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.

• The partial order h<f=g makes the above formula true, hence it is satisfiable.

• The problem is NP-complete.

))()(()( fhhggf

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.

)()( 241321 xxxxxx

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.

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).

Partial Order Constraints Examples

• Let F = {f, g, h}

• The following are partial order constraints on F:

))()(()(

)()()(

))()(()(

3

2

1

hfghgf

fhhggf

hghfgf

Partial Order Constraints Examples (cont.)

• Φ1 can be satisfied by the orders:

– g < f < h

– g = h < f

• Φ2 cannot be satisfied.

• Can Φ3 be satisfied?

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

Partial Order Constraints Satisfiabilty

• 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 Φ is true when atoms in Φ are 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

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)

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

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 Φ is 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 Φ on a set of symbols F is encoded by a propositional formula Φ’ such that Φ is satisfiable if and only if Φ’ is.

• A good encoding yields short formulae with a small number of variables.

“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.

a

Encoding of Axioms:Examples

For a set F• Reflexivity of = is encoded by:

• Transitivity of < is encoded by:

ffFf

hfhggfFhgf

)(,,

“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. (just to encode transitivity, for example).

Integer assignment and solution

Let Φ be a partial order constraint on F, and let |F |=n.

• An integer assignment for Φ is a mapping μ : F→{1,…,n}

• An integer solution for Φ is an assignment θ which makes Φ true under the natural interpretations of < and = on the natural numbers.

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.

))()(()(1 hghfgf

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.

A symbol-based propositional encoding

• Encoding: A partial order constraint Φ on a set of symbols F is encoded by a propositional formula Φ’ such that Φ is satisfiable if and only if Φ’ is.

• Our approach: Symbol-based. we encode the values of the symbols in F in an integer solution.

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

nk 2log

A symbol-based propositional encoding, cont.

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

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

1,..., fff k

)(||)(||1

ii

k

ik gfgf

A symbol-based propositional encoding, cont.

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

for k > 1

111||)(|| gfgf

1

|| ( ) ||

( ) (( ) || ( ) || )

k

k k k k k

f g

f g f g f g

A symbol-based propositional encoding, cont.

• We encode a constraint Φ by 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:

)()( hggf

121212 ,,,,, hhggff

23log

3

2 k

n

222)()( hggf

A symbol-based encoding Example, cont.

2

2 2 2 2 1

|| ( ) ||

( ) (( ) || ( ) || )

f g

g f f g f g

111||)(|| gfgf

)()(||)(|| 22112 hghghg

A symbol-based encoding Example, cont.

Putting it all together:

A satisfying assignment:

))()((

)))()(()((

1122

112222

hghg

gfgfgf

1

0

1222

11

ffhg

hg

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

1 1

2 2 2 1

0

1

g h

g h f f

Theorem 2

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

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.

What if Φ is long? Alternative encoding

• 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.

Alternative encodingExample

• The constraint:

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

))()(()( gfhggf

21

))(())((

)(

22212

1211

hgagfa

aaa

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

Motivation Term Rewrite Systems

• A term rewrite system is a set of rules of the form

where l and r are terms constructed from function symbols and variables.

rl

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.

Term Rewrite SystemsExample

• Fibonacci:

1

2

3

4

5

: 0

: ( ) ( )

: (0) 0

: ( (0)) (0)

: ( ( ( ))) ( ( )) ( )

r x x

r s x y s x y

r fib

r fib s s

r fib s s x fib s x fib x

Term Rewrite SystemsExample, cont.

Computing fib(3):• fib(s(s(s(0))) →5 fib(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 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?

LPO Termination of 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 Term Rewrite Systems, cont.

• LPO termination problem:

is there a partial order >F over function symbols, such that for every rule it hold that Where is the order induced by >F

• LPO termination implies termination of the rewrite system.

rl lpo

lpo

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):

)()()( syxsyxs Flpo

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.

Experimentation results

poSATTTT

Total8.983647.48

Average0.0120.86

Max0.477317.63

• Running time in seconds for strict-LPO termination

Thank you!