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Solving Partial Order Constraints for LPO termination - PowerPoint PPT Presentation

Solving Partial Order Constraints for LPO termination. 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.

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Solving Partial Order Constraints for LPO termination

• 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.

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.

• 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.

• 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).

Partial Order ConstraintsExamples

• Let F = {f, g, h}

• The following are partial order constraints on F:

Partial Order ConstraintsExamples (cont.)

• Φ1can be satisfied by the orders:

• g < f < h

• g = h < f

• Φ2 cannot be satisfied.

• Can Φ3 be satisfied?

≤ 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

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

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)

• 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

• 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.

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.

• 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.

Encoding of Axioms:Examples

For a set F

• Reflexivity of = is encoded by:

• Transitivity of < is encoded by:

“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).

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.

• 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.

• 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.

• 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.

• 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

• Fora symbol f in F, the k-bit representation of its value is:

• A constraint of the form (f = g) is encoded by:

• A constraint of the form (f > g) is encoded recursively by: for k > 1

• We encode a constraint Φby replacing each atom (e.g. (f < g)) with its k-bit encoding.

A symbol-based encodingExample

The constraint on F={g,f,h}:

The propositional variables are:

A symbol-based encoding Example, cont.

A symbol-based encoding Example, cont.

Putting it all together:

A satisfying assignment:

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

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

• 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.

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

Alternative encoding encoding).Example

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

Alternative symbol-based encoding encoding).Size

• 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

Motivation encoding).Term 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.

Term Rewrite Systems encoding).

• 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.

Term Rewrite Systems encoding).Example

• Fibonacci:

Term Rewrite Systems encoding).Example, 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))

Term Rewrite Systems encoding).Importance

• 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?

LPO Termination of encoding).Term 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.

LPO Termination of encoding).Term 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.

Partial Order Constraints encoding).and 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):

Experimentation encoding).

• 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.

Experimentation results encoding).

• Running time in seconds for strict-LPO termination

Thank you! encoding).