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1 Perfect Matchings in Bipartite Graphs An undirected graph G=(U V,E) is bipartite if U V= and E U V. A 1-1 and onto function f:U V is a perfect matching if for any u U, (u,f(u)) E.

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2 Finding Perfect Matchings is Easy Matching as a flow problem

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3 What About Counting Them? § Let A=(a(i,j)) 1 i,j n be the adjacency matrix of a bipartite graph G=({u 1,...,u n } {v 1,...,v n },E), i.e. - permanent sum over the permutations of {1,...,n} §The number of perfect matchings in the graph is

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4 Cycle-Covers Given an undirected bipartite graph G=({u 1,...,u n } {v 1,...,v n },E), the corresponding directed graph is G’=({w 1,...,w n },E), where (w i,w j ) E iff (u i,v j ) E. Definition: Given a directed graph G=(V,E), a set of node-disjoint cycles that together cover V is called a cycle-cover of G. Observation: Every perfect matching in G corresponds to a cycle-cover in G’ and vice- versa.

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5 Three Ways To View Our Problem 1) Counting the number of Perfect Matchings in a bipartite graph. 2) Computing the Permanent of a 0-1 matrix. 3) Counting the number of Cycle-Covers in a directed graph.

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6 #P - A Complexity Class of Counting Problems L NP iff there is a polynomial time decidable binary relation R, s.t. f #P iff f(x)=| { y | R(x,y) } | where R is a relation associated with some NP problem. We say a #P function is #P-Complete, if every #P function Cook-reduces to it. It is well known that #SAT (i.e - counting the number of satisfying assignments) is #P- Complete. some polynomial

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7 On the Hardness of Computing the Permanent Claim [Val79]: Counting the number of cycle- covers in a directed graph is #P-Complete. Proof: By a reduction from #SAT to a generalization of the problem.

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8 The Generalization: Integer Permanent 2 3 1 2 §Activity: an integer weight attached to each edge (u,v) E, denoted (u,v). §The activity of a matching M is (M)= (u,v) M (u,v). §The activity of a set of matchings S is (M)= M S (M). §The goal is to compute the total activity. 2 2 3 1

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9 Integer Permanent Reduces to 0-1 Permanent 2 the rest of the graph 1 1 We would have loved to do something of this sort...

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10 Integer Permanent Reduces to 0-1 Permanent the rest of the graph So instead we do:

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11 But this is really cheating! The integers may be exponentially large, but we are forbidden to add an exponential number of nodes!

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12 The Solution the rest of the graph...

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13 What About Negative Numbers? § W.l.og, let us assume the only negative numbers are -1’s. § We can reduce the problem to calculating the Permanent modulo (big enough) N of a 0-1 matrix by replacing each -1 with (N-1). § Obviously, Perm mod N is efficiently reducible to calculating the Permanent.

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14 Continuing With The Hardness Proof § We showed that computing the permanent of an integer matrix reduces to computing the permanent of a 0-1 matrix. § It remains to prove the reduction from #SAT to integer Permanent. § We start by presenting a few gadgets.

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15 The Choice Gadget Observation: in any cycle-cover the two nodes must be covered by either the left cycle (true) or the right cycle (false). x= truex= false

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16 The Clause Gadget Observation: §no cycle-cover of this graph contains all three external edges. §However, for every proper subset of the external edges, there is exactly one cycle- cover containing it. each external edge corresponds to one literal

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17 The Exclusive-Or Gadget § The Perm. of the whole matrix is 0. § The Perm. of the matrix resulting if we delete the first (last) row and column is 0. § The Perm. of the matrix resulting if we delete the first (last) row and the last (first) column is 4. 2 3

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18 Plugging in the XOR-Gadget Observe a cycle-cover of the graph with a XOR-gadget plugged as in the below figure. §If e is traversed but not t (or vice versa), the Perm. is multiplied by 4. § Otherwise, the Perm. is added 0. e t

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19 Putting It All Together § One choice gadget for every variable. § One Clause gadget for every clause. x= truex= false if the literal is x x= truex= false if the literal is x

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20 Sum Up § Though finding a perfect matching in a bipartite graph can be done in polynomial time, §counting the number of perfect matchings is #P-Complete, and hence believed to be impossible in polynomial time.

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