Nº 80 - Universidad Nacional de Colombia

Transcripción

FCE
Econografos
Nº 80
Agosto 2015
BASIC ANALYSIS OF THE HEX GAME
ANÁLISIS BÁSICO DEL JUEGO HEX
Mike Woodcock, Fernando Uscategui
y David Corrales
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Econografos Escuela de Economía Nº 80
Agosto 2015
BASIC ANALYSIS OF THE HEX GAME
Mike Woodcock1, Fernando Uscategui2, David Corrales3
Abstract
The objective of this paper is to analyze the game of Hex through the use of Game Theory
and Graph Theory. Hex is a game where each player must connect two opposite sides by a
continuous path of pieces in a hexagonal grid within a rhombus-shaped board. The size of
the board is usually 14×14 but the game can be found in a wide range of sizes such as 11×11
and 17×17. Although the strategy is not as deep as in chess, it is still complex, and just like
chess, Hex is a no-chance game, and thus it is a perfect candidate to be examined using some
formal tools. This game has some interesting features that make it more interesting; unlike
chess, the game will never end in a tie, second, the number of possible movements is finite
and third, the second player will always win, through this paper, we will show some of these
features.
Keywords: Game theory, Hex, Strategy, Hex Theorem, Strategy-stealing
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JEL Classification: C72, C65
1
Estudiante de sexto semestre de Economía de la Universidad Nacional de Colombia. ([email protected])
Estudiante de quinto semestre de Economía de la Universidad Nacional de Colombia.
([email protected])
3 Estudiante de quinto semestre de Economía de la Universidad Nacional de Colombia. ([email protected])
2
Universidad Nacional de Colombia Sede Bogotá - Facultad de Ciencias Económicas
Mike Woodcock, Fernando Uscategui, David Corrales
ANÁLISIS BÁSICO DEL JUEGO HEX
Resumen
El objetivo de este documento es analizar el juego del Hex a través del uso de algunas
herramientas de la Teoría de Juegos y la Teoría de Grafos. Hex es un juego en el cual cada
jugador debe conectar dos extremos opuestos de un tablero en forma de rombo a través de
un camino continuo de fichas. El tamaño del tablero es usualmente 14×14 pero también se
puede encontrar en tamaños como 11×11 y 17×17. Las estrategias usadas en el Hex, aunque
no tan complicadas como las del ajedrez, son complejas y no existe ningún elemento de
suerte, esto hace al Hex un candidato ideal para ser analizado a través de herramientas
matemáticas formales. El Hex tiene algunas características interesantes: contrario al ajedrez
el juego nunca puede terminar en tablas, el número de movimientos es finito y por último el
primer jugador siempre tendrá una estrategia ganadora. Demostraremos algunas de estas
cosas a lo largo de este documento. Análisis
Palabras clave: Teoría de Juegos, Hex, Estrategia, Teorema Hex, Robo de estrategias.
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Clasificación JEL: C72, C65
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Director Centro Editorial-FCE
Álvaro Zerda Sarmiento
Equipo Centro Editorial-FCE
Nadeyda Suárez Morales
Pilar Ducuara López
Yuly Rocío Orjuela Rozo
Contacto: Centro Editorial FCE-CID
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Este documento puede ser reproducido citando la fuente. El contenido y la forma del presente
material es responsabilidad exclusiva de sus autores y no compromete de ninguna manera a la
Escuela de Economía, ni a la Facultad de Ciencias Económicas, ni a la
Universidad Nacional de Colombia.
Rector
Ignacio Mantilla Prada
Vicerector General
Jorge Iván Bula Escobar
Facultad de Ciencias Económicas
Decano
José Guillermo García Isaza
Vicedecano
Rafael Suárez
Escuela de Economía
Director
Álvaro Martín Moreno Rivas
Coordinador Programa Curricular de
Economía
Raúl Chamorro Narváez
Centro de Investigaciones para
El Desarrollo CID
Director
Manuel José Antonio Muñoz Conde
Subdirectora
Vilma Narváez
FACULTAD DE CIENCIAS ECONÓMICAS
CENTRO DE INVESTIGACIONES PARA EL DESARROLLO - CID
Escuela de Economía
Contents
Introduction ............................................................................................................................6
Basic Strategies ......................................................................................................................8
Long Term Strategy .............................................................................................................11
The Symmetry of Attacking and Defending ....................................................................11
Hex Theorem ....................................................................................................................12
Strategy-Stealing Argument .............................................................................................16
The Extent of Hex Theorem: Several proofs .......................................................................17
Uniqueness of every path .................................................................................................17
Every set is its own subset ............................................................................................17
Two winning paths can’t exist at the same time ...........................................................18
The no return of every path ..............................................................................................18
Closing Notes .......................................................................................................................21
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References ............................................................................................................................23
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Introduction
Hex was invented independently by the mathematicians: Peit Hein in 1942 and John Nash
in 1947, but it wasn’t until 1952 that it started being sold commercially under the name of
Hex. The game is played on a hexagonal grid within a rhombus-shaped board and the
objective is to connect two parallel sides marked with the player’s color through a continuous
path of chips of that same color. The rules are as follows: each player can place only one
chip per turn and the players alternate turns to place a chip in any empty hexagon on the
board, the first player is decided randomly through a coin toss and after the first movement
has been made the second player has the option to either switch colors with his adversary or
to keep his own. This is done to lessen any potential first turn advantage4. The game of Hex
can be analyzed as a non-cooperative zero-sum game with perfect information. It is noncooperative because given the payouts ((0,1) or (1,0)) where the winning player receives one
and the losing player gets zero. The match can never end in a tie and given that attacking is
the equivalent of defending5, there will never be any incentives to cooperate. It is zero-sum
because for one player to win the other must lose, and it has perfect information because
each player knows every possible movement and every previous movement during each
match. The game of Hex is considered to be ultra-weakly solved6 as John Nash proved in
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1949 that the first player to move always has a winning strategy.
4
This rule is not taken account during the development of the formal reasoning of this paper.
We will prove this later on in the document.
6
There are 3 levels on which a game can be solved: the first consists of knowing the existence of a winning
strategy, the second, knowing which strategy is the winning one and the third is when the final move of the
game can be determined given any starting position.
5
Figure 1: Random Board of Hex
Self-Made Graph
In figure 1, we observe the first 6 movements of a random game of Hex. Here the player
who is playing black would most likely play E5 as their next move with the intention of
connecting all of their pieces, alternatively if it were the other player's turn the next logical
move would be to also play E5 as to avoid black from connecting their pieces. The act of
connecting two pieces of the same color together is called ‘building a bridge’. These bridges
may be classified as weak or strong, depending on how easy it is for the other player to block
the connection. The move we just reviewed is an example of a weak bridge, as the player
playing white could have easily blocked the connection by placing one of their chips in E5.
Figure 2a is another example of a weak bridge, as the player with the white pieces can easily
stop the other from connecting A and C by placing one of their own chips in B. Figure 2b
represents an example of a strong bridge. This means that if the player playing black were
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A and D with a single move. Hence if a player were to connect their sides of the board with
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to place a piece in D the other player would have no way of blocking the connection between
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only strong bridges the game could be considered as over because the player would only
need to 'fill in the blanks' to win. This shows just how strong these types of bridges are.
Figure 2: Types of Bridges
(A) Weak Bridge
(B) Strong Bridge
Self-Made Graph
Basic Strategies
The first thing that we must understand is the construction of bridges; that means connecting
two positions which are one hex away from each other. Let us call connecting two or more
close chips a basic connection; these are important because the outcome of the game will
depend on how good they are. Figure 3 shows all the positions where the player playing
black can place a chip in order establish a basic connection with A. It also represents every
other possible move that can be played7 because any other position on the board could be
interpreted as a rotation or reflection of figure 3. For instance, let's suppose that player 2
wants to connect A with the Hex adjacent to E and D that is missing in the figure, then we
would just need to rotate figure 3 by -45 degrees and the result would be the same as if player
2 wanted to connect A with C, thus it is possible to analyze every move that creates a basic
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connection through the rotation of this figure.
7
Player 2 could play on a hex far away from A, however in this section we just analyze the movements that
connect two close positions
Figure 3: Basic Connections
Self-Made Graph
Given any position in the board, for instance, the black chip standing alone in figure 1, the
next logical movement could be interpreted 'locally' in figure 3. Let us suppose that the black
player was trying to connect the black borders as fast as possible, thus they would use a
straight line similar to the one in figure 1. It is as if, in the figure 3, the black player was
trying to connect A with C, then they would need to place a chip in the B and C to connect
them in order to win. The problem here is that if Black placed a chip in B then White would
block them by playing C and if Black plays C, they will be blocked in B, thus making them
unable to connect through a straight line. Using some basic game theory to explain it, we
can use the extended form diagrammed in figure 4.
Figure 4: Weak Bridge in Extended Form
(A) Player 1 plays next
(B) Player 2 plays next
Self-Made Graph
or strictly dominated strategy, the outcome in both cases (when the black player plays first
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and second) is that they will not be able to connect. That means that playing a straight line
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We can see that neither figure 4a or 4b represent a Nash Equilibrium (NE) nor a dominated
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is very poor strategy, of course it is necessary to keep in mind that it is possible that the other
player does not block the line due to several factors (e.g. trying to connect their own path or
realizing that there is no point in blocking that path). This means that the strategy of a straight
line or a weak bridge should be one of the last options to consider but it should not be
completely tossed aside.
Figure 5: Strong Bridge in Extended Form
Self-Made Graph
On the other hand, we have a strong bridge, which consists of going in a diagonal line. In
the figure 3 that means connecting A with D through either B or E. The analysis of the
strategy is represented in the figure 5. Here there are zero Nash Equilibriums just as with the
weak bridge, but in this case, there exists a non-strictly dominated strategy. To connect with
D the black player has three different options: he can start by placing a chip in either B or E,
and thus he would be blocked in D, the third option is to play a strong bridge, that is to say
to play D so that the other player has only two options, play B or E, if either of these happen
the Black player in his next turn can play the other hex, thus connecting D with A. In this
last scenario, (first move is D) the White player cannot stop the Black player from connecting
the two hexes. As we can see in figure 5, by following the path 1D we will always get a
payout of 1, meaning that we were able to connect those hexes, and the other player will
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always get a negative score (the opposite meaning), so a rational player will not try to block
a strong bridge8, meaning that if there is a path made by strong bridges9 the game has already
ended and the owner of that path won, only if there is no other shorter strong-bridges path.
So as we can see, a strategy that uses some weak bridges will be easily blocked. Therefore,
we can conclude that most of the cases the strong bridges will be used.
Long Term Strategy
So far, we have talked about short-term strategy, as it is very useful and it is something every
casual and advanced Hex player should know, however those strategies were very basic. In
this section will briefly talk about long term strategies, these strategies require the use
mathematical processes to analyze due to the exponential increase of possible moves. One
of the most interesting results that has been found about Hex, is the existence of a winning
strategy for the first player. It states that there is a strategy that if played by the first player
to act, it assures that they will win, no matter what the second player does. Meaning that
there is an advantage over the player who plays second, that is the reason why the second
player has the option of switching colors (and become the first player) after the first move.
In order to prove the existence of a winning strategy we will need the Hex Theorem, but this
just proves the existence of a winning strategy, it will not tell us which player will win. To
prove that the player one will always win, we will need to understand the concept of stealing
strategies, which will be explained later on in the paper along with the symmetry between
attacking and defending.
The Symmetry of Attacking and Defending
For this section we will consider an attacking move as any move made with the intention of
getting closer to connecting the opposite sides of the board and a defensive move will be any
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a player will attack when they feel confident that they are not at risk of losing the game and
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move made with the intention of preventing the opponent from doing so. Understanding this,
8
A player can still play a chip within an adversary's strong bridge if they have a goal other than blocking in
mind such as connecting their own hexes.
9
The path will lack the hexes to complete the strong bridge i.e. B and E in figure 2b, so the path will not be
completely full of hexes.
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they will defend when their opponent is either close to winning or stands to gain a lot from
a certain move.
The reason that the strategy of attacking and the strategy of defending overlap is that any
chip placed by a player cannot be harmful to them in any way; this means that if one player
plays a chip, whether it be attacking or defending, that chip will open up possibilities for
new bridges and connections that previously were not possible. Even if a player were to play
a chip in a hex on the board that does not permit any new bridges or connections, they will
still be better off than before as that chip will have blocked some of the other player’s
connections.
Thus if a player focuses only on defending they will eventually have enough weak and strong
bridges built so that they could threaten their opponent with winning should they not switch
their strategy.
Hex Theorem
Basically, the Hex Theorem sustains that, if the hex board is completely full (one chip in
every hex), there is always a path connecting the player 1's borders or the player 2's borders,
i.e. there's a winning path for one of the two players.
Figure 6: Edges, Vertices and Faces in Hex
.
(A) One vertex and three edges colored
(B) Edges colored
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Self-Made Graph
There are different ways to prove this, here we will use the same method that Matthew Calvin
(Calvin, 2013) used to explain David Gale´s paper of 1979, as we consider it a more
straightforward approach to it, also it allows for easier understanding. To fully understand
the demonstration of the Hex theorem, we have to think of the Hex board as a graph 10, in
order to do that it is necessary to know the very basic concepts of graph theory edge, vertex
and face. A vertex in the Hex board would look like the blue point in figure 6a; we can
understand it as any corner of any hex, so in the figure 6a there are 13 vertices. In the same
figure we can also see three edges in red; so an edge will be any border of any hex, thus in
the figure there are a total of 15 edges. It is necessary to keep in mind that the border of the
Hex board is not the joint of the edges of the hexes in the border; that is the black or white
area in the figure 1. Any edge will have two adjoint hexes except for the edges in the border,
as they will have only one adjoint hex and one adjoint border; for instance in the figure 6b
the red edge has the A hex and the 𝐶 hex as adjoint hexes, notice that the 𝐵 and 𝐷 hexes
have only one vertex that has the red edge as its edge, but 𝐵 and 𝐷 are not considered as
adjoint hexes for the red edge. Finally, a face will be the hex itself. It can be a white hex
(with a white chip in it), a black hex or an empty hex (no chip in it). With that in mind, we
could define 𝐻 as any Hex board of n dimensions (ℝ𝑛 ) completely full (a chip in every hex).
Here we will work with 𝑛 = 2, as there is no real reason for working with higher dimensions,
̂ = (𝑉, 𝐸), where
it would just mean the use of heavier topological concepts. Let's define 𝐻
V are all the vertices of 𝐻, and 𝐸 are all the edges in H, whose adjoint hexes are not of the
same color, and all edges that are in the border of the board, only if the border and its adjoint
hex are opposite colors. To make it clearer let's look at the figure 6b, the red edge will be
contained in E because 𝐴 and 𝐶 (its adjoint hexes) are opposite colors, but the blue edge is
not contained in E because D and C are both white. The figure 7 represents the board of a
random game of Hex, where we can observe 𝑉 in blue (i.e. all the vertices of 𝐻) and 𝐸 in
̂ , that means removing the board from figure 7 and keeping 𝐸
red; while figure 8 is just 𝐻
and 𝑉. Notice that every edge in E is together with another edge in E, it is impossible for an
edge to be alone (later on we will be prove this), and every element in E is part of a cycle11
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or a path12.
10
A set of vertices and edges
Also known as loop because it returns to its starting point
12
A path will not return over any edge it has already passed
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Figure 7: Random Hex Game
Self-Made Graph
̂ ) will have three types of degree; 0, 1 and 2. It is
Vertices under this construction (𝐻
impossible to have a vertex with a greater degree than 2 or for one to have a degree of 1
while not being in any corner. In a Graph made by adjoint hexes, as the Hex board (leaving
no free spaces in it, as the Hex board does), 𝐺 = (𝑉′, 𝐸′), with V all the vertices and E all
̂ , so If 𝐺 and 𝐻
̂ are in the same hex board then |V ′ | ≥ |V|, it is a broader
the edges (unlike 𝐻
definition) the maximum degree of any vertex in V will be 3. Now we apply the conditions
of 𝑉 to those vertices, that means that the maximum degree is reduced to 2. Let's take a
̂ , with 𝑁(𝑉2 ) = 3, therefore it has 3 incident edges 𝑒𝑖 𝜖 E with 𝑖 = 1, 2, 3,
vertex 𝑉2 ∈ 𝑉 in 𝐻
𝐵′ must be different colors, and given that the 𝐶 edge is in 𝐸, then 𝐶′ and 𝐵′ are opposite
colors, therefore if 𝐴′ and 𝐶′ are opposite colors from 𝐵′, then they are the same color,
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it would look like the blue point in figure 6a; given that the edge 𝐵 is in 𝐸 therefore 𝐴′ and
therefore the edge 𝐴 would not belong to 𝐸 so 𝑁(𝑉2 ) = 2 therefore if any vertex 𝑉2 ∈ 𝑉 in
̂ cannot have degree 3 or higher.
𝐻
To prove that the edges in the corners of the board (in 𝐸) are the only ones that can have a
degree equal to 1, let's use the figure 6a again, now let's imagine any vertex 𝑣 with a degree
of 1 and is not in any corner, as the blue one on the figure, but let's suppose that the edges 𝐶
and 𝐵 are not there, the only incident edge to 𝑣 is A, therefore the chip in 𝐴′ and 𝐶 ′ are
opposites colors, and due to 𝐶 does not exist, 𝐶′ and 𝐵′ are the same color, the same for 𝐴′
and 𝐵′, then if 𝐴′ is the same color of 𝐵′, and 𝐶′ is the same color of 𝐵′, then 𝐴′ and 𝐶′ must
be the same color, but this a contradiction, so the only way there is to make sense of it is that
𝐴′ or 𝐶′ don't exist (because it is the border). Any vertex in 𝑉 can belong to either a cycle, a
path or it can be an isolated vertex. Being part of a cycle means that its degree is two (no
other option).
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Self-Made Graph
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̂ of a Random Hex Game
Figure 8: 𝑯
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Because we know that the only vertices which 𝑁 (𝑉) = 1 are the ones in the corners, then
a path has to start in a corner and finish in one of them too. In addition, we know that every
vertex in a path is at least 1-vertex (i.e. vertex of degree 1), that means that a path cannot be
finished in any part of the board that is not a corner, therefore every path is a corner-corner
path. We can go even further, given that there are only four corners, then there are only four
1-vertices, therefore, there will always be two paths (in any 𝐻) that never intersect each other
(if they do it would mean there is a vi which 𝑁(𝑣𝑖 ) > 2)13, and due to there are two nonintersected paths they must start and end in borders of the same color and also we can notice
that a path has two sides, which are opposite colors (either white or black) by following one
of them we will connect the white borders of the board or the black borders. Therefore, if
the board is full there is a winning path14 for one of the two players.
Strategy-Stealing Argument
Now we just need the Strategy-stealing argument to be able to prove that the first to play can
always win. This strategy is very simple and straightforward. Let us suppose there exists a
winning strategy for the second player, but if there is such a strategy (which can be only be
played by the second player), the first player will be aware of it (because the players are
completely rational and have perfect prevision) and would try to use it. Thus, player 1 faces
the problem that this strategy can only be played by whoever goes second in the game. The
way to solve this is that Player 1 will place their first chip in any random hex within the
board and then will ignore it by imagining that the board is completely empty. This means
that when the second player places what effectively is the second chip of the game, it would
be as if they had just placed the very first chip and as player 1 is next to move, in their mind
they will be going second, thus placing the second chip of the game and giving them access
to player 2s winning strategy. The difference is that the real first player has one chip more,
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the one that is being ignored, and because no chip in Hex can harm the one that played it,
this will not matter in the long run. If at any point in the game, the next move requires placing
a chip in the hex that contains the very first chip (the one that was being ignored), they can
put a chip in any other hex and ignore it and it would be as if the chip that was being ignored
13
14
See the section “The no return of every path”
One winning path made by chips, or two winning paths made by edges.
was just played. Therefore, if there is a winning strategy only for the second player, the
player one can steal it and win.
The Extent of Hex Theorem: Several proofs
In this section, we will use Hex Theorem to prove several important things to give a basic
mathematical description of the Hex game. In general, the aim of this section is to try to
assimilate the nature and constitution of Hex, while we discover the features that this game
has to offer. For this reason, this section will use the tools described above. These various
proofs are based on the book “Game Theory” (Maschler, Solan, & Zamir, 2013).
Uniqueness of every path
Proving the uniqueness of every path is akin to proving that every line within the board can
be continued in a unique way. To do this we must first describe a graph 𝐻 (as we did when
we were discussing the Hex Theorem) with a set of vertices 𝑉 and a set of edges 𝐸. Here a
path will be any continuous connection between the two borders of one of the players within
𝐻. Now we will need two basic concepts: the first one derives from Set Theory and the
second one is a logical deduction made by taking into account the existence of the second
player.
Every set is its own subset
Let 𝑆 be the set of hexes in the board; this set contains every possible move that can be
played and every possible combination of those moves. Let 𝑊𝑖 ⊆ 𝑃(𝑆) be all the subsets of
𝑆 where 𝑖 are all the possible board positions that contain a winning bridge. The complement
of 𝑊𝑖 is 𝑊𝑖′ ⊆ P(S), this subset will contain all the possible board combinations that a losing
player can get. Thus if we combine every possible board position that contains a winning
bridge and every possible board position that does not we will obtain 𝑆 like so 𝑊𝑖 ∪ Wi′ =
S. For this let’s suppose that paths are not unique thus there exist two winning bridges made
from the exact same chips and movements, 𝑊1 = 𝑊2 , according to the Set Theory if two
opposite also holds true, that means that 𝑊2 is at least as good or ‘easy’ as 𝑊1 , given that
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𝑊2 ⊆ 𝑊1 . This tells us that the path 𝑊1 is at least as good or ‘easy’ as 𝑊2 and that the
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sets are equal then each set is contained within the other, more formally, 𝑊1 ⊆ 𝑊2 and
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only one chip can occupy a Hex at a time then it must hold true that 𝑊1 and 𝑊2 are exactly
the same path. 𝑊1 ≡ 𝑊2 Thus 𝑊1 is unique.
Two winning paths can’t exist at the same time
The second definition is the following: given two winning paths, 𝑊𝑖 , one for each player
where 𝑖 can be white (𝑤) or black (𝑏); the black wining path is equal (using the second
definition of equal =̈ ) to the white winning path, 𝑊𝑏 =̈ 𝑊𝑤 , if and only if 𝑊𝑖 = 𝑇(𝑊−𝑖 )
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where 𝑖 is 𝑤 or 𝑏 and −𝑖 is the complement; and 𝑇 is a transformation from a winning path
to another winning path of the same dimensions (𝑇 ∶ 𝑊1 → 𝑊2 ) where 𝑇 is a 𝜋(2𝑛 +
1)/2 rotation16 where 𝑛 𝜖 ℤ. Using this definition, it is impossible for two winning paths to
exist at the same time, because they will have to cross at one point or another and there can
only be one chip per hex. That chip will only belong to one of the two paths, so the other
path will lack a chip, therefore that path will not be a winning path, and thus we infer that it
is impossible for two paths to share the same hexes.
The no return of every path
Here, we will prove that a path will never return to a vertex through which it previously
passed. In figure 9, we can see a Hex board, which has geographic direction for each one of
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the 4 corners: North (N), South (S), West (W) and East (E).
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16
Where 𝑊1 is equal to 𝑊−𝑖 using the first definition.
It is a rotation 90 or 270 degrees and so on.
Figure 9: Hex Board with Cardinal Directions
Self-Made Graph
So far, we know that there exists a bridge (path) or broken line that describes the “victory”
for one of the players, and we also know that every vertex in a path is at least 1 degree
because, as we described before, every path is a corner-corner path. Every time that the game
ends (full board) there are two paths17 that the reader can follow in order to find the winning
player18. This is true based on the proofs stated above. If Black player wins, the paths are
connecting the (W, S) and the (N, E) corners; otherwise, if White player wins, the paths are
connecting (W, N) and (S,E) corners19.
Therefore, there is no possible for the way for a link between (N, S) and (W, E) to exist. The
problem resides in the face or hex20 placed in the corners that could depicts both colors 21, if
and (W, E).
17
Made by edges and not chips
Both paths are winning paths for the same player
19
Depending on who made the first move
20
Hex is the game and hex is the hexagon
21
These four hexes are directly connected to both borders, black and white
18
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have the same color or different color there is no way to build a path which connects (N, S)
19
we combine any possibility, it means that if there is a full board, whether both corner hexes
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The first possibility is represented in Figure 10A and that is that two opposite corner chips
have the same color 22 and if one wants to make a Black winning bridge23 and the path start
in the left side down (beginning red path in the corner S), as we can see in the figure 10, in
one moment the path has to cross the bridge to connect with the other side of the path that
started in the top right side (beginning red path in N) what in this context it would be a
contradiction because if we suppose a black winning bridge that blocks any possibility that
there would be both opposite chips that create a vertex who (which) cross to connect with
the another side of the path in the top (corner N)24. The other case can be seen in figure 10B
in which opposite corner chips have different colors, in this case what happens is similar to
the previous case, let’s suppose that there will be a black winning bridge and keep in mind
that the left side down chip is white and black to the top. In this case, the difference resides
that there will be a black chip in the (S, E) border (to create a winning black bridge), in this
sense, the path will have the same obstruction as the previous case with the difference that
the bridge does not start in the corner S but in one of the black border hexes (S, E). Thus, we
reject the possibility of linking together two opposite corners of the board.
Figure 10: Opposite Corner Path
20
(A) Same color
(B) Different colors
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Self-Made Graph
Let’s suppose both black
A bridge is a path made by chips and not edges
24
See the section “Hex Theorem”
22
23
In this sense, if every path (broken line) connects adjacent corners, it always has to contain
at least 1-degree vertex, so a “victory” cannot be shaped as a cycle (2-degree vertex set).
Moreover, it is impossible that a cycle leads us to a victory. We can prove it through a
different method than the one we used before. We know that every edge of the winning path
in E rely on being around the two opposites colors, so we know that a player wins if they
made a bridge that links their two borders, this means that the first chip from each border is
always of the same color, if we iterate that process we will see that both broken line are a
corner-corner path (adjacent corner for both paths according to the winning graph), so it is
impossible that this forms a cycle given that one of the players has to win in order for a
broken line to exist and the bridge had to link the winning color, thus in that bridge there is
not another color that would have made it so that the broken line returns at the same vertex
as a cycle does. This proof is similar to the one previously explained where two opposite
corners cannot be connected because no path can cross over a winning bridge.
Closing Notes
We have proven up to this point, that the game of hex is indeed a complex one, and that it
can be analyzed both through the usage of inductive and deductive methods and advanced
mathematical tools. However the theorems demonstrated in this paper are but an initial
approach to fully understanding the game, many things remain to be proven such as what is
exactly the winning strategy that leads the first player to always win. Until this is proven,
we will not be able to say that the game is fully solved like other games such as Checkers or
Limit Texas Hold'em so there is still room for more study and investigation on the topic.
We have also shown through the use of game theory the existence of weakly dominated and
strongly dominated strategies within the game, and perhaps one of the main conclusions we
can draw is that the extent of the rationality of the players is what will
in most cases determine the outcome of the game, as more rational agents will be able to
precisely because the precise strategy that leads to victory is still unknown, so the player that
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construction of stronger bridges and ultimately to victory. We can draw this conclusion
21
develop stronger long term strategies as opposed to weak short term ones, leading to the
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can predict and adjust to the other's play better will always have the advantage, regardless
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22
of going first or second.
References
Calvin, M. (2013, September). The joy of hex and brouwers fixed point theorem. Retrieved
2013-09-30, from https://vigoroushandwaving.wordpress.com/2013/09/30/the-joy-of-hexand-brouwers-fixed-point-theorem/
Gale, D. (1979). The game of hex and the brouwer fixed-point theorem. The American
Mathematical
Monthly,
86
(10),
pp.
818-827.
Retrieved
from
http://www.jstor.org/stable/2320146 Maschler, M., Solan, E., & Zamir, S. (2013). Game
theory. Cambridge University Press.
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Maschler, M., Solan, E., & Zamir, S. (2013). Game theory. Cambridge University Press.

Nº 80 - Universidad Nacional de Colombia

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