a topic on algorithm design in computer science

1. Lower Bounds on
Ramsey Number
Presented by:- Smit Raj
Guide:- Prof. S.P. Pal
IIT Kharagpur

2. 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 𝐾 𝑛 yields a monochromatic
(red/blue) 𝑲 𝒔 or 𝑲 𝒕 .
“ Ramsey's theorem states that such a number
exists for all s and t .”
• It’s obvious that, R(s , t) = R(t , s).

3. R(4 , 2) = 4 = R(2 , 4)
𝐾4
For every s ≥ 2
R(s , 2) = s = R(2 , s)
• Either the graph has one red/blue edge or a
monochromatic 𝐾 𝑠 .

4. R(3 , 3) ≠ 5
The value of R(3 , 3) = 6.
𝐾5
R(3 , 3) ≠ 5 because we had a counter example with no
red or blue 𝐾3 in it.
.

5. 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.

6. 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

7. • 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]

8. 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
𝟐 𝒙 (here x is no. of edges) has either a red 𝐾 𝑠
or blue 𝐾 𝑡 .
e.g. R(4 , 6) = 40 ??

9. 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.

10. 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 ∈ 𝐾 𝑠 ) is
1 ̷ 2.
Probability of red color 𝐾 𝑠 is 2;𝑠(𝑠;1)/2 .
Probability of blue color 𝐾 𝑠 is 2;𝑠(𝑠;1)/2 .
Probability of monochromatic 𝐾 𝑠 is 21;𝑠(𝑠;1)/2 .

11. We choose n large enough such that there exist a
possible coloring of 𝐾 𝑛 such that no monochromatic 𝐾 𝑠 is
there.
Probability that there exist monochromatic 𝐾 𝑠 in 𝐾 𝑛 is
𝑛 1;𝑠(𝑠;1)/2
P(𝐾 𝑠 ) = 𝑠 2 .
We want to choose n as large as possible so that P(𝐾 𝑠 ) is
less than 1.
We take n = 2(s−1)/2

12. 𝑠
𝑛
P(𝐾 𝑠 ) ≤ 21;𝑠(𝑠;1)/2
𝑠!
< 2;𝑠(𝑠;1)/2 ⨯ 2 𝑠(𝑠;1)/2
<1
Which proves that there exist a coloring for large
enough n so that no monochromatic 𝐾 𝑠 is found.

13. R(𝐾1,2 ; 2) = 3 R(𝐾1,2 ; 3) = 4
𝑘 𝑡−1 +1 𝑖𝑓 𝑘 ≡ 𝑡 ≡ 0(𝑚𝑜𝑑 2)
R(𝐾1,𝑡 ; k) =
𝑘 𝑡−1 +2 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

14. A lower bound proof for R(𝑲 𝒔,𝒕 ;k) :- [6]
Theorem :-
R(𝐾 𝑠,𝑡 ;k) > (2 π 𝑠𝑡)1/(𝑠:𝑡) ((s + t)/𝑒 2 )𝑘 (𝑠𝑡;1)/(𝑠:𝑡)
(e denotes the base for natural logarithms)
let {𝑘1 , 𝑘2 , 𝑘3 ………, 𝑘 𝑘 } are the set of colors
Proof :- Probability that edge x (x ∈ 𝐾 𝑠,𝑡 ) is of color 𝑘1 =
1/k.
Probability of 𝐾 𝑠,𝑡 to be of 𝑘1 = 𝑘 ;𝑠𝑡

15. Probability of 𝐾 𝑠,𝑡 to be monochromatic = k. 𝑘 ;𝑠𝑡
= 𝑘 1;𝑠𝑡
𝑛 𝑠:𝑡
Number of 𝐾 𝑠,𝑡 that can be selected from 𝐾 𝑛 = 𝑠:𝑡 𝑠
Probability that there exist a monochromatic 𝐾 𝑠,𝑡 in 𝐾 𝑛 is
𝑛 𝑠:𝑡
P(𝐾 𝑠,𝑡 ) = 𝑠:𝑡 𝑠
𝑘 1;𝑠𝑡
elementary calculations have shown that when we choose
1/(𝑠:𝑡)
n ≤ (2 π 𝑠𝑡) ((s + t)/𝑒 2 )𝑘 (𝑠𝑡;1)/(𝑠:𝑡)

16. then , P(𝐾 𝑠,𝑡 ) < 1
Which means for
1/(𝑠:𝑡)
n ≤ (2 π 𝑠𝑡) ((s + t)/𝑒 2 )𝑘 (𝑠𝑡;1)/(𝑠:𝑡)
we have a coloring in which no monochromatic 𝐾 𝑠,𝑡 exists.
Hence, R(𝐾 𝑠,𝑡 ; k) > n

17. 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)