This page is for Go players interested in mathematics (or abstract reasoning). It is also a bit of a joke, because what makes go so special is its unbelievable complexity. And here, we actually analyze, understand, (dominate even) the situation the way we do tic-tac-toe! WOW!
We'll be considering 2 possible sets of rules, American and Mathematical rules, most frequently just the second because of its simplicity.
Go on an Empty Board - - - White wins.
Black cannot play and loses.Go on a 1 by 1 - - - White wins.
Next we examine Go on a 1 by 1. Three static positions are possible, one where the Board is empty, one where the board has a black stone and one where the board has a white stone. Please note that only the first of these 3 is a legal Go position. In each of the last two, the stone on the board has no liberty and must be removed as soon as it hits the board. 67% of all possible positions are illegal.- In American rules, both players pass and the result is a draw on the board and usually a win for white because of komi.
- In Mathematical rules, the player who cannot play loses. Since black cannot play, he loses.
Go on a 1 by 2 - - - White wins.
Let's examine Go on a 1 by 2. Nine static positions are possible: The five legal ones: one where the Board is empty, two where the board has a black stone, two where the board has a white stone. The four illegal ones: one with two black stones, one with two white stones and two with a black stone and a white stone. 44% of all possible positions are illegal.With Mathematical rules, black plays first, gets captured and cannot play again. He loses.
Go on a 1 by 3 - - - Black wins.
Let's examine Go on a 1 by 3. 27 static positions are possible:- 0 stone: one legal position.
- 1 stone: 6 legal positions.
- 2 stones: 8 legal positions, 4 illegal ones.
- 3 stones: 8 illegal positions.
With Mathematical rules, black plays first in the middle and wins because white cannot play.
Go on a 1 by 4 - - - Black wins.
Let's examine Go on a 1 by 4. 81 static positions are possible:- 0 stone: one legal position.
- 1 stone: 8 legal positions.
- 2 stones: 20 legal positions, 4 illegal ones.
- 3 stones: 12 legal positions, 20 illegal positions.
- 4 stones: 16 illegal positions.
With Mathematical rules, black plays first in the second point and white can only play on the third or fourth. Either way black captures. White cannot play and loses.
Go on a 2 by 2 - - - Black wins.
Number of static positions
How many static positions exist and how many are legal? This is purely an exercise in counting (mathematicians call it combinatorics). You can safely skip the section, although I would want my students in Probability or Statistics to do it.There are 4 locations on the board and each of them can have a black stone, a white stone or nothing. Therefore there are 3 to the power 4 = 81 static positions for a 2 by 2 board. Let's break them down by the number of stones on the board and into legal versus illegal positions.
- There is only one position with zero stones (legal).
- There are 4 positions with one black stone and 4 positions with one white stone (all legal).
- All positions with 2 stones will be legal. They can be broken down into
2 black stones, 2 white stones and one of each.
- 2 black stones: 2 positions on the diagonals and 4 positions where one side is occupied, hence 6.
- By symmetry, there are 6 positions with 2 white stones
- a black stone and a white stone: there are 12 such positions: Black can be on any of the 4 corners and white on any of the remaining 3.
- positions with 3 stones: there are 36. Let's count them: 4 legal positions have 3 black stones. 4 illegal and 8 legal positions with 2 black stones. 4 illegal and 8 legal positions with 1 black stones. 4 legal positions have 0 black stones. This adds up to 32 postions with 3 stones, 8 illegal and 24 legal.
- positions with 4 stones (all illegal since there are no liberties left!) The hard way is to count 1 with all 4 black, 4 with 3 black, 6 with 2 black, 4 with 1 black and 1 with no black stone. This adds up to 16. The easy way is to notice that each of the 4 spots can have a black or a white stone, hence 2 to the power 4 = 16 different illegal positions.
The Second Chetrit Conjecture:
As n increases, the percentage of illegal positions among all static positions on an n by n Go board decreases.
The play by mathematical rules
- Move 1:
Without much loss of generality, we can assume that black plays here.
- Move 2:
If White answers next to Black, she gets captured and cannot play again,
hence loses. Therefore, we'll assume that she plays on the opposite corner.
- Move 3:
Again without much loss of generality (don't you hate it when mathematicians
wave their hands like that to skip explaining what's going on?), we can
assume that Black plays here.
- Move 4:
Black was also in atari and white captures. Please note that
the static position now has two white stones on the left of the board.
- Move 5:
Black must now choose between the top right and the bottom right.
Playing the top right leads to a loss and
playing the bottom right leads to a win. Let us prove the first statement: If black
plays the top right, white will capture and black cannot recapture without
the board looking exactly like it did after the first
move. Therefore black has no legal move and loses.
Let's assume that black plays the bottom right move, Atari!
- Move 6:
White has no choice but to capture.
- Move 7:
Black captures all tree white stones.
- Move 8:
White must play on the opposite corner for the same reason as in move 2.
- Move 9:
Black can play on the bottom left and prolong the game (to be explored by the serious
student) or play on the top right and win instantly:
- White cannot play Move 10: White's only possible move is the bottom left which captures Black, but this move would recreate exactly the static position which followed Move 4 and is illegal. White has lost.
Therefore, if he plays well, Black can win in 9 moves on a 2 by 2 and the game has no more interest than a tic tac toe game.
When we can do a similar absolute proof that Black wins on a 19 by 19, the game will lose some of its interest! In fact, maybe under the influence, I have found a marvelous proof of this, but my margin is not wide enough...
If you are more serious than I am, please be the one to solve Go once and for all for the next available sizes (2 by 3, 2 by 4, 3 by 3, etc...) If you do, please send me your complete solution and I will append it here (and give you credit, naturally).
