In general, the space between two lines on the goban is about 24mm in the lengthwise
direction and 22mm across. The goban is thus not quite square. In the case of the
stones, they are biconvex and have a diameter of about 22mm.
| Here is a goban of 19x19 lines. Note that some points are emphasised. These are called hoshi. |
 | Diag. 1: 'a' and 'b' are adjacent intersections, but 'b' and 'c' are not. |
Two stones are adjacent if they occupy adjacent intersections.
A chain is a group of one or several stones adjacent one to another of the same
colour.
The liberties of a chain are the empty intersections adjacent to the stones of that
chain.
 | Diag. 2: The four white stones marked X are adjacent one to another. They form a chain that has five liberties: the intersections marked by the letters 'a', 'b', 'c', 'd' and 'e'. |
 | Diag. 3: The black stones border a territory of 7 intersections. Note that the edge of the grid forms a natural border to the territory, but it is perfectly possible to have a territory that does not touch the edge of the board (imagine that the grid is a continent surrounded by the sea, that the edge of the board represents the shore, and that the stones represent the frontiers between the countries on this continent). |
| Go is played by two players. The player who starts the game plays with the black stones and the other with the white stones. Taking turns, the players place a stone of their own colour on an empty intersection of the goban, or else they pass. |
Passing is essentially a way to indicate to the other player that you consider the game to be over.
| When a player fills in the last liberty of an enemy chain, he captures it by removing the stones in that chain from the goban. Further, when placing a stone, a player must not form a chain with no liberties, except if, in doing so, he captures an enemy chain. |
When a chain has only one liberty, it is said to be in atari.
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| Diag. 4: The three white stones X form a chain which is in atari (because it has only one liberty, at 'a'). | Diag. 5: If Black plays at 1, he fills in the last liberty of the white stones... | Diag. 6: ... and so Black captures the white stones and removes them from the goban. |
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| Diag. 7: According to the capture rule, White can play at 'a' and take Black. In this case it is said that Black has only one eye (the intersection 'a') and that he is dead. | Diag. 8: White is unable to play at either 'b' or 'c' and so he can never capture Black. Black is said to have two eyes (the intersections 'b' and 'c') and to be alive. | Diag. 9: If Black plays at 'd' (or 'e'), White will play at 'e' ( or 'd') and capture him. Likewise, if White plays at 'd' (or 'e') Black will capture him. In other words, neither player has any interest in playing at 'd' or 'e'. In this case, the stones X are said to be alive in seki, and 'd' and 'e' are said to be neutral intersections. |
| When placing a stone, a player must not restore the goban to a state identical to any state he had created before. |
The following diagrams show the simplest and commonest case of repetition, which happens to be called ko.
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Diag. 10: If Black plays at 'a', he captures the white stone X which is in atari. |
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Diag. 11: White cannot play back at 'b' immediately and take the black stone 1, even though it is in atari, because otherwise he would be reproducing the position of Diagram 10. He must therefore play elsewhere. The shrewd thing for White to do, with his move elsewhere, is to try to create a threat sufficiently serious for Black to be keen to respond to it immediately and so not have time to play at 'b' himself. If Black responds to the threat, White will now be able to play at 'b', since his previous move will have altered the state on the goban. Then it will be Black's turn to find a threat, and so on, so long as neither player connects. |
| The game stops when the two players pass one after the other. The points are then
counted up. Each intersection in a player's territory earns him a point, as does each
of his stones still present on the goban. It so happens that it is an advantage for Black to play first. Therefore, in an even game, White also receives, by way of compensation, points called komi. Komi is normally 7 and a half points (the half point is used to avoid drawn games). The winner is the one with more points. |
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Diag. 12: At this stage, all the territories have been closed off, and none of
the borders can be captured by the opponent. This is the time to pass and to
count up the points. * Black has 8 points of territory in the lower left and 2 in the upper right. He also has 33 stones on the board. His total is 43 points. * White has 2 points of territory in the upper left and 9 in the lower right. He also has 27 stones on the board. His total is 38 points. * Black thus has 5 points more than White. But if account is taken of the komi, White wins by 2 and a half. |
In practice, in order to shorten games without altering the score, the players may, by mutal agreement, remove dead enemy stones from the goban just before counting up the points, without having to add the moves necessary to capture them. In case of disagreement (which is almost always exceptional), it is sufficient to continue the game until all the possible disputes are settled.
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Diag. 13: If Black plays at 'a', he captures the white stones X. If White tries to save them by playing at 'a' himself, Black plays at 'b' and captures them anyway. Since, as it happens, all the territories have been closed off, the two players pass. Black then removes the stones X from the board, and the points are counted up. Confirm that Black wins by a point and a half. |
Important note: in practice, it is possible to use a quick method of counting up which avoids having to determine the number of stones that are on the board. This method is described later in this document.
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Here is the start of a 9-stone handicap game. Black begins by placing 9 stones on the board. Only then does White place his first stone (move 1 in this example). By tradition, Black places the handicap stones on the hoshis. |
* during the game, to keep the stones you have captured and to give a stone to
the opponent whenever you pass, as if it had been captured;
* at the end of the game, if Black has played last, force White to give him
an extra stone;
* just before counting up the points, place the enemy stones you hold inside
the other player's territories.
In this way, in an even game, each player will at the end have used the same number of stones, which will all be on the goban; there will be no need to count them.
In a game with n handicap stones, the total of black stones on the goban will be equal to the total of white stones plus the n-1 extra points. So again there will be no need to count up the stones.
In these two cases, the winner will be whoever has the most empty intersections, not forgetting, in even games, to add the komi to the White total, and in handicap games likewise to add a half point to the White total.
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Diag. 14: This even game has just finished, and it is Black who has made the last move. During the game, Black captured 5 white stones, and White captured 2 black stones. The stones marked X are removed from the board because they are bound to be captured sooner or later. |
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Diag. 15: First method of counting up in an even game The empty intersections and the stones that are on the board are counted up, the captured stones being ignored. * Black has 23 intersections in his territory and 24 stones on the board. * White has 14 intersections in his territory and 20 stones on the board. White receives in addition a komi of 7 points and a half. * Black wins by (23+24)-(14+20+7.5), or 5 points and a half. |
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Diag. 16: Second method of counting up in an even game
Each player puts the enemy stones in his possession inside the territory of the
other (the stones marked X). Black has 9 white stones to place (5 stones
captured during the game, 2 stones removed from the board after the two
passes signifying the end of the game, 1 stone that White has given him because
Black played last, and 1 stone for the pass by White). White has 5 (2 stones
captured during the game, 2 stones removed from the board after the two passes
signifying the end of the game, and 1 stone for the pass by Black).
* Black has 18 intersections inside his territory, and White 5. |
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Diag. 17: This game, which was at three stones handicap, has just finished. White played the last move. During the game, White captured 5 stones. Black captured no white stones. After the two passes signifying the end of the game, White removes the 2 black stones X from the board because he is bound to be able to capture them. The points can then be counted up. |
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Diag. 18: First method of counting up in a handicap game The empty intersections and the stones that are on the board are counted up, the captured stones being ignored.
* Black has 28 empty intersections in his territory and 16 stones on the
board. |
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Diag. 19: Second method of counting up in a handicap game Each player puts the enemy stones in his possession inside the territory of the other (the stones marked X). Black has 1 white stone to place (which corresponds to the pass by White at the end of the game). White has 8 (5 stones captured in the course of the game, 2 stones removed from the board after the two passes signifying the end of the game, and 1 stone for the pass by Black). Then only the remaining empty intersections are counted up, with no need to know how many handicap stones there were nor how many stones remain on the board.
* Black has 20 empty intersections inside his territory, and White 15. |