Solving Partial Order Constraints for LPO termination

1. Solving Partial Order Constraints for LPO termination

2. Outline • We are given a partial order constraint, which is a formula of the form • The constraint is satisfiable if there exists a partial order on the symbols of the formula (f, g, h in the example above) that makes the formula true. • The partial order h<f=g makes the above formula true, hence it is satisfiable. • The problem is NP-complete.

3. SAT formulaeand SAT-solver • Logic formulae in propositional logic: • Satisfiability: Is there an assignment of truth values to the variables in the formula, which makes the formula true? • SAT solvers test for satisfiability of a formula. • Can handle hundreds of thousands of clauses. • A practical way to handle NP-hard problems, up to some extent.

4. Outline, cont. • To decide satisfiability of a constraint Φ: • Encode it as a SAT formula Φ’; • Use SAT-solver to the satisfiability of Φ’ . • Running time of a SAT solver depends upon the formula’s length and number of variables. • We want to encode the constraint in a compact way – to reduce the running time of the SAT solver.

5. Partial order constraints • A partial order constraint is just like a formula in propositional logic, except that propositions are atoms of the form(f < g) or (f = g).

6. Partial Order ConstraintsExamples • Let F = {f, g, h} • The following are partial order constraints on F:

7. Partial Order ConstraintsExamples (cont.) • Φ1can be satisfied by the orders: • g < f < h • g = h < f • Φ2 cannot be satisfied. • Can Φ3 be satisfied?

8. Total Order ≤ is a total order if it is: • Transitive. • Reflexive. • Anti-symmetric. • Total on a set F: for every a,b in F, a ≤b or b ≤a. Example: g<f=d<h=e

9. Partial Order ConstraintsSatisfiabilty • The definition of satisfiabilty is changed. • A constraint Φ on a set F is satisfiable if there exists an total order ≤ on F such that Φis true when atoms in Φare interpreted as follows: • (a < b) in Φ is true iff (a ≤ b) and ¬(b ≤ a) in the total order • (a = b) in Φ is true iff (a ≤ b) and (b ≤ a) in the total order

10. Axioms on the derived relations < and = For every f, g and h in F: • Reflexivity: (f = f) • Symmetry: (f = g) implies (g = f) • Asymmetry: ¬((f > g) and (g < f)) • Transitivity:((f > g) and (g > h)) implies (f > h)((f = g) and (g = h)) implies (f = h)

11. More Axioms on < and = • Identity: ((f > g) and (g = h)) implies (f > h) ((f = g) and (g > h)) implies (f > h) • Comparability:(f > g) or (f < g) or (f = g) • Comment: Comparability is required from a total order

12. Partial and Total orders • Recall that any partial order can be extended into a total order. • For negation-free constraints: If there exists an partial order such that Φis satisfied, then there exists a total order as well. • We are concerned with the question of satisfiability of partial order constraints. Not, for example, the number of possible partial orders satisfying Φ.

13. Solving Partial Order ConstraintsUsing SAT-solvers • Encoding: A partial order constraint Φon a set of symbols F is encoded by a propositional formula Φ’ such that Φis satisfiable if and only if Φ’ is. • A good encoding yields short formulae with a small number of variables.

14. “Atom-based” Encoding • First approach: atom-based encoding. Encode each atom as a propositional variable. • Notation: denotes the propositional variable corresponding to atom a. • Enforce the axioms of order and equality by adding propositions to the formula.

15. Encoding of Axioms:Examples For a set F • Reflexivity of = is encoded by: • Transitivity of < is encoded by:

16. “Atom-based” encoding:Size of the resulting formulae • For |F| = n, the resulting formula involves: • O(n2) variables. • O(n3) connectives. • This bound is always met. • Even if the formula is short, we still need • A variable for every pair of symbols in F. • A few connectives for every three symbols in F. (justto encode transitivity, for example).

17. Integer assignment and solution Let Φbe a partial order constraint on F, and let |F |=n. • An integer assignment for Φis a mapping μ : F→{1,…,n} • An integer solution for Φis an assignment θwhich makes Φtrue under the natural interpretations of < and = on the natural numbers.

18. Integer assignment and solutionExample • Consider again the constraint: • The assignment f = 2, g=1, h=1 is an integer solution for Φ1. • So is: f=2, g=1, h=3.

19. Theorem 1 • A partial order constraint is satisfiable if and only if it has an integer solution. • Reminder: a constraint is satisfiable if there exists a total order that makes it true.

20. A symbol-based propositional encoding • Encoding: A partial order constraint Φon a set of symbols F is encoded by a propositional formula Φ’ such that Φis satisfiable if and only if Φ’ is. • Our approach: Symbol-based. we encode the values of the symbols in F in an integer solution.

21. A symbol-based propositional encoding, cont. • Each symbol is modeled using propositional variables that encode the binary representation of its value. • Constraints of the form (f < g) or (f = g) are interpreted as constraints on integers, encoded in k-bit arithmetic. • Denote the encoding by: ||(f < g)||k

22. A symbol-based propositional encoding, cont. • Fora symbol f in F, the k-bit representation of its value is: • A constraint of the form (f = g) is encoded by:

23. A symbol-based propositional encoding, cont. • A constraint of the form (f > g) is encoded recursively by: for k > 1

24. A symbol-based propositional encoding, cont. • We encode a constraint Φby replacing each atom (e.g. (f < g)) with its k-bit encoding.

25. A symbol-based encodingExample The constraint on F={g,f,h}: The propositional variables are:

26. A symbol-based encoding Example, cont.

27. A symbol-based encoding Example, cont. Putting it all together: A satisfying assignment:

28. A symbol-based encodingSolution of the Example • We found the satisfying assignment: • The corresponding integer solution is:f=3, h=2, g=2 • The corresponding total order is: h=g<f

29. Theorem 2 • A partial order constraint Φ on symbols F is satisfiable if and only if its symbol-based propositional encoding is satisfiable.

30. A symbol-based encoding:Size • The encoding of a constraint Φ with n symbols involves: • O(nlog(n)) variables. • O(|Φ|log(n)) connectives. • Recall that in the first approach, the encoding involved: • O(n2) variables. • O(n3) connectives.

31. Encode symbols as k-bit representation (as in symbol-based encoding). Encode each atom that appears in Φ as a special variable. Add propositions to connect atoms of the formula (just once!)with the symbol-based encoding of their meanings. What if Φ is long?Alternative encoding

32. Alternative encodingExample • The constraint:is encoded as follows: (a1, a2 are the special variables)

33. Alternative symbol-based encodingSize • The encoding of a constraint Φ with on a set F with n symbols involves • O(n2) variables.(for each unique atom in Φ) • O(|Φ|+ n2log(n)) connectives. • Good for long constraints

34. MotivationTerm Rewrite Systems • A term rewrite system is a set of rules of the form where l and r are terms constructed from functionsymbols and variables.

35. Term Rewrite Systems • The rules of a TRS are used to rewrite expressions. • A rule s→t can be applied on an expression e if e matches a sub-term of s.

36. Term Rewrite SystemsExample • Fibonacci:

37. Term Rewrite SystemsExample, cont. Computing fib(3): • fib(s(s(s(0))) →5fib(s(s(0))+fib(s(o)) • fib(s(s(0))+fib(s(o)) →4 fib(s(s(0))+s(0) • fib(s(s(0))+s(0) →5 fib(s(0))+fib(0)+s(0) • fib(s(0))+fib(0)+s(0) →3 fib(s(0))+0+s(0) • fib(s(0))+0+s(0) →1 fib(s(0))+s(0) • fib(s(0))+s(0) →4 s(0)+s(0) • s(0)+s(0) →2 s(s(0))

38. Term Rewrite SystemsImportance • A subject of theoretical interest. • Many applications, for example: • Automatic theorem proving. • Verifications of programs and chips. • Termination of a TRS: does the rewrite stop after a finite number of steps, for every expression?

39. LPO Termination ofTerm Rewrite Systems • In general, termination of terms rewrite systems is undecidable. • Lexicographic path order (LPO) is an order on the terms, that is induced by a partial order on the function symbols. • The partial order on functions can be thought of as “precedence” order.

40. LPO Termination ofTerm Rewrite Systems, cont. • LPO termination problem: is there a partial order >F over function symbols, such that for every rule it hold thatWhere is the order induced by >F • LPO termination implies termination of the rewrite system.

41. Partial Order Constraintsand LPO Termination • The lexicographic path order (LPO) is induced by a partial order on the function symbols. • The rules of the TRS unfolds a partial order constrain on the function symbols. • Example: consider the rule s(x)+y → s(x+y):

42. Experimentation • Termination tools are tested against benchmarks. • Benchmarks used: 751 rewrite systems. • Our method, poSAT, is compared with another method called TTT. • TTT: a fast and powerful tool with a convenient interface.

43. Experimentation results • Running time in seconds for strict-LPO termination

44. Thank you!