counting perfect matchings in planar graphs

1FKT Algorithm by PlusOne

Counting perfect matchings in planar graphsWith the FKT algorithm

Perfect Matching

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FKT Algorithm by PlusOne

Given a graph G = (V,E), a matching M in G is a set of pairwise non-adjacent edges. M is said to be perfect if every vertex of G is included in M.

Def


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Def Cycle Cover


A set of disjoint cycles in G that contain all vertices of G.

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Def Cycle Cover


A cycle cover that satisfies…• All cycles have even number of vertices.• Every cycle has a direction (clockwise or counter-clockwise).

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Non-even Directed Cycle Cover

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Even Directed Cycle Cover

DefEven Directed Cycle Cover (EDCC)


M1

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M2

ObsEvery ordered pair of perfect matchings can be uniquely mapped to an EDCC.

<M1, M2>

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ObsEvery ordered pair of perfect matchings can be uniquely mapped to an EDCC.

It’s easy to see that every EDCC can also be decomposed into an ordered pair of perfect matchings.

<M1, M2>

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Counting Perfect Matchings


|𝑃𝑀|2 = |𝐸𝐷𝐶𝐶|

The number of perfect matchings

The number of even directed cycle

covers

So counting perfect matchings can be converted into counting EDCCs.


𝒊 1 2 3 4 5 6 7 8

𝜋(𝑖) 6 3 4 5 2 7 8 1

Recall Discrete Mathematics II and Linear Algebra.

Tip Permutations

The parity of a permutation = the parity of # swaps from the original sequence

sgn(𝜋): sign of a permutation +1 (𝑒𝑣𝑒𝑛 𝑝𝑒𝑟𝑚𝑢𝑡𝑎𝑡𝑖𝑜𝑛)−1 (𝑜𝑑𝑑 𝑝𝑒𝑟𝑚𝑢𝑡𝑎𝑡𝑖𝑜𝑛)


A: adjacency matrix 𝐴𝑖𝑗 =

+1 (if edge i → j)−1 (if edge j → i)

0 (if no edge between i, j)

sgn(𝜋): sign of a permutation +1 (偶排列)

−1 (奇排列)val 𝜋 ∶ 𝐴𝑖,𝜋(𝑖)

Def Determinant of Oriented GraphIs the determinant of the adjacency matrix!

𝐷𝑒𝑡 𝐴 = 𝜋𝑆𝑔𝑛 𝜋 ∙ 𝑉𝑎𝑙(𝜋)

By definition,


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𝒊 1 2 3 4 5 6 7 8

𝜋(𝑖) 6 3 4 5 2 7 8 1

ObsEvery Directed Cycle Cover (DCC) corresponds to a permutation.

𝑖, 𝜋 𝑖 ↔ 𝑒𝑑𝑔𝑒 < 𝑖, 𝜋 𝑖 >


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𝒊 1 2 3 4 5 6 7 8

𝜋(𝑖) 6 3 4 5 2 7 8 1

ObsFor a permutation 𝜋, if ∀ 𝑖, 𝜋 𝑖 ∈ 𝐸, we can map 𝜋 to a DCC.


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𝒊 1 2 3 4 5 6 7 8

𝜋(𝑖) 7 3 4 5 2 1 8 6


ObsFor a permutation 𝜋, if ∀ 𝑖, 𝜋 𝑖 ∈ 𝐸, we can map 𝜋 to a DCC.


The inverse proposition : for a permutation 𝜋 , if it cannot be mapped to a DCC, then ∃ 𝑖, 𝜋 𝑖 ∉ 𝐸.

⇒ 𝐴𝑖,𝜋 𝑖 = 0

⇒ val 𝜋 = 0

Det A =

𝜋

𝑆𝑔𝑛 𝜋 𝑉𝑎𝑙(𝜋) =

𝐷𝐶𝐶

𝑆𝑔𝑛 𝜋 𝑉𝑎𝑙(𝜋)

Non-even DCCs will be cancelled out.


You can always find another DCC with an opposite direction for the odd cycle.

𝜋

𝜋′

𝑠𝑔𝑛 𝜋 = 𝑠𝑔𝑛(𝜋′)

Remember that 𝐴𝑖𝑗 = −𝐴𝑗𝑖

So 𝑉𝑎𝑙 𝜋 = −1 𝑘 ∙ 𝑉𝑎𝑙 𝜋′

(k is the number of edges in the cycle)

Therefore: 𝑠𝑔𝑛 𝜋 𝑣𝑎𝑙 𝜋 + 𝑠𝑔𝑛 𝜋 𝑣𝑎𝑙 𝜋′ = 0

Obs

Only EDCCs contribute to Det(A)


Det A

=

𝜋


Our goal: Find an orientation that makes all 𝑠𝑔𝑛 𝜋 𝑣𝑎𝑙 𝜋 = 1(then 𝑃𝑀2 = 𝐸𝐷𝐶𝐶 = 𝐷𝑒𝑡(𝐴))

Permutations that are not DCC, val(𝜋)=0

This depends on the orientation!

≤ |𝐸𝐷𝐶𝐶|This is equal when all

𝑠𝑔𝑛(𝜋)𝑣𝑎𝑙(𝜋)=1

=

𝐷𝐶𝐶


=

𝐸𝐷𝐶𝐶

𝑆𝑔𝑛 𝜋 𝑉𝑎𝑙(𝜋) +

𝑁𝑜𝑡 𝑒𝑣𝑒𝑛 𝐷𝐶𝐶


=

𝐸𝐷𝐶𝐶



Alg FKT Algorithm: Finding an Orientation

Planar Graph: a graph that can be drawn in the 2D plane so that its edges intersect only at its vertices.

Not Planar Planar

FKT Algorithm by PlusOne


Planar Graph: a graph that can be drawn in the 2D plane so that its edges intersect only at its vertices.

Face: a planar graph can be seen as a mesh of faces.

Face!

1. Find a spanning tree for G.Call this tree T1.



1. Find a spanning tree for G.Call this tree T1.

2. Orient the edges arbitrarily.

3. Construct a second tree T2, whose vertices are the faces of T1.Put an edge between faces that share an edge that is not in T1.4. Starting with the leaves of T2, orient these edges of G such that each face has an odd number of edges oriented clockwise. 19FKT Algorithm by PlusOne



Euler’s Formula

For any cycle C, e = v + f − 1, where e is the number of edges inside C, v is the number of vertices inside C, and f is the number of faces inside C.

𝑣 = 4

𝑓 = 5

Lemma

If all faces have an odd number of clockwise edges then all cycles in an EDCC are oddly oriented.

Cycle 𝐶, 𝑒 = 8.

𝑘𝑖: the number of clockwise lines on the boundary of face.

𝑘: the number of clockwise lines on Cycle (C).

Proof• Let 𝐶 be a nice cycle, let 𝑐𝑖 be the number of clockwise lines on

the boundary of face i in 𝐶, and 𝑐 be the number of clockwise lines on 𝐶.

• We oriented each face to have an odd number of clockwise lines,

so 𝑐𝑖 ≡ 1𝑚𝑜𝑑 2, so 𝑓 ≡ 𝑖=1𝑓

𝑐𝑖 𝑚𝑜𝑑 2.

• But also 𝑖=1𝑓

𝑐𝑖 = 𝑐 + 𝑒 (each interior line is counted as clockwise once).

• So 𝑓 ≡ 𝑐 + (𝑣 + 𝑓 − 1) 𝑚𝑜𝑑 2, so 𝑐 ≡ (𝑣 − 1) 𝑚𝑜𝑑2.

• But 𝑣 ≡ 0 𝑚𝑜𝑑 2 ,as C is a nice cycle.

• So 𝐶, and hence every nice cycle, is oddly oriented.


Euler’s Formula

For any cycle C, e = v + f − 1, where e is the number of edges inside C, v is the number of vertices inside C, and f is the number of faces inside C.

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= ((−1) × (−1))𝑘

= 1


𝑠𝑔𝑛 𝜋 𝑣𝑎𝑙 𝜋

Lemma

If all cycles in an EDCC are oddly oriented then 𝑠𝑔𝑛 𝜋 𝑣𝑎𝑙 𝜋 = 1 for all corresponding 𝜋.

Even cycle

sgn(𝜋): sign of a permutation +1 (偶排列)

−1 (奇排列)

val 𝜋 ∶ 𝐴𝑖,𝜋(𝑖)

= 𝑠𝑔𝑛 𝐶𝑖 ∙ 𝑣𝑎𝑙 𝐶𝑖

= 𝑠𝑔𝑛 𝐶𝑖 𝑣𝑎𝑙(𝐶𝑖)

C1 C2

C3

Oddly oriented

The FKT algorithms finds a feasible orientation!

FKT Algorithm by PlusOne 23

Lemma

If all cycles in an EDCC are oddly oriented then 𝑠𝑔𝑛 𝜋 𝑣𝑎𝑙 𝜋 = 1 for all corresponding 𝜋.

LemmaIf all faces have an odd number of clockwise edges then all cycles in an EDCC are oddly oriented.

Det (A) = |EDCC|

FKT algorithm

FKT Algorithm by PlusOne 24

Graph GRun FKT to find

orientation GCompute det 𝐴 PM = det(𝐴)

How to count perfect matchings?

counting perfect matchings in planar graphs

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