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Mathematical Bounds of Komi
Has any upper or lower limits for the 'correct' komi been proven mathematically?
SAS: It's easy to prove that 0 is a lower bound for the correct komi and 361 an upper bound. It doesn't seem likely that any useful bounds could be proved. But for practical purposes we know the correct komi anyway - as Roger Clegg once said " Arno: I agree that 361 is an easy to prove upper bound, but how does the easy proof for 0 as lower bound look like? That assumes that having the first move is an advantage. How do you prove that? [1] SAS: You don't have to prove that being Black is an advantage, only that it isn't a disadvantage. It isn't a disadvantage, because Black is allowed to pass his first move. For simplicity I will assume that two consecutive passes end the game, but other rules should give the same result. Let's suppose that playing the first move (as opposed to merely passing) leads to an n point loss. First suppose the komi is 0. Then Black will pass and White will pass, and the position on the board is a jigo, and so the game is drawn. Now suppose the komi is -n (reverse komi of n). Then Black will pass. If White now passes he loses by the n reverse komi, but if he plays then he loses by the n reverse komi plus the n point disadvantage of first move. So either way, the -n komi (or any other negative komi) gives a win for Black. Arno: The proof for the upper bound goes something like this: if Black starts and White plays no move, Black can win by at most 361 points as he doesn't make any captures. Thus the upper bound for komi is 361, otherwise White would always win. My guess is that the correct komi is somewhere between 0-12 points. But that I cannot prove. Miz: After actually thinking about it a bit I see that 0 and 361 are of course obvious bounds. Improving on the 361 might not be too hard though. We'd just have to prove that White can live with some points. :) I agree that komi 7 feels about right for us humans, anyway. [1] removed my "thinko" of negative komi. I was thinking along the lines of not being allowed to pass on the first move, which of course is not what go rules are about. --Arno Well, set a lower bound at 0, and an upper bound at 361. So it's somewhere between that. As I see it, the object of komi is to get the winning percentage down close to 50%. So then as proof I present that komi 12 is surely sufficient to give White more than a 50% winning percentage. Now considering that White is supposed to be the slightly stronger player, if both players are of equal strength, then Black, starting with sente, should win, due to the proverb that you should give up a stone to get sente. Komi should then be at least 1, so I propose 1 as a new lower bound. Now also considering that you should add a move and then sacrifice, a lower bound for komi should be 2. To calculate the optimal value of komi, theoretically allow draws in the ruleset, and then choose komi to maximize the possibility of a draw. You should get a line graph with a point in the middle, and then values of komi indicating percentage along that line. White is on the left, Black on the right. Which side has the greater % of draw based on +-0.5? (I.e. there should be some statistical uncertanity at 50%). Give jigos to that player. To make it more clear, if you get a value of x for optimal komi based on maximum probability of draw, and a 52% win (for white) at 0.5 but a loss of 47% for -0.5, then there is a 2%-3% discrepancy around x komi. Since it is more difficult for white to win with x komi, then White should win jigos. The same statistical analysis can be used to determine the value of x handicap stones, for x-handicap-stone games only. Just analyze x stone handicap games and recalculate the komi based on maximizing the probability of a draw. Does this sound reasonable? Why doesen't someone do it? It might be PhD material. Get a grant to run a go club/server? I'm there ;) But to prove it mathematically, using bounds, you might want to come up with an equation for a limit. How you structure it would be the key. -kungfu SAS: Yes, you can examine statistics for games using various komi, preferably between very strong players (as they are less erratic). The statistics that I've seen indicate that the correct komi is 7. (See, for example, the link that I've added for the Roger Clegg quote at the top of the page.) But this is just statistics. However convincing it may be, it is not a mathematical proof of anything. As I said above, it doesn't seem likely that any useful bounds could be proved. lavalyn: You assume that the rulesets can always end the game in a scoring position - and thus ignore situations such as triple ko or eternal life. By some odd fluke, there could be White perfect play leading to eternal life or can force a void position optimal (i.e. to avoid it is to let White win). Komi can actually be zero without logical fault under standard Japanese or Chinese rulesets (superko notwithstanding). SAS: But in order to prove things about go, we need a logically consistent and complete ruleset. No such ruleset that I know of allows a game to end in no result. (Japanese rules are not complete and consistent. I don't know about Chinese rules - see my question on the Chinese Rules page.) This is a copy of the living page "Mathematical Bounds of Komi" at Sensei's Library. (C) the Authors, published under the OpenContent License V1.0. |