ICALP g Mingji Xia

Transcription

1 : : ICALP 2010 g Institute of Software Chinese Academy of Sciences

2 : 1 2 3

3 Matrix multiplication : Product of vectors and matrices. AC = e [n] a ec e A = (a e), C = (c e). ABC = e,e [n] a eb ee c e A = (a e), B = (b ee ), C = (c e ). Vector A (resp. C) is also a unary function e [n] a e Matrix B is also binary function (e, e ) [n] 2 b ee

4 #F H : The input of #F H is a bipartite graph G(U, V, E). Each vertex v U V has a function F v. The value of this problem is F v (e v,1,..., e v,dv ), e 1,...,e t [n] v U V where e v,i is the ith edge of v. If v U, F v F. If v V, F v H.

5 Gadgets in #F H : Except defining the value (a function of arity 0) of an instance, we can also define the function of a gadget. Example (AB)(x 1, x 2 ) = e a x 1,eb e,x2 A = (a x1,e), B = (b e,x2 ). Given a gadget Γ, we can define its function F Γ in its external edges {x i }, by a summation over its internal edges {e i }, where e v,i is the ith edge of v. F Γ (x 1,..., x s ) = e 1,...,e t [n] v U V F v (e v,1,..., e v,dv ).

6 More examples : M N (M N) x1 x 2,x 3 x 4 = M x1,x 3 N x2,x 4 We also use a vector of length [n] k to denote a function of arity k. F (M M M) Also denoted as F (M 3 ).

7 : AB = AEB = AMM 1 B = (AM)(M 1 B). (E is identity matrix.) Theorem (Valiant 2004) #{f} {h} and #{F } {H} have the same value. F = fm 3, H = (M 1 ) 2 h.

8 An example: Fibonacci function : A symmetric function F of arity k in Boolean variables is denoted as [f 0, f 1,..., f k ], where f i is the value of F on inputs of weights i. F 1 = [a 0, a 1 ] = [0, 1], F 2 = [a 0, a 1, a 2 ] = [0, 1, 1], F 3 = [a 0, a 1, a 2, a 3 ] = [0, 1, 1, 2],... F k = [a 0, a 1,..., a k ]. ({a i } is Fibonacci sequence.) = k denotes the equivalent relation of arity k. (= k ) = [1, 0,..., 0, 1]. (= a,b ) = [a, 0,..., 0, b]. k For any d, there is between #{= 1, 1 1,..., = ( 1, 1 d } {= ϕ,ϕ 1 2 ) } and #{F 1,..., F d } {= 2 }. 1 ϕ The base M is, where ϕ = ϕ 2.

9 An example: Fibonacci function : For any d, there is between #{= 1, 1 1,..., = ( 1, 1 d } {= ϕ,ϕ 1 2 ) } and #{F 1,..., F d } {= 2 }. 1 ϕ The base M is, where ϕ = ϕ 2. Because 5 a k = ϕ k (1 ϕ) k, 5 Fk = (1, ϕ) k (1, 1 ϕ) k. Because (= 1, 1 k ) = (1, 0) k (0, 1) k, (= 1, 1 k )M k = 5 F k. (M 1 ) 2 (= ϕ,ϕ 1 2 ) = (= 2 )

10 size : F k = [a 0, a 1,..., a k ], for any d, #{F 1, F 2,..., F d } {= 2 } is polynomial time computable. A problem is #P-hard, if all problems in #P can be reduced to it. Lemma P k+1 = [a 0, a 1,..., a k, 2a k ], for any d 2, #{F 1, F 2,..., F d, P d+1 } {= 2 } is #P-hard. Theorem For any d 2, there exists a complex binary function H d such that #{= 1, = 2,..., = d+1 } {H d } is #P-hard, but #{= 1, = 2,..., = d } {H d } is polynomial time computable. Only need to build a from #{F 1, F 2,..., F d, P d+1 } {= 2 } (domain is [2]) to #{= 1, = 2,..., = d+1 } {H d } (domain is [m]). (This is also a from #{F 1, F 2,..., F d } {= 2 } to #{= 1, = 2,..., = d } {H d }.)

11 Maximum degree and complexity : Many hardness results are strengthened to maximum degree bounded version. SAT 3SAT (NP-hard) #SAT #2SAT (#P-hard) Lots in [Vadhan 2001]... #CSP(F)(#{= 1, = 2,...} F) #{= 1, = 2, = 3 } F (each variables occurs at most 3 times), where F is composed of complex functions in Boolean variables. [Cai,Lu,Xia 2009] By this result, our theorem does not hold for Boolean domain. By our theorem, we can not get this kind of strong degree bounded #P hardness results for general counting problem classes. If #CSP({G}) is hard, there exists some d (depends on G) such that #{= 1,..., = d } {G} is hard, where G is a 0-1 weighted undirected graph. [Dyer,Greenhill 2000]

12 : (another version) Matrix multiplication satisfies associative law. (AB)C = A(BC). Theorem (Valiant 2004) #{F M 3 } {H} and #{F } {M 2 H} have the same value.

13 : From #{F 1, F 2,..., F d } {= 2 } to #{= 1, = 2,..., = d } {H d } Find M such that F k = (= k )M k. Recall, Theorem: there exists H d (= M M),... Two small tricks for constructing such a M. 1 [a 0, a 1, a 2..., a k ] = k c i i=0 c [1, r i, ri 2,..., rk i ]. If r i are different integers, there is an integer solution for c i and c. c i [1, r i, ri 2,..., rk i ] can be realized in (= k)m k by setting c i rows of (1, r i ) in M. If c i is negative, we utilize k s=1 ri s = 1 for i = 1, 2,..., k, where r s = e 2πs k+1.

14 Remark : In this, just some problems examples are constructed. There is no requirement on domain size m and function H d. Under these weak requirements, we realize a strong transformation on one side by. Open problem: Assume FP #P. Is there some class of problems, such that predicating the complexity of these problems is uncomputable? In some papers about dichotomy, it is considered whether their dichotomy are computable, and some of them are open.

15 : Conjecture Under some conditions, the of holds. For example, #{f} {h} and #{F } {H} have the same value, where f and F are ternary functions, and h and H are binary functions. Then, there exists nonsingular matrix M such that, F = fm 3, H = (M 1 ) 2 h. It does not holds for some special cases. For example, #{(1, 0), (1, 0)} {(4, 1)} and #{(1, 1), (1, 2)} {(4, 0)}.

16 some cases : Theorem (Lovász 1967, Dyer,Goldberg,Paterson 2006) Suppose H 1 and H 2 are directed acyclic graphs. If for all directed acyclic graphs G, the number of homomorphisms from G to H 1 is equal to the number of homomorphisms from G to H 2, then H 1 and H 2 are isomorphic. If there is a between #R = {H 1 } and #R = {H 2 }, it can be proved that the base M is a permutation matrix. R = denotes {= 1, = 2,... = k,...}.

17 some cases : In the following, we only consider #F {= 2 }. Range is real number. Domain is arbitrary [d]. F is composed of unary functions. F is composed of one symmetric binary function. F is composed of two symmetric binary functions F 1 and F 2. All eigenvalues of F 1 are equal. F is composed of two symmetric binary functions F 1 and F 2. All eigenvalues of F 1 are different, so is F 2. Domain is [2]. F is composed of two symmetric binary functions. F is composed of one unary function and one symmetric binary function. F is composed of one symmetric ternary function.

18 Proof sketch Unary functions : Domain is arbitrary [d]. F is composed of unary functions. #{F 1,..., F k } {= 2 } and #{H 1,..., H k } {= 2 } have the same value. That is, F F = H H, where matrices F = (F 1, F 2,..., F k ) and H = (H 1, H 2,..., H k ). The base of is simply HF 1. Because F F = H H, M M = E. M change = 2 into = 2.

19 Proof sketch Two symmetric binary functions : Domain is arbitrary [d]. F is composed of two symmetric binary functions F 1 and F 2. All eigenvalues of F 1 are different, so is F 2. #{F 1, F 2 } {= 2 } and #{H 1, H 2 } {= 2 }. By the result for one symmetric function, we only need to prove for #{F 1, MF 2 M } {= 2 } and #{F 1, T F 2 T } {= 2 }, where F 1 and F 2 are diagonal matrix with different diagonal entries, and M, T are orthogonal matrices. Using polynomial interpolation method, we can realize E i by F 1 and F 2. E i (x, y) is always zero, except E i (i, i) = 1. Because tr(e i ME j M ) = tr(e i T E j T ), M 2 ij = T 2 ij.

20 Proof sketch Two symmetric binary functions : ( M ij T ij ) 2 = ±1, assume T ij is not zero. Because tr(e i ME j M E k ME l M ) = tr(e i T E j T E k T E l T ), M kl T kl = 1. M ij T ij M il T il M kj T kj Consider matrix R = ( M ij T ij ). This means, the 2 2 submatrix R ik,jl has rank 1. We can prove R has rank 1. (If there are zero T ij, the proof is quite a few more complicated.)

21 Proof sketch Two symmetric binary functions : R = uv, then UT V = M. u, v are column vector composed of 1 and 1. U (resp. u) is the diagonal matrix whose diagonal is u (resp. v). Compose U with F 1. UF 1 U = F 1. V F 2 V = F 2.

22 Proof sketch One symmetric ternary functions : Domain is [2]. #{F } {= 2 } and #{H} {= 2 }. We construct some gadgets to prove this case. This does not work for F = [x, y, x, y]. The value of #{F } {= 2 } is always zero on non-bipartite graphs. There is a direct proof for [x, y, x, y] case.

23 Open problems : Whether the of holds? If it does not hold, the condition is not a necessary condition. Can we find some other sufficient conditions and s? If it does hold, can we utilize it to prove complexity results?

24 : Thank you œ