aa r X i v : . [ m a t h . C O ] F e b HEX IMPLIES Y
TOMASZ PRYTU LA
Abstract.
We give a simple and short proof of the fact that the board gameof Y cannot end in a draw. Our proof, based on the analogous result for thegame of Hex (the so-called ‘Hex Theorem’), is purely topological and does notdepend on the shape of the board. We also include a simplified version ofGale’s proof of Hex Theorem. Introduction
Games of Hex and Y are turn-based, two player, abstract strategy games, be-longing to a family of connection games. Both games, in their original form, areplayed on a subset of a hexagonal tessellation of the plane, where the players (com-monly denoted by red and blue) take turns to place stones on unoccupied hexagonalcells. Once the stone is placed it cannot be moved or removed. Hex is played on arectangular board, whose two pairs of opposite sides are denoted by red, and bluerespectively. The goal for the red player is to create a path of red stones joiningtwo red sides, and the goal for the blue player is to create a blue path joining twoblue sides. The game of Y is played on a triangular board, and the goal for bothplayers is to create a connected chain of stones joining all three sides of the board(such a chain will generically have a shape of the letter Y, hence the name of thegame). Figure 1 shows exemplary boards and positions for Hex and Y.
Figure 1.
Boards for games Hex and Y. On both boards theposition is won by red.Hex was first described by Hein in 1942, and rediscovered by Nash in 1948,and it has been quite extensively studied since (see [Gar59] for a brief history ofHex). The game of Y is an interesting variation on the theme, and was discoveredindependently by Milnor and Shannon in 1950s, and rediscovered by Schensted and
Mathematics Subject Classification.
Key words and phrases.
Game of Hex, Game of Y, Hex Theorem, connection game.The author was supported by the EU Horizon 2020 program under the Marie Sk lodowska-Curiegrant agreement no. 713683 (COFUNDfellowsDTU).
Titus in 1953 [Gar08, Nas52, ST75]. An appealing feature of Y is that the objectiveof the game for both players is the same, as opposed to Hex.One of the main mathematical properties of Hex is that it cannot end in a draw.More precisely, once the Hex board is completely filled with stones, then there iseither a red path joining the red sides, or a blue path joining the blue sides, butnot both. This is known as ‘Hex Theorem’. The same is true for the game of Y;we will refer to it as ‘Y Theorem’. Over the years multiple proofs of Hex Theoremappeared (see e.g., [BBC00, Gal79, Ber76]), however, there are few arguments for YTheorem, and all the proofs known to us are based on the combinatorial propertiesof the board and use some sort of induction or recursion [vR06, HvR06]. The mainpurpose of this note is to show that Y Theorem is in fact equivalent to Hex Theorem,and thus obtain a new, very simple and purely topological proof of Y Theorem.
Theorem.
Hex Theorem and Y Theorem are equivalent.
The fact that Y Theorem implies Hex Theorem is an easy trick, as a game ofHex can be seen as a continuation of a game of Y from a certain position. This wasobserved by Schensted [ST75], [vR06, Section 4.7]. Our efforts thus go into provingthe converse implication. The key idea is to ‘double’ the Y board by reflecting italong one of its sides, and then treat the resulting 4–gon as a Hex board.We present both theorems in a generalized form, where the game is played onan arbitrary triangulation of a 2–disk. This generality, besides the obvious benefitof obtaining a more general result, also allows us to use precise language of graphtheory.For the sake of completeness we also include the proof of Hex Theorem. We claimno originality for the ideas behind this proof, apart from a minor simplification atthe very end, which allows us to keep the proof concise. This note may thus alsoserve as a short and self-contained proof of Hex Theorem in a generalized setting.2.
Hex and Y in a generalized form
We present games of Hex and Y with boards being triangulations of a disk, withlabels on the boundary vertices, such that players color vertices. To obtain thisview from the classical one, one takes a dual triangulation to the Hex or the Yboard, and then one labels the sides accordingly; see Figure 2.
Figure 2.
Dual triangulations to Hex and Y boards.
Board B : Let B be a triangulation of a disk. A path between two vertices, a and b in B is a sequence of vertices a = v , v , . . . , v k = b such that any twoconsecutive vertices are connected by an edge. The boundary cycle of B splits intoa concatenation of k paths α , . . . , α k . That is, the end vertex of path α i is thebeginning vertex of path α i +1 (indices taken mod k ). We call paths α i the sides EX IMPLIES Y 3 of B . To avoid certain degenerate cases, we assume that every α i contains at leastone edge (and thus at least two vertices). Games of Hex and Y:
Players take turns to color the vertices of the trian-gulation of board B with red and blue. One can color only non-colored vertices,and once a vertex is colored, it stays colored until the end of the game. The gamesdiffer in the shape of the board and the winning condition. A subset of vertices A of B is connected if for any two vertices in A there is a path in A joining these twovertices. A chain is a connected subset of vertices having the same color. Game of Hex:
The board has 4 boundary paths, called R , B , R , B . Thegoal for the red player is to create a red chain with at least one vertex in R and atleast one vertex in R . The goal for the blue player is to create a blue chain withat least one vertex in B and at least one vertex in B . Game of Y:
The board has 3 boundary paths l , l , l . The goal for eitherplayer is to form a chain of their color, containing at least one vertex in l , l , and l . 3. Equivalence of Hex Theorem and Y Theorem Y ⇒ Hex.
This observation appears throughout the literature and is attributedto Schensted [ST75]. Consider a Hex board B . Add one vertex r and join it byedges to all vertices of R . Add one vertex b and join it by edges to all vertices of B . Color r red, and color b blue. Now we view the obtained board B ′ as a Y game, with l = { b }∪{ ( B ∩ R ) }∪{ r } , l = B ∪{ r } , and l = R ∪{ b } . (Sinceevery side of B contains at least one edge, it is clear that B ′ is a triangulation of adisk.) Boards B and B ′ are presented in Figure 3. B R R B b r l l l Figure 3.
Extending a Hex board to a position on a Y board.The reader easily sees that playing Y from this position onwards is the same asplaying Hex on the original board. By Y Theorem, this particular position has aunique winner, thus so does the corresponding Hex game.
Hex ⇒ Y. First we prove that there exists at least one winner. Consider aY board B with sides l , l , and l . Assume that every vertex of B is colored.Take another copy B ′ of B (together with the coloring) and glue it to B along theidentity map on l . Since l has at least one edge, the resulting space B ∪ l B ′ isa triangulation of a disk with 4 boundary paths. Let l ′ and l ′ denote the sides of B ∪ l B ′ opposite to l and l respectively. Treat B ∪ l B ′ as a Hex board, where l = R , l = B , l ′ = B and l ′ = R . The board B ∪ l B ′ is shown in Figure 4. By T. PRYTU LA
Hex Theorem there is either a red chain joining R and R or a blue chain joining B and B . Assume that it is a red chain and call it C ′ . The case of the blue chainis done analogously.Note that since l is a disconnecting subset of B ∪ l B ′ , chain C ′ has to have atleast one vertex in l . Now let C be a subset of B which is the image of C ′ underthe map p : B ∪ l B ′ → B which folds B ′ onto B . Note that p is simplicial (i.e., iftwo vertices are connected by an edge, then so are their images), color-preservingand it restricts to the identity map on B . Therefore C = p ( C ′ ) is a chain as well.Since C ′ has a vertex in R = l ′ , and since p ( l ′ ) = l , we get that C has verticeson all three sides of B , thus ensuring a win for red. l = B l = R l ′ = R l ′ = B l BB ′ C ′ C p
Figure 4.
Doubling the Y board, and folding it back onto itself.Chain C ′ is pink. Chain C is dashed red.Now we show that there is a unique winner. Let B, B ′ and B ∪ l B ′ be as aboveand let s : B ∪ l B ′ → B ∪ l B ′ be the reflection across l . Now suppose we havea chain C connecting all three sides of B . Then the union C ∪ s ( C ) is a chainconnecting all four sides of B ∪ l B ′ ; see Figure 5. Thus, having two winners of the Y game on B leads to two winners of the corresponding Hex game on B ∪ l B ′ .4. Proof of Hex Theorem
We present essentially the same proof as Gale [Gal79], with one nuance: ratherthan defining a fixed-point free map of a disk to itself, we use our generalizedrepresentation of a board to directly construct a retraction from a disk onto itsboundary circle.Consider a Hex board B , with sides R , B , R , B . Add four vertices r − , r + , b − , b + and connect r − to both b − and b + , and r + to both b − and b + . Then connectevery vertex of R to r − , every vertex of R to r + , every vertex of B to b − , andevery vertex of B to b + . Thus we get a triangulation of a slightly larger disk D ,which is presented in Figure 6. The original board and the game are expressed interms of the new board as follows. The boundary vertices r − and r + are (from thebeginning of the game) colored red, and the vertices b − and b + are colored blue.Hex Theorem for B is equivalent to the following statement for D : there is eithera red chain form r − to r + , or a blue chain from b − to b + , but not both. EX IMPLIES Y 5 l = B l = R l ′ = R l ′ = B l BB ′ s ( C ) C s
Figure 5.
A winner on a Y board leads to a winner on a Hexboard. Chain C is pink. Chain s ( C ) is dashed red. Proof.
First we prove that there is at least one winner. Assume that every vertex of D is colored. Let V + denote the subset of all red vertices of D which are connectedby a red chain to r + . Let V − be defined as all the red vertices that are not in r − . Define subsets of W + and W − for the blue vertices analogously. By definition,there is no edge between a vertex in V + and a vertex of V − , and the same holdsfor W + and W − . Note that also by definition we have that r + is in V + and b + isin W + .Suppose by contrary that there is no winner. This implies that r − is not in V + and thus it is in V − . For the same reason we have that b − is in W − . Let S = ( r − , b − , r + , b + ) denote the boundary cycle of D . We define a map D → S onvertices by sending the above subsets as follows: V − r − ,V + r + ,W − b − ,W + b + . Observe that this assignment gives a simplicial map. Therefore it induces acontinuous map D → S with the topology coming from the standard Euclideanmetric on every triangle. One easily sees that this map is the identity on S , whichgives a retraction of the disk onto its boundary circle, a contradiction [Hat02,Corollary 2.15].It remains to show that there cannot be two winners. Assume the contrary. Thuswe have chains of respective color joining r − to r + and b − to b + , which by definitionlie inside of D and do not intersect each other. By taking such chains with minimalnumber of vertices we can assume that they are embedded paths. Embed D intothe plane, add one more vertex c on the exterior of D , and connect all vertices r − , r + , b − , b + to c by edges such that they do not intersect one another (the readereasily sees that it can always be done, we present one such configuration in Figure 6). T. PRYTU LA b − b + r + r + r + r + r + r + r − r − r − r − r − r − cD Figure 6.
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Department of applied mathematics and computer science, Technical University ofDenmark, Lyngby, Denmark
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