Lower Bounds on Ramsey Number

Presented by:- Smit Raj

Guide:- Prof. S.P. Pal

IIT Kharagpur

Ramsey Number For Two Colors :- R(s , t) [4]

• We define Ramsey number R(s , t) as the

smallest value of n (n ∈ N) for which every red–blue coloring of 𝐾𝑛 yields a monochromatic (red/blue) 𝑲𝒔 or 𝑲𝒕.

“ Ramsey's theorem states that such a number exists for all s and t .” • It’s obvious that, R(s , t) = R(t , s).

𝐾4 For every s ≥ 2

R(s , 2) = s = R(2 , s) • Either the graph has one red/blue edge or a

monochromatic 𝐾𝑠.

R(4 , 2) = 4 = R(2 , 4)

R(3 , 3) ≠ 5 because we had a counter example with no red or blue 𝐾3 in it. .

𝐾5

R(3 , 3) ≠ 5 The value of R(3 , 3) = 6.

𝐾7 𝐾8 R(3 , 4) ≠ 7 R(3 , 4) ≠ 8

In this specific coloring of 𝐾7 & 𝐾8 we didn’t find any red 𝐾3 or blue 𝐾4 as a sub graph.

Theorem : R(s , t) ≤ R(s , t-1) + R(s-1 , t) [3]

e.g. R(3 ,4) ≤ R(3 , 3) + R(2 , 4) ≤ 6 + 4 ≤ 10 but R(3 , 4) = 9 R(4 , 4) ≤ R(3 , 4) + R(4 , 3) ≤ 9 + 9 ≤ 18 and R(4 , 4) = 18

• No exact formula is known to calculate such a number.

• Very few Ramsey numbers have been computed so far

3 4 5 6 7 8 9 3 6 9 14 18 23 28 26 4 18 25 35/41 49/61 55/84 69/115 5 43/49 58/87 80/143 95/216 116/316 6 102/165 109/298 122/495 153/780

Table from [3]

Why calculation of R(s , t) is so difficult ? • There is no algorithm to calculate R(s , t).

• Brute force method- To select successive no.

of vertices & check weather every combination 𝟐𝒙 (here x is no. of edges) has either a red 𝐾𝑠 or blue 𝐾𝑡 . e.g. R(4 , 6) = 40 ??

Let v = 40 , then e in 𝐾40 is 780 → 2780 number of possible bi coloring is there computing weather the particular coloring have monochromatic 𝐾4 or 𝐾6 takes 10;9 sec.

→ 2780 ⨯ 10;9 ⨯ 10;9 years to calculate weather

R(4 , 6) = 40. & is more than 200 years.

Erdős lower bound on Ramsey number R(s , s) [3]

Theorem :- R(s , s) > 2(𝑠;1)/2 Proof :- Probability that edge x is of red color (x ∈ 𝐾𝑠) is 1 ̷ 2.

Probability of red color 𝐾𝑠 is 2;𝑠(𝑠;1)/2.

Probability of blue color 𝐾𝑠 is 2;𝑠(𝑠;1)/2.

Probability of monochromatic 𝐾𝑠 is 21;𝑠(𝑠;1)/2.

We choose n large enough such that there exist a possible coloring of 𝐾𝑛 such that no monochromatic 𝐾𝑠 is there. Probability that there exist monochromatic 𝐾𝑠 in 𝐾𝑛 is

P(𝐾𝑠) = 𝑛𝑠

21;𝑠(𝑠;1)/2.

We want to choose n as large as possible so that P(𝐾𝑠) is less than 1.

We take n = 2(s−1)/2

P(𝐾𝑠) ≤ 21;𝑠(𝑠;1)/2 𝑛

𝑠

𝑠!

< 2;𝑠(𝑠;1)/2 ⨯ 2𝑠(𝑠;1)/2

< 1 Which proves that there exist a coloring for large enough n so that no monochromatic 𝐾𝑠 is found.

R(𝐾1,2 ; 2) = 3 R(𝐾1,2 ; 3) = 4

R(𝐾1,𝑡 ; k) =

𝑘 𝑡 − 1 + 1 𝑖𝑓 𝑘 ≡ 𝑡 ≡ 0(𝑚𝑜𝑑 2)

𝑘 𝑡 − 1 + 2 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

A lower bound proof for R(𝑲𝒔,𝒕;k) :- Theorem :- R(𝐾𝑠,𝑡;k) > (2 π 𝑠𝑡)1/(𝑠:𝑡)((s + t)/𝑒2)𝑘(𝑠𝑡;1)/(𝑠:𝑡) (e denotes the base for natural logarithms) let {𝑘1, 𝑘2, 𝑘3………, 𝑘𝑘} are the set of colors Proof :- Probability that edge x (x ∈ 𝐾𝑠,𝑡) is of color 𝑘1 = 1/k. Probability of 𝐾𝑠,𝑡 to be of 𝑘1 = 𝑘;𝑠𝑡

[6]

Probability of 𝐾𝑠,𝑡 to be monochromatic = k. 𝑘;𝑠𝑡 = 𝑘1;𝑠𝑡

Number of 𝐾𝑠,𝑡 that can be selected from 𝐾𝑛 = 𝑛𝑠:𝑡

𝑠:𝑡𝑠

Probability that there exist a monochromatic 𝐾𝑠,𝑡 in 𝐾𝑛 is

P(𝐾𝑠,𝑡) = 𝑛𝑠:𝑡

𝑠:𝑡𝑠

𝑘1;𝑠𝑡

elementary calculations have shown that when we choose

n ≤ (2 π 𝑠𝑡)1/(𝑠:𝑡)

((s + t)/𝑒2)𝑘(𝑠𝑡;1)/(𝑠:𝑡)

then , P(𝐾𝑠,𝑡) < 1 Which means for

n ≤ (2 π 𝑠𝑡)1/(𝑠:𝑡)

((s + t)/𝑒2)𝑘(𝑠𝑡;1)/(𝑠:𝑡) we have a coloring in which no monochromatic 𝐾𝑠,𝑡 exists. Hence, R(𝐾𝑠,𝑡; k) > n

References 1. http://mathworld.wolfram.com/RamseyNumber.html 2. http://en.wikipedia.org/wiki/Ramsey%27s_theorem#

Ramsey_numbers 3. Introduction to graph theory , by Douglas Brent West 4. Modern Graph Theory, by Bela, Bollobas 5. http://theoremoftheweek.wordpress.com/2010/05/0

2/theorem-25-erdoss-lower-bound-for-the-ramsey-numbers

6. On multicolor Ramsey number for complete Bipartite graphs, by Fan R K Chung & R L Graham (June 1974)