SAT and the Polya Permanent Problemcsoliver/Artikel/kullmann-sat07-talk.pdfA clause-set is lean iff...

Oliver Kullmann

Main resultsComplement invariance

Lean clause-sets

Minimal unsatisfiability

SAT decision

Hypergraph2-colouring

AutarkiesConstructing autarkysystems

L-matrices andSNS-matrices

Qualitative matrixanalysisOrigins

SNS-matrices

The SAT view

The solution (of manyproblems)

Conclusions andOutlook

SAT and the Polya Permanent Problem

Oliver KullmannComputer Science Department

Swansea University

SAT 2007, Lisbon, May 30, 2007

SAT: Connecting combinatorics and linear algebra

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Overview

1 Main results

2 Hypergraph 2-colouring

3 Autarkies

4 Qualitative matrix analysis

5 Conclusions and Outlook

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



A few general notations

c(F ) for the number of clauses

n(F ) for the number of variables

> is the empty clause-set

⊥ is the empty clause

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Complement-invariant clause-setsFor clauses C and clause-sets F :

C := {x : x ∈ C}F := {C : C ∈ F}.

F is called complement-invariant if F = F .

F is complement-invariant iffthere is a clause-set F0 with F = F0 ∪ F0.

Such a F0 with c(F0) = c(F )2 is called a core half.

For exampleF =

{{a, b}, {b, c},{a, b}, {b, c}

}is complement-invariant, with core halfs

{{a, b}, {b, c}

}or

{{a, b}, {b, c}

}for example.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



The reduced deficiencyFor a clause-set F the deficiency is

δ(F ) := c(F )− n(F ).

The reduced deficiency is

δr(F ) :=12(δ(F )− n(F )).

For a core half F0 of F we have

δr(F ) = δ(F0).

For the example F ={{a, b}, {b, c}, {a, b}, {b, c}

}we

haveδ(F ) = 4− 3 = 1

δr(F ) =12(1− 3) = −1 = (2− 3).

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Autarkies and lean clause-setsRecall:

An autarky for a clause-set F is a partial assignmentwhich satisfies every clause it touches.

A clause-set is lean iff it has no non-trivial autarkies.

The only satisfiable lean clause-set is >.

Every minimally unsatisfiable clause-set is lean.

Other examples of lean clause-sets are obtain byTseitin-extensions of lean clause-sets.

The lean kernel of a clause-set F is the largest(subsumption-wise) lean sub-clause-set of F . It canalso be obtained by repeated autarky reduction(actually also in one step).

The lean kernel of a clause-set consists of exactly all theclauses which can be used in some resolution refutation.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Computing the lean kernel in generalIn general:

1 Deciding whether a clause-set is lean iscoNP-complete.

2 The best algorithm for computing the lean kernel(thus also deciding leanness) seems to be given by

using a SAT solver which returns the set of variablesused in a resolution refutation it foundthis can be computed efficiently without muchoverhead; for example the OKsolver always does it(for intelligent backtracking); see [Kullmann, Silva,Lynce; SAT06].

Applications of the computation of the lean kernel:1 for computing a MU sub-clause-set (which must be

contained in the lean kernel);2 for computing a MAXSAT sub-clause-set (which must

be contained in the complement of the lean kernel).

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Lean complement-invariant clause-sets

Deciding leanness is coNP-complete also forcomplement-invariant clause-sets. However we have:

If the complement-invariant F is lean, then δr(F ) ≥ 0.

This is similar to the general situation, where wehave δ(F ) ≥ 1 for lean clause-sets.

Actually, in the general situation, matchingleanness suffices to establish δ(F ) ≥ 1, while forcomplement-invariant linear leanness suffices.

So we can ask for deciding leanness ofcomplement-invariant clause-sets F with δr(F ) = 0. Now,transferring [Robertson, Seymour, Thomas 1999;McCuaig 2004], we get

Theorem Deciding whether a complement-invariant Fwith δr(F ) = 0 is lean is decidable in polynomial time.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Generalising minimal unsatisfiabilityConsidering an arbitrary autarky systems (which allowfor specialisations of autarkies) we have the followingnotions for clause-sets F :

1 F is A-satisfiable iff the A-lean kernel of F is >.2 F is minimally A-unsatisfiable iff F isA-unsatisfiable, while every strict sub-clause-set of Fis A-satisfiable.

3 F is barely A-lean iff A is A-lean, but after removalof any clause this is no longer the case.

4 F is A-autarky indecomposable iff F cannot beobtained by a variable-disjoint union of two A-leanclause-sets F1, F2, where to the clauses of F2 literalsx with var(x) ∈ var(F1) can be added (arbitrarily).

For the full autarky system, A-satisfiability is satisfiability,and minimal A-unsatisfiability is minimal unsatisfiability.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Characterising minimal A-unsatisfiability

Trivially, every minimally unsatisfiable clause-set is lean.We can get an equivalence, and this for arbitrary autarkysystems A, as follows:

Lemma Consider a normal autarky system A. Then aclause-set F is minimally A-unsatisfiable if and only if

1 F is barely A-lean, and2 F is A-indecomposable.

This lemma might be useful even for the full autarkysystem. We will apply it to the autarky system given bylinear autarkies.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Deciding minimal unsatisfiability

Deciding whether a general F or a complement-invariantF is minimally unsatisfiable is DP-complete.

Theorem Deciding whether a complement-invariantclause-set F with δr(F ) = 0 is minimally unsatisfiable canbe done in polynomial time.

Remark: While in general performing reduction by linearautarkies needs the power of linear programming, hereactually we only need to consider balanced linearautarkies, which can be handled more efficiently by justconsidering systems of linear equations.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



SAT for complement-invariant clause-setsFinally, what about SAT decision for (arbitrary)complement-invariant clause-sets?

Obviously, still SAT decision is NP-complete.

Specialisation to δr(F ) = 0 doesn’t help (padding).

Recall:1 The same problem arises for arbitrary clause-sets F

and the (normal) deficiency.2 Cure: the maximal deficiency δ∗(F ).

Analogously, we introduce the maximal reduceddeficiency

δ*r (F ) := max

F ′⊆Fδr(F ).

For a core half F0 of F we have

δ*r (F ) = δ∗(F0).

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Polynomial time SAT decision

Constructivising [Robertson, Seymour, Thomas;McCuaig] we should be able to prove:

Conjecture The following functional computation problemcan be solved in polynomial time: Given a square matrixA over {−1, 0,+1}, if A is not an SNS-matrix, then asingular matrix A′ ∈ Q(A) over Z can be computed.

Theorem Assume that the conjecture holds true. Thenfor complement-invariant clause-sets F with δ*

r (F ) = 0the satisfiability problem is decidable in polynomial time(providing also a satisfying assignment).

It follows that if F is unsatisfiable, then a minimallyunsatisfiable sub-clause-set can be computed inpolynomial time. (Open: Can we also compute aminimum minimally unsatisfiable sub-clause-set?!?)

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



The main conjectureConjecture For fixed k ∈ N0, the satisfiability problem forcomplementation-invariant clause-sets F with δ*

r (F ) ≤ kis decidable in polynomial time.

The underlying combinatorial structures is of goodinterest to the graph theory community and thecombinatorics community; results in this direction couldhave two bearings:

embedding the (narrow) graph theoretical /combinatorial problems into the (richer) satisfiabilityproblem makes certain operations much moretransparent — hopefully a better understanding alsofor the original problems is finally reached;

embedding the (rich) graph theoretical /combinatorial problem into the (slim) satisfiabilityproblem adds combinatorial structure to SAT.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Hypergraph 2-colouring

A (finite) hypergraph is a pair G = (V , E), where V isthe (finite) vertex set, and E is a set of subsets of V .

A 2-colouring of G is a map f : V → {0, 1} such thatno hyperedge is monochromatic, i.e., for every H ∈ Ethere are v , w ∈ H with f (v) = 0 and f (w) = 1.

Translating the hypergraph 2-colouring problem into aSAT problem is done trivially by creating a “positive copy”and a “negative copy” of the hyperedges, for example thehypergraph (given by its hyperedges){

{a, b}, {b, c}}

is translated into the clause-set{{va, vb}, {vb, vc}, {va, vb}, {vb, vc}

}.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Hypergraphs vs. complement-invariance

So the hypergraph 2-colouring problem is exactly thesame as the SAT-problem for complement-invariantclause-sets which are also PN clause-sets, that is, everyclause is either positive or negative.

How much more general is SAT for (general)complement-invariant clause-sets compared to PNcomplement-invariant clause-sets?

1 The standard translation of arbitrary clause-sets intoPN clause-sets (by introducing new variables) can besymmetrised, and then yields a translation ofcomplement-invariant clause-sets into PNcomplement-invariant clause-sets.

2 This translation is not only SAT-preserving, but alsoMU-, autarky- and #SAT-preserving.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Why not just hypergraphs?

So, modulo a very simple (AC0) and very well behavedreduction, complement-invariant clause-sets and PNcomplement-invariant clause-sets can be identified w.r.t.SAT and variations.However, there is one central asset of SAT, namelyresolution:

Resolution, symmetrised, does not leave the classof complement-invariant clause-sets.

However, it leaves the class of PNcomplement-invariant clause-sets!

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Using special satisfying assignmentsConsider some restricted form of satisfying assignments,via a predicate S(ϕ, F ), which is true if the partialassignment ϕ satisfies F and fulfils the specialrequirement, e.g.,

ϕ is matching satisfying (recall Stefan Szeider’s talk)

ϕ is NAESAT-satisfying, i.e., in every clause there isnot only a satisfied literal, but also a falsified one.

You obtain an autarky system: A “S-autarky” for F is apartial assignment ϕ such that ϕ satisfies F [var(ϕ)],where for a set V of variables F [V ] denotes the restrictionof F to V .

From matching satisfying assignments we getmatching autarkies.

From NAESAT-satisfying assignments we getbalanced autarkies.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Balanced autarkies

Consider a clause-set F and its “symmetrised” versionF := F0 ∪ F0:

1 F0 is satisfiable w.r.t. balanced autarkies, i.e.NAESAT-satisfiable, iff F is satisfiable.

2 F0 is lean w.r.t. balanced autarkies iff F is lean.3 F0 is minimally unsatisfiable w.r.t. balanced autarkies

iff F is minimally unsatisfiable.

Thus the symmetrisation operation translates between

the general property for complement-invariant F0∪F0

the balanced version of the property for general F0.

In this way we can make use of “both worlds”: The“balanced SAT-properties” of arbitrary clause-sets vs. the“general SAT-properties” of complement-invariantclause-sets.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Using the clause-variable matrix

For a clause-set F let M(F ) denote the clause-variablematrix:

1 As was first noticed by [Davydov/Dovydova], that F issatisfiable can be expressed in a natural way byM(F ) using “matrix speech”.

2 Various versions can be given, exploiting duality fromthe realm of linear programming.

All these properties are best expressed using the notionsof qualitative matrix analysis:

That a variable-clause matrix is balanced lean isexactly the same as being an L-matrix,

where the simplest case, where the matrix is square(deficiency = 0) is called SNS-matrix.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



The origins of qualitative matrix analysis

“Qualitative matrix analysis” (QMA) is matrix analysismodulo the equivalence relation between matrices given

by having the same sign pattern.

[Brualdi, Shader: Matrices of sign-solvable linear system(1995)] describes the foundations of QMA in qualitativeeconomics:

Qualitative economics is usually considered tohave originated with the work of Samuelson whodiscussed the possibility of determiningunambiguously the qualitative behavior ofsolution values of a system of equations. In hispioneering paper Lancaster put it this way:

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Origins continued

Economists believed for a very long time,and most economists would still hope it to be so,that a considerable body of sensible economicpropositions could be expressed in a qualitativeway, that is, in a form in which the algebraic signof some effect is predicted from a knowledge ofthe signs, only, of the relevant structuralparameters of the system.

We consider here the fundamental notion, that a squarereal matrix A has a sign-invariant determinant, that is,for all matrices A′ with the same sign pattern as A wehave sgn(det(A′)) = sgn(det(A)).

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Sign patterns and zero patternsThe sign pattern sgn(M) of matrix M is the{−1, 0,+1}-matrix sgn(M) of the same dimensiongiven by entrywise sgn-formation.The qualitative class of M,Q(M) := {M ′ : sgn(M ′) = sgn(M)}, is the set of allmatrices with the same sign pattern.The null pattern of M is sgn(|M|) (a {0, 1}-matrix),where |M| denotes entrywise absolute-valueformation.

For example: M =

(4 −50 2

)sgn(M) =

(1 −10 1

)sgn(|M|) =

(1 10 1

).

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



SNS-matrices

By definition, A is has sign-invariant determinant if andonly if

(i) either ∀A′ ∈ Q(A) : det(A′) = 0

(ii) or ∀A′ ∈ Q(A) : det(A′) > 0

(iii) or ∀A′ ∈ Q(A) : det(A′) < 0.

For every matrix Q(A) is connected, while det : Q(A) → Ris continuous, and so det(Q(A)) is connected.

Thus cases (ii) or (iii) hold iff ∀A′ ∈ Q(A) : det(A′) 6= 0, inwhich case A is called an SNS-matrix(“sign-non-singular”).

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Diagonals

For a square matrix A of order n:

The diagonal corresponding to permutation π ∈ Sn

is the vector (Ai,π(i))i∈{1,...,n}.

The main diagonal is the diagonal corresponding tothe identity map.

A diagonal is non-null if all entries are non-zero.

For example, the permutation(

1 2 32 3 1

)∈ S3

corresponds to the diagonal ••

•

.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



The Leibniz determinant expansion

For a square matrix A of order n we have

det(A) =∑π∈Sn

sgn(π) ·n∏

i=1

Ai,π(i).

In other words, we run through all diagonals of A andsum up their products, weighted by the sign of thecorresponding permutation.

Clearly we need to consider only non-null diagonals(they correspond to the non-null terms in thedeterminant expansion).

Remark: The determinant expansion does not yield apoly-time computation of the determinant, this howevercan be achieved by Gaussian elimination.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



The determinant expansion for order 3

The three even diagonals of a square matrix of order 3are •

••

,

••

•

,

••

•

,

and the three odd diagonals are ••

•

,

••

•

,

••

•

.

(Only the non-null diagonals need to be considered for agiven matrix, that is, where all entries are not zero.)

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



First characterisation

Lemma For a square matrix A we have

(i) A has sign-invariant determinant with sign ε ∈ {±1}if and only if every non-null term in the determinantexpansion has sign ε, and there is one such term.

(ii) A has sign-invariant determinant with sign 0 iff thereis no non-null term in the determinant expansion.

Proof: For Part (i) use Laplace determinant expansionafter a row, where two terms with positive resp. negativesign differ, to see that in Q(A) a positive and a negativesign of the determinant can be realised.

Do the same also for Part (ii), using that the existence ofa non-null term implies the existence of a positive as wellas a negative term (if the determinant is zero).

√

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



An exampleLet’s consider n = 3. The even resp. odd permutations inS3 are

123, 231, 312

321, 132, 213.

Now −1 −1 11 −1 0−1 −1 −1

is an SNS-matrices (all four non-null term are −1), while1 −1 0

0 1 11 0 1

is not.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Associated bipartite graphsFor a {0, 1}-matrix A of dimension n ×m let bip(A) bethe bipartite graph with bipartition ({1, . . . , n}, {1, . . . , m})and edges {i , j} for Ai,j = 1.

So the non-null diagonals of a square matrix Acorrespond 1-1 to the perfect matchings of bip(sgn(|A|)),

for example

1 1 00 0 11 1 0

yields 1

>>>>

>>>>

1

2

====

=== 2

3

��

��3

with two perfect

matchings {{1, 1}, {2, 3}, {3, 2}

},{

{1, 2}, {2, 3}, {3, 1}}.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Sign-singular matrices

By the Lemma we get that a matrix A is sign-singularif and only if bip(A) does not have a perfect matching.

This can be checked in polynomial time. It remainsthe (much) harder task of deciding whether a matrixA is SNS.

BTW, the generalisation of property SNS to matrices ofarbitrary dimension (“L-matrices”) is coNP-complete.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



The Permanent

For a square matrix A of order n we define

per(A) =∑π∈Sn

n∏i=1

Ai,π(i).

So the permanent is defined by an expansion like thedeterminant, only that we do not take the sign of thepermutations into account.

The permanent of a square {0, 1}-matrix A is thenumber of perfect matchings of bip(A) (a#P-complete problem).

By the previous lemma we get:

A square matrix A is SNS if and only ifdet(A) 6= 0 and per(|A|) = |det(A)|.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



{0, 1}-matrices as adjacency-matrices

For a square {0, 1}-matrix A of order n let dgl(A) denotethe directed graph with vertex set {1, . . . , n} and an edge(i , j) if Ai,j = 1, while by dg(A) we denote the irreflexiveversion of dgl(A) (ignoring the entries of A on the maindiagonal). For example

A =

1 1 0 1 00 1 0 0 11 1 1 0 00 0 1 1 01 0 0 0 1

7→ dg(A) = 1

��

��

��...

....

....

...

4

��

5

^^>>>>>>>>

3

GG��// 2

OO

.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Even cycles in digraphs

Every square matrix A with det(A) 6= 0 must have anon-zero diagonal. So by column permutation, w.l.o.g. forSNS-decision we can assume a non-zero main diagonal.Now for a square {0, 1}-matrix A with non-zero diagonalthe following assertions are equivalent:

1 A is SNS.2 per(A) = det(A).3 Each term in the determinant expansion is ≥ 0.4 There is no odd non-zero diagonal.5 dgl(A) (dg(A)) has no even cycle.

Proof: A cyclic permutation (a1, . . . , an) is odd iff n iseven; n is the length of the corresponding cycle in dgl(A).A general permutation is odd iff the cycle decompositionhas an odd number of odd cyclic permutations.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Counter-examples as autarkiesFor a square matrix A:

A is SNS iff∀A′ ∈ Q(A) ¬∃ x 6= 0 : A′x = 0. Thus

A is not SNS iff∃A′ ∈ Q(A) ∃ x 6= 0 : A′x = 0.

W.l.o.g.: A = sgn(A), i.e., A is {−1, 0,+1}-matrix. What isa {−1, 0,+1}-matrix ?!

A (multi-)clause-set !

A is the clause-variable matrix of a clause-set F .x is interpreted as a partial assignment (zero entriesmean unassigned), and yields an autarky — every

clause which is touched is satisfied.

Moreover, in every touched clause there must be also afalsified literal — so its a balanced autarky.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



Example

Consider the non-SNS matrix

1 −1 00 1 11 0 1

. The

clause-set F0 with this clause-variable matrix is

F0 :={{a, b}, {b, c}, {a, c}

},

which has exactly two balanced autarkies, namelyϕ1 := 〈c → 1, a, b → 0〉 and ϕ2 := 〈c → 0, a, b → 1〉. Thecorresponding solution to M(F ) · x = 0 is1 −1 0

0 1 11 0 1

·

−1−11

= 0.

(Remark: Here we can use M(F0) itself, and thus ϕ1, ϕ2are balanced linear autarkies.)

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



On the importance of SNS-matrix decision

As we have seen now, the following decision problemsare all equivalent to the SNS-matrix decision problem bysimple reductions:

1 Is a complement-invariant clause-set F withδr(F ) = 0 lean?

2 Is a clause-set F with δ(F ) = 0 balanced lean?3 Is a PN clause-set F with δ(F ) = 0 balanced lean?4 Given a directed graph, does it has no directed

circuits of even length?

We have also seen that leanness is closely related tobeing minimally unsatisfiable (here), which forhypergraphs translates into the property of beingminimally non-2-colourable.

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



The original Pólya problem

Expanding the arguments we have given, one can alsoshow the close relation to the following problem posed byPólya (1913):

Given a {0, 1}-square matrix A, when can some of the 1’sbe changed to −1’s in such a way that the permanent of

A equals the determinant of the modified matrix?

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



The solution

In 1996 independently by Robertson, Seymour,Thomas and McCuaig a polynomial time algorithmfor the Pólya problem was demonstrated.

Thus also SNS-matrix decision can be done inpolynomial time.

(Since the argumentation is quite complicated (40pages of dense argumentation), it is not really clearwhether, in case the matrix was found not to be anSNS-matrix, also a witness for this can be computedefficiently.)

Oliver Kullmann


Lean clause-sets


SAT decision





SNS-matrices

The SAT view



SummaryThe class of complementation-invariantclause-sets has been introduced, together with thenotion of reduced deficiency.Poly-time time SAT decision was shown (relative to abelievable conjecture) for complementation-invariantclause-sets of maximal reduced deficiency zero.Many relations to graph theory, hypergraph theory,combinatorics, matrix analysis — the link provided byautarky theory.The main conjecture should provide a fertile groundfor interesting investigations into SAT algorithms,which should find the attention of the widercombinatorics community.

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