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

Colourful Problems n We’ll look at a class of problems on colourings of graphs and hypergraphs. Sunday, 9 June, 13

Folkman Graphs Sunday, 9 June, 13

Folkman Graphs n Let G and H be graphs. We say that G is a k -folkman graph for H if for any assignment of one of k colours to each vertex of G , there is a monochromatic copy induced copy of H . 2-folkman graph for P 3 Sunday, 9 June, 13

Folkman Graphs n Let G and H be graphs. We say that G is a k -folkman graph for H if for any assignment of one of k colours to each vertex of G , there is a monochromatic copy induced copy of H . 2-folkman graph for P 3 use both red and blue Sunday, 9 June, 13

Folkman Graphs n Let G and H be graphs. We say that G is a k -folkman graph for H if for any assignment of one of k colours to each vertex of G , there is a monochromatic copy induced copy of H . 2-folkman graph for P 3 use both use both use both red and blue red and blue red and blue Sunday, 9 June, 13

Folkman Graphs n Let G and H be graphs. We say that G is a k -folkman graph for H if for any assignment of one of k colours to each vertex of G , there is a monochromatic copy induced copy of H . red or blue ? 2-folkman graph for P 3 use both use both use both use both use both red and blue red and blue red and blue red and blue red and blue Sunday, 9 June, 13

Folkman Graphs n For graphs H and G , we can form a hypergraph F on the vertices of G whose edges are those subsets of 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). Sunday, 9 June, 13

Folkman Graphs n For a graph H of order n , how small an order can a k-folkman graph have? G = K n n For or the smallest k -folkman graph K n k ( n − 1) + 1 has order n k n Substitution operation gives about . Can we do better? Sunday, 9 June, 13

Folkman Graphs n Theorem (JIB-VR) Fix k. Then for any graph H of order n , there is a k -folkman graph for H with O ( n 2 log 2 n ) vertices. Sunday, 9 June, 13

Folkman Graphs α ∈ (0,1/ 2] n Sketch of proof: Fix . We’ll construct a graph G of order at most a constant times n 2 log 2 n for which every induced subgraph of α | V ( G )| ⎢ ⎥ order has an induced copy of H . (For ⎣ ⎦ α = 1/ k this implies that G is a k -folkman graph for H ). Sunday, 9 June, 13

Folkman Graphs n Sketch of proof: From the fact that for z > 1 there is a prime between z and 2 z, we pick a prime p such that α n log n ≤ p + 1 ≤ 8 4 α n log n . p + 1 ⎢ ⎥ Set and q to be the remainder when x = ⎢ ⎥ ⎣ ⎦ n you divide p + 1 by n (so x is about a constant times log n ). Sunday, 9 June, 13

Folkman Graphs t = p + 1 n Sketch of proof: Set and take a projective N = p 2 + p + 1 plane of order p: has points and P P lines, with every line having t points. n For every line l , we take a random partition of l l 1 , … , l n into n parts , with each part of size x or x + 1, and join every vertex of to every vertex of l i iff is an edge of H. l j v i v j Sunday, 9 June, 13

Folkman Graphs 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. Sunday, 9 June, 13

Folkman Graphs n Sketch of proof: l projective plane of order p Sunday, 9 June, 13

Folkman Graphs n Sketch of proof: l α proportion of vertices projective plane of order p Sunday, 9 June, 13

Folkman Graphs n Sketch of proof: Let X be a fixed subset of points α N ⎢ ⎥ of cardinality . Let E be the event that the ⎣ ⎦ subgraph induced by X does contain an induced copy of H . Sunday, 9 June, 13

Folkman Graphs n Sketch of proof: Let X be a fixed subset of points α N ⎢ ⎥ of 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 x l ∑ − x x l t Prob( E ) ≤ n N e l ∈ . P Sunday, 9 June, 13

Folkman Graphs ∑ x l n Sketch of proof: We need to estimate . l ∈ P Sunday, 9 June, 13

Folkman Graphs ∑ x l n Sketch of proof: We need to estimate . l ∈ P L P n Lemma: Let be a collection of lines of with L = | L | ≥ N 1/2 + ε ε > 0 for some fixed . Then ∑ ~ α . x l N L l ∈ L Sunday, 9 June, 13

Folkman Graphs P n Sketch of proof: Let the points and lines of p 1 , … , p N l 1 , … , l N be and, respectively, with X = { p i 1 , … , p i m } . . Sunday, 9 June, 13

Folkman Graphs N × N n Sketch of proof: We form an matrix B whose rows and columns are indexed by the P points and lines of with ⎧ λ if p i ∈ l j ⎪ B i , j = ⎨ − 1 ⎪ otherwise, ⎩ λ λ = p + where , so that is a root of p y 2 − 2 py + p 2 − p = 0. Sunday, 9 June, 13

Folkman Graphs n Let denote the i -th row of B . Then from the b i properties of a projective plane, for i ≠ j , λ 2 + 2 p λ ( − 1) + ( p 2 + p + 1 − 2 p − 1)( − 1) 2 b i ⋅ b j = λ 2 − 2 p λ + p 2 − p = = 0. Sunday, 9 June, 13

Folkman Graphs n Let denote the i -th row of B . Then from the b i properties of a projective plane, for i ≠ j , λ 2 + 2 p λ ( − 1) + ( p 2 + p + 1 − 2 p − 1)( − 1) 2 b i ⋅ b j = λ 2 − 2 p λ + p 2 − p = = 0. λ n That is, with this choice of , the rows of B are orthogonal! Sunday, 9 June, 13

Folkman Graphs p i 1 , … , p i m m = α N n Let be the points of X ( ), and ⎢ ⎥ ⎣ ⎦ 2 b i 1 + ... + b i m consider . Sunday, 9 June, 13

Folkman Graphs p i 1 , … , p i m m = α N n Let be the points of X ( ), and ⎢ ⎥ ⎣ ⎦ 2 b i 1 + ... + b i m consider . n By the orthogonality of the rows of B we see that 2 + ... + b i m 2 2 b i 1 + ... + b i m = b i 1 m (( p + 1) λ 2 + p 2 ) = α N 5/2 ~ Sunday, 9 June, 13

Folkman Graphs n Let be the restriction of to the columns b i j c i j L c i 1 + ... + c i m = ( ψ 1 , … , ψ L ) of . Let . , so that ψ l = λ x l − ( m − x l ) . Sunday, 9 June, 13

Folkman Graphs n Let be the restriction of to the columns b i j c i j L c i 1 + ... + c i m = ( ψ 1 , … , ψ L ) of . Let . , so that ψ l = λ x l − ( m − x l ) . n By the Cauchy-Schwartz inequality, we find that 2 ⎛ λ + 1 ⎞  c 1 +…+ c m  2 ≥ ∑ − m L x l . ⎜ ⎟ ⎝ ⎠ L l ∈ L Sunday, 9 June, 13

Folkman Graphs n Now 2 ⎛ λ + 1 ⎞  c 1 +…+ c m  2 ≥ ∑ − m L x l . ⎜ ⎟ ⎝ ⎠ L l ∈ L 2 = m (( p + 1) λ 2 + p 2 ) ~ α N 5/2 b i 1 + ... + b i m Sunday, 9 June, 13

Folkman Graphs n Now 2 ⎛ λ + 1 ⎞  c 1 +…+ c m  2 ≥ ∑ − m L x l ⎜ ⎟ ⎝ ⎠ L l ∈ L 2 = m (( p + 1) λ 2 + p 2 ) ~ α N 5/2 b i 1 + ... + b i m 2 ≥ c i 1 + ... + c i m 2 b i 1 + ... + b i m so as clearly we find that λ + 1 m (( p + 1)( p 2 + 2 ∑ − m L ≤ p + p ) + p 2 ) ~ α N 5/4 . x l L l ∈ L Sunday, 9 June, 13

Folkman Graphs λ + 1~ L ≥ N (1/2) + ε n As and , we have N ∑ ~ α x l N L . l ∈ L In particular, ∑ ≥ ( α / 2) N 3/2 . x l l ∈ P Sunday, 9 June, 13

Folkman Graphs n Thus the probability of is bounded above by E ⎛ ⎞ − α x N t N 3/2 ⎟ n N e ⎜ 2 α N ⎢ ⎥ ⎣ ⎦ ⎝ ⎠ ⎛ ⎞ N α − α log α + log n − α 2 ⋅ 3 α log n ⋅ N ⎜ ⎟ ⎝ ⎠ ≤ e t = o(1) Sunday, 9 June, 13

Folkman Graphs n Thus for large n there is a graph G of order N = p 2 + p + 1 = O ( n 2 log 2 n ) for which every α N ⎢ ⎥ induced subgraph of order contains H as an ⎣ ⎦ α = 1/ k induced subgraph. For G is a k -folkman graph for H of order O ( n 2 log 2 n ). Sunday, 9 June, 13

Folkman Graphs n Even for small graphs and k = 2, finding the smallest folkman graphs is difficult! graph smallest order of a 2-folkman graph 6 P 3 9 K 1,3 9–11 P 4 9 C 4 9 3 ∪ K 1 P 9 K 4 − e Sunday, 9 June, 13