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13 Ramsey Theory The Ramsey number R(r,s) is the smallest n for which any 2-coloring of K n contains a monochromatic red K r or a monochromatic blue K s where r,s 2.

14 Examples R(2,2) = 2 R(3,3) = 6 R(4,4) = 18 R(5,5) = 58 R(6,6) 102

15 Propositions of Ramsey Theory R(a,b) = R(b,a) R(2,m) = m R(3,m) = 3m!

16 Generalized Ramsey Theory Ramsey Theory can be generalized to have r colorings. Define for natural numbers I 1,...,I r the Ramsey number R(I 1, I 2,..., I r ) to be the least number n for which any r-coloring of K n contains a monochromatic r color K I r.

17 Chromatic number The chromatic number χ(h) is defined to be the least number of colors needed to color the vertices of H so that the endpoints of each edge are not all the same color. If χ(h) > 2 then every 2-coloring of K n contains a monochromatic K m

18 Overview Helpful to think about combinatorial problems in terms of games. 1. Achievement game 2. Avoidance game 3. Pekec s game (combination of Achievement game and Avoidance game) 1 / 9

19 Achievement Game 1. Start with K n and 2 players, each with a unique color. 2. Give each player a graph (P1 gets G, P2 gets H). 3. Players alternate turns, coloring an edge on K n each turn. 4. Aim: Each player aims to color a subgraph isomorphic to their graph in their color. 2 / 9

20 Achievement Game 1. Start with K n and 2 players, each with a unique color. 2. Give each player a graph (P1 gets G, P2 gets H). 3. Players alternate turns, coloring an edge on K n each turn. 4. Aim: Each player aims to color a subgraph isomorphic to their graph in their color. Imagine we are exploring the case of R(3, 3) = 6. We have two players trying to color monochromatic K 3 s. If n = 6, so the players are coloring K 6, then there will be a winner. Because we have shown R(3, 3) = 6. 3 / 9

21 Achievement Game 1. Start with K n and 2 players, each with a unique color. 2. Give each player a graph (P1 gets G, P2 gets H). 3. Players alternate turns, coloring an edge on K n each turn. 4. Aim: Each player aims to color a subgraph isomorphic to their graph in their color. Imagine we are exploring the case of R(3, 3) = 6. We have two players trying to color monochromatic K 3 s. If n = 6, so the players are coloring K 6, then there will be a winner. Because we have shown R(3, 3) = 6. If n = 5, so the players are coloring K 5, then there may be tie. If n = 3, then the players must tie (because neither can color a K 3 ). 4 / 9

22 Achievement Game Theorem Theorem In the achievement game, if all players are trying to achieve the same graph (i.e. G = H) and there is a winning strategy, then it belongs to the first player. Proof idea. The proof uses a strategy stealing argument. If the second player had a winning stategy, it would be a strategy on K n 1, then the first player could use that strategy on K n and win. Another way to think about it is that it s never bad for a player to color an edge, so the first person to go must have the winning strategy - just like tic-tac-toe. Corollary Suppose the players are trying to build the same graph. There exists some n such that for all m n, if the players are coloring K m, then the first player wins with best play. Proof: n = R(G, G). 5 / 9

23 Avoidance Game Same setup, except that the players want to avoid building a monochromatic subgraph isomorphic to their graph. If they do make one, then they lose. The theorem above does not hold for the avoidance game. Depending on the circumstance, coloring an edge could either help or hurt a player. Say the player can force the other player to build the graph by coloring an edge, then it s good to color an edge. Say if the player colors any edge, they build their monochromatic subgraph, then it s bad to color an edge. For two players playing on K 6, exhaustive computer search shows that player 2 wins with best play. The Ramsey number R(G, H) gives an upper bound on n for which if they play on K n, at least one player loses. 6 / 9

24 Pekec s Game Now we combine the Achievement and Avoidance game. One player (achiever) tries to create a monochromatic G in their color. Other player (avoider) tries to prevent the creation of a monochromatic G in the opponent s color. Lemma If Pekec s game is played on K m and the achiever is trying to create a monochromatic G where m 2R(G, G), then the achiever wins with best play. 7 / 9

25 Pekec s Game Lemma Lemma If Pekec s game is played on K m and the achiever is trying to create a monochromatic G where m 2R(G, G), then the achiever wins with best play. Proof. We prove by giving a strategy for the achiever. 1. Split edges into three sets, two of which form the complete subgraph on R(H, H) vertices, and the other has the remaining edges. 2. Achiever assigns an isomorphism between the two K R(H,H) subgraphs. 3. Whenever the opponent colors an edge in one subgraph, the achiever colors the corresponding edge in the other subgraph. 4. If the opponent colors an edge in the third set, so does the achiever until there are more edges. (technically determines are ordering, but not really important) 5. By definition of R(H, H), if the achiever s monochromatic H does not appear their color is the first set of edges, then it will appear in the second set of edges in the achiever s color. 6. So the acheiver wins. 8 / 9

26 Explanation of Proof of Lemma Say G = K 3. Say m = 15. Note that 15 =m 2R(G, G) = 2R(3, 3) = 12. Then set1 = K 6 K 1 5. Then set2 = K 6 K 1 5. Then set3 = remaining 3 edges. After coloring, set1 and set2 will look identical, except that colors will be flipped. i.e. e set1 has color c f (e) set2 has color c. So there will be a monochromatic K 3 in set1 and set2, each of a different color. Hence there is a monochromatic K 3 of the achiever s color, so they win! 9 / 9

27 Vertex Coloring A legal vertex coloring on a simple graph f : V (G) [r] is created by assigning one of r colors to each vertex in the graph G such that no adjacent vertices are the same color. The chromatic number χ(g) is the smallest number of colors, k, where a legal vertex coloring is possible.

28 Coloring the Plane The Hadwiger?Nelson Problem: What is the chromatic number of the plane, χ(r 2 )? Setup: Define the infinite unit distance graph of R 2 as R 2, the graph generated by taking each point in R 2 as a vertex and marking vertices as adjacent (i.e connecting them with an edge), iff they are unit distance apart. Theorem (Bruijn-Erdős Theorem) For every infinite graph G and every finite integer k, G can be colored with k colors iff all of its finite subgraphs can be colored with k colors.

29 Upper and Lower Bounds for χ(r 2 ) The best known bounds are 4 χ(r 2 ) 7. How do we find these bounds?

30 Upper and Lower Bounds for χ(r 2 ) The best known bounds are 4 χ(r 2 ) 7. How do we find these bounds? It s easy to show that if we try to color the plane with one or two colors, there will be always be a case where two points unit distance apart are assigned the same color.

31 Upper and Lower Bounds for χ(r 2 ) The best known bounds are 4 χ(r 2 ) 7. How do we find these bounds? It s easy to show that if we try to color the plane with one or two colors, there will be always be a case where two points unit distance apart are assigned the same color. The Moser Spindle shows that χ(r 2 ) 4

32 Upper and Lower Bounds for χ(r 2 ) The best known bounds are 4 χ(r 2 ) 7. How do we find these bounds? It s easy to show that if we try to color the plane with one or two colors, there will be always be a case where two points unit distance apart are assigned the same color. The Moser Spindle shows that χ(r 2 ) 4 If we tesselate the plane using regular hexagons (of diameter slightly less than unit distance) we can show that it is possible to color the plane with 7 colors, which yields the upper bound χ(r 2 ) 7

33 Homework Problems 1. Draw another unit distance graph, G with χ(g) = Show that R(2, m) = m. * Hint: One possible answer uses 10 vertices.