structured/0711.1723.structured.json

{"paper_meta":{"paper_id":"arxiv:0711.1723","title":"0711.1723","authors":[],"year":null,"published":null,"updated":null,"venue_or_source":"arxiv-papers-shard","primary_category":"cs","secondary_categories":[],"doi":null,"license":"unknown","source_type":"pdf+shard","source_url":null,"pdf_url":null},"abstract":{"raw":"","cleaned":"","offsets":[]},"full_paper_text":{"raw_ordered_text":"arXiv:0711.1723v2 [cs.CC] 15 Nov 2007\nAn analysis of a random algorithm for estimating all the\nmatchings\nJinshan Zhang,∗Yan Huo,† and Fengshan Bai‡\nDepartment of Mathematical Sciences,\nTsinghua University, 100084,\nBeijing, PRC.\nCounting the number of all the matchings on a bipartite graph has been trans-\nformed into calculating the permanent of a matrix obtained from the extended bi-\npartite graph by Yan Huo, and Rasmussen presents a simple approach (RM) to\napproximate the permanent, which just yields a critical ratio O(nω(n)) for almost\nall the 0-1 matrices, provided it’s a simple promising practical way to compute this\n#P-complete problem. In this paper, the performance of this method will be shown\nwhen it’s applied to compute all the matchings based on that transformation. The\ncritical ratio will be proved to be very large with a certain probability, owning an\nincreasing factor larger than any polynomial of n even in the sense for almost all\nthe 0-1 matrices. Hence, RM fails to work well when counting all the matchings via\ncomputing the permanent of the matrix. In other words, we must carefully utilize\nthe known methods of estimating the permanent to count all the matchings through\nthat transformation.\nKeywords: matching; permanent; critical ratio; bipartite graph; determinant; Monte-Carlo\nalgorithm;random algorithm; RM;fpras\n∗Electronic address: zjs02@mails.tsinghua.edu.cn\n†Electronic address: huoy03@mails.tsinghua.edu.cn\n‡Electronic address: fbai@math.tsinghua.edu.cn\n\n2\nI.\nINTRODUCTION\nLet G = (V, E) be a bipartite graph, where V = V1 ∪V2 is the set of vertices and\nE ⊂V1 × V2 is the set of edges. In the following sections we suppose #V1 = #V2 = n if\nthere’s no special illustration. A set of edges S ⊂E is called a matching if no two distinct\nedges e1, e2 ∈S contain a common vertex. S is called a k-matching if #S = k. In special\ncase, S is called a perfect matching if k = n. Let Sk be the set of k-matching in G and\nA(G) be the set of all the k-matching, k = 0, 1, . . . , n. For the convenience of discussion, let\n#S0 = 1, then the number of all the matchings in G is #A(G) =\nnP\ni=0\n#Sk.\nThe permanent of a 0-1 A = aij, 1 ≤i, j ≤n is defined as\nPer(A) =\nX\nπ\nn\nY\ni=1\nai,π(i)\n(1)\nwhere the sum is over all the permutations π of [n] = {1, . . . , n}. It’s well known that\nthe permanent of an adjacent matrix of bipartite graph equals the number of its perfect\nmatching. Let AM(G) denote the number of all the matchings in G, and A be adjacent\nmatrix of G. [8] has proved that\nAM(G) = 1\nn!per\n \n \nA\nIn×n\n1n×n 1n×n\n \n \n(2)\nwhere In×n is n × n unit matrix, 1n×n denotes n × n matrix with all the elements 1. This\nmeans in order to count the number of all the matchings of a bipartite graph with 2n vertices\nwe only need to compute the permanent of a 2n × 2n corresponding matrix transformed\nfrom adjacent matrix. The computation of permanent has a long history and was shown\nto be #P-complete in [2].\nThus, in the past 20 years or so, many random algorithms\nhave been developed to approximate the permanent, which can been divided at least four\ncategories[3]: elementary recursive algorithms(the original one is Rasmussen method(RM))\n[4]; reductions to determinants [5, 7, 9, 11]; iterative balancing [12]; and Markov chain\nMonte Carlo [13, 16, 19].\nAll these methods try to find a fully-polynomial randomized\napproximation scheme fpras for computing the permanent. fpras is such a scheme which,\nwhen given ε and inputs matrix A, outputs a estimator(usually a unbiased estimator)Y of\nthe permanent such that\nPr((1 −ε)per(A) ≤Y ≤(1 + ε)per(A)) ≥3\n4\n(3)\n\n3\nand runs in polynomial time in n and ε−1, here 3/4 may be boosted to 1 −δ for any\ndesired δ > 0 by running the algorithm O(log(δ−1)) and taking the median of the trials\n[10]. Then a straightforward application of Chebychev’s inequality shows that running the\nalgorithm O( E(Y 2)\nE2(Y )ε−2) times and taking the mean of the results can make the probability\nmore than 3/4(e.g. running 4 E(Y 2)\nE2(Y )ε−2 times). Hence, if the critical ratio E(Y 2)\nE2(Y ) is bounded\nby a polynomial of inputs A, we’ll get an fpras for the permanent of A. Another modified\nscheme called fpras for almost all inputs means: choose a matrix from A(n, 1/2)(A(n, 1/2)\ndenotes a probability space of n × n 0-1 matrices where each entry is chosen to be 1 or 0\nwith the same probability 1/2), or equivalently choose a matrix u.a.r. from A(n) (A(n)\nrepresents the set of n × n 0-1 matrices), and the following\nPr(critical ratio of A is bounded by a polynomial of the input A )= 1 −o(1) as n −→∞\nholds.(Note that this is a much weaker requirement than that of an fpras). If a proposition P\nrelating to n satisfies Pr(P is true)= 1 −o(n), we say P holds whp(whp is the abbreviation\nof ”with high probability”). Thus, that there is an fpras for almost all the matrix means the\ncritical ratio of A is bounded by a polynomial of the input A whp. A exciting result, that\nMarkov Chain approach led to the first fpras for the permanent of any 0-1 matrix(actually\nof any matrix with nonnegative entry) was shown by[16]. However, its high exponent of\npolynomial running time makes it difficult to be a practical method to approximate the\npermanent. RM and reductions to determinants seem to be two practical approaches esti-\nmating permanent due to their simply feasibility, and both of them have been proved to be\nan fpras for almost all the 0-1 matrices. besides, [3] promises a good prospect on computing\npermanent via clifford algebra if some difficulties can be conquered. RM also has developed\nto be a kind of approaches called sequential importance sampling way, which is widely used\nin statistical physics, see[14].\nIn this paper, we’ll, by RM, compute the number of all the matchings based on the\nabove transformation and give its performance theoretically, say, an analysis of critical\nratio in the sense ”for almost all the 0-1 matrix” of that matrix with a special structure. In\nsection II, A new alternative estimator operating directly on the adjacent matrix without any\ntransformation will be presented and proved to be equivalent to approximation performing\non the transformed matrix by RM. In section III, a low bound of the critical ratio for almost\nall the matrices will be presented, which is larger than any polynomial of n with a certain\nprobability. Hence, RM does not perform well in computing the number of all the matchings\n\n4\nas in computing the number of perfect matching. In section IV we’ll propose some analytic\nresults w.r.t. the expectation and variance of the number of all the matchings of a matrix\nselected u.a.r from G(m, n)(G(m, n) denotes the set of bipartite graph with #V1 = #V2 = n\nas its vertices and exact m edges). These results seem likely to contribute to the upper\nbound of critical ratio for almost all matrices, but the calculations are more arduous and\nwill be left for latter paper.\nII.\nAN EQUIVALENT ESTIMATOR\nAll the notations have the same meanings as those in the previous section without\nspecial illustration. Let A an n × n 0-1 matrix be an adjacent matrix of a bipartite graph\nG = (V, E), (V = V1\nS V2). Set YA a random variable. Then RM can be stated as follows:\ninputs: A an n × n 0-1 matrix;\noutputs: YA the estimator of permanent A;\nif n=0; then\nYA = 1\nelse\nW = {j : a1j = 1}\nif W = ∅then\nYA = 0\nelse\nChoose J u.a.r. from W\nYA = |W|Y1J\nY1j denotes the submatrix obtained from A by removing the 1st row and the jth\ncolumn.\nNote this heuristic idea comes from the Laplace’s expansion.\nOur following\nalgorithm(for easy discussion, call it AMM) is also inspired by another expansion. we first\npresents our algorithm for the number of all the matchings, and then give the explanation\nand proof of equivalence between AMM and RM on the transformed matrix:\ninputs: A an n × n 0-1 adjacent matrix of G;\n\n5\noutputs: YA the estimator of the number of all the matchings of G;\nif n=0; then\nYA = 1\nelse\nW = {j : a1j = 1} S{0}\nChoose J u.a.r. from W\nYA = |W|Y1J\nY10 denotes a submatrix of A by removing the 1st row(of course, it’s not necessarily a square\nmatrix). Define a new terminology AM on the matrix. let B = {bij, 1 ≤i ≤m, 1 ≤j ≤n}\nan m × n matrix, m ≤n. let AM(∅) = 1, by induction on m.\nAM(B) := AM(B10) +\nn\nX\nj=1\nb1,jB1j\n(4)\nThen we have the following theorem.\nTheorem 1.\nLet A be an n × n adjacent matrix of a bipartite graph G, Then\nAM(A)is the number of all the matchings of G.\nProof:\nIt’s easy to check, when k\n≥\n1, the number of k-matching of G equals\nP\ni1,··· ,ik\nP\nπ\nai1,π(i1) · · · aik,π(ik), where i1\n< i2 · · · < ik chosen from {1, 2, · · · , n}, π de-\nnotes the permutation of{i1, i2, · · · , ik}.\nThus,\nthe number of all the matchings\nis\nnP\nk=1\nP\ni1<···<ik⊆{1,··· ,n}\nP\nπ\nai1,π(i1) · · ·aik,π(ik) + 1, where 1 denotes the number of 0-matching.\nNote that if the AM(A) is written in terms of sum of elements of the matrix A, then it’s\nclearly to see AM(A) =\nnP\nk=1\nP\ni1<···<ik⊆{1,··· ,n}\nP\nπ\nai1,π(i1) · · ·aik,π(ik) + 1.□\nCorollary1.\nLet A = {aij1 ≤i, j ≤n} be an n × n 0-1 matrix and YA is obtained\nby above AMM. Then YA is unbiased for AM(A), E(YA) = AM(A)\nProof: We prove for any m × n 0-1 matrix A, 1 ≤i ≤m, 1 ≤j ≤n}, which will be widely\nused in the following proves. AMM is unbiased for AM(A). For any fixed n, by induction\non m, k=0,∀1 ≤l ≤n, ∀a k × l 0-1 matrix A, the equation E(YA) = AM(A) is trivial.\nNow suppose ∀k ≤m, k ≤l ≤n, a k × l 0-1 matrix A has E(YA) = AM(A). Then when\nk = m, let |W| = q, we have\n\n6\nE(YA) =\nX\nj∈W\nE(YA|J = j)Pr(J = j)\n=\nX\nj∈W\nE(qYA1j|J = j)q−1\n=\nX\nj∈W\nE(YA1j)\n=\nX\nj∈W\nAM(A1j)\n= AM(A).\n□\nAnother simple corollary can also be obtained.\nTo estimate the number of all the\nmatching in G, by RM operating on B =\n \n \nA\nIn×n\n1n×n 1n×n\n \n divided by n! is equivalent to\noperating on A by AMM, in precise words, which can be stated as follows.\nCorollary2.\nLet XA be the output of RM operating on A , YB be the output of\nAMM operating on transformed matrix B divided by n!. Then XA and YB has the same\ndistribution.\nProof: Note that by RM after n-th step operating on B, YB = Sn ∗Y1n×n/n!, where Sn is a\nnumber obtained from the first n steps, and obviously Y1n×n ≡n!. Hence, we have YB = Sn.\nThe same distribution of Sn and XA can be verified step by step.□\nCorollary3. AM(A) = 1\nn!per\n \n \nA\nIn×n\n1n×n 1n×n\n \n .\nProof: This is a direct deduction of corollary2. Let XA be the output of RM operating on\nA , YB be the output of AMM operating on transformed matrix B divided by n!.\nAM(A) = E(XA) = E(YB) = 1\nn!per\n \n \nA\nIn×n\n1n×n 1n×n\n \n \n□\nSo in the following section, we’ll use AMM to compute all the matchings instead of RM\nsince some methodologies similar to Rasmussen can be utilized. Another small advantage\n\n7\nby AMM is that the critical ratio is smaller than that directly obtained from RM. The\ncritical ratio by RM would be (2n)!, see Theorem 2.2[4], while the critical ratio by AMM\nwould be (n + 1)n.\nTheorem2.\nLet A = {aij, 1 ≤i, j ≤n} be an n × n adjacent matrix of a bipartite\ngraph G, and let XA be the output of AMM. Then E(XA)2\nE(X2\nA) ≤(n + 1)n. Generally, Let A be\nan m × n 0-1 matrix, m ≤n. XA be the output of AMM. Then E(XA)2\nE(X2\nA) ≤(n + 1)m\nProof: Induction on m, For any fixed n. k = 0,∀1 ≤l ≤n, ∀a k × l 0-1 matrix A, the\ninequation is trivial. In the case k = m, let |W| = q, we have\nE(X2\nA) =\nX\nj∈W\nE(X2\nA|J = j)Pr(J = j)\n=\nX\nj∈W\nE(q2X2\nA1j|J = j)q−1\n=\nX\nj∈W\nE(X2\nA1j)q\n≤\nX\nj∈W\nE(XA1j)2(n + 1)m−1q\n≤(\nX\nj∈W\nE(XA1j))2(n + 1)m−1q\n= E(XA)2(n + 1)m\n□\nIII.\nA LOWER BOUND OF CRITICAL RATIO FOR ALMOST ALL THE\nMATRICES\nRasmussen shows that although the critical ratio of RM is factorial in n, it does indeed\nprovide an fpras for almost all the matrix. However, the similar result can not be anticipated\nwhen computing all the matchings by RM. In fact the critical ratio for almost all the matrix\nwould be more than n\n√n/2−1 with a certain probability. To prove this, we need to define\nsome new denotations. Since there’re two probability spaces, we use the subscript σ denote\nthe calculus w.r.t. the probability space the algorithm lies in, say, coin-tosses, and subscript\nA represent the calculus w.r.t. the space probability the random matrices lie in. A(m, n, p)\ndenotes the probability space of all m × n 0-1 random matrices where each entry is chosen\n\n8\nto be 1 with probability p, andA(m, n) denotes the set of all m × n 0-1 matrices .\nTo obtain the mean and variance of the output of AMM on average under probability\nmeasure PrA, we need the following lemma.\nLemma1 Let f(m, n) defined as f(m, n) = anf(m−1, n) + cnf(m−1, n−1), where m ≤n\nare two nonnegative integers, an and cn are two infinite positive series w.r.t. n. And ∀\n0 ≤l ≤n, f(0, l) = 1. Then\nf(m, n) =\nm\nP\nk=1\nP\ns0+s1+···+sk=m−k\ns0,···sk≥0\ncn · · · cn−k+1as0\nn · · · ask\nn−k + am\nn\nProof: By induction on m. Obviously, the case p=0 is trivial. Suppose when p ≤m −1\n∀p ≤l ≤n, f(p, l) =\npP\nk=1\nP\ns0+s1+···+sk=p−k\ncl · · · cl−k+1as0\nl · · · ask\nl−k + ap\nl holds, then when p=m,\nwe have\nanf(m −1, n) =\nm−1\nX\nk=1\nX\ns0+s1+···+sk=m−1−k\ncn · · ·cn−k+1as0+1\nn\n· · · ask\nn−k + am\nn\n=\nm−1\nX\nk=1\nX\ns0+s1+···+sk=m−k\ns0≥1\ncn · · · cn−k+1as0\nn · · · ask\nn−k + am\nn\nand\ncnf(m −1, n −1) =\nm−1\nX\nk=1\nX\ns0+s1+···+sk=m−1−k\ncn · · · cn−kas0\nn−1 · · · ask\nn−1−k + cnam−1\nn−1\n=\nm−1\nX\nk=1\nX\ns1+s2+···+sk+1=m−1−k\ncn · · · cn−kas1\nn−1 · · · ask+1\nn−1−k + cnam−1\nn−1\n=\nm\nX\nk=2\nX\ns1+s2+···+sk=m−k\ncn · · · cn−k+1as1\nn−1 · · · ask\nn−k + cnam−1\nn−1\n=\nm\nX\nk=1\nX\ns1+s2+···+sk=m−k\ncn · · · cn−k+1as1\nn−1 · · · ask\nn−k\n=\nm\nX\nk=1\nX\ns0+s1+···+sk=m−k\ns0=0\ncn · · · cn−k+1as0\nn as1\nn−1 · · · ask\nn−k\nFrom the above two equation, there holds\n\n9\nf(m, n) = anf(m −1, n) + cnf(m −1, n −1)\n=\nm\nX\nk=1\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\ncn · · · cn−k+1as0\nn · · · ask\nn−k + am\nn\nThe previous n can be replaced by any l, where m ≤l ≤n □\nUsing lemma1 we can easily obtain two following Theorems.\nTheorem3.\nChoose Am,n u.a.r.\nfrom A(m, n), m ≤n, or equivalently let Am,n\nfrom A(m, n, 1/2). Then\nEA(AM(Am,n)) =\nm\nX\nk=0\nCk\nm\nP k\nn\n2k\nwhere Ck\nm =\nm!\nk!(m−k)! and P k\nn =\nn!\n(n−k)!\nProof: Induction on m. The case p=0, EA(AM(A)) = 1 is trivial. Suppose ∀p ≤m −1,\np ≤l ≤n EA(AM(Ap,l)) =\npP\nk=0\nCk\np\nP p−k\nl\n2p−k =\npP\nk=0\nCk\np\nP k\nl\n2k\nwhen p=m, ∀m ≤l ≤n, we have\nEA(AM(Am,l)) = EA(AM(A1,0\nm,l) +\nn\nX\nj=1\na1,jAM(A1,j\nm,l))\n= EA(AM(Am−1,l)) +\nn\nX\nj=1\nEA(a1,j)EA(AM(Am−1,l−1)\n= EA(AM(Am−1,l)) + n\n2EA(AM(Am−1,l−1)\nUsing lemma1, here al ≡1, andcl = l\n2 then\nEA(AM(Am,l)) =\nm\nX\nk=1\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\ncl · · · cl−k+1 + 1\n=\nm\nX\nk=1\nP k\nl\n2k\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\n1 + 1\n=\nm\nX\nk=1\nP k\nl\n2k Ck\nm + 1\n=\nm\nX\nk=0\nP k\nl\n2k Ck\nm\n\n10\n□\nTheorem4 Choose Am,n u.a.r. from A(m, n), m ≤n, and let XAm,n be the output by\nAMM. Then\nEA(Eσ(XAm,n)) =\nm\nX\nk=0\nCk\nm\nP k\nn\n2k\nand\nEA(Eσ(X2\nAm,n)) =\nm\nX\nk=0\nP k\nnP k\nn+3\n2m+k\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\n(n + 2)s0(n + 2 −1)s1 · · · (n + 2 −k)sk\nProof: The first equation is is trivial since Eσ(X2\nAm,n) = AM(Am,l). For the second one,\nwe use induction on m. The case p=0 is obvious. Suppose ∀Ap,l where 0 ≤p ≤m −1,\np ≤l ≤n the second equation holds. When p = m, noting the fact M = |W| −1 is a\nbinomial variable with parameter l and 1/2(recall W/{0} is the set of column indices with\na 1 in the first row), then\nEA(Eσ(X2\nAm,l)) =\nl\nX\nq=0\nEA(Eσ(X2\nAm,l)|M = q)PrA(M = q)\n=\nl\nX\nq=0\nEA((q + 1)\nX\nj∈W\nEσ(X2\nA1j\nm,l)|M = q)PrA(M = q)\n=\nl\nX\nq=0\nEA((q + 1)Eσ(X2\nAm−1,l) + q(q + 1)Eσ(X2\nAm−1,l−1))PrA(M = q)\n= (EA(M) + 1)EA(Eσ(X2\nAm−1,l)) + (EA(M2) + EA(M))EA(Eσ(X2\nAm−1,l−1))\n= (l + 2\n2\n)EA(Eσ(X2\nAm−1,l)) + (l2 + 3l\n4\n)EA(Eσ(X2\nAm−1,l−1))\nUsing lemma1, here al = l+2\n2 , and cl = l2+3l\n4 .Then\nEA(Eσ(X2\nAm,l)) =\nm\nX\nk=1\nP k\nl P k\nl+3\n4k\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\n(l + 2\n2\n)s0(l + 2 −1\n2\n)s1 · · · (l + 2 −k\n2\n)sk + (l + 2\n2\n)m\n=\nm\nX\nk=0\nP k\nl P k\nl+3\n2k+m\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\n(l + 2)s0(l + 2 −1)s1 · · · (l + 2 −k)sk\n□\nTheorem5 Choose An,n u.a.r. from A(n, n), and let XAn,n be the output by AMM. Then\nwhp h(n) ≤EA(Eσ(XAn,n)) ≤nh(n), where h(n) = (n!)2\n2n\n2k∗\n(n−k∗)!(k∗)2, k∗= ⌊−1 + √2n + 3⌋.\n\n11\nwhere ⌊∗⌋denotes the largest integer no more than ∗.\nProof:\nEA(Eσ(XAn,n)) =\nn\nX\nk=0\nCk\nn\nP k\nn\n2k\n=\nn\nX\nk=0\nCk\nn\nP n−k\nn\n2n−k\n= (n!)2\n2n\nn\nX\nk=0\n2k\n(n −k)!(k!)2\nand let bk =\n2k\n(n−k)!(k!)2, then\nbk\nbk−1 = 2(n−k+1)\nk2\n, set\nbk\nbk−1 ≥1 we have k ≤−1 + √2n + 3, thus,\nbk∗= max\nk=0,··· ,n bk. Thus, obviously\n(n!)2\n2n bk∗≤EA(Eσ(XAn,n)) ≤n(n!)2\n2n bk∗\n□\nTheorem6 Choose An,n u.a.r. from A(n, n), and let XAn,n be the output by AMM. Then\nwhp\nEA(Eσ(X2\nAn,n))\nE2\nA(Eσ(XAn,n)) ≥n(√n/2)\nProof: Numerical experiment shows the above result.\nhowever the theoretical analysis\nseems so hard than until now I haven’t thought out the way to show the comparably tight\nfor EA(Eσ(X2\nAn,n)) since the order of\nP\ns0+s1+···+sk=n−k\n(n + 2)s0(n + 2 −1)s1 · · · (n + 2 −k)sk is\ntoo difficult to gain a good lower bound. The following bound is easy to check and the best\none among methods I thought out,\nEA(Eσ(X2\nAn,n)) ≥\nn\nX\nk=0\n(n!)2(n + 3)!\n22n\n2k(k + 2)k\n(k!)2(k + 3)!(n −k)!\nHowever it still can’t reach the goal. Therefore, the proof of this theorem will be left for the\nfuture.\nEven if Theorem6 has been proved, unfortunately, the critical ratio for almost all the\nmatrices can not obtained from this theorem since two random variables are not independent.\nIn order to accomplish the ultimate result, we need to calculate the EA(E2\nσ(X2\nAn,n)). Using\nthe induction similar to theorem4, we can obtain the recursion of EA(E2\nσ(X2\nAm,n))(recall M\nis a binomial variable with parameter n and 1\n2).\nEA(E2\nσ(X2\nAm,n)) = 2(EA(M3) + 2EA(M2) + EA(M))EA(Eσ(X2\nAm−1,n)Eσ(X2\nAm−1,n−1))\n+(EA(M2)+2EA(M)+1)EA(E2\nσ(X2\nAm−1,n))+(EA(M4)+2EA(M3)+EA(M2))EA(E2\nσ(X2\nAm−1,n−1))\n\n12\nComparing EA(E2\nσ(X2\nAm,n)) with E2\nA(Eσ(X2\nAn,n)) and computing their ratio have to be\ndone.\nOur main aim of doing this is to find the matrices satisfying Eσ(X2\nAm,n) ≤\nEA(E2\nσ(X2\nAm,n))g(n), where g(n) is a polynomial of n. However, the ratio of\nEA(E2\nσ(X2\nAm,n))\nE2\nA(Eσ(X2\nAn,n)) is\nso large that it can’t accomplish our goal. Thus we deduce our requirement whp to with\na certain probability p > 0, and in our results p = 1\n2 −ε where ε is no more than 0.02. To\nprove the theorem, we need the following lemma, which will be proved in section IV.\nLemma2 Let B(m, n) denote the set of all n × n 0-1 matrices with exact m 1’s, m ≫n.\nChoose B u.a.r. from B(m, n). Then\nE(AM(B)) =\nn\nX\nk=0\n(Ck\nn)2k!Cm−k\nn2−k\nCm\nn2\nand\nE(AM2(B))\nE2(AM(B)) = 1 + o(1), n →∞\nTheorem7 Choose An,n u.a.r. from A(n, n), and let XAn,n be the output by AMM. Then\nPr(\nEσ(X2\nAn,n)\nE2σ(XAn,n) ≥n\n√n/2−1) ≥\nn2\nP\ni=(1/2+ε)n2 Ck\nn2\n2n2\nwhere c is a constant no more 10, andε ≤0.02.\nProof: From lemma2 we know if we set m = (1/2 + ε)n2 and q =\nCm−k\nn2−k\nCm\nn2 . When n goes to\ninfinity, noting k ≤n ≪m, n2, there holds\nq = Cm−k\nn2−k\nCm\nn2\n= m(m −1) · · ·(m −k)\nn2(n2 −1) · · · (n2 −k)\nand\nln(q) =\nk−1\nX\ni=0\n[ln(m −i) −ln(n2 −i)]\n= kln( m\nn2) +\nk−1\nX\ni=0\n[ln(1 −i\nm) −ln(1 −i\nn2)]\n= kln( m\nn2) −\nk−1\nX\ni=0\n[ i\nm −i\nn2 + O( i2\nm2)]\n= kln( m\nn2) −k(k −1)\n2\n( 1\nm −1\nn2) + O( k3\nm2)\n\n13\nThus, noting that km−1 ≤2nm−1 = O(n3m−2)\nq = ( m\nn2)kexp[−k2\n2 ( 1\nm −1\nn2) + O( n3\nm2)]\n= ((1/2 + ε)n2\nn2\n)kexp[−k2\n2 (\n1\n(1/2 + ε)n2 −1\nn2) + O(\nn3\n((1/2 + ε)n2)2)]\n≤e−1(1/2 + ε)k\nLet B selected u.a.r. from B(m, n) Since E(AM2(B))\nE2(AM(B)) = 1 + o(1), as n →∞\nthen Pr(AM(B) < 5\n6E(AM(B))) →0, as n →∞. So, if m ≥(1/2 + ε)n2 and ε ≤0.02, we\nhave whp\nEσ(X2\nB) ≥E2\nσ(XB)\n= AM2(B)\n≥(5\n6E(AM(B)))2\n= (5\n6\nn\nX\nk=0\n(Ck\nn)2k!Cm−k\nn2−k\nCm\nn2\n)2\n≥(\nn\nX\nk=0\n(Ck\nn)2k!5e−1\n6 (1/2 + ε)k)2\n≥\nn\nX\nk=0\nP k\nnP k\nn+3\n2n+k\nX\ns0+s1+···+sk=n−k\ns0,···sk≥0\n(n + 2)s0(n + 2 −1)s1 · · · (n + 2 −k)sk\n= EA(Eσ(XAn,n)).\nNoting Pr(A ∈\nS\nm≥(1/2+ε)n2 B(m, n)) =\nn2\nP\ni=(1/2+ε)n2\nCk\nn2\n2n2\n,\nthus Pr(Eσ(X2\nAn,n) ≥EA(Eσ(XAn,n))) ≥\nn2\nP\ni=(1/2+ε)n2\nCk\nn2\n2n2\n.\nUsing Markov’s inequality,\nPr(Eσ(X2\nAn,n) ≥nEA(Eσ(XAn,n))) ≤1\nn →0\nthen whp Eσ(X2\nAn,n) ≤nEA(Eσ(XAn,n)). Finally, we have\nPr(\nEσ(X2\nAn,n)\nE2σ(XAn,n) ≥1\nn\nEA(Eσ(X2\nAn,n))\nEA(Eσ(XAn,n))) ≥\nn2\nP\ni=(1/2+ε)n2 Ck\nn2\n2n2\n\n14\nApply theorem6 to the above formula, we have\nPr(\nEσ(X2\nAn,n)\nE2\nσ(XAn,n) ≥n\n√n/2−1) ≥\nn2\nP\ni=(1/2+ε)n2 Ck\nn2\n2n2\nIV.\nTHE NUMBER OF ALL THE MATCHINGS ON RANDOM GRAPH.\nIn this section, we consider the expectation and variance of the number of all the\nmatchings on G selected u.a.r. from G(m, n). We have the following theorem.\nTheorem8 Choose G u.a.r.\nfrom G(m, n), where G(m, n) denotes the set of bipar-\ntite graph with #V1 = #V2 = n as its vertices and exact m edges, m ≫n, and let AM(G)\ndenotes the number of all the matchings in G. Then we have\nE(AM(G)) =\nn\nX\nk=0\n(Ck\nn)2k!E(XM(k))\nand\nE(AM2(G)) =\nn\nX\nk=0\nk\nX\ni=0\n(Ck\nn)2k!\nmin(i,n−k)\nX\np=0\nCp\nn−kCi−p\nk\nP p\nn−i+p\ni−p\nX\nj=0\nCj\ni−p[Fn−j(i−p−j)]E(XM(k+i−j))\n+\nn\nX\nk=1\nk−1\nX\ni=0\n(Ck\nn)2k!\nmin(i,n−k)\nX\np=0\nCp\nn−kCi−p\nk\nP p\nn−i+p\ni−p\nX\nj=0\nCj\ni−p[Fn−j(i −p −j)]E(XM(k+i−j))\nwhere E(XM(k)) = Cm−k\nn2−k/Cm\nn2 and Fn(p) =\npP\nr=0\n(−1)rCr\npP p−r\nn−r\nProof: we’ll use the methodology in [6]; Let M(k) be a k-matching on V1 + V2, For G ∈\nG(m, n), define the random variable XM(G) to be 1 if M(k) is contained in G, and otherwise\n0. The expectation and second moment of AM(G) is as follows.\nE(AM(G)) = E(\nn\nX\nk=0\nX\nM(k)\nXM(k)) =\nn\nX\nk=0\nX\nM(k)\nE(XM(k))\nand\nE(AM2(G)) = E((\nn\nX\nk=0\nX\nM(k)\nXM(k))2) =\nn\nX\nk=0\nn\nX\ni=0\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\nwhere ∀0 ≤k ≤n, M(k) and M\n′(k) range over all (Ck\nn)2k! k-matching’s on V1 + V2. Note\nthat\nE(XM(k)) = Cm−k\nn2−k\nCm\nn2\n\n15\nThe first equation follows quickly. For the second, in order to compute E(XM(k)X\n′\nM(i)),\nwe have to calculate the number of pairs of M(k) and M\n′(i) as a function of the overlap\nj = |M(k) T M\n′(i)|. For any fixed k, suppose i ≤k, we need to compute the number of the\npairs of M(k) and M\n′(i), where i = 0, · · · , k, and M\n′(i) ranges over all (Ci\nn)2i! i-matching’s\non V1 + V2. The problem can be equivalently stated as follows: There’re n different letters\nand n different envelopes. Among these letters, there’re exact k(0 ≤k ≤n) labeled letters,\neach of which has only one ’mother envelope’ among envelopes. Different labeled letters\nhave different mother envelopes. We call a j-fit if there’re exact j labeled letters put into\nits own mother envelope. Now choose i(0 ≤i ≤k)letters from these n letters, then put\nthem into i envelopes, and each letter can only be put into one envelope. ∀possible j,\nhow many circumstances of j-fit are there? We can solve this problem like this: Suppose\nthere’re p letters unlabeled and i −p labeled letters among the selected letters, obviously,\n0 ≤p ≤min(n −k, i), the number of ways of choosing letters is Cp\nn−kCi−p\nk\n. If the labeled\nletters has been laid, then the number of the ways of putting p unlabeled letters is P p\nn−(i−p).\nFor any j(0 ≤j ≤i−p), there’re Cj\ni−p ways putting exact j labeled letters in its own mother\nenvelope. The last one we need to deal with is how many ways to put i−p−j labeled letters\ninto n −j envelopes which contain all these i −p −j letters’ mother envelopes, satisfying\n0-fit. By the principle of inclusion-exclusion see[1], we can easily obtain the number of the\nways is Fn−j(i −p −j), where Fn(p) =\npP\nr=0\n(−1)rCr\npP p−r\nn−r. Noting that p ranges over 0 to\nmin(i, n −k), and j ranges over 0 to i −p, for each k and i ≤k. Then\nX\nM′(i)\nE(XM(k)X\n′\nM(i)) =\nmin(i,n−k)\nX\np=0\nCp\nn−kCi−p\nk\nP p\nn−i+p\ni−p\nX\nj=0\nCj\ni−p[Fn−j(i −p −j)]E(XM(k+i−j))\n\n16\nwhere E(XM(k)) = Cm−k\nn2−k/Cm\nn2 and Fn(p) =\npP\nr=0\n(−1)rCr\npP p−r\nn−r.\nConsider,\nn\nX\nk=0\nn\nX\ni=0\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i)) = (\nn\nX\nk=0\nk\nX\ni=0\n+\nn−1\nX\nk=0\nn\nX\ni=k+1\n)\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\nk\nX\ni=0\n+\nn−1\nX\nk=0\nn\nX\ni=k+1\n)\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\nk\nX\ni=0\n+\nn\nX\ni=1\ni−1\nX\nk=0\n)\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\nk\nX\ni=0\n+\nn\nX\nk=1\nk−1\nX\ni=0\n)\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\nk\nX\ni=0\n+\nn\nX\nk=1\nk−1\nX\ni=0\n)\nX\nM(k)\nX\nM′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\n(Ck\nn)2k!\nk\nX\ni=0\n+\nn\nX\nk=1\n(Ck\nn)2k!\nk−1\nX\ni=0\n)\nX\nM′(i)\nE(XM(k)X\n′\nM(i))\nReplace P\nM′(i)\nE(XM(k)X\n′\nM(i)) by\nmin(i,n−k)\nP\np=0\nCp\nn−kCi−p\nk\nP p\nn−i+p\ni−p\nP\nj=0\nCj\ni−p[Fn−j(i −p −j)]E(XM(k+i−j)), then the second equation is\nachieved.\n□\nRemark: To complete the proof of theorem7, we also need to know whether the ratio\nE(AM2(G))\nE2(AM(G)) goes to 1 as n goes to infinity, adding the condition such as m2n−3 →∞as n\n→∞. We guess such a result is right, however the calculus seems very difficult. And this\nresult also contributes to the upper bound of critical ratio for almost all the matrices.\nAcknowledgments\n[1] M. Hall JR. Combinatorial Theory, Blaisdell, Waltham Massachusetts (1967).\n\n17\n[2] Valiant.\nThe complexity of computing the permanent,\nTheoretical Computer Science. 8,\n189-201 (1979).\n[3] S. Chien, L. Rasmussen and A. Sinclair. Clifford Algebras and approximating the permanent,\nProceedings of the 34th Annual Symposium on Theory of Computing (STOC). 222C231 (2002).\n[4] L. E. Rasmussen. Approximating the Permanent: a Simple Approach, Random Structures\nand Algorithms. 5, 349-361 (1994).\n[5] C. Godsil and I. Gutman. On the matching polynomial of a graph, Algebraic Methods in\nGraph Theory. 241-249 (1981).\n[6] A. Frieze and M. Jerrum.\nAn Analysis of A Monte Carlo Algorithm For Estimating the\nPermanent, Combinatorica. 15(1), 67-83 (1995).\n[7] N. Karmarkar, R. Karp, R. Lipton, L. Lov ́asz, and M. Luby,. A Monte-Carlo algorithm for\nestimating the permanent, SIAM Journal on Computing. 22, 284-293 (1993).\n[8] Yan Huo, Heng Liang, Siqi Liu and Fengshan Bai. Approximating the monomer-dimer con-\nstants through matrix permanent, arXiv:0708.1641v2:cond-mat.stat-mech . (2007).\n[9] A. Barvinok. Polynomial time algorithms to approximate permanents and mixed discriminants\nwithin a simply exponential factor, Random Struct. Algorithms. 14, 29-61 (1999).\n[10] H. Chernoff.\nA measure ofasymptotic efficiencyfor tests ofa hypothesis based on the sum\nofobservations, Ann. Math. Stat. 23, 493-509 (1952).\n[11] A. Barvinok. New permanent estimators via non-commutative determinants, Preprint. (2000).\n[12] N. Linial, A. Samorodnitsky and A. Wigderson. A deterministic strongly polynomial algorithm\nfor matrix scaling and approximate permanents, Combinatorica. 20, 545-568 (2000).\n[13] M. Jerrum and A. Sinclair. Approximating the permanent,, SIAM Journal on Computing.\n18, 1149-1178 (1989).\n[14] I.Beichl and F. Sullivan. Approximating the Permanent via Importance Sampling with Ap-\nplication to the Dimer Covering Problem, Journal of Computational Physics. 149, 128-147\n(1999).\n[15] Shmuel Friedland and Daniel Levy. A polynomial-time approximation algorithm for the num-\nber of k-matchings in bipartite graphs, arXiv:cs.CC/0607135 v1 . 392-401 (2006).\n[16] M. Jerrum, A. Sinclair and E. Vigoda. A polynomial-time approximation algorithm for the\npermanent of a matrix with non-negative entries, Proceedings of the 33rd ACM Symposium\non Theory of Computing. 712C721 (2001).\n\n18\n[17] A. Frieze and S. Suen. Counting the number of hamilton cycles in random digraphs, Random\nStructrues and Algorithms.(1992).\n[18] B. Bollob ́as. Random Graph, Cambridge University Press. Second Edition (2001).\n[19] C. Kenyon, D. Randall, and A. Sinclair. Approximating the number of dimer coverings of a\nlattice, Journal of Statistical Physics. 83, 637-659 (1996).\n[20] George E. Andrews, Richard Askey and Ranjan Roy. Special Function, Cambridge University\nPress. (1999).","paragraphs":[{"paragraph_id":"p1","order":1,"text":"arXiv:0711.1723v2 [cs.CC] 15 Nov 2007\nAn analysis of a random algorithm for estimating all the\nmatchings\nJinshan Zhang,∗Yan Huo,† and Fengshan Bai‡\nDepartment of Mathematical Sciences,\nTsinghua University, 100084,\nBeijing, PRC.\nCounting the number of all the matchings on a bipartite graph has been trans-\nformed into calculating the permanent of a matrix obtained from the extended bi-\npartite graph by Yan Huo, and Rasmussen presents a simple approach (RM) to\napproximate the permanent, which just yields a critical ratio O(nω(n)) for almost\nall the 0-1 matrices, provided it’s a simple promising practical way to compute this\n#P-complete problem. In this paper, the performance of this method will be shown\nwhen it’s applied to compute all the matchings based on that transformation. The\ncritical ratio will be proved to be very large with a certain probability, owning an\nincreasing factor larger than any polynomial of n even in the sense for almost all\nthe 0-1 matrices. Hence, RM fails to work well when counting all the matchings via\ncomputing the permanent of the matrix. In other words, we must carefully utilize\nthe known methods of estimating the permanent to count all the matchings through\nthat transformation.\nKeywords: matching; permanent; critical ratio; bipartite graph; determinant; Monte-Carlo\nalgorithm;random algorithm; RM;fpras\n∗Electronic address: zjs02@mails.tsinghua.edu.cn\n†Electronic address: huoy03@mails.tsinghua.edu.cn\n‡Electronic address: fbai@math.tsinghua.edu.cn"},{"paragraph_id":"p2","order":2,"text":"2\nI.\nINTRODUCTION\nLet G = (V, E) be a bipartite graph, where V = V1 ∪V2 is the set of vertices and\nE ⊂V1 × V2 is the set of edges. In the following sections we suppose #V1 = #V2 = n if\nthere’s no special illustration. A set of edges S ⊂E is called a matching if no two distinct\nedges e1, e2 ∈S contain a common vertex. S is called a k-matching if #S = k. In special\ncase, S is called a perfect matching if k = n. Let Sk be the set of k-matching in G and\nA(G) be the set of all the k-matching, k = 0, 1, . . . , n. For the convenience of discussion, let\n#S0 = 1, then the number of all the matchings in G is #A(G) =\nnP\ni=0\n#Sk.\nThe permanent of a 0-1 A = aij, 1 ≤i, j ≤n is defined as\nPer(A) =\nX\nπ\nn\nY\ni=1\nai,π(i)\n(1)\nwhere the sum is over all the permutations π of [n] = {1, . . . , n}. It’s well known that\nthe permanent of an adjacent matrix of bipartite graph equals the number of its perfect\nmatching. Let AM(G) denote the number of all the matchings in G, and A be adjacent\nmatrix of G. [8] has proved that\nAM(G) = 1\nn!per"},{"paragraph_id":"p3","order":3,"text":"A\nIn×n\n1n×n 1n×n"},{"paragraph_id":"p4","order":4,"text":"(2)\nwhere In×n is n × n unit matrix, 1n×n denotes n × n matrix with all the elements 1. This\nmeans in order to count the number of all the matchings of a bipartite graph with 2n vertices\nwe only need to compute the permanent of a 2n × 2n corresponding matrix transformed\nfrom adjacent matrix. The computation of permanent has a long history and was shown\nto be #P-complete in [2].\nThus, in the past 20 years or so, many random algorithms\nhave been developed to approximate the permanent, which can been divided at least four\ncategories[3]: elementary recursive algorithms(the original one is Rasmussen method(RM))\n[4]; reductions to determinants [5, 7, 9, 11]; iterative balancing [12]; and Markov chain\nMonte Carlo [13, 16, 19].\nAll these methods try to find a fully-polynomial randomized\napproximation scheme fpras for computing the permanent. fpras is such a scheme which,\nwhen given ε and inputs matrix A, outputs a estimator(usually a unbiased estimator)Y of\nthe permanent such that\nPr((1 −ε)per(A) ≤Y ≤(1 + ε)per(A)) ≥3\n4\n(3)"},{"paragraph_id":"p5","order":5,"text":"3\nand runs in polynomial time in n and ε−1, here 3/4 may be boosted to 1 −δ for any\ndesired δ > 0 by running the algorithm O(log(δ−1)) and taking the median of the trials\n[10]. Then a straightforward application of Chebychev’s inequality shows that running the\nalgorithm O( E(Y 2)\nE2(Y )ε−2) times and taking the mean of the results can make the probability\nmore than 3/4(e.g. running 4 E(Y 2)\nE2(Y )ε−2 times). Hence, if the critical ratio E(Y 2)\nE2(Y ) is bounded\nby a polynomial of inputs A, we’ll get an fpras for the permanent of A. Another modified\nscheme called fpras for almost all inputs means: choose a matrix from A(n, 1/2)(A(n, 1/2)\ndenotes a probability space of n × n 0-1 matrices where each entry is chosen to be 1 or 0\nwith the same probability 1/2), or equivalently choose a matrix u.a.r. from A(n) (A(n)\nrepresents the set of n × n 0-1 matrices), and the following\nPr(critical ratio of A is bounded by a polynomial of the input A )= 1 −o(1) as n −→∞\nholds.(Note that this is a much weaker requirement than that of an fpras). If a proposition P\nrelating to n satisfies Pr(P is true)= 1 −o(n), we say P holds whp(whp is the abbreviation\nof ”with high probability”). Thus, that there is an fpras for almost all the matrix means the\ncritical ratio of A is bounded by a polynomial of the input A whp. A exciting result, that\nMarkov Chain approach led to the first fpras for the permanent of any 0-1 matrix(actually\nof any matrix with nonnegative entry) was shown by[16]. However, its high exponent of\npolynomial running time makes it difficult to be a practical method to approximate the\npermanent. RM and reductions to determinants seem to be two practical approaches esti-\nmating permanent due to their simply feasibility, and both of them have been proved to be\nan fpras for almost all the 0-1 matrices. besides, [3] promises a good prospect on computing\npermanent via clifford algebra if some difficulties can be conquered. RM also has developed\nto be a kind of approaches called sequential importance sampling way, which is widely used\nin statistical physics, see[14].\nIn this paper, we’ll, by RM, compute the number of all the matchings based on the\nabove transformation and give its performance theoretically, say, an analysis of critical\nratio in the sense ”for almost all the 0-1 matrix” of that matrix with a special structure. In\nsection II, A new alternative estimator operating directly on the adjacent matrix without any\ntransformation will be presented and proved to be equivalent to approximation performing\non the transformed matrix by RM. In section III, a low bound of the critical ratio for almost\nall the matrices will be presented, which is larger than any polynomial of n with a certain\nprobability. Hence, RM does not perform well in computing the number of all the matchings"},{"paragraph_id":"p6","order":6,"text":"4\nas in computing the number of perfect matching. In section IV we’ll propose some analytic\nresults w.r.t. the expectation and variance of the number of all the matchings of a matrix\nselected u.a.r from G(m, n)(G(m, n) denotes the set of bipartite graph with #V1 = #V2 = n\nas its vertices and exact m edges). These results seem likely to contribute to the upper\nbound of critical ratio for almost all matrices, but the calculations are more arduous and\nwill be left for latter paper.\nII.\nAN EQUIVALENT ESTIMATOR\nAll the notations have the same meanings as those in the previous section without\nspecial illustration. Let A an n × n 0-1 matrix be an adjacent matrix of a bipartite graph\nG = (V, E), (V = V1\nS V2). Set YA a random variable. Then RM can be stated as follows:\ninputs: A an n × n 0-1 matrix;\noutputs: YA the estimator of permanent A;\nif n=0; then\nYA = 1\nelse\nW = {j : a1j = 1}\nif W = ∅then\nYA = 0\nelse\nChoose J u.a.r. from W\nYA = |W|Y1J\nY1j denotes the submatrix obtained from A by removing the 1st row and the jth\ncolumn.\nNote this heuristic idea comes from the Laplace’s expansion.\nOur following\nalgorithm(for easy discussion, call it AMM) is also inspired by another expansion. we first\npresents our algorithm for the number of all the matchings, and then give the explanation\nand proof of equivalence between AMM and RM on the transformed matrix:\ninputs: A an n × n 0-1 adjacent matrix of G;"},{"paragraph_id":"p7","order":7,"text":"5\noutputs: YA the estimator of the number of all the matchings of G;\nif n=0; then\nYA = 1\nelse\nW = {j : a1j = 1} S{0}\nChoose J u.a.r. from W\nYA = |W|Y1J\nY10 denotes a submatrix of A by removing the 1st row(of course, it’s not necessarily a square\nmatrix). Define a new terminology AM on the matrix. let B = {bij, 1 ≤i ≤m, 1 ≤j ≤n}\nan m × n matrix, m ≤n. let AM(∅) = 1, by induction on m.\nAM(B) := AM(B10) +\nn\nX\nj=1\nb1,jB1j\n(4)\nThen we have the following theorem.\nTheorem 1.\nLet A be an n × n adjacent matrix of a bipartite graph G, Then\nAM(A)is the number of all the matchings of G.\nProof:\nIt’s easy to check, when k\n≥\n1, the number of k-matching of G equals\nP\ni1,··· ,ik\nP\nπ\nai1,π(i1) · · · aik,π(ik), where i1\n< i2 · · · < ik chosen from {1, 2, · · · , n}, π de-\nnotes the permutation of{i1, i2, · · · , ik}.\nThus,\nthe number of all the matchings\nis\nnP\nk=1\nP\ni1<···<ik⊆{1,··· ,n}\nP\nπ\nai1,π(i1) · · ·aik,π(ik) + 1, where 1 denotes the number of 0-matching.\nNote that if the AM(A) is written in terms of sum of elements of the matrix A, then it’s\nclearly to see AM(A) =\nnP\nk=1\nP\ni1<···<ik⊆{1,··· ,n}\nP\nπ\nai1,π(i1) · · ·aik,π(ik) + 1.□\nCorollary1.\nLet A = {aij1 ≤i, j ≤n} be an n × n 0-1 matrix and YA is obtained\nby above AMM. Then YA is unbiased for AM(A), E(YA) = AM(A)\nProof: We prove for any m × n 0-1 matrix A, 1 ≤i ≤m, 1 ≤j ≤n}, which will be widely\nused in the following proves. AMM is unbiased for AM(A). For any fixed n, by induction\non m, k=0,∀1 ≤l ≤n, ∀a k × l 0-1 matrix A, the equation E(YA) = AM(A) is trivial.\nNow suppose ∀k ≤m, k ≤l ≤n, a k × l 0-1 matrix A has E(YA) = AM(A). Then when\nk = m, let |W| = q, we have"},{"paragraph_id":"p8","order":8,"text":"6\nE(YA) =\nX\nj∈W\nE(YA|J = j)Pr(J = j)\n=\nX\nj∈W\nE(qYA1j|J = j)q−1\n=\nX\nj∈W\nE(YA1j)\n=\nX\nj∈W\nAM(A1j)\n= AM(A).\n□\nAnother simple corollary can also be obtained.\nTo estimate the number of all the\nmatching in G, by RM operating on B ="},{"paragraph_id":"p9","order":9,"text":"A\nIn×n\n1n×n 1n×n"},{"paragraph_id":"p10","order":10,"text":"divided by n! is equivalent to\noperating on A by AMM, in precise words, which can be stated as follows.\nCorollary2.\nLet XA be the output of RM operating on A , YB be the output of\nAMM operating on transformed matrix B divided by n!. Then XA and YB has the same\ndistribution.\nProof: Note that by RM after n-th step operating on B, YB = Sn ∗Y1n×n/n!, where Sn is a\nnumber obtained from the first n steps, and obviously Y1n×n ≡n!. Hence, we have YB = Sn.\nThe same distribution of Sn and XA can be verified step by step.□\nCorollary3. AM(A) = 1\nn!per"},{"paragraph_id":"p11","order":11,"text":"A\nIn×n\n1n×n 1n×n"},{"paragraph_id":"p12","order":12,"text":".\nProof: This is a direct deduction of corollary2. Let XA be the output of RM operating on\nA , YB be the output of AMM operating on transformed matrix B divided by n!.\nAM(A) = E(XA) = E(YB) = 1\nn!per"},{"paragraph_id":"p13","order":13,"text":"A\nIn×n\n1n×n 1n×n"},{"paragraph_id":"p14","order":14,"text":"□\nSo in the following section, we’ll use AMM to compute all the matchings instead of RM\nsince some methodologies similar to Rasmussen can be utilized. Another small advantage"},{"paragraph_id":"p15","order":15,"text":"7\nby AMM is that the critical ratio is smaller than that directly obtained from RM. The\ncritical ratio by RM would be (2n)!, see Theorem 2.2[4], while the critical ratio by AMM\nwould be (n + 1)n.\nTheorem2.\nLet A = {aij, 1 ≤i, j ≤n} be an n × n adjacent matrix of a bipartite\ngraph G, and let XA be the output of AMM. Then E(XA)2\nE(X2\nA) ≤(n + 1)n. Generally, Let A be\nan m × n 0-1 matrix, m ≤n. XA be the output of AMM. Then E(XA)2\nE(X2\nA) ≤(n + 1)m\nProof: Induction on m, For any fixed n. k = 0,∀1 ≤l ≤n, ∀a k × l 0-1 matrix A, the\ninequation is trivial. In the case k = m, let |W| = q, we have\nE(X2\nA) =\nX\nj∈W\nE(X2\nA|J = j)Pr(J = j)\n=\nX\nj∈W\nE(q2X2\nA1j|J = j)q−1\n=\nX\nj∈W\nE(X2\nA1j)q\n≤\nX\nj∈W\nE(XA1j)2(n + 1)m−1q\n≤(\nX\nj∈W\nE(XA1j))2(n + 1)m−1q\n= E(XA)2(n + 1)m\n□\nIII.\nA LOWER BOUND OF CRITICAL RATIO FOR ALMOST ALL THE\nMATRICES\nRasmussen shows that although the critical ratio of RM is factorial in n, it does indeed\nprovide an fpras for almost all the matrix. However, the similar result can not be anticipated\nwhen computing all the matchings by RM. In fact the critical ratio for almost all the matrix\nwould be more than n\n√n/2−1 with a certain probability. To prove this, we need to define\nsome new denotations. Since there’re two probability spaces, we use the subscript σ denote\nthe calculus w.r.t. the probability space the algorithm lies in, say, coin-tosses, and subscript\nA represent the calculus w.r.t. the space probability the random matrices lie in. A(m, n, p)\ndenotes the probability space of all m × n 0-1 random matrices where each entry is chosen"},{"paragraph_id":"p16","order":16,"text":"8\nto be 1 with probability p, andA(m, n) denotes the set of all m × n 0-1 matrices .\nTo obtain the mean and variance of the output of AMM on average under probability\nmeasure PrA, we need the following lemma.\nLemma1 Let f(m, n) defined as f(m, n) = anf(m−1, n) + cnf(m−1, n−1), where m ≤n\nare two nonnegative integers, an and cn are two infinite positive series w.r.t. n. And ∀\n0 ≤l ≤n, f(0, l) = 1. Then\nf(m, n) =\nm\nP\nk=1\nP\ns0+s1+···+sk=m−k\ns0,···sk≥0\ncn · · · cn−k+1as0\nn · · · ask\nn−k + am\nn\nProof: By induction on m. Obviously, the case p=0 is trivial. Suppose when p ≤m −1\n∀p ≤l ≤n, f(p, l) =\npP\nk=1\nP\ns0+s1+···+sk=p−k\ncl · · · cl−k+1as0\nl · · · ask\nl−k + ap\nl holds, then when p=m,\nwe have\nanf(m −1, n) =\nm−1\nX\nk=1\nX\ns0+s1+···+sk=m−1−k\ncn · · ·cn−k+1as0+1\nn\n· · · ask\nn−k + am\nn\n=\nm−1\nX\nk=1\nX\ns0+s1+···+sk=m−k\ns0≥1\ncn · · · cn−k+1as0\nn · · · ask\nn−k + am\nn\nand\ncnf(m −1, n −1) =\nm−1\nX\nk=1\nX\ns0+s1+···+sk=m−1−k\ncn · · · cn−kas0\nn−1 · · · ask\nn−1−k + cnam−1\nn−1\n=\nm−1\nX\nk=1\nX\ns1+s2+···+sk+1=m−1−k\ncn · · · cn−kas1\nn−1 · · · ask+1\nn−1−k + cnam−1\nn−1\n=\nm\nX\nk=2\nX\ns1+s2+···+sk=m−k\ncn · · · cn−k+1as1\nn−1 · · · ask\nn−k + cnam−1\nn−1\n=\nm\nX\nk=1\nX\ns1+s2+···+sk=m−k\ncn · · · cn−k+1as1\nn−1 · · · ask\nn−k\n=\nm\nX\nk=1\nX\ns0+s1+···+sk=m−k\ns0=0\ncn · · · cn−k+1as0\nn as1\nn−1 · · · ask\nn−k\nFrom the above two equation, there holds"},{"paragraph_id":"p17","order":17,"text":"9\nf(m, n) = anf(m −1, n) + cnf(m −1, n −1)\n=\nm\nX\nk=1\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\ncn · · · cn−k+1as0\nn · · · ask\nn−k + am\nn\nThe previous n can be replaced by any l, where m ≤l ≤n □\nUsing lemma1 we can easily obtain two following Theorems.\nTheorem3.\nChoose Am,n u.a.r.\nfrom A(m, n), m ≤n, or equivalently let Am,n\nfrom A(m, n, 1/2). Then\nEA(AM(Am,n)) =\nm\nX\nk=0\nCk\nm\nP k\nn\n2k\nwhere Ck\nm =\nm!\nk!(m−k)! and P k\nn =\nn!\n(n−k)!\nProof: Induction on m. The case p=0, EA(AM(A)) = 1 is trivial. Suppose ∀p ≤m −1,\np ≤l ≤n EA(AM(Ap,l)) =\npP\nk=0\nCk\np\nP p−k\nl\n2p−k =\npP\nk=0\nCk\np\nP k\nl\n2k\nwhen p=m, ∀m ≤l ≤n, we have\nEA(AM(Am,l)) = EA(AM(A1,0\nm,l) +\nn\nX\nj=1\na1,jAM(A1,j\nm,l))\n= EA(AM(Am−1,l)) +\nn\nX\nj=1\nEA(a1,j)EA(AM(Am−1,l−1)\n= EA(AM(Am−1,l)) + n\n2EA(AM(Am−1,l−1)\nUsing lemma1, here al ≡1, andcl = l\n2 then\nEA(AM(Am,l)) =\nm\nX\nk=1\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\ncl · · · cl−k+1 + 1\n=\nm\nX\nk=1\nP k\nl\n2k\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\n1 + 1\n=\nm\nX\nk=1\nP k\nl\n2k Ck\nm + 1\n=\nm\nX\nk=0\nP k\nl\n2k Ck\nm"},{"paragraph_id":"p18","order":18,"text":"10\n□\nTheorem4 Choose Am,n u.a.r. from A(m, n), m ≤n, and let XAm,n be the output by\nAMM. Then\nEA(Eσ(XAm,n)) =\nm\nX\nk=0\nCk\nm\nP k\nn\n2k\nand\nEA(Eσ(X2\nAm,n)) =\nm\nX\nk=0\nP k\nnP k\nn+3\n2m+k\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\n(n + 2)s0(n + 2 −1)s1 · · · (n + 2 −k)sk\nProof: The first equation is is trivial since Eσ(X2\nAm,n) = AM(Am,l). For the second one,\nwe use induction on m. The case p=0 is obvious. Suppose ∀Ap,l where 0 ≤p ≤m −1,\np ≤l ≤n the second equation holds. When p = m, noting the fact M = |W| −1 is a\nbinomial variable with parameter l and 1/2(recall W/{0} is the set of column indices with\na 1 in the first row), then\nEA(Eσ(X2\nAm,l)) =\nl\nX\nq=0\nEA(Eσ(X2\nAm,l)|M = q)PrA(M = q)\n=\nl\nX\nq=0\nEA((q + 1)\nX\nj∈W\nEσ(X2\nA1j\nm,l)|M = q)PrA(M = q)\n=\nl\nX\nq=0\nEA((q + 1)Eσ(X2\nAm−1,l) + q(q + 1)Eσ(X2\nAm−1,l−1))PrA(M = q)\n= (EA(M) + 1)EA(Eσ(X2\nAm−1,l)) + (EA(M2) + EA(M))EA(Eσ(X2\nAm−1,l−1))\n= (l + 2\n2\n)EA(Eσ(X2\nAm−1,l)) + (l2 + 3l\n4\n)EA(Eσ(X2\nAm−1,l−1))\nUsing lemma1, here al = l+2\n2 , and cl = l2+3l\n4 .Then\nEA(Eσ(X2\nAm,l)) =\nm\nX\nk=1\nP k\nl P k\nl+3\n4k\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\n(l + 2\n2\n)s0(l + 2 −1\n2\n)s1 · · · (l + 2 −k\n2\n)sk + (l + 2\n2\n)m\n=\nm\nX\nk=0\nP k\nl P k\nl+3\n2k+m\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\n(l + 2)s0(l + 2 −1)s1 · · · (l + 2 −k)sk\n□\nTheorem5 Choose An,n u.a.r. from A(n, n), and let XAn,n be the output by AMM. Then\nwhp h(n) ≤EA(Eσ(XAn,n)) ≤nh(n), where h(n) = (n!)2\n2n\n2k∗\n(n−k∗)!(k∗)2, k∗= ⌊−1 + √2n + 3⌋."},{"paragraph_id":"p19","order":19,"text":"11\nwhere ⌊∗⌋denotes the largest integer no more than ∗.\nProof:\nEA(Eσ(XAn,n)) =\nn\nX\nk=0\nCk\nn\nP k\nn\n2k\n=\nn\nX\nk=0\nCk\nn\nP n−k\nn\n2n−k\n= (n!)2\n2n\nn\nX\nk=0\n2k\n(n −k)!(k!)2\nand let bk =\n2k\n(n−k)!(k!)2, then\nbk\nbk−1 = 2(n−k+1)\nk2\n, set\nbk\nbk−1 ≥1 we have k ≤−1 + √2n + 3, thus,\nbk∗= max\nk=0,··· ,n bk. Thus, obviously\n(n!)2\n2n bk∗≤EA(Eσ(XAn,n)) ≤n(n!)2\n2n bk∗\n□\nTheorem6 Choose An,n u.a.r. from A(n, n), and let XAn,n be the output by AMM. Then\nwhp\nEA(Eσ(X2\nAn,n))\nE2\nA(Eσ(XAn,n)) ≥n(√n/2)\nProof: Numerical experiment shows the above result.\nhowever the theoretical analysis\nseems so hard than until now I haven’t thought out the way to show the comparably tight\nfor EA(Eσ(X2\nAn,n)) since the order of\nP\ns0+s1+···+sk=n−k\n(n + 2)s0(n + 2 −1)s1 · · · (n + 2 −k)sk is\ntoo difficult to gain a good lower bound. The following bound is easy to check and the best\none among methods I thought out,\nEA(Eσ(X2\nAn,n)) ≥\nn\nX\nk=0\n(n!)2(n + 3)!\n22n\n2k(k + 2)k\n(k!)2(k + 3)!(n −k)!\nHowever it still can’t reach the goal. Therefore, the proof of this theorem will be left for the\nfuture.\nEven if Theorem6 has been proved, unfortunately, the critical ratio for almost all the\nmatrices can not obtained from this theorem since two random variables are not independent.\nIn order to accomplish the ultimate result, we need to calculate the EA(E2\nσ(X2\nAn,n)). Using\nthe induction similar to theorem4, we can obtain the recursion of EA(E2\nσ(X2\nAm,n))(recall M\nis a binomial variable with parameter n and 1\n2).\nEA(E2\nσ(X2\nAm,n)) = 2(EA(M3) + 2EA(M2) + EA(M))EA(Eσ(X2\nAm−1,n)Eσ(X2\nAm−1,n−1))\n+(EA(M2)+2EA(M)+1)EA(E2\nσ(X2\nAm−1,n))+(EA(M4)+2EA(M3)+EA(M2))EA(E2\nσ(X2\nAm−1,n−1))"},{"paragraph_id":"p20","order":20,"text":"12\nComparing EA(E2\nσ(X2\nAm,n)) with E2\nA(Eσ(X2\nAn,n)) and computing their ratio have to be\ndone.\nOur main aim of doing this is to find the matrices satisfying Eσ(X2\nAm,n) ≤\nEA(E2\nσ(X2\nAm,n))g(n), where g(n) is a polynomial of n. However, the ratio of\nEA(E2\nσ(X2\nAm,n))\nE2\nA(Eσ(X2\nAn,n)) is\nso large that it can’t accomplish our goal. Thus we deduce our requirement whp to with\na certain probability p > 0, and in our results p = 1\n2 −ε where ε is no more than 0.02. To\nprove the theorem, we need the following lemma, which will be proved in section IV.\nLemma2 Let B(m, n) denote the set of all n × n 0-1 matrices with exact m 1’s, m ≫n.\nChoose B u.a.r. from B(m, n). Then\nE(AM(B)) =\nn\nX\nk=0\n(Ck\nn)2k!Cm−k\nn2−k\nCm\nn2\nand\nE(AM2(B))\nE2(AM(B)) = 1 + o(1), n →∞\nTheorem7 Choose An,n u.a.r. from A(n, n), and let XAn,n be the output by AMM. Then\nPr(\nEσ(X2\nAn,n)\nE2σ(XAn,n) ≥n\n√n/2−1) ≥\nn2\nP\ni=(1/2+ε)n2 Ck\nn2\n2n2\nwhere c is a constant no more 10, andε ≤0.02.\nProof: From lemma2 we know if we set m = (1/2 + ε)n2 and q =\nCm−k\nn2−k\nCm\nn2 . When n goes to\ninfinity, noting k ≤n ≪m, n2, there holds\nq = Cm−k\nn2−k\nCm\nn2\n= m(m −1) · · ·(m −k)\nn2(n2 −1) · · · (n2 −k)\nand\nln(q) =\nk−1\nX\ni=0\n[ln(m −i) −ln(n2 −i)]\n= kln( m\nn2) +\nk−1\nX\ni=0\n[ln(1 −i\nm) −ln(1 −i\nn2)]\n= kln( m\nn2) −\nk−1\nX\ni=0\n[ i\nm −i\nn2 + O( i2\nm2)]\n= kln( m\nn2) −k(k −1)\n2\n( 1\nm −1\nn2) + O( k3\nm2)"},{"paragraph_id":"p21","order":21,"text":"13\nThus, noting that km−1 ≤2nm−1 = O(n3m−2)\nq = ( m\nn2)kexp[−k2\n2 ( 1\nm −1\nn2) + O( n3\nm2)]\n= ((1/2 + ε)n2\nn2\n)kexp[−k2\n2 (\n1\n(1/2 + ε)n2 −1\nn2) + O(\nn3\n((1/2 + ε)n2)2)]\n≤e−1(1/2 + ε)k\nLet B selected u.a.r. from B(m, n) Since E(AM2(B))\nE2(AM(B)) = 1 + o(1), as n →∞\nthen Pr(AM(B) < 5\n6E(AM(B))) →0, as n →∞. So, if m ≥(1/2 + ε)n2 and ε ≤0.02, we\nhave whp\nEσ(X2\nB) ≥E2\nσ(XB)\n= AM2(B)\n≥(5\n6E(AM(B)))2\n= (5\n6\nn\nX\nk=0\n(Ck\nn)2k!Cm−k\nn2−k\nCm\nn2\n)2\n≥(\nn\nX\nk=0\n(Ck\nn)2k!5e−1\n6 (1/2 + ε)k)2\n≥\nn\nX\nk=0\nP k\nnP k\nn+3\n2n+k\nX\ns0+s1+···+sk=n−k\ns0,···sk≥0\n(n + 2)s0(n + 2 −1)s1 · · · (n + 2 −k)sk\n= EA(Eσ(XAn,n)).\nNoting Pr(A ∈\nS\nm≥(1/2+ε)n2 B(m, n)) =\nn2\nP\ni=(1/2+ε)n2\nCk\nn2\n2n2\n,\nthus Pr(Eσ(X2\nAn,n) ≥EA(Eσ(XAn,n))) ≥\nn2\nP\ni=(1/2+ε)n2\nCk\nn2\n2n2\n.\nUsing Markov’s inequality,\nPr(Eσ(X2\nAn,n) ≥nEA(Eσ(XAn,n))) ≤1\nn →0\nthen whp Eσ(X2\nAn,n) ≤nEA(Eσ(XAn,n)). Finally, we have\nPr(\nEσ(X2\nAn,n)\nE2σ(XAn,n) ≥1\nn\nEA(Eσ(X2\nAn,n))\nEA(Eσ(XAn,n))) ≥\nn2\nP\ni=(1/2+ε)n2 Ck\nn2\n2n2"},{"paragraph_id":"p22","order":22,"text":"14\nApply theorem6 to the above formula, we have\nPr(\nEσ(X2\nAn,n)\nE2\nσ(XAn,n) ≥n\n√n/2−1) ≥\nn2\nP\ni=(1/2+ε)n2 Ck\nn2\n2n2\nIV.\nTHE NUMBER OF ALL THE MATCHINGS ON RANDOM GRAPH.\nIn this section, we consider the expectation and variance of the number of all the\nmatchings on G selected u.a.r. from G(m, n). We have the following theorem.\nTheorem8 Choose G u.a.r.\nfrom G(m, n), where G(m, n) denotes the set of bipar-\ntite graph with #V1 = #V2 = n as its vertices and exact m edges, m ≫n, and let AM(G)\ndenotes the number of all the matchings in G. Then we have\nE(AM(G)) =\nn\nX\nk=0\n(Ck\nn)2k!E(XM(k))\nand\nE(AM2(G)) =\nn\nX\nk=0\nk\nX\ni=0\n(Ck\nn)2k!\nmin(i,n−k)\nX\np=0\nCp\nn−kCi−p\nk\nP p\nn−i+p\ni−p\nX\nj=0\nCj\ni−p[Fn−j(i−p−j)]E(XM(k+i−j))\n+\nn\nX\nk=1\nk−1\nX\ni=0\n(Ck\nn)2k!\nmin(i,n−k)\nX\np=0\nCp\nn−kCi−p\nk\nP p\nn−i+p\ni−p\nX\nj=0\nCj\ni−p[Fn−j(i −p −j)]E(XM(k+i−j))\nwhere E(XM(k)) = Cm−k\nn2−k/Cm\nn2 and Fn(p) =\npP\nr=0\n(−1)rCr\npP p−r\nn−r\nProof: we’ll use the methodology in [6]; Let M(k) be a k-matching on V1 + V2, For G ∈\nG(m, n), define the random variable XM(G) to be 1 if M(k) is contained in G, and otherwise\n0. The expectation and second moment of AM(G) is as follows.\nE(AM(G)) = E(\nn\nX\nk=0\nX\nM(k)\nXM(k)) =\nn\nX\nk=0\nX\nM(k)\nE(XM(k))\nand\nE(AM2(G)) = E((\nn\nX\nk=0\nX\nM(k)\nXM(k))2) =\nn\nX\nk=0\nn\nX\ni=0\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\nwhere ∀0 ≤k ≤n, M(k) and M\n′(k) range over all (Ck\nn)2k! k-matching’s on V1 + V2. Note\nthat\nE(XM(k)) = Cm−k\nn2−k\nCm\nn2"},{"paragraph_id":"p23","order":23,"text":"15\nThe first equation follows quickly. For the second, in order to compute E(XM(k)X\n′\nM(i)),\nwe have to calculate the number of pairs of M(k) and M\n′(i) as a function of the overlap\nj = |M(k) T M\n′(i)|. For any fixed k, suppose i ≤k, we need to compute the number of the\npairs of M(k) and M\n′(i), where i = 0, · · · , k, and M\n′(i) ranges over all (Ci\nn)2i! i-matching’s\non V1 + V2. The problem can be equivalently stated as follows: There’re n different letters\nand n different envelopes. Among these letters, there’re exact k(0 ≤k ≤n) labeled letters,\neach of which has only one ’mother envelope’ among envelopes. Different labeled letters\nhave different mother envelopes. We call a j-fit if there’re exact j labeled letters put into\nits own mother envelope. Now choose i(0 ≤i ≤k)letters from these n letters, then put\nthem into i envelopes, and each letter can only be put into one envelope. ∀possible j,\nhow many circumstances of j-fit are there? We can solve this problem like this: Suppose\nthere’re p letters unlabeled and i −p labeled letters among the selected letters, obviously,\n0 ≤p ≤min(n −k, i), the number of ways of choosing letters is Cp\nn−kCi−p\nk\n. If the labeled\nletters has been laid, then the number of the ways of putting p unlabeled letters is P p\nn−(i−p).\nFor any j(0 ≤j ≤i−p), there’re Cj\ni−p ways putting exact j labeled letters in its own mother\nenvelope. The last one we need to deal with is how many ways to put i−p−j labeled letters\ninto n −j envelopes which contain all these i −p −j letters’ mother envelopes, satisfying\n0-fit. By the principle of inclusion-exclusion see[1], we can easily obtain the number of the\nways is Fn−j(i −p −j), where Fn(p) =\npP\nr=0\n(−1)rCr\npP p−r\nn−r. Noting that p ranges over 0 to\nmin(i, n −k), and j ranges over 0 to i −p, for each k and i ≤k. Then\nX\nM′(i)\nE(XM(k)X\n′\nM(i)) =\nmin(i,n−k)\nX\np=0\nCp\nn−kCi−p\nk\nP p\nn−i+p\ni−p\nX\nj=0\nCj\ni−p[Fn−j(i −p −j)]E(XM(k+i−j))"},{"paragraph_id":"p24","order":24,"text":"16\nwhere E(XM(k)) = Cm−k\nn2−k/Cm\nn2 and Fn(p) =\npP\nr=0\n(−1)rCr\npP p−r\nn−r.\nConsider,\nn\nX\nk=0\nn\nX\ni=0\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i)) = (\nn\nX\nk=0\nk\nX\ni=0\n+\nn−1\nX\nk=0\nn\nX\ni=k+1\n)\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\nk\nX\ni=0\n+\nn−1\nX\nk=0\nn\nX\ni=k+1\n)\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\nk\nX\ni=0\n+\nn\nX\ni=1\ni−1\nX\nk=0\n)\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\nk\nX\ni=0\n+\nn\nX\nk=1\nk−1\nX\ni=0\n)\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\nk\nX\ni=0\n+\nn\nX\nk=1\nk−1\nX\ni=0\n)\nX\nM(k)\nX\nM′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\n(Ck\nn)2k!\nk\nX\ni=0\n+\nn\nX\nk=1\n(Ck\nn)2k!\nk−1\nX\ni=0\n)\nX\nM′(i)\nE(XM(k)X\n′\nM(i))\nReplace P\nM′(i)\nE(XM(k)X\n′\nM(i)) by\nmin(i,n−k)\nP\np=0\nCp\nn−kCi−p\nk\nP p\nn−i+p\ni−p\nP\nj=0\nCj\ni−p[Fn−j(i −p −j)]E(XM(k+i−j)), then the second equation is\nachieved.\n□\nRemark: To complete the proof of theorem7, we also need to know whether the ratio\nE(AM2(G))\nE2(AM(G)) goes to 1 as n goes to infinity, adding the condition such as m2n−3 →∞as n\n→∞. We guess such a result is right, however the calculus seems very difficult. And this\nresult also contributes to the upper bound of critical ratio for almost all the matrices.\nAcknowledgments\n[1] M. Hall JR. Combinatorial Theory, Blaisdell, Waltham Massachusetts (1967)."},{"paragraph_id":"p25","order":25,"text":"17\n[2] Valiant.\nThe complexity of computing the permanent,\nTheoretical Computer Science. 8,\n189-201 (1979).\n[3] S. Chien, L. Rasmussen and A. Sinclair. Clifford Algebras and approximating the permanent,\nProceedings of the 34th Annual Symposium on Theory of Computing (STOC). 222C231 (2002).\n[4] L. E. Rasmussen. Approximating the Permanent: a Simple Approach, Random Structures\nand Algorithms. 5, 349-361 (1994).\n[5] C. Godsil and I. Gutman. On the matching polynomial of a graph, Algebraic Methods in\nGraph Theory. 241-249 (1981).\n[6] A. Frieze and M. Jerrum.\nAn Analysis of A Monte Carlo Algorithm For Estimating the\nPermanent, Combinatorica. 15(1), 67-83 (1995).\n[7] N. Karmarkar, R. Karp, R. Lipton, L. Lov ́asz, and M. Luby,. A Monte-Carlo algorithm for\nestimating the permanent, SIAM Journal on Computing. 22, 284-293 (1993).\n[8] Yan Huo, Heng Liang, Siqi Liu and Fengshan Bai. Approximating the monomer-dimer con-\nstants through matrix permanent, arXiv:0708.1641v2:cond-mat.stat-mech . (2007).\n[9] A. Barvinok. Polynomial time algorithms to approximate permanents and mixed discriminants\nwithin a simply exponential factor, Random Struct. Algorithms. 14, 29-61 (1999).\n[10] H. Chernoff.\nA measure ofasymptotic efficiencyfor tests ofa hypothesis based on the sum\nofobservations, Ann. Math. Stat. 23, 493-509 (1952).\n[11] A. Barvinok. New permanent estimators via non-commutative determinants, Preprint. (2000).\n[12] N. Linial, A. Samorodnitsky and A. Wigderson. A deterministic strongly polynomial algorithm\nfor matrix scaling and approximate permanents, Combinatorica. 20, 545-568 (2000).\n[13] M. Jerrum and A. Sinclair. Approximating the permanent,, SIAM Journal on Computing.\n18, 1149-1178 (1989).\n[14] I.Beichl and F. Sullivan. Approximating the Permanent via Importance Sampling with Ap-\nplication to the Dimer Covering Problem, Journal of Computational Physics. 149, 128-147\n(1999).\n[15] Shmuel Friedland and Daniel Levy. A polynomial-time approximation algorithm for the num-\nber of k-matchings in bipartite graphs, arXiv:cs.CC/0607135 v1 . 392-401 (2006).\n[16] M. Jerrum, A. Sinclair and E. Vigoda. A polynomial-time approximation algorithm for the\npermanent of a matrix with non-negative entries, Proceedings of the 33rd ACM Symposium\non Theory of Computing. 712C721 (2001)."},{"paragraph_id":"p26","order":26,"text":"18\n[17] A. Frieze and S. Suen. Counting the number of hamilton cycles in random digraphs, Random\nStructrues and Algorithms.(1992).\n[18] B. Bollob ́as. Random Graph, Cambridge University Press. Second Edition (2001).\n[19] C. Kenyon, D. Randall, and A. Sinclair. Approximating the number of dimer coverings of a\nlattice, Journal of Statistical Physics. 83, 637-659 (1996).\n[20] George E. Andrews, Richard Askey and Ranjan Roy. Special Function, Cambridge University\nPress. (1999)."}],"pages":[{"page":1,"text":"arXiv:0711.1723v2 [cs.CC] 15 Nov 2007\nAn analysis of a random algorithm for estimating all the\nmatchings\nJinshan Zhang,∗Yan Huo,† and Fengshan Bai‡\nDepartment of Mathematical Sciences,\nTsinghua University, 100084,\nBeijing, PRC.\nCounting the number of all the matchings on a bipartite graph has been trans-\nformed into calculating the permanent of a matrix obtained from the extended bi-\npartite graph by Yan Huo, and Rasmussen presents a simple approach (RM) to\napproximate the permanent, which just yields a critical ratio O(nω(n)) for almost\nall the 0-1 matrices, provided it’s a simple promising practical way to compute this\n#P-complete problem. In this paper, the performance of this method will be shown\nwhen it’s applied to compute all the matchings based on that transformation. The\ncritical ratio will be proved to be very large with a certain probability, owning an\nincreasing factor larger than any polynomial of n even in the sense for almost all\nthe 0-1 matrices. Hence, RM fails to work well when counting all the matchings via\ncomputing the permanent of the matrix. In other words, we must carefully utilize\nthe known methods of estimating the permanent to count all the matchings through\nthat transformation.\nKeywords: matching; permanent; critical ratio; bipartite graph; determinant; Monte-Carlo\nalgorithm;random algorithm; RM;fpras\n∗Electronic address: zjs02@mails.tsinghua.edu.cn\n†Electronic address: huoy03@mails.tsinghua.edu.cn\n‡Electronic address: fbai@math.tsinghua.edu.cn"},{"page":2,"text":"2\nI.\nINTRODUCTION\nLet G = (V, E) be a bipartite graph, where V = V1 ∪V2 is the set of vertices and\nE ⊂V1 × V2 is the set of edges. In the following sections we suppose #V1 = #V2 = n if\nthere’s no special illustration. A set of edges S ⊂E is called a matching if no two distinct\nedges e1, e2 ∈S contain a common vertex. S is called a k-matching if #S = k. In special\ncase, S is called a perfect matching if k = n. Let Sk be the set of k-matching in G and\nA(G) be the set of all the k-matching, k = 0, 1, . . . , n. For the convenience of discussion, let\n#S0 = 1, then the number of all the matchings in G is #A(G) =\nnP\ni=0\n#Sk.\nThe permanent of a 0-1 A = aij, 1 ≤i, j ≤n is defined as\nPer(A) =\nX\nπ\nn\nY\ni=1\nai,π(i)\n(1)\nwhere the sum is over all the permutations π of [n] = {1, . . . , n}. It’s well known that\nthe permanent of an adjacent matrix of bipartite graph equals the number of its perfect\nmatching. Let AM(G) denote the number of all the matchings in G, and A be adjacent\nmatrix of G. [8] has proved that\nAM(G) = 1\nn!per\n \n \nA\nIn×n\n1n×n 1n×n\n \n \n(2)\nwhere In×n is n × n unit matrix, 1n×n denotes n × n matrix with all the elements 1. This\nmeans in order to count the number of all the matchings of a bipartite graph with 2n vertices\nwe only need to compute the permanent of a 2n × 2n corresponding matrix transformed\nfrom adjacent matrix. The computation of permanent has a long history and was shown\nto be #P-complete in [2].\nThus, in the past 20 years or so, many random algorithms\nhave been developed to approximate the permanent, which can been divided at least four\ncategories[3]: elementary recursive algorithms(the original one is Rasmussen method(RM))\n[4]; reductions to determinants [5, 7, 9, 11]; iterative balancing [12]; and Markov chain\nMonte Carlo [13, 16, 19].\nAll these methods try to find a fully-polynomial randomized\napproximation scheme fpras for computing the permanent. fpras is such a scheme which,\nwhen given ε and inputs matrix A, outputs a estimator(usually a unbiased estimator)Y of\nthe permanent such that\nPr((1 −ε)per(A) ≤Y ≤(1 + ε)per(A)) ≥3\n4\n(3)"},{"page":3,"text":"3\nand runs in polynomial time in n and ε−1, here 3/4 may be boosted to 1 −δ for any\ndesired δ > 0 by running the algorithm O(log(δ−1)) and taking the median of the trials\n[10]. Then a straightforward application of Chebychev’s inequality shows that running the\nalgorithm O( E(Y 2)\nE2(Y )ε−2) times and taking the mean of the results can make the probability\nmore than 3/4(e.g. running 4 E(Y 2)\nE2(Y )ε−2 times). Hence, if the critical ratio E(Y 2)\nE2(Y ) is bounded\nby a polynomial of inputs A, we’ll get an fpras for the permanent of A. Another modified\nscheme called fpras for almost all inputs means: choose a matrix from A(n, 1/2)(A(n, 1/2)\ndenotes a probability space of n × n 0-1 matrices where each entry is chosen to be 1 or 0\nwith the same probability 1/2), or equivalently choose a matrix u.a.r. from A(n) (A(n)\nrepresents the set of n × n 0-1 matrices), and the following\nPr(critical ratio of A is bounded by a polynomial of the input A )= 1 −o(1) as n −→∞\nholds.(Note that this is a much weaker requirement than that of an fpras). If a proposition P\nrelating to n satisfies Pr(P is true)= 1 −o(n), we say P holds whp(whp is the abbreviation\nof ”with high probability”). Thus, that there is an fpras for almost all the matrix means the\ncritical ratio of A is bounded by a polynomial of the input A whp. A exciting result, that\nMarkov Chain approach led to the first fpras for the permanent of any 0-1 matrix(actually\nof any matrix with nonnegative entry) was shown by[16]. However, its high exponent of\npolynomial running time makes it difficult to be a practical method to approximate the\npermanent. RM and reductions to determinants seem to be two practical approaches esti-\nmating permanent due to their simply feasibility, and both of them have been proved to be\nan fpras for almost all the 0-1 matrices. besides, [3] promises a good prospect on computing\npermanent via clifford algebra if some difficulties can be conquered. RM also has developed\nto be a kind of approaches called sequential importance sampling way, which is widely used\nin statistical physics, see[14].\nIn this paper, we’ll, by RM, compute the number of all the matchings based on the\nabove transformation and give its performance theoretically, say, an analysis of critical\nratio in the sense ”for almost all the 0-1 matrix” of that matrix with a special structure. In\nsection II, A new alternative estimator operating directly on the adjacent matrix without any\ntransformation will be presented and proved to be equivalent to approximation performing\non the transformed matrix by RM. In section III, a low bound of the critical ratio for almost\nall the matrices will be presented, which is larger than any polynomial of n with a certain\nprobability. Hence, RM does not perform well in computing the number of all the matchings"},{"page":4,"text":"4\nas in computing the number of perfect matching. In section IV we’ll propose some analytic\nresults w.r.t. the expectation and variance of the number of all the matchings of a matrix\nselected u.a.r from G(m, n)(G(m, n) denotes the set of bipartite graph with #V1 = #V2 = n\nas its vertices and exact m edges). These results seem likely to contribute to the upper\nbound of critical ratio for almost all matrices, but the calculations are more arduous and\nwill be left for latter paper.\nII.\nAN EQUIVALENT ESTIMATOR\nAll the notations have the same meanings as those in the previous section without\nspecial illustration. Let A an n × n 0-1 matrix be an adjacent matrix of a bipartite graph\nG = (V, E), (V = V1\nS V2). Set YA a random variable. Then RM can be stated as follows:\ninputs: A an n × n 0-1 matrix;\noutputs: YA the estimator of permanent A;\nif n=0; then\nYA = 1\nelse\nW = {j : a1j = 1}\nif W = ∅then\nYA = 0\nelse\nChoose J u.a.r. from W\nYA = |W|Y1J\nY1j denotes the submatrix obtained from A by removing the 1st row and the jth\ncolumn.\nNote this heuristic idea comes from the Laplace’s expansion.\nOur following\nalgorithm(for easy discussion, call it AMM) is also inspired by another expansion. we first\npresents our algorithm for the number of all the matchings, and then give the explanation\nand proof of equivalence between AMM and RM on the transformed matrix:\ninputs: A an n × n 0-1 adjacent matrix of G;"},{"page":5,"text":"5\noutputs: YA the estimator of the number of all the matchings of G;\nif n=0; then\nYA = 1\nelse\nW = {j : a1j = 1} S{0}\nChoose J u.a.r. from W\nYA = |W|Y1J\nY10 denotes a submatrix of A by removing the 1st row(of course, it’s not necessarily a square\nmatrix). Define a new terminology AM on the matrix. let B = {bij, 1 ≤i ≤m, 1 ≤j ≤n}\nan m × n matrix, m ≤n. let AM(∅) = 1, by induction on m.\nAM(B) := AM(B10) +\nn\nX\nj=1\nb1,jB1j\n(4)\nThen we have the following theorem.\nTheorem 1.\nLet A be an n × n adjacent matrix of a bipartite graph G, Then\nAM(A)is the number of all the matchings of G.\nProof:\nIt’s easy to check, when k\n≥\n1, the number of k-matching of G equals\nP\ni1,··· ,ik\nP\nπ\nai1,π(i1) · · · aik,π(ik), where i1\n< i2 · · · < ik chosen from {1, 2, · · · , n}, π de-\nnotes the permutation of{i1, i2, · · · , ik}.\nThus,\nthe number of all the matchings\nis\nnP\nk=1\nP\ni1<···<ik⊆{1,··· ,n}\nP\nπ\nai1,π(i1) · · ·aik,π(ik) + 1, where 1 denotes the number of 0-matching.\nNote that if the AM(A) is written in terms of sum of elements of the matrix A, then it’s\nclearly to see AM(A) =\nnP\nk=1\nP\ni1<···<ik⊆{1,··· ,n}\nP\nπ\nai1,π(i1) · · ·aik,π(ik) + 1.□\nCorollary1.\nLet A = {aij1 ≤i, j ≤n} be an n × n 0-1 matrix and YA is obtained\nby above AMM. Then YA is unbiased for AM(A), E(YA) = AM(A)\nProof: We prove for any m × n 0-1 matrix A, 1 ≤i ≤m, 1 ≤j ≤n}, which will be widely\nused in the following proves. AMM is unbiased for AM(A). For any fixed n, by induction\non m, k=0,∀1 ≤l ≤n, ∀a k × l 0-1 matrix A, the equation E(YA) = AM(A) is trivial.\nNow suppose ∀k ≤m, k ≤l ≤n, a k × l 0-1 matrix A has E(YA) = AM(A). Then when\nk = m, let |W| = q, we have"},{"page":6,"text":"6\nE(YA) =\nX\nj∈W\nE(YA|J = j)Pr(J = j)\n=\nX\nj∈W\nE(qYA1j|J = j)q−1\n=\nX\nj∈W\nE(YA1j)\n=\nX\nj∈W\nAM(A1j)\n= AM(A).\n□\nAnother simple corollary can also be obtained.\nTo estimate the number of all the\nmatching in G, by RM operating on B =\n \n \nA\nIn×n\n1n×n 1n×n\n \n divided by n! is equivalent to\noperating on A by AMM, in precise words, which can be stated as follows.\nCorollary2.\nLet XA be the output of RM operating on A , YB be the output of\nAMM operating on transformed matrix B divided by n!. Then XA and YB has the same\ndistribution.\nProof: Note that by RM after n-th step operating on B, YB = Sn ∗Y1n×n/n!, where Sn is a\nnumber obtained from the first n steps, and obviously Y1n×n ≡n!. Hence, we have YB = Sn.\nThe same distribution of Sn and XA can be verified step by step.□\nCorollary3. AM(A) = 1\nn!per\n \n \nA\nIn×n\n1n×n 1n×n\n \n .\nProof: This is a direct deduction of corollary2. Let XA be the output of RM operating on\nA , YB be the output of AMM operating on transformed matrix B divided by n!.\nAM(A) = E(XA) = E(YB) = 1\nn!per\n \n \nA\nIn×n\n1n×n 1n×n\n \n \n□\nSo in the following section, we’ll use AMM to compute all the matchings instead of RM\nsince some methodologies similar to Rasmussen can be utilized. Another small advantage"},{"page":7,"text":"7\nby AMM is that the critical ratio is smaller than that directly obtained from RM. The\ncritical ratio by RM would be (2n)!, see Theorem 2.2[4], while the critical ratio by AMM\nwould be (n + 1)n.\nTheorem2.\nLet A = {aij, 1 ≤i, j ≤n} be an n × n adjacent matrix of a bipartite\ngraph G, and let XA be the output of AMM. Then E(XA)2\nE(X2\nA) ≤(n + 1)n. Generally, Let A be\nan m × n 0-1 matrix, m ≤n. XA be the output of AMM. Then E(XA)2\nE(X2\nA) ≤(n + 1)m\nProof: Induction on m, For any fixed n. k = 0,∀1 ≤l ≤n, ∀a k × l 0-1 matrix A, the\ninequation is trivial. In the case k = m, let |W| = q, we have\nE(X2\nA) =\nX\nj∈W\nE(X2\nA|J = j)Pr(J = j)\n=\nX\nj∈W\nE(q2X2\nA1j|J = j)q−1\n=\nX\nj∈W\nE(X2\nA1j)q\n≤\nX\nj∈W\nE(XA1j)2(n + 1)m−1q\n≤(\nX\nj∈W\nE(XA1j))2(n + 1)m−1q\n= E(XA)2(n + 1)m\n□\nIII.\nA LOWER BOUND OF CRITICAL RATIO FOR ALMOST ALL THE\nMATRICES\nRasmussen shows that although the critical ratio of RM is factorial in n, it does indeed\nprovide an fpras for almost all the matrix. However, the similar result can not be anticipated\nwhen computing all the matchings by RM. In fact the critical ratio for almost all the matrix\nwould be more than n\n√n/2−1 with a certain probability. To prove this, we need to define\nsome new denotations. Since there’re two probability spaces, we use the subscript σ denote\nthe calculus w.r.t. the probability space the algorithm lies in, say, coin-tosses, and subscript\nA represent the calculus w.r.t. the space probability the random matrices lie in. A(m, n, p)\ndenotes the probability space of all m × n 0-1 random matrices where each entry is chosen"},{"page":8,"text":"8\nto be 1 with probability p, andA(m, n) denotes the set of all m × n 0-1 matrices .\nTo obtain the mean and variance of the output of AMM on average under probability\nmeasure PrA, we need the following lemma.\nLemma1 Let f(m, n) defined as f(m, n) = anf(m−1, n) + cnf(m−1, n−1), where m ≤n\nare two nonnegative integers, an and cn are two infinite positive series w.r.t. n. And ∀\n0 ≤l ≤n, f(0, l) = 1. Then\nf(m, n) =\nm\nP\nk=1\nP\ns0+s1+···+sk=m−k\ns0,···sk≥0\ncn · · · cn−k+1as0\nn · · · ask\nn−k + am\nn\nProof: By induction on m. Obviously, the case p=0 is trivial. Suppose when p ≤m −1\n∀p ≤l ≤n, f(p, l) =\npP\nk=1\nP\ns0+s1+···+sk=p−k\ncl · · · cl−k+1as0\nl · · · ask\nl−k + ap\nl holds, then when p=m,\nwe have\nanf(m −1, n) =\nm−1\nX\nk=1\nX\ns0+s1+···+sk=m−1−k\ncn · · ·cn−k+1as0+1\nn\n· · · ask\nn−k + am\nn\n=\nm−1\nX\nk=1\nX\ns0+s1+···+sk=m−k\ns0≥1\ncn · · · cn−k+1as0\nn · · · ask\nn−k + am\nn\nand\ncnf(m −1, n −1) =\nm−1\nX\nk=1\nX\ns0+s1+···+sk=m−1−k\ncn · · · cn−kas0\nn−1 · · · ask\nn−1−k + cnam−1\nn−1\n=\nm−1\nX\nk=1\nX\ns1+s2+···+sk+1=m−1−k\ncn · · · cn−kas1\nn−1 · · · ask+1\nn−1−k + cnam−1\nn−1\n=\nm\nX\nk=2\nX\ns1+s2+···+sk=m−k\ncn · · · cn−k+1as1\nn−1 · · · ask\nn−k + cnam−1\nn−1\n=\nm\nX\nk=1\nX\ns1+s2+···+sk=m−k\ncn · · · cn−k+1as1\nn−1 · · · ask\nn−k\n=\nm\nX\nk=1\nX\ns0+s1+···+sk=m−k\ns0=0\ncn · · · cn−k+1as0\nn as1\nn−1 · · · ask\nn−k\nFrom the above two equation, there holds"},{"page":9,"text":"9\nf(m, n) = anf(m −1, n) + cnf(m −1, n −1)\n=\nm\nX\nk=1\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\ncn · · · cn−k+1as0\nn · · · ask\nn−k + am\nn\nThe previous n can be replaced by any l, where m ≤l ≤n □\nUsing lemma1 we can easily obtain two following Theorems.\nTheorem3.\nChoose Am,n u.a.r.\nfrom A(m, n), m ≤n, or equivalently let Am,n\nfrom A(m, n, 1/2). Then\nEA(AM(Am,n)) =\nm\nX\nk=0\nCk\nm\nP k\nn\n2k\nwhere Ck\nm =\nm!\nk!(m−k)! and P k\nn =\nn!\n(n−k)!\nProof: Induction on m. The case p=0, EA(AM(A)) = 1 is trivial. Suppose ∀p ≤m −1,\np ≤l ≤n EA(AM(Ap,l)) =\npP\nk=0\nCk\np\nP p−k\nl\n2p−k =\npP\nk=0\nCk\np\nP k\nl\n2k\nwhen p=m, ∀m ≤l ≤n, we have\nEA(AM(Am,l)) = EA(AM(A1,0\nm,l) +\nn\nX\nj=1\na1,jAM(A1,j\nm,l))\n= EA(AM(Am−1,l)) +\nn\nX\nj=1\nEA(a1,j)EA(AM(Am−1,l−1)\n= EA(AM(Am−1,l)) + n\n2EA(AM(Am−1,l−1)\nUsing lemma1, here al ≡1, andcl = l\n2 then\nEA(AM(Am,l)) =\nm\nX\nk=1\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\ncl · · · cl−k+1 + 1\n=\nm\nX\nk=1\nP k\nl\n2k\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\n1 + 1\n=\nm\nX\nk=1\nP k\nl\n2k Ck\nm + 1\n=\nm\nX\nk=0\nP k\nl\n2k Ck\nm"},{"page":10,"text":"10\n□\nTheorem4 Choose Am,n u.a.r. from A(m, n), m ≤n, and let XAm,n be the output by\nAMM. Then\nEA(Eσ(XAm,n)) =\nm\nX\nk=0\nCk\nm\nP k\nn\n2k\nand\nEA(Eσ(X2\nAm,n)) =\nm\nX\nk=0\nP k\nnP k\nn+3\n2m+k\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\n(n + 2)s0(n + 2 −1)s1 · · · (n + 2 −k)sk\nProof: The first equation is is trivial since Eσ(X2\nAm,n) = AM(Am,l). For the second one,\nwe use induction on m. The case p=0 is obvious. Suppose ∀Ap,l where 0 ≤p ≤m −1,\np ≤l ≤n the second equation holds. When p = m, noting the fact M = |W| −1 is a\nbinomial variable with parameter l and 1/2(recall W/{0} is the set of column indices with\na 1 in the first row), then\nEA(Eσ(X2\nAm,l)) =\nl\nX\nq=0\nEA(Eσ(X2\nAm,l)|M = q)PrA(M = q)\n=\nl\nX\nq=0\nEA((q + 1)\nX\nj∈W\nEσ(X2\nA1j\nm,l)|M = q)PrA(M = q)\n=\nl\nX\nq=0\nEA((q + 1)Eσ(X2\nAm−1,l) + q(q + 1)Eσ(X2\nAm−1,l−1))PrA(M = q)\n= (EA(M) + 1)EA(Eσ(X2\nAm−1,l)) + (EA(M2) + EA(M))EA(Eσ(X2\nAm−1,l−1))\n= (l + 2\n2\n)EA(Eσ(X2\nAm−1,l)) + (l2 + 3l\n4\n)EA(Eσ(X2\nAm−1,l−1))\nUsing lemma1, here al = l+2\n2 , and cl = l2+3l\n4 .Then\nEA(Eσ(X2\nAm,l)) =\nm\nX\nk=1\nP k\nl P k\nl+3\n4k\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\n(l + 2\n2\n)s0(l + 2 −1\n2\n)s1 · · · (l + 2 −k\n2\n)sk + (l + 2\n2\n)m\n=\nm\nX\nk=0\nP k\nl P k\nl+3\n2k+m\nX\ns0+s1+···+sk=m−k\ns0,···sk≥0\n(l + 2)s0(l + 2 −1)s1 · · · (l + 2 −k)sk\n□\nTheorem5 Choose An,n u.a.r. from A(n, n), and let XAn,n be the output by AMM. Then\nwhp h(n) ≤EA(Eσ(XAn,n)) ≤nh(n), where h(n) = (n!)2\n2n\n2k∗\n(n−k∗)!(k∗)2, k∗= ⌊−1 + √2n + 3⌋."},{"page":11,"text":"11\nwhere ⌊∗⌋denotes the largest integer no more than ∗.\nProof:\nEA(Eσ(XAn,n)) =\nn\nX\nk=0\nCk\nn\nP k\nn\n2k\n=\nn\nX\nk=0\nCk\nn\nP n−k\nn\n2n−k\n= (n!)2\n2n\nn\nX\nk=0\n2k\n(n −k)!(k!)2\nand let bk =\n2k\n(n−k)!(k!)2, then\nbk\nbk−1 = 2(n−k+1)\nk2\n, set\nbk\nbk−1 ≥1 we have k ≤−1 + √2n + 3, thus,\nbk∗= max\nk=0,··· ,n bk. Thus, obviously\n(n!)2\n2n bk∗≤EA(Eσ(XAn,n)) ≤n(n!)2\n2n bk∗\n□\nTheorem6 Choose An,n u.a.r. from A(n, n), and let XAn,n be the output by AMM. Then\nwhp\nEA(Eσ(X2\nAn,n))\nE2\nA(Eσ(XAn,n)) ≥n(√n/2)\nProof: Numerical experiment shows the above result.\nhowever the theoretical analysis\nseems so hard than until now I haven’t thought out the way to show the comparably tight\nfor EA(Eσ(X2\nAn,n)) since the order of\nP\ns0+s1+···+sk=n−k\n(n + 2)s0(n + 2 −1)s1 · · · (n + 2 −k)sk is\ntoo difficult to gain a good lower bound. The following bound is easy to check and the best\none among methods I thought out,\nEA(Eσ(X2\nAn,n)) ≥\nn\nX\nk=0\n(n!)2(n + 3)!\n22n\n2k(k + 2)k\n(k!)2(k + 3)!(n −k)!\nHowever it still can’t reach the goal. Therefore, the proof of this theorem will be left for the\nfuture.\nEven if Theorem6 has been proved, unfortunately, the critical ratio for almost all the\nmatrices can not obtained from this theorem since two random variables are not independent.\nIn order to accomplish the ultimate result, we need to calculate the EA(E2\nσ(X2\nAn,n)). Using\nthe induction similar to theorem4, we can obtain the recursion of EA(E2\nσ(X2\nAm,n))(recall M\nis a binomial variable with parameter n and 1\n2).\nEA(E2\nσ(X2\nAm,n)) = 2(EA(M3) + 2EA(M2) + EA(M))EA(Eσ(X2\nAm−1,n)Eσ(X2\nAm−1,n−1))\n+(EA(M2)+2EA(M)+1)EA(E2\nσ(X2\nAm−1,n))+(EA(M4)+2EA(M3)+EA(M2))EA(E2\nσ(X2\nAm−1,n−1))"},{"page":12,"text":"12\nComparing EA(E2\nσ(X2\nAm,n)) with E2\nA(Eσ(X2\nAn,n)) and computing their ratio have to be\ndone.\nOur main aim of doing this is to find the matrices satisfying Eσ(X2\nAm,n) ≤\nEA(E2\nσ(X2\nAm,n))g(n), where g(n) is a polynomial of n. However, the ratio of\nEA(E2\nσ(X2\nAm,n))\nE2\nA(Eσ(X2\nAn,n)) is\nso large that it can’t accomplish our goal. Thus we deduce our requirement whp to with\na certain probability p > 0, and in our results p = 1\n2 −ε where ε is no more than 0.02. To\nprove the theorem, we need the following lemma, which will be proved in section IV.\nLemma2 Let B(m, n) denote the set of all n × n 0-1 matrices with exact m 1’s, m ≫n.\nChoose B u.a.r. from B(m, n). Then\nE(AM(B)) =\nn\nX\nk=0\n(Ck\nn)2k!Cm−k\nn2−k\nCm\nn2\nand\nE(AM2(B))\nE2(AM(B)) = 1 + o(1), n →∞\nTheorem7 Choose An,n u.a.r. from A(n, n), and let XAn,n be the output by AMM. Then\nPr(\nEσ(X2\nAn,n)\nE2σ(XAn,n) ≥n\n√n/2−1) ≥\nn2\nP\ni=(1/2+ε)n2 Ck\nn2\n2n2\nwhere c is a constant no more 10, andε ≤0.02.\nProof: From lemma2 we know if we set m = (1/2 + ε)n2 and q =\nCm−k\nn2−k\nCm\nn2 . When n goes to\ninfinity, noting k ≤n ≪m, n2, there holds\nq = Cm−k\nn2−k\nCm\nn2\n= m(m −1) · · ·(m −k)\nn2(n2 −1) · · · (n2 −k)\nand\nln(q) =\nk−1\nX\ni=0\n[ln(m −i) −ln(n2 −i)]\n= kln( m\nn2) +\nk−1\nX\ni=0\n[ln(1 −i\nm) −ln(1 −i\nn2)]\n= kln( m\nn2) −\nk−1\nX\ni=0\n[ i\nm −i\nn2 + O( i2\nm2)]\n= kln( m\nn2) −k(k −1)\n2\n( 1\nm −1\nn2) + O( k3\nm2)"},{"page":13,"text":"13\nThus, noting that km−1 ≤2nm−1 = O(n3m−2)\nq = ( m\nn2)kexp[−k2\n2 ( 1\nm −1\nn2) + O( n3\nm2)]\n= ((1/2 + ε)n2\nn2\n)kexp[−k2\n2 (\n1\n(1/2 + ε)n2 −1\nn2) + O(\nn3\n((1/2 + ε)n2)2)]\n≤e−1(1/2 + ε)k\nLet B selected u.a.r. from B(m, n) Since E(AM2(B))\nE2(AM(B)) = 1 + o(1), as n →∞\nthen Pr(AM(B) < 5\n6E(AM(B))) →0, as n →∞. So, if m ≥(1/2 + ε)n2 and ε ≤0.02, we\nhave whp\nEσ(X2\nB) ≥E2\nσ(XB)\n= AM2(B)\n≥(5\n6E(AM(B)))2\n= (5\n6\nn\nX\nk=0\n(Ck\nn)2k!Cm−k\nn2−k\nCm\nn2\n)2\n≥(\nn\nX\nk=0\n(Ck\nn)2k!5e−1\n6 (1/2 + ε)k)2\n≥\nn\nX\nk=0\nP k\nnP k\nn+3\n2n+k\nX\ns0+s1+···+sk=n−k\ns0,···sk≥0\n(n + 2)s0(n + 2 −1)s1 · · · (n + 2 −k)sk\n= EA(Eσ(XAn,n)).\nNoting Pr(A ∈\nS\nm≥(1/2+ε)n2 B(m, n)) =\nn2\nP\ni=(1/2+ε)n2\nCk\nn2\n2n2\n,\nthus Pr(Eσ(X2\nAn,n) ≥EA(Eσ(XAn,n))) ≥\nn2\nP\ni=(1/2+ε)n2\nCk\nn2\n2n2\n.\nUsing Markov’s inequality,\nPr(Eσ(X2\nAn,n) ≥nEA(Eσ(XAn,n))) ≤1\nn →0\nthen whp Eσ(X2\nAn,n) ≤nEA(Eσ(XAn,n)). Finally, we have\nPr(\nEσ(X2\nAn,n)\nE2σ(XAn,n) ≥1\nn\nEA(Eσ(X2\nAn,n))\nEA(Eσ(XAn,n))) ≥\nn2\nP\ni=(1/2+ε)n2 Ck\nn2\n2n2"},{"page":14,"text":"14\nApply theorem6 to the above formula, we have\nPr(\nEσ(X2\nAn,n)\nE2\nσ(XAn,n) ≥n\n√n/2−1) ≥\nn2\nP\ni=(1/2+ε)n2 Ck\nn2\n2n2\nIV.\nTHE NUMBER OF ALL THE MATCHINGS ON RANDOM GRAPH.\nIn this section, we consider the expectation and variance of the number of all the\nmatchings on G selected u.a.r. from G(m, n). We have the following theorem.\nTheorem8 Choose G u.a.r.\nfrom G(m, n), where G(m, n) denotes the set of bipar-\ntite graph with #V1 = #V2 = n as its vertices and exact m edges, m ≫n, and let AM(G)\ndenotes the number of all the matchings in G. Then we have\nE(AM(G)) =\nn\nX\nk=0\n(Ck\nn)2k!E(XM(k))\nand\nE(AM2(G)) =\nn\nX\nk=0\nk\nX\ni=0\n(Ck\nn)2k!\nmin(i,n−k)\nX\np=0\nCp\nn−kCi−p\nk\nP p\nn−i+p\ni−p\nX\nj=0\nCj\ni−p[Fn−j(i−p−j)]E(XM(k+i−j))\n+\nn\nX\nk=1\nk−1\nX\ni=0\n(Ck\nn)2k!\nmin(i,n−k)\nX\np=0\nCp\nn−kCi−p\nk\nP p\nn−i+p\ni−p\nX\nj=0\nCj\ni−p[Fn−j(i −p −j)]E(XM(k+i−j))\nwhere E(XM(k)) = Cm−k\nn2−k/Cm\nn2 and Fn(p) =\npP\nr=0\n(−1)rCr\npP p−r\nn−r\nProof: we’ll use the methodology in [6]; Let M(k) be a k-matching on V1 + V2, For G ∈\nG(m, n), define the random variable XM(G) to be 1 if M(k) is contained in G, and otherwise\n0. The expectation and second moment of AM(G) is as follows.\nE(AM(G)) = E(\nn\nX\nk=0\nX\nM(k)\nXM(k)) =\nn\nX\nk=0\nX\nM(k)\nE(XM(k))\nand\nE(AM2(G)) = E((\nn\nX\nk=0\nX\nM(k)\nXM(k))2) =\nn\nX\nk=0\nn\nX\ni=0\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\nwhere ∀0 ≤k ≤n, M(k) and M\n′(k) range over all (Ck\nn)2k! k-matching’s on V1 + V2. Note\nthat\nE(XM(k)) = Cm−k\nn2−k\nCm\nn2"},{"page":15,"text":"15\nThe first equation follows quickly. For the second, in order to compute E(XM(k)X\n′\nM(i)),\nwe have to calculate the number of pairs of M(k) and M\n′(i) as a function of the overlap\nj = |M(k) T M\n′(i)|. For any fixed k, suppose i ≤k, we need to compute the number of the\npairs of M(k) and M\n′(i), where i = 0, · · · , k, and M\n′(i) ranges over all (Ci\nn)2i! i-matching’s\non V1 + V2. The problem can be equivalently stated as follows: There’re n different letters\nand n different envelopes. Among these letters, there’re exact k(0 ≤k ≤n) labeled letters,\neach of which has only one ’mother envelope’ among envelopes. Different labeled letters\nhave different mother envelopes. We call a j-fit if there’re exact j labeled letters put into\nits own mother envelope. Now choose i(0 ≤i ≤k)letters from these n letters, then put\nthem into i envelopes, and each letter can only be put into one envelope. ∀possible j,\nhow many circumstances of j-fit are there? We can solve this problem like this: Suppose\nthere’re p letters unlabeled and i −p labeled letters among the selected letters, obviously,\n0 ≤p ≤min(n −k, i), the number of ways of choosing letters is Cp\nn−kCi−p\nk\n. If the labeled\nletters has been laid, then the number of the ways of putting p unlabeled letters is P p\nn−(i−p).\nFor any j(0 ≤j ≤i−p), there’re Cj\ni−p ways putting exact j labeled letters in its own mother\nenvelope. The last one we need to deal with is how many ways to put i−p−j labeled letters\ninto n −j envelopes which contain all these i −p −j letters’ mother envelopes, satisfying\n0-fit. By the principle of inclusion-exclusion see[1], we can easily obtain the number of the\nways is Fn−j(i −p −j), where Fn(p) =\npP\nr=0\n(−1)rCr\npP p−r\nn−r. Noting that p ranges over 0 to\nmin(i, n −k), and j ranges over 0 to i −p, for each k and i ≤k. Then\nX\nM′(i)\nE(XM(k)X\n′\nM(i)) =\nmin(i,n−k)\nX\np=0\nCp\nn−kCi−p\nk\nP p\nn−i+p\ni−p\nX\nj=0\nCj\ni−p[Fn−j(i −p −j)]E(XM(k+i−j))"},{"page":16,"text":"16\nwhere E(XM(k)) = Cm−k\nn2−k/Cm\nn2 and Fn(p) =\npP\nr=0\n(−1)rCr\npP p−r\nn−r.\nConsider,\nn\nX\nk=0\nn\nX\ni=0\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i)) = (\nn\nX\nk=0\nk\nX\ni=0\n+\nn−1\nX\nk=0\nn\nX\ni=k+1\n)\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\nk\nX\ni=0\n+\nn−1\nX\nk=0\nn\nX\ni=k+1\n)\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\nk\nX\ni=0\n+\nn\nX\ni=1\ni−1\nX\nk=0\n)\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\nk\nX\ni=0\n+\nn\nX\nk=1\nk−1\nX\ni=0\n)\nX\nM(k),M′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\nk\nX\ni=0\n+\nn\nX\nk=1\nk−1\nX\ni=0\n)\nX\nM(k)\nX\nM′(i)\nE(XM(k)X\n′\nM(i))\n= (\nn\nX\nk=0\n(Ck\nn)2k!\nk\nX\ni=0\n+\nn\nX\nk=1\n(Ck\nn)2k!\nk−1\nX\ni=0\n)\nX\nM′(i)\nE(XM(k)X\n′\nM(i))\nReplace P\nM′(i)\nE(XM(k)X\n′\nM(i)) by\nmin(i,n−k)\nP\np=0\nCp\nn−kCi−p\nk\nP p\nn−i+p\ni−p\nP\nj=0\nCj\ni−p[Fn−j(i −p −j)]E(XM(k+i−j)), then the second equation is\nachieved.\n□\nRemark: To complete the proof of theorem7, we also need to know whether the ratio\nE(AM2(G))\nE2(AM(G)) goes to 1 as n goes to infinity, adding the condition such as m2n−3 →∞as n\n→∞. We guess such a result is right, however the calculus seems very difficult. And this\nresult also contributes to the upper bound of critical ratio for almost all the matrices.\nAcknowledgments\n[1] M. Hall JR. Combinatorial Theory, Blaisdell, Waltham Massachusetts (1967)."},{"page":17,"text":"17\n[2] Valiant.\nThe complexity of computing the permanent,\nTheoretical Computer Science. 8,\n189-201 (1979).\n[3] S. Chien, L. Rasmussen and A. Sinclair. Clifford Algebras and approximating the permanent,\nProceedings of the 34th Annual Symposium on Theory of Computing (STOC). 222C231 (2002).\n[4] L. E. Rasmussen. Approximating the Permanent: a Simple Approach, Random Structures\nand Algorithms. 5, 349-361 (1994).\n[5] C. Godsil and I. Gutman. On the matching polynomial of a graph, Algebraic Methods in\nGraph Theory. 241-249 (1981).\n[6] A. Frieze and M. Jerrum.\nAn Analysis of A Monte Carlo Algorithm For Estimating the\nPermanent, Combinatorica. 15(1), 67-83 (1995).\n[7] N. Karmarkar, R. Karp, R. Lipton, L. Lov ́asz, and M. Luby,. A Monte-Carlo algorithm for\nestimating the permanent, SIAM Journal on Computing. 22, 284-293 (1993).\n[8] Yan Huo, Heng Liang, Siqi Liu and Fengshan Bai. Approximating the monomer-dimer con-\nstants through matrix permanent, arXiv:0708.1641v2:cond-mat.stat-mech . (2007).\n[9] A. Barvinok. Polynomial time algorithms to approximate permanents and mixed discriminants\nwithin a simply exponential factor, Random Struct. Algorithms. 14, 29-61 (1999).\n[10] H. Chernoff.\nA measure ofasymptotic efficiencyfor tests ofa hypothesis based on the sum\nofobservations, Ann. Math. Stat. 23, 493-509 (1952).\n[11] A. Barvinok. New permanent estimators via non-commutative determinants, Preprint. (2000).\n[12] N. Linial, A. Samorodnitsky and A. Wigderson. A deterministic strongly polynomial algorithm\nfor matrix scaling and approximate permanents, Combinatorica. 20, 545-568 (2000).\n[13] M. Jerrum and A. Sinclair. Approximating the permanent,, SIAM Journal on Computing.\n18, 1149-1178 (1989).\n[14] I.Beichl and F. Sullivan. Approximating the Permanent via Importance Sampling with Ap-\nplication to the Dimer Covering Problem, Journal of Computational Physics. 149, 128-147\n(1999).\n[15] Shmuel Friedland and Daniel Levy. A polynomial-time approximation algorithm for the num-\nber of k-matchings in bipartite graphs, arXiv:cs.CC/0607135 v1 . 392-401 (2006).\n[16] M. Jerrum, A. Sinclair and E. Vigoda. A polynomial-time approximation algorithm for the\npermanent of a matrix with non-negative entries, Proceedings of the 33rd ACM Symposium\non Theory of Computing. 712C721 (2001)."},{"page":18,"text":"18\n[17] A. Frieze and S. Suen. Counting the number of hamilton cycles in random digraphs, Random\nStructrues and Algorithms.(1992).\n[18] B. Bollob ́as. Random Graph, Cambridge University Press. Second Edition (2001).\n[19] C. Kenyon, D. Randall, and A. Sinclair. Approximating the number of dimer coverings of a\nlattice, Journal of Statistical Physics. 83, 637-659 (1996).\n[20] George E. Andrews, Richard Askey and Ranjan Roy. Special Function, Cambridge University\nPress. (1999)."}]},"section_tree":[],"assets":{"figures":[],"tables":[],"images":[]},"math_expressions":{"inline":[],"block":[{"equation_id":"eq1","equation_number":null,"raw_text":"Let G = (V, E) be a bipartite graph, where V = V1 ∪V2 is the set of vertices and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq2","equation_number":null,"raw_text":"E ⊂V1 × V2 is the set of edges. In the following sections we suppose #V1 = #V2 = n if","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq3","equation_number":null,"raw_text":"edges e1, e2 ∈S contain a common vertex. S is called a k-matching if #S = k. In special","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq4","equation_number":null,"raw_text":"case, S is called a perfect matching if k = n. Let Sk be the set of k-matching in G and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq5","equation_number":null,"raw_text":"A(G) be the set of all the k-matching, k = 0, 1, . . . , n. For the convenience of discussion, let","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq6","equation_number":null,"raw_text":"#S0 = 1, then the number of all the matchings in G is #A(G) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq7","equation_number":null,"raw_text":"i=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq8","equation_number":null,"raw_text":"The permanent of a 0-1 A = aij, 1 ≤i, j ≤n is defined as","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq9","equation_number":null,"raw_text":"Per(A) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq10","equation_number":null,"raw_text":"i=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq11","equation_number":null,"raw_text":"where the sum is over all the permutations π of [n] = {1, . . . , n}. It’s well known that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq12","equation_number":null,"raw_text":"AM(G) = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq13","equation_number":null,"raw_text":"Pr(critical ratio of A is bounded by a polynomial of the input A )= 1 −o(1) as n −→∞","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq14","equation_number":null,"raw_text":"relating to n satisfies Pr(P is true)= 1 −o(n), we say P holds whp(whp is the abbreviation","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq15","equation_number":null,"raw_text":"selected u.a.r from G(m, n)(G(m, n) denotes the set of bipartite graph with #V1 = #V2 = n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq16","equation_number":null,"raw_text":"G = (V, E), (V = V1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq17","equation_number":null,"raw_text":"if n=0; then","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq18","equation_number":null,"raw_text":"YA = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq19","equation_number":null,"raw_text":"W = {j : a1j = 1}","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq20","equation_number":null,"raw_text":"if W = ∅then","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq21","equation_number":null,"raw_text":"YA = 0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq22","equation_number":null,"raw_text":"YA = |W|Y1J","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq23","equation_number":null,"raw_text":"if n=0; then","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq24","equation_number":null,"raw_text":"YA = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq25","equation_number":null,"raw_text":"W = {j : a1j = 1} S{0}","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq26","equation_number":null,"raw_text":"YA = |W|Y1J","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq27","equation_number":null,"raw_text":"matrix). Define a new terminology AM on the matrix. let B = {bij, 1 ≤i ≤m, 1 ≤j ≤n}","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq28","equation_number":null,"raw_text":"an m × n matrix, m ≤n. let AM(∅) = 1, by induction on m.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq29","equation_number":null,"raw_text":"AM(B) := AM(B10) +","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq30","equation_number":null,"raw_text":"j=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq31","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq32","equation_number":null,"raw_text":"clearly to see AM(A) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq33","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq34","equation_number":null,"raw_text":"Let A = {aij1 ≤i, j ≤n} be an n × n 0-1 matrix and YA is obtained","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq35","equation_number":null,"raw_text":"by above AMM. Then YA is unbiased for AM(A), E(YA) = AM(A)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq36","equation_number":null,"raw_text":"on m, k=0,∀1 ≤l ≤n, ∀a k × l 0-1 matrix A, the equation E(YA) = AM(A) is trivial.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq37","equation_number":null,"raw_text":"Now suppose ∀k ≤m, k ≤l ≤n, a k × l 0-1 matrix A has E(YA) = AM(A). Then when","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq38","equation_number":null,"raw_text":"k = m, let |W| = q, we have","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq39","equation_number":null,"raw_text":"E(YA) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq40","equation_number":null,"raw_text":"E(YA|J = j)Pr(J = j)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq41","equation_number":null,"raw_text":"E(qYA1j|J = j)q−1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq42","equation_number":null,"raw_text":"= AM(A).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq43","equation_number":null,"raw_text":"matching in G, by RM operating on B =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq44","equation_number":null,"raw_text":"Proof: Note that by RM after n-th step operating on B, YB = Sn ∗Y1n×n/n!, where Sn is a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq45","equation_number":null,"raw_text":"number obtained from the first n steps, and obviously Y1n×n ≡n!. Hence, we have YB = Sn.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq46","equation_number":null,"raw_text":"Corollary3. AM(A) = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq47","equation_number":null,"raw_text":"AM(A) = E(XA) = E(YB) = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq48","equation_number":null,"raw_text":"Let A = {aij, 1 ≤i, j ≤n} be an n × n adjacent matrix of a bipartite","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq49","equation_number":null,"raw_text":"Proof: Induction on m, For any fixed n. k = 0,∀1 ≤l ≤n, ∀a k × l 0-1 matrix A, the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq50","equation_number":null,"raw_text":"inequation is trivial. In the case k = m, let |W| = q, we have","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq51","equation_number":null,"raw_text":"A) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq52","equation_number":null,"raw_text":"A|J = j)Pr(J = j)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq53","equation_number":null,"raw_text":"A1j|J = j)q−1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq54","equation_number":null,"raw_text":"= E(XA)2(n + 1)m","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq55","equation_number":null,"raw_text":"Lemma1 Let f(m, n) defined as f(m, n) = anf(m−1, n) + cnf(m−1, n−1), where m ≤n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq56","equation_number":null,"raw_text":"0 ≤l ≤n, f(0, l) = 1. Then","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq57","equation_number":null,"raw_text":"f(m, n) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq58","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq59","equation_number":null,"raw_text":"s0+s1+···+sk=m−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq60","equation_number":null,"raw_text":"Proof: By induction on m. Obviously, the case p=0 is trivial. Suppose when p ≤m −1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq61","equation_number":null,"raw_text":"∀p ≤l ≤n, f(p, l) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq62","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq63","equation_number":null,"raw_text":"s0+s1+···+sk=p−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq64","equation_number":null,"raw_text":"l holds, then when p=m,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq65","equation_number":null,"raw_text":"anf(m −1, n) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq66","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq67","equation_number":null,"raw_text":"s0+s1+···+sk=m−1−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq68","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq69","equation_number":null,"raw_text":"s0+s1+···+sk=m−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq70","equation_number":null,"raw_text":"cnf(m −1, n −1) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq71","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq72","equation_number":null,"raw_text":"s0+s1+···+sk=m−1−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq73","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq74","equation_number":null,"raw_text":"s1+s2+···+sk+1=m−1−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq75","equation_number":null,"raw_text":"k=2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq76","equation_number":null,"raw_text":"s1+s2+···+sk=m−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq77","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq78","equation_number":null,"raw_text":"s1+s2+···+sk=m−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq79","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq80","equation_number":null,"raw_text":"s0+s1+···+sk=m−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq81","equation_number":null,"raw_text":"s0=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq82","equation_number":null,"raw_text":"f(m, n) = anf(m −1, n) + cnf(m −1, n −1)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq83","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq84","equation_number":null,"raw_text":"s0+s1+···+sk=m−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq85","equation_number":null,"raw_text":"EA(AM(Am,n)) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq86","equation_number":null,"raw_text":"k=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq87","equation_number":null,"raw_text":"m =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq88","equation_number":null,"raw_text":"n =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq89","equation_number":null,"raw_text":"Proof: Induction on m. The case p=0, EA(AM(A)) = 1 is trivial. Suppose ∀p ≤m −1,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq90","equation_number":null,"raw_text":"p ≤l ≤n EA(AM(Ap,l)) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq91","equation_number":null,"raw_text":"k=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq92","equation_number":null,"raw_text":"2p−k =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq93","equation_number":null,"raw_text":"k=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq94","equation_number":null,"raw_text":"when p=m, ∀m ≤l ≤n, we have","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq95","equation_number":null,"raw_text":"EA(AM(Am,l)) = EA(AM(A1,0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq96","equation_number":null,"raw_text":"j=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq97","equation_number":null,"raw_text":"= EA(AM(Am−1,l)) +","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq98","equation_number":null,"raw_text":"j=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq99","equation_number":null,"raw_text":"= EA(AM(Am−1,l)) + n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq100","equation_number":null,"raw_text":"Using lemma1, here al ≡1, andcl = l","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq101","equation_number":null,"raw_text":"EA(AM(Am,l)) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq102","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq103","equation_number":null,"raw_text":"s0+s1+···+sk=m−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq104","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq105","equation_number":null,"raw_text":"s0+s1+···+sk=m−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq106","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq107","equation_number":null,"raw_text":"k=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq108","equation_number":null,"raw_text":"EA(Eσ(XAm,n)) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq109","equation_number":null,"raw_text":"k=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq110","equation_number":null,"raw_text":"Am,n)) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq111","equation_number":null,"raw_text":"k=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq112","equation_number":null,"raw_text":"s0+s1+···+sk=m−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq113","equation_number":null,"raw_text":"Am,n) = AM(Am,l). For the second one,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq114","equation_number":null,"raw_text":"we use induction on m. The case p=0 is obvious. Suppose ∀Ap,l where 0 ≤p ≤m −1,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq115","equation_number":null,"raw_text":"p ≤l ≤n the second equation holds. When p = m, noting the fact M = |W| −1 is a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq116","equation_number":null,"raw_text":"Am,l)) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq117","equation_number":null,"raw_text":"q=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq118","equation_number":null,"raw_text":"Am,l)|M = q)PrA(M = q)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq119","equation_number":null,"raw_text":"q=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq120","equation_number":null,"raw_text":"m,l)|M = q)PrA(M = q)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq121","equation_number":null,"raw_text":"q=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq122","equation_number":null,"raw_text":"Am−1,l−1))PrA(M = q)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq123","equation_number":null,"raw_text":"= (EA(M) + 1)EA(Eσ(X2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq124","equation_number":null,"raw_text":"= (l + 2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq125","equation_number":null,"raw_text":"Using lemma1, here al = l+2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq126","equation_number":null,"raw_text":"2 , and cl = l2+3l","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq127","equation_number":null,"raw_text":"Am,l)) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq128","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq129","equation_number":null,"raw_text":"s0+s1+···+sk=m−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq130","equation_number":null,"raw_text":"k=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq131","equation_number":null,"raw_text":"s0+s1+···+sk=m−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq132","equation_number":null,"raw_text":"whp h(n) ≤EA(Eσ(XAn,n)) ≤nh(n), where h(n) = (n!)2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq133","equation_number":null,"raw_text":"(n−k∗)!(k∗)2, k∗= ⌊−1 + √2n + 3⌋.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq134","equation_number":null,"raw_text":"EA(Eσ(XAn,n)) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq135","equation_number":null,"raw_text":"k=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq136","equation_number":null,"raw_text":"k=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq137","equation_number":null,"raw_text":"= (n!)2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq138","equation_number":null,"raw_text":"k=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq139","equation_number":null,"raw_text":"and let bk =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq140","equation_number":null,"raw_text":"bk−1 = 2(n−k+1)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq141","equation_number":null,"raw_text":"bk∗= max","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq142","equation_number":null,"raw_text":"k=0,··· ,n bk. Thus, obviously","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq143","equation_number":null,"raw_text":"s0+s1+···+sk=n−k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq144","equation_number":null,"raw_text":"k=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq145","equation_number":null,"raw_text":"Am,n)) = 2(EA(M3) + 2EA(M2) + EA(M))EA(Eσ(X2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq146","equation_number":null,"raw_text":"a certain probability p > 0, and in our results p = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq147","equation_number":null,"raw_text":"E(AM(B)) 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