan. • Example (from the previous slides): •  can be satisfied with move(r1,l1,l2,0) = true • Thus move(r1,l1,l2,0) is a solution for (P,0) • It’s the only solution - no other way to satisfy 

  14. Planning • How to find an assignment of truth values that satisfies ? • Use a satisfiability algorithm • Example: the Davis-Putnam algorithm • First need to put  into conjunctive normal form e.g.,  = D  (D  A B)  (D A B)  (D A  B)  A • Write as a set of clauses (disjuncts of literals)  = {D, (D  A B), (D A B), (D A  B), A} • Two special cases: •  = {} is a formula that’s always true •  = {…, (), …} is a formula that’s always false

  15. The Davis-Putnam Procedure Depth-first backtracking through combinations of truth values • Select a variable P in  • Recursive call on   P Simplify: remove alloccurrences of P If  = {…, (), …}then backtrack Else if  = {} thenhave a solution • Recursive callon   P Simplify: remove alloccurrences of P If  = {…, (), …}then backtrack Else if  = {} then have a solution

  16. Local Search • Let u be an assignment of truth values to all of the variables • cost(u,) = number of clauses in  that aren’t satisfied by u • flip(P,u) = u with the truth value of P reversed • Local search: • Select a random assignment u • while cost(u,) ≠ 0 • if there is a P such that cost(flip(P,u),) < cost(u,) then • randomly choose any such P • u flip(P,u) • else return failure • Local search is sound • If it finds a solution it will find it very quickly • Local search is not complete: can get trapped in local minima

  17. GSAT • Basic-GSAT: • Select a random assignment u • while cost(u,) ≠ 0 • choose the P that minimizes cost(flip(P,u),) • Not guaranteed to terminate • GSAT: • restart after a max number of flips • return failure after a max number of restarts • The book discusses several other stochastic procedures • One is Walksat • works better than both local search and GSAT • I’ll skip the details

  18. Discussion • Recall the overall approach: • for n = 0, 1, 2, …, • encode (P,n) as a satisfiability problem  • if  is satisfiable, then • From the set of truth values that satisfies , extract a solution plan and return it • How well does this work?

  19. Discussion • Recall the overall approach: • for n = 0, 1, 2, …, • encode (P,n) as a satisfiability problem  • if  is satisfiable, then • From the set of truth values that satisfies , extract a solution plan and return it • How well does this work? • By itself, not very practical (takes too much memory and time) • But it can be combined with other techniques • e.g., planning graphs

  20. BlackBox • The BlackBox procedure combines planning-graph expansion and satisfiability checking • It is roughly as follows: • for n = 0, 1, 2, … • Graph expansion: • create a “planning graph” that contains n “levels” • Check whether the planning graph satisfies a necessary(but insufficient) condition for plan existence • If it does, then • Encode (P,n) as a satisfiability problem  but include only the actions in the planning graph • If  is satisfiable then return the solution

  21. More about BlackBox • Memory requirement still is combinatorially large, but less than satisfiability alone • It was one of the two fastest planners in the 1998 planning competition

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