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### hex

Course: MATH 550, Fall 2009
School: McGill
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Word Count: 1283

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on Notes HEX Tibor Szab o MATH 550, Winter 2009 Graph Theory and Combinatorics The board of the game HEX consists of hexagonal cells arranged in a rhombus shaped lattice of size 11 11. The game was invented by the Danish mathematician Piet Hein in 1942. There are two players RED and BLUE. The players alternately occupy cells of the board with stones of their color. The goal of player RED is to connect the top...

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on Notes HEX Tibor Szab o MATH 550, Winter 2009 Graph Theory and Combinatorics The board of the game HEX consists of hexagonal cells arranged in a rhombus shaped lattice of size 11 11. The game was invented by the Danish mathematician Piet Hein in 1942. There are two players RED and BLUE. The players alternately occupy cells of the board with stones of their color. The goal of player RED is to connect the top side to the bottom side with red stones, while player BLUE wants to connect the left side to the right side with blue stones. The winner is the one who succeeds. (Remark: The Jordan Curve Theorem implies that there is no position with all cells occupied such that both players succeeded.) While it is not immediate from the denition, one can prove that HEX is Maker/Breaker-type game. For this, consider the Maker/Breaker game that has the set of cells as its board X and all the paths connecting top to bottom as the family F of winning sets (See the notes on Positional Games for denitions). Consider the game where RED is Maker, who tries to occupy a winning set, that is, a path from top to bottom. If RED wins as Maker then he clearly also wins as RED in HEX. On the other hand if BLUE wins as Breaker, that only means he prevented RED in occupying a top bottom RED path. Did he also win in HEX? That is, did he create a left-right blue path? According to a theorem John Nash, the answer is yes. Hence HEX is indeed a Maker/Breaker game. Theorem 1 (Nash, 1947) HEX cannot end in a draw. The fact that after a complete game of HEX one of the players must have won, eventually follows from Sperners Lemma (that is in turn equivalent to Brouwers xed point theorem). Let C be a plane graph that is isomorphic to a cycle. We say that the plane graph G is a triangulation of C if C is a subgraph of G, all other vertices of G are in the interior face of C (there is an interior by the Jordan Curve Theorem), and all the faces of G except the innite face is a triangle. 1 Lemma 2 (Sperners Lemma) Let us be given a triangulation G of the cycle C with three designated vertices v1 , v2 , v3 V (C) and a coloring c : V (G) {1, 2, 3} of the vertices with three colors having the following properties: c(vi ) = i for i = 1, 2, 3. For any pair i, j {1, 2, 3}, i = j, the vertices on the segment of C between vi and vj use only colors i and j. Then G contains a polychromatic triangle face (a triangle face whose vertices have three distinct colors). Proof. The proof follows from the Handshaking Lemma. We create an auxiliary graph H from the colored triangulation. The vertex set V (H) consists of the faces of G. Two faces F and F are adjacent in H if they have a common edge of G on their boundary whose end points x, y have distinct colors: c(x) = c(y). Let us look at the degrees in H. Let F be a triangle face of G. If the vertices of F all have the same color, then F has degree 0. If there are two identical and one dierent color among the three vertices, then F has degree 2 in H. A polychromatic triangle, should G contain one, would have degree three. What is the degree of the innite face? For any pair i, j {1, 2, 3}, i = j, the segment of C between vi and vj contains an odd number of color ips between i and j, since the starting and ending points have dierent colors. Hence there are an odd number of edges going the from innite region to the triangles faces that are along one segment of C. Hence the degree of the innite region is the sum of three odd numbers, odd itself. The auxiliary graph H we constructed must have an even number of odd vertices so there must be an odd number of polychromatic triangles. Hence there is at least one. Remark Sperners Lemma can be extended analogously to higher dimensions, to appropriately colored triangulations of the n-dimensional simplex. Let G be a triangulation of cycle C with three distinct designated vertices v1 , v2 , v3 . These vertices dene three path segments of C: C1 , C2 , C3 , such that for every i, j {1, 2, 3}, i = j, we have that V (Ci ) V (Cj ) = {vk }, where k {1, 2, 3}, k = i, k = j. A connector of G is a subgraph consisting of a vertex v V (G) and three paths P1 , P2 , and P3 that connect v to C1 , C2 , and C3 , respectively. 2 Lemma 3 (Connector Lemma) (Hochberg-McDiarmid-Saks, 1995) Let G be a triangulation of cycle C with three distinct designated vertices v1 , v2 , v3 . For any blue/red-coloring of V (G) there exists a monochromatic connector. Proof. We assume there is no monochromatic connector. We create a threecoloring of V (G) that satises the conditions of Sperners Lemma. For each vertex v there is some segment of C that cannot be reached from v via a monochromatic path. For c(v) we select the smallest index i for which there is no monochromatic path from v to segment Ci . This coloring clearly satises the properties of Sperners Lemma. Hence there is monochromatic triangle face T . Two of the vertices of T have the same color in the red/bluecoloring of our Lemma. Let us say that vertices x and y of T are both colored red. Let us assume w.l.o.g. that c(x) < c(y). On the one hand we have no red path from x to segment Cc(x) . On the other hand, by the minimality of index c(y), there is a red path from y to segment Cc(x) , since c(x) < c(y). This is obviously a contradiction, since going rst from x to y and then ...

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