Rainbow matchings Existence and countingweb.mat.bham.ac.uk/G.Perarnau/slides/budapest.pdf · 2015....

Transcript of Rainbow matchings Existence and countingweb.mat.bham.ac.uk/G.Perarnau/slides/budapest.pdf · 2015....

Rainbow matchingsExistence and counting

Guillem Perarnau

Universitat Politècnica de CatalunyaDepartament de Matemàtica Aplicada IV

2nd September 2011Budapest

joint work with Oriol Serra
Outline

1 The problem

2 Counting with the Local Lemma

3 Our Approach

4 Random Models

Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 2 / 19
Outline

1 The problem


3 Our Approach

4 Random Models

Rainbow matchings and Latin transversals

Edge coloring. C : E(Kn,n) −→ N

Perfect matching: M = {ei indep}

Rainbow matching:no repeated colors in M.

Integer square matrix A = {aij}

Transversal Tσ = {aiσ(i)}

Latin Transversal:no repeated entries in Tσ.

0BB@1 5 4 27 2 6 35 4 2 13 5 3 1

1CCA








0BB@1 5 4 27 2 6 35 4 2 13 5 3 1

1CCA








0BB@1 5 4 27 2 6 35 4 2 13 5 3 1

1CCA








0BB@1 5 4 27 2 6 35 4 2 13 5 3 1

1CCA

0BB@1 5 4 27 2 6 35 4 2 13 5 3 1

1CCA








0BB@1 5 4 27 2 6 35 4 2 13 5 3 1

1CCA








0BB@1 5 4 27 2 6 35 4 2 13 5 3 1

1CCA

Open problems on Latin squares - Existence

Every latin square of odd order admits a latin transversal.

Conjecture (Ryser, 1967)

Every latin square admits a partial latin transversal of size n − 1.

Conjecture (Brualdi, 1975)

Every latin square admits a partial latin transversal of size n −O(log2 n).

Theorem (Hatami and Shor, 2008)

For every integer matrix, if no entry appears more than n4e times, then ithas a latin transversal.

Proposition (Erdős and Spencer, 1991)

Open problems on Latin squares - Counting

Let zn be the number of latin transversals of the cyclic group of order n.Then there exists two constants 0 < c1 < c2 < 1 such that

cn1 n! < zn < cn2 n!

Conjecture (Vardi, 1991)

Let zn be the number of latin transversals of the cyclic group of order n.Then

an < zn < bn√

nn!

where a = 3.246 and b = 0.614.

Theorem (McKay, McLeod and Wanless, 2006 / Cavenagh and Wan-less, 2010)

Open problems on Latin squares - Counting

Let zn be the number of latin transversals of the cyclic group of order n.Then there exists two constants 0 < c1 < c2 < 1 such that

cn1 n! < zn < cn2 n!

Conjecture (Vardi, 1991)

Let zn be the number of latin transversals of the cyclic group of order n.Then

an < zn < bn√

nn!

where a = 3.246 and b = 0.614.

Theorem (McKay, McLeod and Wanless, 2006 / Cavenagh and Wan-less, 2010)

Outline

1 The problem


3 Our Approach

4 Random Models

Poisson Paradigm

A1, . . . ,Am bad events Pr(Ai) = pi ,

Pr

m\

i=1

Ai

!?

1 If Ai are mutually independent

Pr

m\

i=1

Ai

!=

mYi=1

(1− pi) ∼ e−µ µ =mX

i=1

pi expected numberof bad events

2 If µ < 1, by the union bound

Pr

m\

i=1

Ai

!≥ 1−

mXi=1

Pr(Ai) = 1− µ > 0

Poisson paradigm: If the dependencies among Ai are rare.

Pr

m\

i=1

Ai

!= (1 + o(1))e−µ

Poisson Paradigm


Pr

m\

i=1

Ai

!?


Pr

m\

i=1

Ai

!=

mYi=1

(1− pi) ∼ e−µ µ =mX

i=1



Pr

m\

i=1

Ai

!≥ 1−

mXi=1

Pr(Ai) = 1− µ > 0


Pr

m\

i=1

Ai

!= (1 + o(1))e−µ

Poisson Paradigm


Pr

m\

i=1

Ai

!?


Pr

m\

i=1

Ai

!=

mYi=1

(1− pi) ∼ e−µ µ =mX

i=1



Pr

m\

i=1

Ai

!≥ 1−

mXi=1

Pr(Ai) = 1− µ > 0


Pr

m\

i=1

Ai

!= (1 + o(1))e−µ

Poisson Paradigm


Pr

m\

i=1

Ai

!?


Pr

m\

i=1

Ai

!=

mYi=1

(1− pi) ∼ e−µ µ =mX

i=1



Pr

m\

i=1

Ai

!≥ 1−

mXi=1

Pr(Ai) = 1− µ > 0


Pr

m\

i=1

Ai

!= (1 + o(1))e−µ

Lovász Local Lemma

dependency graph H,

V (H) = {A1, . . . ,Am}

E(H) = {dependencies among events}, Pr(Ai |T

j∈S Aj) = Pr(Ai)

∃x1, . . . , xm ∈ (0, 1) such that

Pr(Ai) ≤ xiY

Aj∈N(Ai )

(1− xj)

Then,

Pr

m\

i=1

Ai

!>

mYi=1

(1− xi) COUNTING (lower bound)

Lopsided version (Erdős and Spencer, 1991)


dependency graph H,

V (H) = {A1, . . . ,Am}E(H) = {dependencies among events}

, Pr(Ai |T

j∈S Aj) = Pr(Ai)

∃x1, . . . , xm ∈ (0, 1) such that

Pr(Ai) ≤ xiY

Aj∈N(Ai )

(1− xj)

Then,

Pr

m\

i=1

Ai

!>

mYi=1




dependency graph H,

V (H) = {A1, . . . ,Am}E(H) = {dependencies among events}, Pr(Ai |

Tj∈S Aj) = Pr(Ai)

∃x1, . . . , xm ∈ (0, 1) such that

Pr(Ai) ≤ xiY

Aj∈N(Ai )

(1− xj)

Then,

Pr

m\

i=1

Ai

!>

mYi=1




dependency graph H,


Tj∈S Aj) = Pr(Ai)

∃x1, . . . , xm ∈ (0, 1) such that

Pr(Ai) ≤ xiY

Aj∈N(Ai )

(1− xj)

Then,

Pr

m\

i=1

Ai

!>

mYi=1




dependency graph H,


Tj∈S Aj) = Pr(Ai)

∃x1, . . . , xm ∈ (0, 1) such that

Pr(Ai) ≤ xiY

Aj∈N(Ai )

(1− xj)

Then,

Pr

m\

i=1

Ai

!> 0 EXISTENCE

Then,

Pr

m\

i=1

Ai

!>

mYi=1




dependency graph H,


Tj∈S Aj) = Pr(Ai)

∃x1, . . . , xm ∈ (0, 1) such that

Pr(Ai) ≤ xiY

Aj∈N(Ai )

(1− xj)

Then,

Pr

m\

i=1

Ai

!>

mYi=1




dependency graph H,


Tj∈S Aj) = Pr(Ai)

∃x1, . . . , xm ∈ (0, 1) such that

Pr(Ai) ≤ xiY

Aj∈N(Ai )

(1− xj)

Then,

Pr

m\

i=1

Ai

!>

mYi=1



Upper bound using local Lemma (Lu and Szekely, 2009)

ε-near dependency graph H,

V (H) = {A1, . . . ,Am}

Pr(Ai ∩ Aj) = 0 if (i , j) ∈ E(H)

for any S ⊆ [m] \ N(Ai)

Pr(Ai |\j∈S

Aj) ≥ (1− ε) Pr(Ai)

Then,

Pr

m\

i=1

Ai

!≤Y

i

(1− (1− ε) Pr(Ai)) COUNTING (upper bound)



V (H) = {A1, . . . ,Am}Pr(Ai ∩ Aj) = 0 if (i , j) ∈ E(H)


Pr(Ai |\j∈S

Aj) ≥ (1− ε) Pr(Ai)

Then,

Pr

m\

i=1

Ai

!≤Y

i






Pr(Ai |\j∈S

Aj) ≥ (1− ε) Pr(Ai)

Then,

Pr

m\

i=1

Ai

!≤Y

i






Pr(Ai |\j∈S

Aj) ≥ (1− ε) Pr(Ai)

Then,

Pr

m\

i=1

Ai

!≤Y

i


Matchings of Kn,n

LetM be a collection of matchings of Kn,n (or Kn).Choose a perfect matching F u.a.r..

For any M ∈M define,

AM = {M ⊆ F}

V (H) = {AM}M∈ME(H) between AM and AN if M and N are in conflict (M ∪ N is not amatching)

LetM be a sparse set of matchings of Kn,n (or Kn).

Then H is both a negative dependency graph and an ε-near dependencygraph.

Theorem (Lu and Szekely, 2009)

• k -cycle free permutations • Latin rectangles n × r • d-regular graphs (configuration model)

Matchings of Kn,n



AM = {M ⊆ F}






Matchings of Kn,n



AM = {M ⊆ F}






Matchings of Kn,n



AM = {M ⊆ F}






Matchings of Kn,n



AM = {M ⊆ F}






Outline

1 The problem


3 Our Approach

4 Random Models

Setting the bad events

Set the following bad events

Given an edge coloring

Ae,f if e and f independent andsame color

In this matching A(11),(34) holds

If no event holds, we have arainbow matching

M = {Ae,f : e, f ∈ E(Kn,n), c(e) = c(f ), e and f independent}

but...

M is not sparse








but...

M is not sparse








but...

M is not sparse








but...

M is not sparse








but...

M is not sparse








but...

M is not sparse

Counting rainbow matchings

Fix an edge–coloring of Kn,n such that no color appears more than n/ktimes. Let µ = |M|/n(n − 1).If k ≥ 12 then there exist two constants γ1(k) < 1 < γ2(k), such that

e−γ2(k)µ ≤ Pr(M rainbow) ≤ e−γ1(k)µ

for a random matching M.

Theorem

Counting rainbow matchings

Fix an edge–coloring of Kn,n such that no color appears more than n/ktimes. Let µ = |M|/n(n − 1).If k ≥ 12 then there exist two constants γ1(k) < 1 < γ2(k), such that

e−γ2(k)µ ≤ Pr(M rainbow) ≤ e−γ1(k)µ

for a random matching M.

Theorem

If zn is the number of latin transversals in a latin square of size n whereeach entry appear at most n/k times (k ≥ 12), then there exist 0 < c1 <c2 < 1 constants such that

cn1 n! ≤ zn ≤ cn2 n!

Corollary

Outline

1 The problem


3 Our Approach

4 Random Models

Ryser conjecture whp

Ryser conjecture is difficult −→ Is it true for almost all latin squares?

First step: settle a model for random latin squares DIFFICULT!Jacobson, M. T. and Matthews, P., Generating uniformly distributed random Latin squares,

Journal of Combinatorial Designs, 4, 1996, 6, 405–437.

McKay, B. D. and Wanless, I. M., Most Latin squares have many subsquares, Journal of

Combinatorial Theory. Series A, 86, 1999, 2, 322–347.

Cavenagh, N. J. and Greenhill, C. and Wanless, I. M., The cycle structure of two rows in a

random Latin square, Random Structures & Algorithms, 33, 2008, 3, 286–309.

Given integer matrix A

Pr(@ latin transversal)� Pr(A latin square) ∼ e−n2



First step: settle a model for random latin squares

DIFFICULT!

Jacobson, M. T. and Matthews, P., Generating uniformly distributed random Latin squares,










First step: settle a model for random latin squares DIFFICULT!

Jacobson, M. T. and Matthews, P., Generating uniformly distributed random Latin squares,








Random models of colorings

Let s = nk the number of colors.

Random Model 1: Uniform modelChoose a color for each edge u.a.r.Prob space: [s]n

2

Random Model 2: Regular model

Uniformly among s-edge-colorings, each color appearing n/k times.




2






2






2






2



Results

Let C be a random edge coloring of Kn,n in the URM with s ≥ n colors.Then,

Pr(M rainbow) = e−c(k)µ,

where µ ∼ n2k and c(k) = 2k“

1− (k − 1) log“

kk−1

””.

Let C be a random edge-coloring of Kn,n in the RRM with s ≥ n colors.Then

Pr(M rainbow) = e−(c(k)+o(1))µ.

Proposition

Every random edge–coloring of Kn,n with s ≥ n colors in the URM (RRM)has a rainbow matching with high probability.

Theorem

In particular

Pr(@ latin transversal) ≤ 1n

Results




1− (k − 1) log“

kk−1

””.



Proposition


Theorem

In particular


Results




1− (k − 1) log“

kk−1

””.



Proposition


Theorem

In particular


Results

Thanks for your attention.


The problemCounting with the Local LemmaOur ApproachRandom Models

Rainbow matchings Existence and countingweb.mat.bham.ac.uk/G.Perarnau/slides/budapest.pdf · 2015....

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Transcript of Rainbow matchings Existence and countingweb.mat.bham.ac.uk/G.Perarnau/slides/budapest.pdf · 2015....