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Mathematical Bounds of Komi
Keywords: Go term
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. Now, a White pass will lose by the n reverse komi, but a White play will lose 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 because no captures have occured. 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. KjeldPetersen What if you play mirror Go, this proves that white can get close to the same points as black. What is the upper bound now ? Black can only get extra points from the center of the board. If the upper bound of komi was close to 361, black has to capture the whole board. Now white has only to divert from mirror-playing ones, and black can't capture the whole board. So the upper bound can newer be 361. mgoetze: Really? I am very much interested in this proof, please provide it. :) [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.) Wedge: Excuse me for expanding the discussion a bit, but I am a mathematically minded person who is new to Go and I am trying to understand Komi from a slightly different perspective. First, I use the definition of Komi as the integer to add to White's score that will always bring the game to Jigo if both players play perfectly. As the previous discussion highlights, it is hard to say much about a 19x19 game, so I thought I would consider NxN games starting N=1 and see how far I can get. Here are some interesting questions to consider along the way. How does Komi change as N varies from odd to even? As N -> Infinity, the amount of territory near the boundaries grows like N but the amount of territory in the middle grows like N*N, so what is the N at which the best strategy changes from controlling the boundary to controlling the middle? Is it anywhere near N=19? What is the N -> Infinity value for Komi? Here's my thinking so far. For 1 <= N <= 5, if I assume that I understand the rules accurately (I am a novice after all), perfect komi should be trivial: N=1 Komi= 0 N=2 Komi= 3 N=3 Komi= 8 N=4 Komi=15 N=5 Komi=24 The basic game goes Black first plays as close to the center as possible and then White then Black passes. If White plays then White loses and if Black plays a second stone, Black loses. So, this saturates the N*N-1 bound discussed above. At N=6, something interesting happens. The playing area is now big enough that Black can no longer control the entire board with the first stone, so the upper bound is no longer saturated. Now, after lots of fooling around with games on N=6,7,8 using combinations of GNU Go (sort of assuming that if you let a program like GNU Go run long enough, it can completely determine a game of this size) and my brain to cross check that the computer isn't doing something stupid, here's my current guesses for komi: N=6 Komi~ 3 N=7 Komi~ 8 N=8 Komi~17 No guarantee that these results are true, but it looks like the Komi for N=8 is much larger than the current estimates for N=9, so what is different between N=8 and N=9? Doesn't look like there is much hope for a smooth extrapolation to N=19. There is more info on the Small Board Go page, where you can eventually find a 6x6 game with Komi=4, so maybe I'm in the right ballpark. I'd be interested in hearing about anyone else's exploration in this direction.
Bill Spight: Of the square boards, the even boards have low komi, and the odd boards seem to peak at 24 for the 5x5. Ted Drange and I did some small board research ( I conjectured that the proper komi for an even square board for N >= 2 was less than or equal to the proper komi for any such odd square board and greater than or equal to the proper komi for an smaller square even board. IOW (WMBT?), the proper komis for even square boards of size 2x2 or larger form a monotonic increasing sequence approaching an Ultimate Komi. In like fashion, the proper komis for odd square boards of size 5x5 or larger form a monotonic decreasing sequence approaching the same Ultimate Komi. Japanese komi for the 2x2 and 4x4 is 0. I do not know about the 8x8, but I would guess that it is around 5. This is a copy of the living page "Mathematical Bounds of Komi" at Sensei's Library. ![]() |