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Go is actually a finite two-person game of perfect information and cannot end in a draw. Then by Zermelo's theorem, it is exactly one of the two has winning strategy, either Black or White.

So my question is which one? the Black or the White?

Update: Despite of the game is too large to calculate directly, my (roughly) idea is to prove inductively, from $3 \times 3$, $4 \times 4$ until $19 \times 19$. If for any $n$, one player(suppose black) has winning strategy, then it seems a conclusion that the player(black) has winning strategy.

Update: It seems the result is also rely on the scoring rule.

So another idea is to consider scoring rules, since the more compensation the White earned, the higher probability the winning strategy White has. Different results may be yielded within area scoring and territory scoring.

Widawensen
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Popopo
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    It can actually end in a draw when the komi is a natural number small enough, though this is uncommon. Do you know that it is a 'finite' game? There are double kos... – Sh4pe Nov 05 '12 at 14:51
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    Is it possible to answer this question without also simultaneously devising a perfect strategy? – Nick Alger Nov 05 '12 at 14:57
  • @Sh4pe Yes, as there is only 181 black and 180 white stones(as soldiers) in game, and removed from the game after they 'fall in battle'. – Popopo Nov 05 '12 at 15:00
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    I thought it was well known that the answer is not known. (And it seems Zermelo might not have called what you call Zermelo's theorem a theorem.) – Did Nov 05 '12 at 15:00
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    I see no reason to think answering this question is within our current capability. – GEdgar Nov 05 '12 at 15:00
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    The induction idea in the Update sounds very strange to me. In fact, I wonder what the inductive step from 3x3 to 4x4, for example, could look like. – Did Nov 05 '12 at 15:06
  • @did Ummm...it seems need much more works. – Popopo Nov 05 '12 at 16:17
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    Go has been being played for thousands of years, and komi is in an increasing tendency, used to be $3.5$, $4.5$, now it is $6.5$. I think finding the right komi is equivalent to your question. And it is still only guessed. – Berci Nov 05 '12 at 16:23
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    Your update two is also not true. The concept of mirror go is well studied and while it can in principle give an advantage, mindless copying certainly does not work. (The 15th chapter of the popular manga Hikaru no go is built upon this.) – Willie Wong Nov 05 '12 at 16:29
  • @Popopo: Okay, after having looked at http://en.wikipedia.org/wiki/Go_and_mathematics, I believe that there is an upper bound to the legal game length :) It's not that simple as stating the number of stones that fit on the board though... – Sh4pe Nov 05 '12 at 16:45
  • @WillieWong ...You are right, it's not a winning strategy even without komi. – Popopo Nov 05 '12 at 17:46
  • Assuming no komi, if player 2 has a winning strategy, then black can simply pass his first turn. –  Nov 06 '12 at 11:41
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    @michael: in which case the other player will also pass and force a draw. War Games, anyone? – Willie Wong Nov 06 '12 at 11:45
  • A rare diplomatic solution :). – michael Nov 06 '12 at 11:48

1 Answers1

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Mini-go results up to 5x6, 4x7, and 3x9 are known, see Solving Go for Rectangular Boards. In general, the first player has the advantage, and can be expected to win by a certain number of points under optimal play. Optimal play on the 3x7 board is extremely different than optimal play on the 3x9 board.

If the Komi is above/below the optimal points, then the first player will lose/win. Minor changes in rules, such as japanese/chinese scoring, superko rules, and so on, can change the optimal play.

Optimal plays for 6x6 and 5x7 are currently unknown, and are believed to be 4 and 9 points for black. However, these have not yet been exhaustively proven.

Based on known results so far, there is no inductive process that can be applied to these results.

Ed Pegg
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