Ramsey Numbers Paper

By Marilyn Titus

Table of ContentsItem Page Number

Table Of Contents 1

Ramsey Number Introduction 2

Common Notation Used Throughout and Other Information 3

Items to Prove 4

External Resources Used 4

Items given at the start 5

Counter graph R(5,5) is greater than 45. 6

S(n): R(n,n) = 2q, q is an integer ≥ 1 7

S(n): R(n,n) = 2[R(n-1,n-1) +2(n-3)] +2, n≥3 8-9

S(n): R(n,n) = 2(2^n – 2n +1), n≥2 10

R(n,n) calculated up until and including R(20,20) 11

S(n) : R3c = c(R3{c-1} -1) + 2, c≥2, R31 = 3 122

Ramsey Number Introduction• In layman’s terms, a Ramsey number is this: What is

the smallest number of people R(n,n) to guarantee that there is a group of n people that exclusively know each other OR a group of n people that exclusively do not know each other? Each person must either “know” or “not know” a person, there is no “middle ground”.

• In terms of graph theory, given a complete graph G, labeling its edges with one of two colors (or properties), what is the smallest size G can have in order to be forced to contain a complete graph of Kn of one color within it? (This can also be R(n,k) where the question is rewritten to ask for Kn of one color and Kk of another color, but this paper doesn’t concern this subset of Ramsey Numbers.)

3

Common Notation Used Throughout and Other Information

• All hyperlinks in this paper are property of their respective author(s). I make no claim to any such proprietary information and wish to give credit to other authors where it is due. The only thing I claim as my own is the logic of the proofs in these slides.

• Kn refers to the complete graph of n vertexes. For more information: See http://en.wikipedia.org/wiki/Complete_graph

• A complete graph G has a size R(n,n) when it has the following properties:– Each edge is labeled with one of two colors.– It is required to have a subgraph of Kn within it with all edges of a single color. – This is the smallest sized graph to have the requirement above.– For example, R(3,3) = 6 because it has to have K3 within it, but 5 is not required to. 7 is not

R(n,n) because every graph of size 6 has K3 within it, given each edge is labeled with one of two colors.

• A complete graph with size of R3C is one that :– Each edge is labeled with one of c colors.– It is required to have K3 within it of a single color. – It is the smallest such graph to meet this requirement.– R3C = R(3,3,3,3,3,3,3,3,…..) c times. R32 = R(3,3) ; R33 = R(3,3,3); R34 = R(3,3,3,3); and so forth.

4

Items to prove

• R(n,n) = 2q, q is an integer, n≥2. • R(n,n) = 2[R(n-1,n-1) + 2(n-3)] + 2, n≥3.• R(n,n) = 2[2^n-2n+1], n≥2.• R3c = c(R3c-1 -1) + 2, c≥2, R31 = 3

External resources used• http://en.wikipedia.org/wiki/Ramsey's_theorem • http://mathworld.wolfram.com/RamseyNumber.html• http://en.wikipedia.org/wiki/Complete_graph• http://en.wikipedia.org/wiki/Ramsey%27s_theorem• http://www.dharwadker.org/clique/

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Items given at the start

• G is a complete graph with unlabeled, un-weighted edges. Each edge is labeled with one of two colors, red and blue, and G has a size of R(n,n) where n ≥ 2.

• R(n,n) > 1 for all n ≥ 2.

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Counter graph for R(5,5) > 45

• Assume R(5,5) ≤ 45. Then all graphs of that size will be required to have K5 within them. But K5 doesn’t exist in the graph in the excel file below, as found by me, and verified by the peer-reviewed program written by A. Dharwadker at http://www.dharwadker.org/clique/, which the reader is invited to use as well to try to find a graph of size 5.

• The graph in Excel and the zip file with the graph in program format and its output where it says there are no cliques of size 5.

Microsoft Excel 97-2003 Worksheet

graphsForProgram.zip

If watching by video, attachments available on request. 7

S(n): R(n,n) = 2q, q is an integer ≥ 1.

• Assume S(n) is false. There exists some R(n,n) that is an odd integer.• From prior given information, examine a vertex, (v0). If R(n,n) is odd then it will have even

number of edges. There will be q edges of each color, meaning R(n,n) = 2q+1.• Unlabel one of the edges. Meaning there will be q edges of one color and q-1 edges of

another color. Without loss of generality we will say q edges of red and q-1 edges of blue.• A complete graph of one color must have edges of only that color. E.G. K^n red has no blue

edges and K^n blue has no red edges. • Assume this v0 is a part of K^n. (If not repeat above steps until n vertexes remaining….Then

one of the vertexes must be.)• Assume K^n is red, then it would be forced in q edges. By the pigeonhole principle, q edges

from a vertex is forced when there are q-1 edges of each color and then adding a edge. This gives 2(q-1) +1 edges, which is a contradiction of the above that R(n,n) is odd.

• Assume K^n is blue, then it would be forced in q-1 blue edges. By the pigeonhole principle, this would give 2(q-2) + 1 edges, which a contraction that R(n,n) is odd.

• If R(n,n) is odd we end up with a contradiction, in that it is not the “smallest” sized graph to contain R(n,n).

• Therefore R(n,n) must be even when n≥2. The reason for the when clause is because we must have at least one edge for this proof to hold.

• S(n) is true by proof by contradiction. 8

S(n): R(n,n) = 2[R(n-1,n-1) +2(n-3)] +2, n≥3.

• We know from prior proof R(n,n) = 2q. • Examine a vertex in R(n,n) called v0. It will have q-1 edges red, q-1

edges blue, and one unlabeled edge that will force either q red edges or q blue edges. Assume this vertex is in Kn.

• Examine the subgraph that is formed from q edges of the same color. We will call these q edges from v0 red without loss of generality.

• This is a subgraph of q edges, graph M. Examine a vertex in this subgraph M. Call it r1. It connects to all other ri with either a red edge or a blue edge, by definition.

• R(n,n) = 2(q-1) +2 = 2(r1 to blue without Kn + r1 to red without Kn + r1) + 2.

• See next page:

9

Continued from prior page…• In this subgraph M, assume r1 has B blue edges and T red edges

between itself and the other ri. • If B is equal to R(n-1,n-1) then we know R(n,n) exists, because if Kn-1

is blue then Kn blue exists with r1 and the n-1 vertexes in Kn-1. If Kn-1 is red then Kn red exists with the n-1 vertexes in Kn-1 and v0.

• Now we have R(n,n) = 2[R(n-1,n-1) -1 + r1 + T] + 2. R(n,n) = 2[R(n-1,n-1) + T] + 2.

• The next question is what is T? By the pigeonhole principle, T is equal to 2(n-3), using prior methods stated. {If the reader wishes this can also be proved by induction…Since I prefer the most elegant proof possible this is not included.}

• This means R(n,n) = 2 [ R(n-1,n-1) + 2(n-3) ] + 2. Since this was our statement to prove. This is proven.

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S(n): R(n,n) = 2(2n – 2n +1), n≥2• Leverage R(n,n) = 2[R(n-1,n-1) + 2(n-3)] + 2, n≥3.• By alegbra R(n,n) = 2[R(n -1,n-1) + 2n -5], n≥3.• Basis steps: (for proof of basis steps see: http://mathworld.wolfram.com/RamseyNumber.html )

– n=3, R(3,3) = 6, 2(8-6+1) = 6– n=4, R(4,4) = 18, 2(16-8+1) = 18

• Induction: Assume S(n) is true for R(n,n) show that it must be true for R(n+1,n+1).• We know R(n+1,n+1) = 2[R(n,n) + 2(n+1) -5], n+1≥3.• Using the assumption R(n,n) = 2(2n – 2n +1), then R(n+1,n+1) = 2[2(2n – 2n +1) + 2(n+1) -5].• Using alegbra: R(n+1,n+1) = 2[2n+1 -4n+2+2n+2-5].• R(n+1,n+1) = 2(2n+1 – 2n -1). • Which is exactly what one gets when one plugs in n+1 into the formula above. But this only works

for n≥3 as that is what our prior formula worked for. To prove it for n=2 we must do the work for that case.

• R(2,2) = 2(4-4+1) = 2. Therefore the formula is true for 2 as well and now it can be expanded for that case.

• Therefore S(n) is proven for all integers greater than or equal to two.

11

R(n,n) calculated up until and including R(20,20)

12

n R(n,n) (as paper states) R(n,n) previously established bounds (see references list for sources)

1 1 1

2 2 2

3 6 6

4 18 18

5 46 [43, 49]

6 106 [102, 165]

7 230 [205, 540]

8 482 [282, 1870]

9 990 [565, 6588]

10 2010 [798, 23556]

11 4054 [1597, 184755]

12 8146 [1637, 705431]

13 16334 [2557, 2704155]

14 32714 [2989, 10400599]

15 65478 [5485, 40116599]

16 131010 [5605, 155117519]

17 262078 [8917, 601080389]

18 524218 [11005, 2333606219]

19 1048502 [17885, 9075135299]

202097074

** Did not see any bounds available in an easily accessible place for higher numbers**

S(n) : R3c = c(R3{c-1} -1) + 2, c≥2, R31 = 3

• We know R3c has at least one vertex. Off of it must come the same number of colored edges and one force edge. (Use same method as described to prove R(n,n) is even.) This demonstrates R3C = 3(q-1) + 2, where q is the max number of edges of each color before R(n,n) is formed.

• Examine q-1 edges from a vertex v0 of a given color, say red. We know that in this subgraph, none of the additional edges may be red. For example, v0 connects to r1,r2,r3,….ri with a “red” edge. If r1 connects to r2,r3,r4….ri with a “red” edge then K3 exists which is what is attempting to be avoided. Repeat for all other rj.

• Now what is being asked in this q-1 edges is when K3 exists given one less color to work with, namely red. This value is R3c-1. So we cannot have R3{c-1} edges of a given color or R3C exists. This means our q-1 is equal to R3{c-1} -1.

• This means R3C = c(R3{c-1} -1) +2. • This can be further expanded on by induction (left to the reader). The

basis steps are: R32 = 2(3-1) + 2 = 6, R33 = 3(6-1) + 2 = 17, which is also true. For basis steps, see: http://en.wikipedia.org/wiki/Ramsey%27s_theorem 13