Perfect Matchings in Bipartite Graphs

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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 fU?V is a perfect
matching if for any u?U, (u,f(u))?E.
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Finding Perfect Matchings is Easy
Matching as a flow problem
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What About Counting Them?

• Let A(a(i,j))1?i,j?n be the adjacency matrix of
  a bipartite graph G(u1,...,un?v1,...,vn,E),
  i.e. -

• The number of perfect matchings in the graph is

permanent
sum over the permutations of 1,...,n
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Cycle-Covers

• Given an undirected bipartite graph
  G(u1,...,un?v1,...,vn,E), the corresponding
  directed graph is G(w1,...,wn,E), where
  (wi,wj)?E iff (ui,vj)?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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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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P - A Complexity Class of Counting Problems

• L?NP iff there is a polynomial time decidable
  binary relation R, s.t.

some polynomial

• 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.

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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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The Generalization Integer Permanent
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.

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1
2
2
3
2
1
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Integer Permanent Reduces to 0-1 Permanent
the rest of the graph
We would have loved to do something of this
sort...
1
2
1
10
Integer Permanent Reduces to 0-1 Permanent
the rest of the graph
So instead we do
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• 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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The Solution
the rest of the graph
...
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What About Negative Numbers?

• W.l.og, let us assume the only negative numbers
  are -1s.
• 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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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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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).

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The Clause Gadget
each external edge corresponds to one literal

• 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.

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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.

-1
-1
2
3
-1
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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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Putting It All Together

• One choice gadget for every variable.
• One Clause gadget for every clause.

if the literal is x
if the literal is ?x
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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.