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]
β’ We define Ramsey number R(s , t) as the
smallest value of n (n β N) for which every redβblue coloring of πΎπ yields a monochromatic (red/blue) π²π or π²π.
β Ramsey's theorem states that such a number exists for all s and t .β β’ Itβs obvious that, R(s , t) = R(t , s).
πΎ4 For every s β₯ 2
R(s , 2) = s = R(2 , s) β’ Either the graph has one red/blue edge or a
monochromatic πΎπ .
R(4 , 2) = 4 = R(2 , 4)
R(3 , 3) β 5 because we had a counter example with no red or blue πΎ3 in it. .
πΎ5
R(3 , 3) β 5 The value of R(3 , 3) = 6.
πΎ7 πΎ8 R(3 , 4) β 7 R(3 , 4) β 8
In this specific coloring of πΎ7 & πΎ8 we didnβt find any red πΎ3 or blue πΎ4 as a sub graph.
Theorem : R(s , t) β€ R(s , t-1) + R(s-1 , t) [3]
e.g. R(3 ,4) β€ R(3 , 3) + R(2 , 4) β€ 6 + 4 β€ 10 but R(3 , 4) = 9 R(4 , 4) β€ R(3 , 4) + R(4 , 3) β€ 9 + 9 β€ 18 and R(4 , 4) = 18
β’ No exact formula is known to calculate such a number.
β’ 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 ? β’ There is no algorithm to calculate R(s , t).
β’ Brute force method- To select successive no.
of vertices & check weather every combination ππ (here x is no. of edges) has either a red πΎπ or blue πΎπ‘ . e.g. R(4 , 6) = 40 ??
Let v = 40 , then e in πΎ40 is 780 β 2780 number of possible bi coloring is there computing weather the particular coloring have monochromatic πΎ4 or πΎ6 takes 10;9 sec.
β 2780 β¨― 10;9 β¨― 10;9 years to calculate weather
R(4 , 6) = 40. & is more than 200 years.
ErdΕs lower bound on Ramsey number R(s , s) [3]
Theorem :- R(s , s) > 2(π ;1)/2 Proof :- Probability that edge x is of red color (x β πΎπ ) is 1 Μ· 2.
Probability of red color πΎπ is 2;π (π ;1)/2.
Probability of blue color πΎπ is 2;π (π ;1)/2.
Probability of monochromatic πΎπ is 21;π (π ;1)/2.
We choose n large enough such that there exist a possible coloring of πΎπ such that no monochromatic πΎπ is there. Probability that there exist monochromatic πΎπ in πΎπ is
P(πΎπ ) = ππ
21;π (π ;1)/2.
We want to choose n as large as possible so that P(πΎπ ) is less than 1.
We take n = 2(sβ1)/2
P(πΎπ ) β€ 21;π (π ;1)/2 π
π
π !
< 2;π (π ;1)/2 β¨― 2π (π ;1)/2
< 1 Which proves that there exist a coloring for large enough n so that no monochromatic πΎπ is found.
R(πΎ1,2 ; 2) = 3 R(πΎ1,2 ; 3) = 4
R(πΎ1,π‘ ; k) =
π π‘ β 1 + 1 ππ π β‘ π‘ β‘ 0(πππ 2)
π π‘ β 1 + 2 ππ‘βπππ€ππ π
A lower bound proof for R(π²π,π;k) :- Theorem :- R(πΎπ ,π‘;k) > (2 Ο π π‘)1/(π :π‘)((s + t)/π2)π(π π‘;1)/(π :π‘) (e denotes the base for natural logarithms) let {π1, π2, π3β¦β¦β¦, ππ} are the set of colors Proof :- Probability that edge x (x β πΎπ ,π‘) is of color π1 = 1/k. Probability of πΎπ ,π‘ to be of π1 = π;π π‘
[6]
Probability of πΎπ ,π‘ to be monochromatic = k. π;π π‘ = π1;π π‘
Number of πΎπ ,π‘ that can be selected from πΎπ = ππ :π‘
π :π‘π
Probability that there exist a monochromatic πΎπ ,π‘ in πΎπ is
P(πΎπ ,π‘) = ππ :π‘
π :π‘π
π1;π π‘
elementary calculations have shown that when we choose
n β€ (2 Ο π π‘)1/(π :π‘)
((s + t)/π2)π(π π‘;1)/(π :π‘)
then , P(πΎπ ,π‘) < 1 Which means for
n β€ (2 Ο π π‘)1/(π :π‘)
((s + t)/π2)π(π π‘;1)/(π :π‘) we have a coloring in which no monochromatic πΎπ ,π‘ exists. Hence, R(πΎπ ,π‘; 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)
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