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What is Ramsey Theory ?

What is Ramsey Theory ?. What is Ramsey Theory ? It might be described as the study of unavoidable regularity in large structures. Complete disorder is impossible. T. Motzkin. What is Ramsey Theory ? It might be described as the study of unavoidable regularity in large structures.

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What is Ramsey Theory ?

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  1. What is Ramsey Theory?

  2. What is Ramsey Theory? It might be described as the study of unavoidable regularity in large structures.

  3. Complete disorder is impossible. T. Motzkin What is Ramsey Theory? It might be described as the study of unavoidable regularity in large structures.

  4. Ramsey’s Theorem(1930) For any k < l and r, there exists R = R(k,l,r) so that for any r-coloring of the k-element sets of an R-element set, there is always some l-element set with all of its k-element subsets having the same color.

  5. Frank Plumpton Ramsey (1903-1930) Ramsey’s Theorem(1930) For any k < l and r, there exists R = R(k,l,r) so that for any r-coloring of the k-element sets of an R-element set, there is always some l-element set with all of its k-element subsets having the same color.

  6. k - finite X Euclidean Ramsey Theory

  7. k - finite X k Cong(X) - family of all X which are congruent to X Euclidean Ramsey Theory (i.e., “copies” of X up to some Euclidean motion)

  8. k - finite X k Cong(X) - family of all X which are congruent to X N C = i ‘ we have X Euclidean Ramsey Theory (i.e., “copies” of X up to some Euclidean motion) X is said to be Ramsey if for all r there exists N = N(X,r) such that for every partition , ‘ Cong(X) and some i. for some X

  9. k - finite X k Cong(X) - family of all X which are congruent to X N C = i ‘ we have X r N X Euclidean Ramsey Theory (i.e., “copies” of X up to some Euclidean motion) X is said to be Ramsey if for all r there exists N = N(X,r) such that for every partition , ‘ Cong(X) and some i. for some X

  10. N N If then there is a finite subset Y such that Y r N r X X Compactness Principle

  11. N N If then there is a finite subset Y such that Y r N r X X Compactness Principle Example 1 |X| = 2 X =

  12. N N If then there is a finite subset Y such that Y r r N a unit simplex in . r X X Compactness Principle Example 1 |X| = 2 X = r to be the r+1 vertices of For a given r, take Y r

  13. N N If then there is a finite subset Y such that Y r r r N a unit simplex in . r X X r Compactness Principle Example 1 |X| = 2 X = r to be the r+1 vertices of For a given r, take Y r Then Y X .

  14. n n Let Q denote the set of 2 vertices of the n-cube.

  15. n n Let Q denote the set of 2 vertices n of the n-cube. Then Q is Ramsey.

  16. n n Let Q denote the set of 2 vertices n of the n-cube. Then Q is Ramsey. Theorem. For any k and r, there exists N = N(k,r) such that any r-coloring of Q contains a monochromatic Q . N k

  17. n n Let Q denote the set of 2 vertices n of the n-cube. Then Q is Ramsey. Theorem. For any k and r, there exists N = N(k,r) such that any r-coloring of Q contains a monochromatic Q . N k Idea of proof: (induction) k = 1 Choose N(1,r) = r + 1

  18. n n Let Q denote the set of 2 vertices n of the n-cube. Then Q is Ramsey. Theorem. For any k and r, there exists N = N(k,r) such that any r-coloring of Q contains a monochromatic Q . N k r+1 …… Idea of proof: (induction) k = 1 Choose N(1,r) = r + 1 Consider the r + 1 points:

  19. (……………,1,……………,0,……………) and (……………,0,……………,1,……………) Since only r colors are used then some pair must have the same color, say 1 This is a monochromatic Q .

  20. (……………,1,……………,0,……………) and (……………,0,……………,1,……………) Since only r colors are used then some pair must have the same color, say 1 This is a monochromatic Q . So far, so good!

  21. N + 2 N 1 r+1 k = 2 Choose N(2,r) = (r + 1) + (r + 1) =

  22. N + 2 N 1 N N 2 1 N N 1 2 r+1 k = 2 Choose N(2,r) = (r + 1) + (r + 1) = Consider the points: (1,0,0,…,0,1,0,0,…,0) (1,0,0,…,0,0,1,0,…,0) …… (1,0,0,…,0,0,0,0,…,1) (0,1,0,…,0,1,0,0,…,0) (0,1,0,…,0,0,1,0,…,0) …… (0,1,0,…,0,0,0,0,…,1) (0,0,1,…,0,1,0,0,…,0) ………

  23. N + 2 N 1 N N 2 1 N N 1 2 r+1 k = 2 Choose N(2,r) = (r + 1) + (r + 1) = Consider the points: (1,0,0,…,0,1,0,0,…,0) (1,0,0,…,0,0,1,0,…,0) …… (1,0,0,…,0,0,0,0,…,1) (0,1,0,…,0,1,0,0,…,0) (0,1,0,…,0,0,1,0,…,0) …… (0,1,0,…,0,0,0,0,…,1) (0,0,1,…,0,1,0,0,…,0) ………

  24. N + 2 N N 1 2 N 1 N N 2 1 N N 1 2 r+1 k = 2 Choose N(2,r) = (r + 1) + (r + 1) = Consider the points: (1,0,0,…,0,1,0,0,…,0) (1,0,0,…,0,0,1,0,…,0) (1,0,0,…,0) …… (1,0,0,…,0,0,0,0,…,1) (0,1,0,…,0,1,0,0,…,0) (0,1,0,…,0,0,1,0,…,0) …… (0,1,0,…,0,0,0,0,…,1) (0,0,1,…,0,1,0,0,…,0) ………

  25. N + 2 N N 1 2 N 1 N N 2 1 N N 1 2 r+1 k = 2 Choose N(2,r) = (r + 1) + (r + 1) = Consider the points: (1,0,0,…,0,1,0,0,…,0) (1,0,0,…,0,0,1,0,…,0) (1,0,0,…,0) …… (1,0,0,…,0,0,0,0,…,1) (0,1,0,…,0,1,0,0,…,0) (0,1,0,…,0,0,1,0,…,0) …… (0,1,0,…,0,0,0,0,…,1) (0,0,1,…,0,1,0,0,…,0) ………

  26. N + 2 N N N N 2 1 2 1 N 1 N N 2 1 N N 2 1 r+1 k = 2 Choose N(2,r) = (r + 1) + (r + 1) = Consider the points: (1,0,0,…,0,1,0,0,…,0) (1,0,0,…,0,0,1,0,…,0) (1,0,0,…,0) …… (1,0,0,…,0,0,0,0,…,1) (0,1,0,…,0,1,0,0,…,0) (0,1,0,…,0,0,1,0,…,0) (0,1,0,…,0) …… (0,1,0,…,0,0,0,0,…,1) (0,0,1,…,0,1,0,0,…,0) ………

  27. N + 2 N N N N 2 1 1 2 N 1 points: N N (1,0,0,…,0,1,0,0,…,0) 2 1 (1,0,0,…,0,0,1,0,…,0) (1,0,0,…,0) …… N N 1 2 (1,0,0,…,0,0,0,0,…,1) (0,1,0,…,0,1,0,0,…,0) (0,1,0,…,0,0,1,0,…,0) (0,1,0,…,0) …… (0,1,0,…,0,0,0,0,…,1) (0,0,1,…,0,1,0,0,…,0) ……… ……… r+1 k = 2 Choose N(2,r) = (r + 1) + (r + 1) = Consider the

  28. N N N N N N 1 1 2 1 2 1 + Since the N points represented by the can be r-colored in at most r ways, then the original r-coloring of Q induces an r - coloring of Q . ‘s 2

  29. N N N N N N N 1 1 1 2 1 1 2 + Since the N points represented by the can be r-colored in at most r ways, then the original r-coloring of Q induces an r - coloring of Q . ‘s 2 r+1 Since N = r + 1 = r + 1, some pair has the same coloring, say 2

  30. N N N N N N N 1 1 2 1 2 1 1 i j 1 1 i 2 + j 2 Since the N points represented by the can be r-colored in at most r ways, then the original r-coloring of Q induces an r - coloring of Q . ‘s 2 r+1 Since N = r + 1 = r + 1, some pair has the same coloring, say 2 (………,1,………0,……) (identical colorings) (………,0,………1,……)

  31. N N N N N N N 1 1 1 2 1 2 1 i j 1 1 i 2 + j 2 Since the N points represented by the can be r-colored in at most r ways, then the original r-coloring of Q induces an r - coloring of Q . ‘s 2 r+1 Since N = r + 1 = r + 1, some pair has the same coloring, say 2 (………,1,………0,……) monochromatic by choice of N 1 (………,0,………1,……)

  32. N N N N N N N 1 1 1 2 1 2 1 i j 1 1 i 2 + j 2 Since the N points represented by the can be r-colored in at most r ways, then the original r-coloring of Q induces an r - coloring of Q . ‘s 2 r+1 Since N = r + 1 = r + 1, some pair has the same coloring, say 2 (………,1,………0,……) Thus, all 4 are monochromatic (………,0,………1,……)

  33. N N N N N N N 1 1 1 2 1 1 2 i j 1 1 i 2 + j 2 Since the N points represented by the can be r-colored in at most r ways, then the original r-coloring of Q induces an r - coloring of Q . ‘s 2 r+1 Since N = r + 1 = r + 1, some pair has the same coloring, say 2 (………,1,………0,……) Thus, all 4 are monochromatic (………,0,………1,……) 2 These 4 points form a monochromatic Q :

  34. N N N N N N N 1 1 2 1 1 2 1 i j 1 1 i 2 + (……,1,………0,......,1,……,0,……) j 2 (……,1,………0,......,0,……,1,……) i i j j 1 2 2 1 (……,0,………1,......,1,……,0,……) (……,0,………1,......,0,……,1,……) Since the N points represented by the can be r-colored in at most r ways, then the original r-coloring of Q induces an r - coloring of Q . ‘s 2 r+1 Since N = r + 1 = r + 1, some pair has the same coloring, say 2 (………,1,………0,……) Thus, all 4 are monochromatic (………,0,………1,……) 2 These 4 points form a monochromatic Q :

  35. For k = 3, we can take N(3,r) = N + N + N 1 3 2 where

  36. For k = 3, we can take N(3,r) = N + N + N 1 3 2 where Continuing this way, the theorem is proved.

  37. For k = 3, we can take N(3,r) = N + N + N 1 3 2 where Continuing this way, the theorem is proved. Note that by this technique, the bounds we get are rather large.

  38. For k = 3, we can take N(3,r) = N + N + N 1 3 2 27 For example, it shows that N(4,2) 2 + 13. where Continuing this way, the theorem is proved. Note that by this technique, the bounds we get are rather large.

  39. For k = 3, we can take N(3,r) = N + N + N 1 3 2 27 For example, it shows that N(4,2) 2 + 13. where Continuing this way, the theorem is proved. Note that by this technique, the bounds we get are rather large. What is the true order of growth here?

  40. With this technique, we can prove the: Product Theorem. If X and Y are Ramsey then the Cartesian product X Y is also Ramsey. x

  41. With this technique, we can prove the: Product Theorem. If X and Y are Ramsey then the Cartesian product X Y is also Ramsey. x Corollary: (Any subset of) the vertices of an n-dimensional rectangular parallelepiped is Ramsey.

  42. With this technique, we can prove the: Product Theorem. If X and Y are Ramsey then the Cartesian product X Y is also Ramsey. x Corollary: (Any subset of) the vertices of an n-dimensional rectangular parallelepiped is Ramsey. For example, any acute triangle is Ramsey.

  43. With this technique, we can prove the: Product Theorem. If X and Y are Ramsey then the Cartesian product X Y is also Ramsey. x Corollary: (Any subset of) the vertices of an n-dimensional rectangular parallelepiped is Ramsey. For example, any acute triangle is Ramsey. What about ?

  44. How can we get obtuse Ramsey triangles?

  45. n How can we get obtuse Ramsey triangles? Example. Choose n = R(7, 9, r) and consider the set S of points x in having all coordinates zero except for 7 coordinates which have in order the values 1, 2, 3, 4, 3, 2, 1.

  46. x = (0 0 0 1 0 2 0 0 0 3 4 0 0 3 0 0 0 0 2 0 1 0 0 0 0) n How can we get obtuse Ramsey triangles? Example. Choose n = R(7, 9, r) and consider the set S of points x in having all coordinates zero except for 7 coordinates which have in order the values 1, 2, 3, 4, 3, 2, 1.

  47. x = (0 0 0 1 0 2 0 0 0 3 4 0 0 3 0 0 0 0 2 0 1 0 0 0 0) n n There are ( ) such points in S. 7 How can we get obtuse Ramsey triangles? Example. Choose n = R(7, 9, r) and consider the set S of points x in having all coordinates zero except for 7 coordinates which have in order the values 1, 2, 3, 4, 3, 2, 1.

  48. x = (0 0 0 1 0 2 0 0 0 3 4 0 0 3 0 0 0 0 2 0 1 0 0 0 0) n n There are ( ) such points in S. 7 How can we get obtuse Ramsey triangles? Example. Choose n = R(7, 9, r) and consider the set S of points x in having all coordinates zero except for 7 coordinates which have in order the values 1, 2, 3, 4, 3, 2, 1. Any r-coloring of S induces an r-coloring of the 7-sets of {1,2,……,n}

  49. By the choice of n = R(7, 9, r), there exists some 9-set with all its 7-sets having the same color.

  50. i i i i i i i i i 3 9 6 5 7 4 1 2 8 By the choice of n = R(7, 9, r), there exists some 9-set with all its 7-sets having the same color. x = (……x ……x ……x ……x ……x ……x ……x ……x ……x ……) A = (…….1…….2…….3……..4…….3…….2…….1……..0…….0……) B = (…….0…….1…….2……..3…….4…….3…….2……..1…….0……) C = (…….0…….0…….1……..2…….3…….4…….3……..2…….1……)

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