Approximate Counting CSPs Hunt for Galois Connection

Transcription

1 Approximate Counting CSPs Hunt for Galois Connection Andrei A. Bulatov Simon Fraser University La Trobe University 203

2 2/36 Constraint Satisfaction Problem Decision: Given a conjunctive formula R decide if it is satisfiable ( 2 x y) R ( z, ) K Counting: Given a conjunctive formula R ( 2 x y) R ( z, ) K find the number of satisfying assignments If all relations are from set Γ, the problem is denoted CSP(Γ), #CSP(Γ)

3 3 / 36 Examples: #SAT, Linear Equations #3-SAT: = #CSP( C 3 ) Instance: A propositional formula Φ= C K C n in 3-CNF. Objective: How many satisfying assignments are there? #Linear Equations: Instance: A system of linear equations a a n x x + K + a L + K + a m nm x x m m = = b b n = #CSP( F ) Objective: How many solutions are there?

4 4/36 CSP and Galois Connection For a relation R an n-ary operation f on the same set is a r r r polymorphism of R if for any a a K, a R, 2, r r r f a, a, K, an ) R ( 2 n f is a polymorphism of Γ if it is a polymorphism of every relation from Γ Theorem (Jeavons; 998) Let Γ and be constraint languages on a set D such that every polymorphism of Γ is also polymorphism of. Then CSP( ) is poly time reducible to CSP(Γ).

5 5/36 More CSPs, More Galois Connections CSP usual polymorphisms QCSP surjective operations CSPs with global const. usual polymorphisms (almost) Counting usual polymorphisms Optimization multi-morphisms fractional polymorphisms weighted polymorphisms

6 6/36 More CSPs, More Galois Connections Approximate optimization fractional polymorphisms 2 QCSP + disjunction surjective hyperoperations Approximate counting????

7 7/36 Outline How Galois connection works Galois connection for counting Partition functions and `Galois connection

8 8/36 How It Works Suppose we have a CSP R ( z, x) R2 ( y) K First rewrite it as follows: Is the sentence y, zkr ( z, x) R2( y) K true? Now suppose that has a `definition Then we can substitute and transform R 2 R2( y) = t Q( t) Q( t, y) y, zkr ( z, x) ( t Q( t) Q( t, y)) K y, z, tkr ( z, x) Q( t) Q( t, y) K Therefore CSP({ R, R2, K}) CSP({ R, Q, R3, K })

9 9/36 How It Works 2 Relation R is pp-definable in Γ if R( x) = y Φ( y), where Φ is a conjunctive formula using relations from Γ is pp-definable in Γ if every its relation is. Theorem (Jeavons; 998) Let Γ and be constraint languages on a set D such that is pp-definable in Γ. Then CSP( ) is poly time reducible to CSP(Γ). r r r r

10 0/36 How It Works 3 Galois Correspondence (Geiger, 968; BKKR, 970; ) Let Γ and be constraint languages on a set D. Then is pp-definable in Γ if and only if every polymorphism of Γ is also a polymorphism of.

11 /36 More Galois Connections conjunctive formulas (no quantifiers) -formulas -formulas partial operations works for any kind of CSP surjective operations works for QCSP unary functions More complicated for fractional polymorphisms, multimorphisms, etc.

12 2/36 G.C. for Counting CSPs Suppose we have a #CSP R z, We cannot rewrite as a sentence Suppose that has a `definition R 2 Then we substitute and transform This is not a counting CSP R ( 2 y x) R ( ) K x y, zkr ( z, x) R ( ) K, 2 y R2( y) = t Q( t) Q( t, y) ( z, x) ( t Q( t) Q( t, )) K y tkr ( z, x) Q( t) Q( t, ) K y

13 3/36 G.C. for Counting CSPs 2 Theorem (B.,Dalmau; 2003) Let Γ and be constraint languages on a set D such that is pp-definable in Γ. Then #CSP( ) is poly time reducible to #CSP(Γ). Reduction is through interpolation

14 4/36 Approximate Counting Algorithm Alg is an ε-approximation algorithm for a counting CSP if it is polynomial time and for any instance P of the problem it outputs number Alg(P) such that # P Alg( P) # P < ε Alg is said to be an FPRAS if take P and ε as input and returns number Alg(P) satisfying the condition above in time polynomial in P and /ε AP-reductions are reductions that preserve approximation

15 5/36 AP-Reductions Definitions by conjunctive formulas are AP-reductions If has a definition R( y, t) = Q( t) Q( t, y), then R 2 CSP({ R}) CSP({ Q}) However, interpolation is not an NP-reduction Moreover, the usual Galois connection does not work for APreductions

16 6/36 Example Let R be the following ternary relation R = and Q = 0 0 I claim CSP({ Q}) CSP({ R}) AP k 2 Algorithm: - Insert k R s for each S - Solve the problem - Divide by (2 k ) s and round down

17 7/36 Max-Quantifiers and AP-Reductions Let Φ( x, K, xn, y, K, ym ) be a conjunctive formula in language Γ, and for an assignment ϕ to x, K, x n let #ϕ denote the number of extensions of ϕ that satisfy Φ. Let M be the maximum of these values. We say that Φ is a max-implementation of R if ϕ ( x r ) R # ϕ = M Theorem (B., Hedayaty, 20) If there is a max-implementation of R in Γ, then #CSP(Γ Γ {R}) is AP-reducible to #CSP(Γ)

18 8/36 Max-Co-Clones We write R( x, K, xn ) = max y, K, ymφ( x, K, xn, y, K, y The max-co-clone Γ of a constraint language Γ consists of all relations that can be expressed using: Relations in Γ The equality relation Conjunction Max-quantification m Problem: Find a Galois correspondence for max-co-clones )

19 9/36 Co-Clones all relations self complement upsets affine downsets =, trivial monotone

20 20/36 Boolean Approximate Counting Theorem (Dyer,Goldberg,Jerrum, 2007) Let Γ be a constraint language over {0,}. Then either #CSP(Γ) is solvable in polynomial time; or it is as hard to approximate as #BIS; or it is as hard to approximate as #SAT.

21 2/36 G.C. for Max-Quantification Problem: Find a Galois correspondence for max-co-clones In the Boolean case every max-co-clone is a regular co-clone It is not the case for larger sets Every max-co-clone is a partial co-clone Therefore G.C. should be with some class of partial functions What class??

22 22/36 What We Would Like To Separate with G.C. Bipartite #k-coloring : Input. A bipartite graph Objective. Find the number of its k-colorings It is believed that the hierarchy FPRAS Bip-3-Col Bip-5-Col Bip-7-Col is strict

23 23/36 k-existential Quantifiers If Φ is a formula with free variables x, K, x n and y, a tuple, satisfies a K,a n Ψ( x, K, xn ) = k yφ( x, K, xn, y) iff Φ( a, K, a, b) is true for at least k values b. n = If D = k then k = The k-existential co-clone Γ k, k Nof Γ consists of all relations definable from Γ and equality by conjunction and k-existential quantifier

24 24/36 K-Surjective Operations A (partial) operation f : D n D is said to be k-surjective if for any A, K, An D with A = = A n = k the set f A, contains at least k elements K K, A ) ( n For a constraint language Γ and a set k N, the set of all k-surjective polymorphisms of Γ is denoted by m(k)-pol(γ) Theorem (B.,Hedayaty; 202) For any constraint language Γ, for any k N, Inv( m( k) Pol( Γ)) = Γ k

25 25 / 36 Weighted #CSP Γ is a set of functions f: D R (natural, real, complex) Instance of #CSP(Γ): R ( 2 x y) R ( z, ) K f y) f ( z, ) ( 2 x K Given an instance I the weight of a mapping σ: V D is computed as w σ ) = f ( σ ( x), σ ( y)) f ( σ ( z), σ ( x), σ ( )) K ( 2 x Then Z( I ) = σ: V D w( σ )

26 26 / 36 Counting Problems: Fourier Coefficients Let f :{0, } R be a function and Fourier coefficient n + S = { i, K, ik} {, K, n} fˆ( S) = 2 n fˆ ( S) x, K, x is given by n f ( x n {0,}, K, x n )( ) x i + L+ x i k

27 27 / 36 ω-clones We can define ω-clones similar to co-clones: - becomes product - becomes summation - also add a `trivial operation of multiplying by a constant Γ denotes the set of all functions that can be obtained from Γ applying the operations above Limits: R R for = lim R, R2,K Γ i i

28 28 / 36 Functional Clones 2 A set of functions closed under multiplication by a constant, products, summation, and limits is said to be an ω-clone The ω-clone generated by a set of functions Γ is denoted Γ Theorem (B.,Dyer,Goldberg,Jerrum; 20) If Γ Γ is finite then #CSP(Γ ) #CSP(Γ) AP

29 29 / 36 Log Supermodular Functions Question 2: Any `morphisms for ω-clones? n + A function f :{0, } R is said to be log supermodular if for n any x r, y r {0, } r r v r r r f ( x) f ( y) f ( x y) f ( x y) Lemma LSM is an ω-clone

30 30 / 36 Log Supermodular Functions 2 Theorem Let Γ be a set of functions n + f : {0,} R containing all nonnegative unary functions. Then either Γ is trivial, or it is ω LSM, or it is the set of all functions on {0,} LSM IMP All functions Trivial functions Very trivial functions

31 3 / 36 Multi Morphisms A k-tuple of k-ary operations ( g, K, g k ), g i :{0,} {0,}, is said to be a multimorphism of function f if for any x, K, xn, K, x kn {0,} k f ( x, K, xn ) K f ( xk, K, x kn ) f g ( x, K, x ), K, g ( x, K, x )) K ( k n kn f ( g ( x, K, xk), K, g( x, K, x n kn Log supermodularity is equivalent to (, ) multimorphism Do multimorphisms preserve products, summation, and limit? ))

32 32 / 36 Multi Morphisms 2 No example of a multimorphism that does not Threshold multimorphism g, K, g ) : ( k g ( x, K, ) = i x k iff at least i arguments equal Preserves products, summation, limit, because f has a threshold multimorphism iff it is log supermodular No direct proof

33 33 / 36 Fourier Coefficients Revisited n Let f :{0, } R be a function and S = { i, K, ik} {, K, n} ˆ S Fourier coefficient f ( ) is given by fˆ( S) = 2 n x, K, x n f ( x n {0,}, K, x ) ( ) + L+ x Let y r S denote tuple with yi = if i S and y otherwise Fourier transform of f is function given by fˆ ( y r i = 0 fˆ S ) = fˆ( S) n x i ik

34 Fourier Coefficients vs Products and Summation 34 / 36 Summation: Let f ( x, K, x, y) be a function and Then g( x, K, xn ) = f ( x, K, xn,0) + f ( x, K, x r r gˆ ( x) = fˆ( 0) n n,) Product: Convolution Theorem. = {,} ( +)

35 35 / 36 ω-clones and Fourier Transform r Let f ( x) = f ( x) and Let C be the set of functions f such that is nonnegative r f * r ( x) = r f ( x) r f ( x) Theorem (McQuillan,Dyer; 20) C is a ω-clone C LSM IMP All functions Trivial functions Very trivial functions

36 Thank you!