Degree Ramsey numbers of trees Bill Kinnersley Department of - - PowerPoint PPT Presentation

Degree Ramsey numbers of trees

Bill Kinnersley

Department of Mathematics University of Illinois at Urbana-Champaign wkinner2@illinois.edu Joint work with

Kevin G. Milans, Douglas B. West

Preliminaries

Graph Ramsey theory: find the smallest n for which every s-edge-coloring

• f Kn has a monochromatic G.

Preliminaries

Graph Ramsey theory: find the smallest n for which every s-edge-coloring

• f Kn has a monochromatic G.

Definition

We write H

s

→ G if every s-edge-coloring of H yields a monochromatic G. (Read: H s-arrows G or H forces G.) R(G; s) = min{n : Kn

s

→ G}. This is “wasteful”; if G is sparse, we don’t really need Kn.

Preliminaries

Graph Ramsey theory: find the smallest n for which every s-edge-coloring

• f Kn has a monochromatic G.

Definition

We write H

s

→ G if every s-edge-coloring of H yields a monochromatic G. (Read: H s-arrows G or H forces G.) R(G; s) = min{n : Kn

s

→ G}. This is “wasteful”; if G is sparse, we don’t really need Kn.

Definition

The s-color degree Ramsey number of a graph G, denoted R∆(G; s), is min{∆(H) : H

s

→ G}.

Stars

Theorem (Burr-Erd˝

• s-Lovàsz, 1976)

R∆(K1,k; 2) =

• 2k − 2,

k even 2k − 1, k odd .

Stars

Theorem (Burr-Erd˝

• s-Lovàsz, 1976)

R∆(K1,k; 2) =

• 2k − 2,

k even 2k − 1, k odd .

Theorem

R∆(K1,k; s) = s(k − 1), k even s(k − 1) + 1, k odd . Monotonicity: if H is a subgraph of G, then R∆(H; s) ≤ R∆(G; s). Hence if ∆(G) = k, then R∆(G; s) ≥ R∆(K1,k; s).

Stars

Theorem

R∆(K1,k; s) = s(k − 1), k even s(k − 1) + 1 k odd .

Proof (upper bound).

K1,s(k−1)+1

s

→ K1,k.

Stars

Theorem

R∆(K1,k; s) = s(k − 1), k even s(k − 1) + 1 k odd .

Proof (upper bound).

K1,s(k−1)+1

s

→ K1,k. We can do better when k is even. Let H be s(k − 1)-regular and have no (k − 1)-factor [Bollobás-Saito-Wormald]. Now H

s

→ K1,k, since avoiding K1,k means finding a (k − 1)-factorization.

Stars

Theorem

R∆(K1,k; s) = s(k − 1), k even s(k − 1) + 1 k odd .

Proof (lower bound).

Suppose ∆(H) < s(k − 1). By Vizing’s Theorem, H is s(k − 1)-edge-colorable, so E(H) decomposes into s(k − 1) matchings. Let each color class be the union of k − 1 matchings; this avoids K1,k.

Stars

Theorem

R∆(K1,k; s) = s(k − 1), k even s(k − 1) + 1 k odd .

Proof (lower bound).

Suppose ∆(H) < s(k − 1). By Vizing’s Theorem, H is s(k − 1)-edge-colorable, so E(H) decomposes into s(k − 1) matchings. Let each color class be the union of k − 1 matchings; this avoids K1,k. We can do better when k is odd. Suppose ∆(H) is s(k − 1)-regular. By Petersen’s Theorem, H decomposes into 2-factors. Let each color class be the union of k−1

2

• f these.

Double-stars, two colors

Definition

The double-star Sa,b is the tree with diameter 3 and central vertices of degrees a and b.

Theorem

R∆(Sa,b; 2) =

• 2b − 2,

b even and a < b 2b − 1

• therwise

.

Double-stars, two colors

Definition

The double-star Sa,b is the tree with diameter 3 and central vertices of degrees a and b.

Theorem

R∆(Sa,b; 2) =

• 2b − 2,

b even and a < b 2b − 1

• therwise

.

Proof (lower bound).

R∆(Sa,b; 2) ≥ R∆(K1,b; 2); this suffices unless a = b and b is even.

Double-stars, two colors

Definition

The double-star Sa,b is the tree with diameter 3 and central vertices of degrees a and b.

Theorem

R∆(Sa,b; 2) =

• 2b − 2,

b even and a < b 2b − 1

• therwise

.

Proof (lower bound).

R∆(Sa,b; 2) ≥ R∆(K1,b; 2); this suffices unless a = b and b is even. For that case, suppose H is (2b − 2)-regular. Alternate red and blue along the edges of an Eulerian circuit. This avoids Sb,b.

Double-stars, two colors

Theorem

R∆(Sa,b; 2) = 2b − 2, b even and a < b 2b − 1

• therwise

.

Proof sketch (upper bound).

Let H be triangle-free and (2b − 1)-regular; we claim H → Sb,b.

Double-stars, two colors

Theorem

R∆(Sa,b; 2) = 2b − 2, b even and a < b 2b − 1

• therwise

.

Proof sketch (upper bound).

Let H be triangle-free and (2b − 1)-regular; we claim H → Sb,b. Suppose some 2-edge-coloring of H avoids Sb,b. Call a vertex majority-red if it lies on b red edges and majority-blue if it lies on b blue edges. WLOG, a plurality of vertices are majority-red.

Double-stars, two colors

Theorem

R∆(Sa,b; 2) = 2b − 2, b even and a < b 2b − 1

• therwise

.

Proof sketch (upper bound).

Let H be triangle-free and (2b − 1)-regular; we claim H → Sb,b. Suppose some 2-edge-coloring of H avoids Sb,b. Call a vertex majority-red if it lies on b red edges and majority-blue if it lies on b blue edges. WLOG, a plurality of vertices are majority-red. No red edge joins majority-red vertices, so each red edge has a majority-blue endpoint. Each majority-red vertex lies on at least b red edges and each majority-blue vertex lies on at most b − 1, hence more majority-blue vertices than majority-red vertices.

Double-stars, two colors

Theorem

R∆(Sa,b; 2) = 2b − 2, b even and a < b 2b − 1

• therwise

.

Proof sketch (upper bound).

If a < b and b even, we can do better. Take C5, replace each vertex with an independent set of size b − 1, and replace each edge with Kb−1,b−1. . . . . . . . . . ... . . .

◮ Each vertex is red, blue, or balanced. ◮ If we avoid Sb−1,b,

◮ no red edges joining red vertices ◮ no red edges joining red and

balanced vertices

◮ no blue edges joining blue vertices ◮ no blue edges joining blue and

balanced vertices

Double-stars, many colors

What about R∆(Sb,b; s) when s ≥ 3?

Double-stars, many colors

What about R∆(Sb,b; s) when s ≥ 3?

Theorem (Jiang)

When T is a tree, we have R∆(T; s) ≤ 2s(∆(T) − 1). Thus R∆(Sb,b; s) ≤ 2s(b − 1). Is this tight?

Double-stars, many colors

What about R∆(Sb,b; s) when s ≥ 3?

Theorem (Jiang)

When T is a tree, we have R∆(T; s) ≤ 2s(∆(T) − 1). Thus R∆(Sb,b; s) ≤ 2s(b − 1). Is this tight? Asymptotically yes, as s tends to infinity.

Double-stars, many colors

Goal: extend upper bound argument for R∆(Sb,b; 2) to R∆(Sb,b; s) for general s.

Definition

Given an s-edge-coloring of H:

◮ vertex v is major in a color if it lies on b edges of that color and minor

• therwise;

“Major in red” generalizes “majority-red” from the 2-color argument.

Double-stars, many colors

Goal: extend upper bound argument for R∆(Sb,b; 2) to R∆(Sb,b; s) for general s.

Definition

Given an s-edge-coloring of H:

◮ vertex v is major in a color if it lies on b edges of that color and minor

• therwise;

◮ a minor edge is an edge whose color is minor at both endpoints.

“Major in red” generalizes “majority-red” from the 2-color argument.

Double-stars, many colors

Lemma

Let H be a triangle-free graph. If an edge-coloring of H having r minor edges avoids Sb,b, then |E(H)| + r =

• v

d∗(v), where d∗(v) denotes the number of edges incident to and minor at v.

Double-stars, many colors

Lemma

Let H be a triangle-free graph. If an edge-coloring of H having r minor edges avoids Sb,b, then |E(H)| + r =

• v

d∗(v), where d∗(v) denotes the number of edges incident to and minor at v.

Proof.

No edge is major at both endpoints, since then it would be the central edge

• f a monochromatic Sb,b. Count edges by the endpoint(s) at which they are

minor; exactly r edges get counted twice.

Double-stars, many colors

Lemma

Let H be a triangle-free graph. If an edge-coloring of H having r minor edges avoids Sb,b, then |E(H)| + r =

• v

d∗(v), where d∗(v) denotes the number of edges incident to and minor at v.

Proof.

No edge is major at both endpoints, since then it would be the central edge

• f a monochromatic Sb,b. Count edges by the endpoint(s) at which they are

minor; exactly r edges get counted twice. Suppose we can find some H in which every edge-coloring avoiding Sb,b has many minor edges.

• v d∗(v) can’t be too large, so we can bound |E(H)|, and hence ∆(H).

If ∆(H) just barely exceeds this bound, then we’ve forced Sb,b.

Double-stars, many colors

Ramanujan graphs have the needed properties.

Definition

A Ramanujan graph is a d-regular graph with smallest eigenvalue at least −2 √ d − 1.

Double-stars, many colors

Ramanujan graphs have the needed properties.

Definition

A Ramanujan graph is a d-regular graph with smallest eigenvalue at least −2 √ d − 1. Let H be a Ramanujan graph. No matter how we color E(H), many edges will join vertices having the same major colors. These edges must be minor.

Double-stars, many colors

Lemma

Let H be a regular triangle-free Ramanujan graph that does not force Sb,b. Then ∆(H) ≤ 2(M + o(1))(b − 1), where M = max

• T⊆{1,...,s}(s − |T|)xT

2 −

T⊆{1,...,s} xT

• 1 −

xT

(

s |T|)

. Here xT represents the fraction of vertices having major colors T.

Double-stars, many colors

Lemma

Let H be a regular triangle-free Ramanujan graph that does not force Sb,b. Then ∆(H) ≤ 2(M + o(1))(b − 1), where M = max

• T⊆{1,...,s}(s − |T|)xT

2 −

T⊆{1,...,s} xT

• 1 −

xT

(

s |T|)

. Here xT represents the fraction of vertices having major colors T. Computation yields:

Theorem

For s ≥ 3, R∆(Sb,b; s) ≤

• (s − 1)s +
• s2 + s + 2 + 4/(s − 1)

s + 1 + 2/s + o(1)

• (b − 1).

Double-stars, lower bound

This is asymptotically tight!

Theorem

For s ≥ 3, R∆(Sb,b; s) ≥

• (s − 1)s +
• s2 + s + 2 + 4/(s − 1)

s + 1 + 2/s − o(1)

• (b − 1).

Double-stars, lower bound

This is asymptotically tight!

Theorem

For s ≥ 3, R∆(Sb,b; s) ≥

• (s − 1)s +
• s2 + s + 2 + 4/(s − 1)

s + 1 + 2/s − o(1)

• (b − 1).

We need to show how to color an arbitrary graph with low enough maximum degree, while avoiding Sb,b. Use a random coloring:

Double-stars, lower bound

This is asymptotically tight!

Theorem

For s ≥ 3, R∆(Sb,b; s) ≥

• (s − 1)s +
• s2 + s + 2 + 4/(s − 1)

s + 1 + 2/s − o(1)

• (b − 1).

We need to show how to color an arbitrary graph with low enough maximum degree, while avoiding Sb,b. Use a random coloring:

◮ randomly choose “planned” major colors for each vertex, distributed

according to the xT from earlier;

Double-stars, lower bound

This is asymptotically tight!

Theorem

For s ≥ 3, R∆(Sb,b; s) ≥

• (s − 1)s +
• s2 + s + 2 + 4/(s − 1)

s + 1 + 2/s − o(1)

• (b − 1).

We need to show how to color an arbitrary graph with low enough maximum degree, while avoiding Sb,b. Use a random coloring:

◮ randomly choose “planned” major colors for each vertex, distributed

according to the xT from earlier;

◮ use the vertex-coloring to guide a random edge-coloring.

Larger trees

We’ve shown:

Theorem

For fixed s, R∆(Sb,b; s) ∼

• (s − 1)s +
• s2 + s + 2 + 4/(s − 1)

s + 1 + 2/s + o(1)

• (b − 1).

R∆(Sb,b; 3) ∼ 3(b − 1); R∆(Sb,b; 4) ∼ 2 + 2

11(1 +

√ 210)(b − 1) ≈ 4.8166(b − 1).

Larger trees

We’ve shown:

Theorem

For fixed s, R∆(Sb,b; s) ∼

• (s − 1)s +
• s2 + s + 2 + 4/(s − 1)

s + 1 + 2/s + o(1)

• (b − 1).

R∆(Sb,b; 3) ∼ 3(b − 1); R∆(Sb,b; 4) ∼ 2 + 2

11(1 +

√ 210)(b − 1) ≈ 4.8166(b − 1). As s grows, the coefficient on (b − 1) tends to 2s. Thus Jiang’s bound R∆(T; s) ≤ 2s(∆(T) − 1) is asymptotically tight.

Larger trees

We’ve shown:

Theorem

For fixed s, R∆(Sb,b; s) ∼

• (s − 1)s +
• s2 + s + 2 + 4/(s − 1)

s + 1 + 2/s + o(1)

• (b − 1).

R∆(Sb,b; 3) ∼ 3(b − 1); R∆(Sb,b; 4) ∼ 2 + 2

11(1 +

√ 210)(b − 1) ≈ 4.8166(b − 1). As s grows, the coefficient on (b − 1) tends to 2s. Thus Jiang’s bound R∆(T; s) ≤ 2s(∆(T) − 1) is asymptotically tight. When s is fixed, can the bound be improved?

Larger trees

Let T (d)

k

denote the maximal d-regular tree of diameter 2k. T (d)

1

is a star; T (d)

2

properly contains Sd,d. To study tightness in Jiang’s bound, study R∆(T (d)

k

; 2) for large k.

Larger trees

Let T (d)

k

denote the maximal d-regular tree of diameter 2k. T (d)

1

is a star; T (d)

2

properly contains Sd,d. To study tightness in Jiang’s bound, study R∆(T (d)

k

; 2) for large k. What is known:

◮ R∆(T (d) 2

; 2) ∼ 8

3(d − 1); ◮ R∆(T (d) 3

; 2) ≥ (3.078... − o(1))(d − 1);

◮ R∆(T (d) 4

; 2) ≥ (3.309... − o(1))(d − 1).

Larger trees

Let T (d)

k

denote the maximal d-regular tree of diameter 2k. T (d)

1

is a star; T (d)

2

properly contains Sd,d. To study tightness in Jiang’s bound, study R∆(T (d)

k

; 2) for large k. What is known:

◮ R∆(T (d) 2

; 2) ∼ 8

3(d − 1); ◮ R∆(T (d) 3

; 2) ≥ (3.078... − o(1))(d − 1);

◮ R∆(T (d) 4

; 2) ≥ (3.309... − o(1))(d − 1). As with double-stars, finding an upper bound (for all k, d, s) reduces to a tricky optimization problem. A solution to this optimization problem yields a lower bound on R∆ via a probabilistic argument.

Thanks

Thank you!