Colourful Problems in Combinatorics Jason I. Brown Dalhousie - - PowerPoint PPT Presentation

Colourful Problems in Combinatorics

Jason I. Brown Dalhousie University

Sunday, 9 June, 13

Colourful Problems

Sunday, 9 June, 13

Colourful Problems

n Eric Mendelsohn’s research papers include a

number on colourings:

Sunday, 9 June, 13

Colourful Problems

n Eric Mendelsohn’s research papers include a

number on colourings:

n On the chromatic index of path decompositions. n Bicolour graphs of Steiner triple systems. n Colouring planar mixed hypergraphs. n On defining numbers of vertex colouring of regular graphs. n On the complexity of coloring areflexive h-ary relations with

given permutation group.

n Computing star chromatic number from related graph

invariants.

n 3-(v, 4, 1) covering designs with chromatic numbers 2 and 3

Sunday, 9 June, 13

n We’ll look at a class of problems on colourings of

graphs and hypergraphs.

Colourful Problems

Sunday, 9 June, 13

Folkman Graphs

Sunday, 9 June, 13

n Let G and H be graphs. We say that G is a

k-folkman graph for H if for any assignment of

• ne of k colours to each vertex of G, there is a

monochromatic copy induced copy of H. 2-folkman graph for

Folkman Graphs

P

3

Sunday, 9 June, 13

n Let G and H be graphs. We say that G is a

k-folkman graph for H if for any assignment of

• ne of k colours to each vertex of G, there is a

monochromatic copy induced copy of H. 2-folkman graph for

Folkman Graphs

P

3

use both red and blue

Sunday, 9 June, 13

n Let G and H be graphs. We say that G is a

k-folkman graph for H if for any assignment of

• ne of k colours to each vertex of G, there is a

monochromatic copy induced copy of H. 2-folkman graph for

Folkman Graphs

P

3

use both red and blue use both red and blue use both red and blue

Sunday, 9 June, 13

n Let G and H be graphs. We say that G is a

k-folkman graph for H if for any assignment of

• ne of k colours to each vertex of G, there is a

monochromatic copy induced copy of H. 2-folkman graph for

Folkman Graphs

P

3

use both red and blue use both red and blue use both red and blue use both red and blue use both red and blue red or blue ?

Sunday, 9 June, 13

n For graphs H and G, we can form a hypergraph F

• n the vertices of G whose edges are those subsets
• f V(G) that induce a copy of H. Then G is a k-

folkman graph for H if the hypergraph F is not k- colourable (in the usual sense).

Folkman Graphs

Sunday, 9 June, 13

n For a graph H of order n, how small an order can

a k-folkman graph have?

n For or the smallest k-folkman graph

has order

n Substitution operation gives about . Can we do

better?

Folkman Graphs

nk G = Kn k(n −1)+1 Kn

Sunday, 9 June, 13

n Theorem (JIB-VR) Fix k. Then for any graph H of

• rder n, there is a k-folkman graph for H with

vertices.

Folkman Graphs

O(n2 log2 n)

Sunday, 9 June, 13

n Sketch of proof: Fix . We’ll construct

a graph G of order at most a constant times for which every induced subgraph of

• rder has an induced copy of H. (For

this implies that G is a k-folkman graph for H).

Folkman Graphs

α ∈(0,1/ 2] n2 log2 n α |V(G)| ⎢ ⎣ ⎥ ⎦ α = 1/ k

Sunday, 9 June, 13

n Sketch of proof: From the fact that for z > 1 there

is a prime between z and 2z, we pick a prime p such that Set and q to be the remainder when you divide p + 1 by n (so x is about a constant times

Folkman Graphs

4 α nlogn ≤ p +1≤ 8 α nlogn. x = p +1 n ⎢ ⎣ ⎢ ⎥ ⎦ ⎥ logn).

Sunday, 9 June, 13

n Sketch of proof: Set and take a projective

plane of order p: has points and lines, with every line having t points.

n For every line l, we take a random partition of l

into n parts , with each part of size x or x + 1, and join every vertex of to every vertex of iff is an edge of H.

Folkman Graphs

t = p +1 P N = p2 + p +1 P

l1,…,ln li l j

vivj

Sunday, 9 June, 13

n Sketch of proof: The properties of projective

planes ensures that the procedure is well defined and that every line induces a graph arising from H by substituting independent sets for the vertices.

Folkman Graphs

Sunday, 9 June, 13

n Sketch of proof:

Folkman Graphs

l projective plane of order p

Sunday, 9 June, 13

n Sketch of proof:

Folkman Graphs

l projective plane of order p α proportion of vertices

Sunday, 9 June, 13

n Sketch of proof: Let X be a fixed subset of points

• f cardinality . Let E be the event that the

subgraph induced by X does contain an induced copy of H.

Folkman Graphs

αN ⎢ ⎣ ⎥ ⎦

Sunday, 9 June, 13

n Sketch of proof: Let X be a fixed subset of points

• f cardinality . Let E be the event that the

subgraph induced by X does contain an induced copy of H.

n Suppose that line l intersects X in points. Then

.

Folkman Graphs

αN ⎢ ⎣ ⎥ ⎦ xl

Prob(E) ≤ nNe

− x t xl

l∈ P

∑

Sunday, 9 June, 13

n Sketch of proof: We need to estimate .

Folkman Graphs

xl

l∈ P

∑

Sunday, 9 June, 13

n Sketch of proof: We need to estimate . n Lemma: Let be a collection of lines of with

for some fixed . Then .

Folkman Graphs

xl

l∈ P

∑

L = | L | ≥ N1/2+ε

L

P

ε > 0

xl

l∈ L

∑

~ α N L

Sunday, 9 June, 13

n Sketch of proof: Let the points and lines of

be and, respectively, with . .

Folkman Graphs P

p1,…, pN l1,…,lN

X = {pi1,…, pim}

Sunday, 9 June, 13

n Sketch of proof: We form an matrix B

whose rows and columns are indexed by the points and lines of with where , so that is a root of

Folkman Graphs

N × N

P

Bi, j = λ if pi ∈l j −1

• therwise,

⎧ ⎨ ⎪ ⎩ ⎪

λ = p + p λ y2 − 2py + p2 − p = 0.

Sunday, 9 June, 13

n Let denote the i-th row of B. Then from the

properties of a projective plane, for i ≠ j, = = = 0.

Folkman Graphs

bi bi ⋅b j λ 2 + 2pλ(−1)+ (p2 + p +1− 2p −1)(−1)2 λ 2 − 2pλ + p2 − p

Sunday, 9 June, 13

n Let denote the i-th row of B. Then from the

properties of a projective plane, for i ≠ j, = = = 0.

n That is, with this choice of , the rows of B are

• rthogonal!

Folkman Graphs

bi bi ⋅b j λ 2 + 2pλ(−1)+ (p2 + p +1− 2p −1)(−1)2 λ 2 − 2pλ + p2 − p λ

Sunday, 9 June, 13

n Let be the points of X ( ), and

consider .

Folkman Graphs

pi1,…, pim

m = αN ⎢ ⎣ ⎥ ⎦

bi1 +...+ bim

2

Sunday, 9 June, 13

n Let be the points of X ( ), and

consider .

n By the orthogonality of the rows of B we see that

= = ~

Folkman Graphs

pi1,…, pim

m = αN ⎢ ⎣ ⎥ ⎦

bi1 +...+ bim

2

bi1 +...+ bim

2

bi1

2 +...+ bim 2

m((p +1)λ 2 + p2) αN 5/2

Sunday, 9 June, 13

n Let be the restriction of to the columns

• f . Let . , so that

.

Folkman Graphs

cij bij

L

ci1 +...+ cim = (ψ 1,…,ψ L) ψl = λxl − (m − xl)

Sunday, 9 June, 13

n Let be the restriction of to the columns

• f . Let . , so that

.

n By the Cauchy-Schwartz inequality, we find that

Folkman Graphs

cij bij

L

ci1 +...+ cim = (ψ 1,…,ψ L) ψl = λxl − (m − xl)  c1 +…+ cm 2 ≥ λ +1 L xl

l∈L

∑

− m L ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

2

.

Sunday, 9 June, 13

n Now

Folkman Graphs

 c1 +…+ cm 2 ≥ λ +1 L xl

l∈L

∑

− m L ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

2

. bi1 +...+ bim

2 = m((p +1)λ 2 + p2) ~ αN 5/2

Sunday, 9 June, 13

n Now

so as clearly we find that

Folkman Graphs

 c1 +…+ cm 2 ≥ λ +1 L xl

l∈L

∑

− m L ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

2

bi1 +...+ bim

2 = m((p +1)λ 2 + p2) ~ αN 5/2

bi1 +...+ bim

2 ≥ ci1 +...+ cim 2

λ +1 L xl

l∈ L

∑

− m L ≤ m((p +1)(p2 + 2 p + p)+ p2) ~ α N 5/4.

Sunday, 9 June, 13

n As and , we have

In particular,

Folkman Graphs

λ +1~ N xl

l∈ L

∑

~ α N L. xl

l∈ P

∑

≥ (α / 2)N 3/2. L ≥ N (1/2)+ε

Sunday, 9 June, 13

n Thus the probability of is bounded above by

= o(1)

Folkman Graphs

E

N αN ⎢ ⎣ ⎥ ⎦ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ nNe

−α 2 x t N 3/2

≤ e

N α−α logα+logn−α 2 ⋅ 3 α logn⋅ N t ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

Sunday, 9 June, 13

n Thus for large n there is a graph G of order

for which every induced subgraph of order contains H as an induced subgraph. For G is a k-folkman graph for H of order

Folkman Graphs

N = p2 + p +1= O(n2 log2 n) αN ⎢ ⎣ ⎥ ⎦ α = 1/ k O(n2 log2 n).

Sunday, 9 June, 13

n Even for small graphs and k = 2, finding the

smallest folkman graphs is difficult!

graph smallest order of a 2-folkman graph

6 9 9–11 9 9 9

Folkman Graphs

P

3

K1,3 P

4

C4 P

3 ∪ K1

K4 − e

Sunday, 9 June, 13