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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 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).
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 .
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.
.
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
    𝟐 𝒙 (here x is no. of edges) has either a red 𝐾 𝑠
or blue 𝐾 𝑡 .

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 ∈ 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 .
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
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.
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
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
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)
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
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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Ramsey number lower bounds

  • 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)