Math 575Problem Set 14

For any two graphs F and H, the generalized Ramsey number r(F, H) is the smallest integer N such that every coloring of the edges of KN contains either a red copy of F or a blue copy

of H. So for instance, r(Pn ,Pn ) is the smallest integer N such that every coloring of the edges

of KN contains a monochromatic path on n vertices.

It follows from this definition that N is also the smallest positive integer such that every graph G on N vertices either has F as a subgraph or the complement of G has H as a subgraph. This is because we may view G as the red graph in some red-blue coloring of the edges ofKN and the complement of G as the blue graph in the same coloring.

Example: r(P4 ,P4 ) ≤ 6 .

Proof. Consider any red - blue coloring of edges of K6 . Since r(3, 3) = 6 , we know that

there is a red or blue copy of K3 in this coloring. Let a, b, c denote the vertices of this

monochromatic triangle and WLOG assume that its edges are red. Let x, y, z denote the remaining three vertices. Then if any edge from {x, y, z} to {a, b, c} is red, we have a red P4 .

On the other hand, if none of the edges between {x, y, z} and {a, b, c} are red, then we have a blue K3,3 , which contains a blue P4 within it.

In fact, r(P4 ,P4 ) ≤ 5 and we can see this via the following argument.

Consider any red - blue coloring of edges of K5 . Then for any vertex v in K5 , some two of its

edges, say va and vb have the same color and hence avb is a monochromatic P3. WLOG,

assume that avb is a red P3. Let x and y be the other two vertices of this K5 . Then if any one

of the edges xa, xb, ya, yb are red then together with avb it would form a red P4 . But if all of

these edge are blue then axby is a blue P4 .

To see that r(P4 ,P4 ) > 4 (and hence that r(P4 , P4 ) = 5 ), just note that if you color any triangle

of K4 red and the other three edges blue, then there is no monochromatic P4. in this coloring.

1. Given that

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r(P4 ,P4 ) = 5 , show that

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r(P5,P5 ) = 6 .Solution: Now consider any red-blue coloring of the edges of

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K6 . Then since

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r(P4 ,P4 ) = 5 , we may assume WLOG that there is a red

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P4 in this coloring. Let this path be a b c d. Let the remaining two vertices be x and y. Then if any of the edges from a, d to x, y are red, we get a red

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P5 right away. So we may assume all these edges are blue and so a, b, x, y forms a blue

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C4 in G. Now if any of the edges from c, d to a, b, x, y are blue, we get a blue

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P5 right away. Hence, c, d a, b, x, y forms a blue

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K2.4 which must contains a blue

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P5 as a subgraph (e.g., a c y d b ). To see that r(P5 ,P5 ) > 5, consider the graph

K4 &∪ K1 - a K4 together with a single

vertex - neither it nor its complement contains a P5 . In terms of colorings, color the edges of

K5 by first coloring all the edges of aK4 red and all the remaining edges blue - so the red

graph is K4 together with an isolated vertex, and the blue graph is K1,4 .

2. Show that r(P4 ,C4 ) = 5 . Solution: First we show r(P4 ,C4 ) ≤ 5 . Consider an arbitrary red-blue coloring of the edges ofK5 . We must show that there is either a red P4 or a blue C4 . Now suppose that v is a vertex in K5 that is incident with at least two red edges – say vu and vw are red. Let x and y be the remaining two vertices. Then either one of xu, xw, yu, yw is red and we have a red P4 or they are all blue and we have a blueC4 .

So it must be that no vertex is incident with two red edges. So that means that in the red graph R, degR (v) ≤ 1 for every vertex v. So, in the blue graph B, degB (v) ≥ 3 for every vertex v.

Now, since there are an odd number of vertices, not all can have degree 3 in B and so there is some vertex u with degB (u) = 4 . But now it is not hard to see that there must be a blueC4 (remember every vertex has blue degree at least 3). The graph

K3 &∪ K1 [i.e., a triangle and an

isolated vertex] shows r(P4 ,C4 ) > 4 . (In terms of colorings, we color the edges ofK4 so that the red graph is K3 together with an isolated vertex, and the blue graph is K1,3 .

3. Recall the theorem of Erdös that states that for 2 ≤ n ≤ m, r(n,m) ≤ r(n −1,m) + r(n,m −1).(a). Use this to show that for any integer n ≥ 2, r(n,n) ≤ 2r(n −1,n) .

Solution: r(n,n) ≤ r(n,n −1) + r(n −1,n) = 2r(n −1,n).

(b). Show that for n ≥ 2, r(3,n) ≤ r(3,n −1) + n.Solution: We have r(3,n) ≤ r(3,n −1) + r(2,n) = r(3,n −1) + n.

(c). Show that for n ≥ 2, r(3,n) ≤n +12

⎛⎝⎜

⎞⎠⎟.

Solution: Use the previous inequality and telescoping. See the presentation and/or your notes for the details.

4. If F is a graph and k is a positive integer, then bykF we mean k disjoint copies of F. So, 2K2 is the graph that consists of two disjoint edges.

(a). Prove:r(2K2 , 2K2 ) = 5 .Solution: Since r(P4 ,P4 ) = 5 , and any P4 contains a 2K2 , we have r(2K2 , 2K2 ) ≤ 5.So see that r(2K2 , 2K2 ) > 4, consider the graph

K3 &∪ K1 .

(b). Prove: r K1,3, K1,3( ) = 6 .

Solution: To see that r K1,3, K1,3( ) ≤ 6, it is enough to note that for any red-blue coloring of the

edges of a K6 and for any vertex v in this K6 , v must be incident with three edges of the same color and thus v is the center of a monochromaticK1,3 . The graph C5 shows that

r K1,3, K1,3( ) > 5 (in terms of coloring, color the edges of C5 red and the other edges blue and a

coloring of the edges of K5 results in which there is no monochromatic K1,3 ).

(c). Prove: r(K1,4 , K1,4 ) = 7 .Solution: Consider a red-blue coloring of the edges of a K7 and suppose that there is no monochromatic K1,4 in this coloring. Consider any vertex v in thisK7 . Since our coloring contains no monochromatic K1,4 , v is not incident with four edges of either color. Thus every vertex is incident with exactly three red edges. But this is impossible since then the red graph would have an odd number of vertices of vertices of odd degree.To see that r K1,4 , K1,4( ) > 6, just note that for any 6-cycle, neither it nor its complement

contains a K1,4 as a subgraph. (in terms of coloring, color the edges of C6 red and the other edges blue and a coloring of the edges of K6 results in which there is no monochromatic K1,4 ).