Rule
The playfield of the HexGo board is divided in 37 hexagonal cells. The game is played by two players. One player takes the red hexagonal stones (chips).
The other player (computer) takes the blue stones. The game proceeds as two players place a stone on an unoccupied cell alternatively. A stone must be
placed each turn, no “passes” or skipping of turn is allowed. The objective of the game is to create an isolated empty cell (territory or “Me”) which is
then claimed to be own cell if it is created in the following configurations (the examples are for the red player).
The other player (computer) takes the blue stones. The game proceeds as two players place a stone on an unoccupied cell alternatively. A stone must be
placed each turn, no “passes” or skipping of turn is allowed. The objective of the game is to create an isolated empty cell (territory or “Me”) which is
then claimed to be own cell if it is created in the following configurations (the examples are for the red player).
(1) For a cell in the inner field
Figs. 1(a)-(c) show cases where a cell, X surrounded by 6 stones. For each case, the “Me” belongs to the red player because the number of the
surrounding red stones is larger than the number of the surrounding blue stones. The “Me” can also be created by forming the configuration shown in Fig.
2(a) with or without a blue stone at the center. If the blue stone has been placed, it is considered to be captured and removed to create a “Me” as shown in
Fig. 2(b). The red player holds the captured blue stone. The cell now belongs to the red player. Note that the surrounding red stones can be placed in
different orientations; therefore, there are three possibilities to surround the stone chip.
surrounding red stones is larger than the number of the surrounding blue stones. The “Me” can also be created by forming the configuration shown in Fig.
2(a) with or without a blue stone at the center. If the blue stone has been placed, it is considered to be captured and removed to create a “Me” as shown in
Fig. 2(b). The red player holds the captured blue stone. The cell now belongs to the red player. Note that the surrounding red stones can be placed in
different orientations; therefore, there are three possibilities to surround the stone chip.
Fig. 1(a)
Fig. 1(b)
Fig. 1(c)
Fig. 2(a)
Fig. 2(b)
(2) For a cell along the edges of the field
There are two different positions at which a “Me” can be created. In Fig. 3(a), a cell is surrounded by 4 stones (a, b, c, d). If the red stones are more than
two (i.e., 3 or 4), the “Me” belongs to the red player. In Figs. 3 (b), a cell is surrounded by three chips (a, b, c). If the red stones are more than one (i.e., 2
or 3), the “Me” belongs to the red player.
two (i.e., 3 or 4), the “Me” belongs to the red player. In Figs. 3 (b), a cell is surrounded by three chips (a, b, c). If the red stones are more than one (i.e., 2
or 3), the “Me” belongs to the red player.
If a cell is surrounded by equal number of red and blue stones as shown in Fig. 4 (a) and (b), the “Me” is considered as a neutral territory. Note that Fig. 4
(a) and (b) are examples and there are different color arrangements which create the neutral “Me”.
(a) and (b) are examples and there are different color arrangements which create the neutral “Me”.
Fig. 3(a)
Fig. 3(b)
Fig. 4(a)
Fig. 4(b)
Once a “Me” is created, the players can only place a stone in the “Me” (either own or opponent’s) when this act creates an own new “Me” or prevents
the opponent to create a new “Me”. Exceptions are the cases in Figs. 1(a), 1(b) and 2(b). In these cases, a stone cannot be placed in the “Me”.
the opponent to create a new “Me”. Exceptions are the cases in Figs. 1(a), 1(b) and 2(b). In these cases, a stone cannot be placed in the “Me”.
When all unoccupied cells except “Me” are filled, the play stops. To determine the winner, first, the neutral “Me” is filled with either red or blue stones
to mask the neutral “Me”. Next, if the player has the opponent’s stones, they are used to fill the opponent “Me” to reduce the number of the opponent’
s “Me”. For this reason, it is a good strategy to capture the opponent’s stones during the game. Finally, the remaining “Me” are counted. The player
with larger number of the “Me” wins.
to mask the neutral “Me”. Next, if the player has the opponent’s stones, they are used to fill the opponent “Me” to reduce the number of the opponent’
s “Me”. For this reason, it is a good strategy to capture the opponent’s stones during the game. Finally, the remaining “Me” are counted. The player
with larger number of the “Me” wins.