Approximating The Permanent

1
Approximating The Permanent
Seminar in Complexity 04/06/2001

• Amit Kagan

2
Topics

• Description of the Markov chain
• Analysis of its mixing time

3
Definitions

• Let G (V1, V2, E) be a bipartite graph on nn
  vertices.
• Let M denote the set of perfect matchings in G.
• Let M(y, z) denote the set of near-perfect
  matchings with holes only at y and z.

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M(u,v)/M Exponentially Large

• Observe the following bipartite graph

u
v
It has only one perfect matching...
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M(u,v)/M Exponentially Large

• But two near-perfect matchings with holes at u
  and v.

u
v
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M(u,v)/M Exponentially Large

• Concatenating another hexagon,
• adds a constant number of vertices,
• but doubles the number of near-perfect matchings,
• while the number of perfect matchings remains 1.

. . .
Thus we can force the ratio M(u,v)/M to be
exponentially large.
7
The Breakthrough

• Jerrum, Sinclair, and Vigoda 2000 introduced an
  additional weight factor.
• Any hole pattern (including that with no holes)
  is equally likely in the stationary distribution
  p.
• p will assign O(1/n2) weight to perfect matchings.

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Edge Weights

• For each edge (y, z) ? E, we introduce a positive
  weight ?(y, z).
• For a matching M, ?(M) ?(i, j)?M?(i, j).
• For a set of matchings S, ?(S) ?M?S?(M).
• We will work with the complete graph on nn
  vertices
• ?(e) 1 for all e ? E
• ?(e) ? 0 for all e ? E

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The Stationary Distribution

• The desired distribution p over O is ?(M) ? ?(M),
  where

w V1 V2 ? ? is the weight function, to be
specified shortly
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The Markov Chain

• Choose an edge e(u,v) uniformly at random.

(i) If M ? M and e ? M, let M M\e, (ii)
if M ? M(u,v), let M M?e,
(iii) if M ? M(u,z) where z ? v, and (y,v) ? M,
let M M?e\(y,v), (iv) if M ? M(y,v)
where y ? u, and (u,z) ? M, let M
M?e\(u,z).
Metropolis rule
3. With probability min1,?(M)/?(M) go to
M otherwise, stay at M.
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The Markov Chain (cont.)

• Finally, we add a self-loop probability of ½ to
  every state.
• This insures the MC is aperiodic.
• We also have irreducibility.

What about the stationary distribution?
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Detailed Balance
P(M,M) gt 0

• Consider two adjacent matchings M and M with
  ?(M) ?(M).

? ?(M)P(M, M) ?(M)P(M, M)
Q(M,M)
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The Ideal Weight

• Recall that ?(M) ? ?(M), where
• Ideally, we would take w w, where

?(M)
?(M)w(u,v)
?(M(u,v))
?(M)
?(M)
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The Concession

• We will content ourselves with weights w
  satisfying

• This perturbation will reduce the relative weight
  of perfect and near-perfect matchings by at most
  a constant factor (4).

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The Mixing Time Theorem

• Assuming the weight function w satisfies the
  above inequality for all (y,z) ? V1 V2 , then
  the mixing time of the MC is bounded above by
  ?(?) O(m6n8(n logn log?-1)), provided the
  initial state is a perfect matching of maximum
  activity.

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Edge Weights Revisited

• We will work with the complete graph on nn
  vetices.
• Think of non-edges e ? E as having a very small
  activity of 1/n!.
• The combined weight of all invalid matchings is
  at most 1.
• We begin with activities ? whose ideal weights w
  are easy to compute, and progress towards our
  target activities.

?(e) 1/n! for all e ? E ?(e) 1/n! for all
e ? E
? 1
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Step I

• We assume at the beginning of the phase w(u,v)
  approximates w(u,v) within ratio 2 for all
  (u,v).
• Before updating an activity, we will find for
  each (u,v) a better approximation, one that is
  within ratio c for some 1 lt c lt 2.
• For this purpose we use the identity

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Step I (cont.)

• The mixing time theorem allows us to sample, in
  polynomial time, from a distribution ? that is
  within variation distance ? of p.
• We choose ? c1/n2, take O(n2 log ?-1) samples
  from ?, and use sample averages.
• Using a few Chernoff bounds, we have, with
  probability 1- (n21)?, approximation within
  ratio c to all of w(u,v).

c1 gt 0 is a sufficiently small constant
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Step I (conclusion)

• Taking c 6/5 and using O(n2 log ?-1) samples,
  we obtain refined estimates w(u,v) satisfying
• 5w(u,v)/6 w(u,v) 6w(u,v)/5

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Step II

• We update the activity of an edge e
• ?(e) ? ?(e) exp(-1/2)
• The ideal weight function w changes by at most a
  factor of exp(1/2).
• Since 6exp(1/2)/5 lt 2, our estimates w after step
  I approximate w within ratio 2 for the new
  activities.

1.978
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Step II (cont.)
?(e) 1/n! for all e ? E ?(e) 1/n! for all e
? E
? 1

• We use the above procedure repeatedly to reduce
  the initial activities to the target activities.

• This requires O(n2 n log n) phases.
• Each phase requires O(n2 log ?-1) samples.
• Each sample requires O(n21 log n) simulation
  steps (mixing time theorem).
• Overall time - O(n26 log2 n log ?-1)

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The ? Error

• We need to set ? so that the overall failure
  probability is strictly less than ?, say ?/2.
• The probability that any phase fails is at most
  O(n3 log n n2?).
• We will take ? c2? / n5 log n .

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Time Complexity
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Conductance

• The conductance of a reversible MC is defined as
  ?min??S???(S), where
• Theorem
• For an ergodic, reversible Markov chain with
  self- loops probabilities P(y,y) ? ½ for all
  states x??,

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Canonical Paths

• We define canonical paths ?I,F from all I ? O to
  all F ? M.
• Denote G ?I,F (I, F) ? O M.
• Certain transitions on a canonical path will be
  deemed chargeable.
• For each transition t denote cp(t) (I, F)
  ?I,F contains t as a chargeable
  transition

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I ? F

• If I ? M, then I ? F consists of a collection of
  alternating cycles.
• If I ? M(y,z), then I ? F consists of a
  collection of alternating cycles together with a
  single alternating path from y to z.

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Type A Path
We assume w.l.g. that the edge (v0, v1) belongs
to I

• Assume I ? M.
• A cycle v0 ? v1 ? ? v2k v0 is unwound by

(i) removing the edge (v0, v1), (ii)
successively, for each 1 i k 1, exchanging
the edge (v2i, v2i1) with (v2i-1, v2i), (iii)
adding the edge (v2k-1, v2k).

• All these transitions are deemed chargeable.

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Type A Path Illustrated
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Type B Path

• Assume I ? M(y,z).
• The alternating path y v0 ? ? v2k1 z is
  unwound by
• (i) successively, for each 1 i k,
  exchanging the edge (v2i-1, v2i) with (v2i-2,
  v2i-1), and
• (ii) adding the edge (v2k, v2k1).
• Here, only the above transitions are deemed
  chargeable.

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Type B Path Illustrated
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Congestion

• We define a notion of congestion of G
• Lemma I
• Assuming the weight w approximates w within
  ratio 2, then t(G) 16m.

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Lemma II

• Let u,y ? V1, v,z ? V2. Then,
• (i) ?(u,v)?(M(u,v)) ?(M), for all vertices
  u,v with u ? v.
• (ii) ?(u,v)?(M(u,z))?(M(y,v)) ?(M)?(M(y,z)),
  for all distinct vertices u,v,y,z with u ? v.
• Observe that Mu,z ? My,v ? (u,v) decomposes
  into a collection of cycles together with an
  odd-length path O joining y and z.

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Corollary III

• Let u,y ? V1, v,z ? V2. Then,
• (i) w(u,v) ?(u,v), for all vertices u,v
  with u ? v.
• (ii) w(u,z)w(y,v) ?(u,v)w(y,z), for all
  distinct vertices u,v,y,z with u ? v.
• (iii) w(u,z)w(y,v) ?(u,v) ?(y,z), for all
  distinct vertices u,v,y,z with u ? v and y ? z.

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Proof of Lemma I

• For any transition t (M,M) and any pair of
  states I, F ? cp(t), we will define an encoding
  ?t(I,F) ? O such that ?t cp(t) ? O is an
  injection, and
• p(I)p(F) 8 minp(M), p(M)p(?t(I,F))
• 16m Q(t)p(?t(I,F))
• Summing over I,F ? cp(t), we get

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The Injection ?t

• For a transition t (M,M) which is involved in
  stage (ii) of unwinding a cycle, the encoding
  is ?t(I,F) I ? F ? (M ? M) \
  (v0, v1).
• Otherwise, the encoding is ?t(I,F) I
  ? F ? (M ? M).

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From Congestion to Conductance

• Corollary IV Assuming the weight
  function w approximates w within ratio 2 for all
  (y,z) ? V1 V2 , then ? 1/100t3n4 1/106m3n4.
• Proof
• Set a 1/10tn2 .
• Let (S,S) be a partition of the state-space.

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Case I

• p(S ? M) / p(S) a and p(S ? M) / p(S) a.
• Just looking at canonical paths of type A we have
  a total flow of p(S ? M)p(S ? M) a2p(S)p(S)
  across the cut.
• Thus, tQ(S,S) a2p(S)p(S), and,
• ?(S) Q(S,S)/p(S)p(S) a2 /t 1/100t3n4.

1/10tn2
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Case II

• Otherwise, p(S ? M) / p(S) lt a .
• Note the following estimates
• p(M) 1/4(n21) 1/5n2
• p(S ? M) lt ap(S) lt a
• p(S \ M) p(S) p(S ? M) gt (1 a)p(S)
• Q(S \ M, S ? M) p(S ? M) lt ap(M)

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Case II (cont.)

• Consider the cut (S \ M, S ? M).
• The weight of canonical paths (all chargeable as
  they cross the cut) is p(S \ M)p(M) (1
  a)p(S)/5n2 p(S)/6n2.

1/10tn2

• Hence, tQ(S \ M,S ? M) p(S)/6n2.
• Q(S,S) p(S)p(S)/15tn2.
• ?(S) Q(S,S)/p(S)p(S) 1/15tn2.

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Summing It Up

• Starting from an initial state X0 of maximum
  activity guarantees p(X0) 1/n!, and hence,
  log(p(X0)-1) O(n log n).
• We showed ?(S) 1/100t3n4, and hence, ?(S)-1
  O(t3n4) O(m3n4).
• Thus, according to the conductance theorem,
  ?x0(?) O(m6n8(n logn log?-1)).