Measuring the Perfection Gap: a case for Integer Komi

This discussion is motivated by the question: How far are top players (human or AI) away from perfect play?

There seems to be mounting statistical evidence, both from professional play and AI analysis, that supports that the correct value of Komi may be 7.

In principle, perfect play, while using the correct integer value of komi, should always lead to a draw. A game that does not end in a draw indicates that at least one of the players must have made imperfect plays. Of course, games could still end in draws even if both players play imperfectly.

Thus, given a set of players, the percentage of games that end in ties is a loose upper bound on the percentage of games played perfectly, or alternatively, the percentage that don’t end in ties is a loose lower bound on the percentage played imperfectly.

Hence, I think it would be interesting for top players (humans and AI) to use integer komi and see what we might observe from the statistics.

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New Zealand agrees with you.

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Interesting proposition! I am not a mathematician myself, but I guess the integer komi would indeed always result in a draw with perfect play.

Now the problem is, that we, poor humans and even the best AI’s are so far from perfect play that this kind of statistic won’t yield reliable results about the perfection gap.

I like integer komi from a different reason: it’s a nice feeling to fight, and not win, not lose, but depart in friendship :slight_smile:

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I made a related thread at Some thoughts about no-komi Go

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Welcome to the forums! I’m glad to see that this post prompted a new person to participate in the discussion here.

I think you make a great point about the best go players still being very far from perfect. I believe that a perfect game of go almost certainly has not yet been played, and that this may remain the case for the foreseeable future. In light of this, it would seem that my original bounding arguments are somewhat vacuous, since we’d be trying to upper bound something that we could reasonably assume to be zero.

While draws do not necessarily indicate perfect play, the absence of them could be used to refute claims about achieving perfect play. For example, if someone creates an amazing new go bot and makes an incredible claim like “this bot plays perfectly in about 50% of its games”, then one could refute this claim by showing that the bot draws far less than 25% of the time when playing against itself.

Another consideration is to model very strong players as having the tendency to find the correct plays, but occasionally and randomly making some point-losing mistakes over the course of a game. A draw would occur if either: 1) both players play perfectly, or 2) their mistakes balance each other out. Even though we might discount case 1 as practically impossible, case 2 would seem more statistically more likely if the players tend to make fewer mistakes. Thus, the observer draw rate (even if assume that perfect games aren’t actually happening) could be statistically related to mistake rate of the strong players under consideration.

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It supports that correct Komi for Chinese should be 7. Supposedly this means that Japanese should be 6 if Black is expected to get last dame in perfect play (thus rendering Chinese a point higher), but also 7 if White is. afaik there is little reason to suspect that one is more common than the other, rendering the average still at 6.5…

Of course, you can come to this same conclusion in a different manner: Say go acts similarly to a coupon stack, with a “coupon” that grants you X amount of points if you take it, followed by more that decrease at a linear rate down to 0 (although CGT will argue down to -1), then the value of the komi for that coupon stack should be half the value of the biggest coupon. If the Komi in Chinese is 7, then the value of the highest coupon (which is a representation of the largest move) would be 14, which would be one more than under Japanese – thus rendering the size of the largest coupon as 13, and likewise a komi of 6.5 for that coupon stack.

But yeah, definitely more data needed, especially if we can manage it for Japanese komi.

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Dame isn’t scored in Japanese, so unlike Chinese it doesn’t matter who plays last

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but in comparison, in Chinese it does matter who plays last, cuz black can get one extra point in comparison to the Japaness, thus making Japense one less in comparison.

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Which is why when they both moved up one step from 5.5 komi, Japanese was able to go to 6.5 but Chinese had to go to 7.5.

Because in Chinese scoring only odd integer values affect the result.

The extra 0.5 is simply a tie breaker. It could be literally any number between 0 and 1, it has nothing to do with the theoretical maximum value move.

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I am aware, but I’m also bringing other ideas into play.

You see, if the optimal Chinese komi is definitely 7, then in the perfect game if b gets the last move, the Chinese has 1 more point than Japanese, making the optimal komi in Japanese 6, but if in that optimal game white gets the last move, the scores would be equal, and both would have optimal komi of 7.

This is the basis of my probabilistic argument.

The basis of the coupon stack argument has more to do with the fact that whoever takes from a coupon stack first has an advantage of half the size of the top coupon. If we treat optimal komi as the exact counter to that first move, then if go acts like a coupon stack, optimal komi is half the value of the first move (a trick used to evaluate points/handicap stone ratio by some)

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I’m implicitly assuming Chinese-like (area scoring) rules (with strict superko) when mentioning a number like 7. However, we don’t know the exact number for certain.

If the correct komi for area rules was known to be definitely 7, then that would be very strong support that the correct komi for Japanese rules is either 6 or 7, but we still cannot say for certain that it is not something else. We do not know if any of peculiarities of the Japanese rules might create a vastly different outcome. Maybe perfect play involves building a seki with an unbalanced number of eyes that don’t score points under Japanese rules. Maybe a bent-four would emerge and be considered dead despite unremovable ko threats (like from a seki) elsewhere on the board. Maybe perfect play under Japanese rules is ill-defined since the game must end in a triple ko that neither player can abandon without losing.

The average between 6 and 7 is 6.5, which makes for a reasonable argument as to why we can use a komi of 6.5 for Japanese rules, given our uncertainty of the situation. However, the correct komi is certainly not 6.5, by definition, since it cannot lead to a draw (by jigo) under perfect play. If an oracle told us that 7 was correct for Chinese rules, and no “funny business” (see above) should happen for perfect play under Japanese rules, then we could only say that the correct value must be either 6 or 7.

I don’t understand this step of your coupon argument. Could you clarify? Do you just mean to say that the largest coupon must be equal to or one less than that for Chinese rules (because of dame filling)? Is this argument just functionally equivalent to the earlier one?

Under Chinese rules, it is possible (although uncommon) for the score difference to be an even integer. This would occur if there is a seki creating an odd number of unfillable dame points. They skipped 6.5 since that was unlikely to make a difference versus 5.5, but in principle it could.

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Indeed, but in the lack of knowledge on the subject, but if my arguments are reasonable it is reasonable to assume that it should be probabilistically accurate komi for providing a 50/50 result among humans.

Again, I’d like to see data with variable-komi bots in that regard for both Chinese and Japanese

EDIT: to resolve the “no ties” and the “probabilistic” idea I have, a margin of .5 for ties might resolve it

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I agree. Given what I know, I might even bet on 6.5 if I had to pick a komi value to give as close to even results for humans under Japanese rules. However, in the context of picking a “correct” komi for hypothetical perfect play, I might not do better than just flipping a coin to pick exactly 6 or 7.

Unfortunately, guessing at either 6 or 7 might not be fair as an even money game. The house (universe) might have things rigged with some funny business around how ko and seki are handled by Japanese rules.

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Depends how you count the error. If you count it directly then the expectation is 1/2 point average error both ways, but if you take squared error (which is usually the common choice), then 6.5 does do better than picking a side, even if the mean is also 6.5

But that means basically nothing anyway

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I was envisioning 0-1 loss, since komi is essentially discrete (and reasonably assumed to reside within a finite set, e.g., {0, 0.5, 1, … , 14.5, 15}).

Squared error has the side-effect assigning different error values to a guess of 6.5 vs 6.66, while they have the same functional impact on the game.

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I was thinking about this while looking at an example in Berlekamps book. If one really wants an even game with Chinese or area scoring rules with integer komi, shouldn’t we in theory also have an even number of points on the board, that is play with an NxN board with N even? In theory you could end up with a game with equal areas under japanese (territory) scoring but with one extra dame point filled in that swings the game on an odd size board in area scoring. (I know there’s more to consider with eyes in seki etc with the scoring)

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(I’ve played this game without firm grasp of the rules as my attitude has been to learn as I encounter and accept punishment for not knowing.)

  • Disregarding komi, is the territoryscore for each player necessarily a whole number?

  • The game might draw without reaching the scoring phase, as mentioned.

  • Is it possible that the result of a game is win/lose without reaching the scoring phase?

  • Putting together @yebellz’s canceling out mistakes argument and @mekliff’s probablistic argument about dame seems to suggest that, in chinese rule, fractional komi is more robust for imperfect players than integer komi, if the correct komi for perfect players is an integer.

  • I haven’t understood what @mekliff’s saying about halving of the value of coupons. ‘Linear’ doesn’t seem to suggest halving, no? Could you explain or link to explanation?

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yes, there is always integer territory, integer captures, and integer stones, meaning it’s always an integer under both Territory (Japanese) and Area (Chinese/NZ) scoring without komi

only if there is a resignation

It has to do with a certain CGT idea. But the idea is simple: if there is a stack of coupons of linearly descending values (say 20,19,18,17,16, etc), and we’ll assume for a second that this is not accompanied by another game, and assuming there is an even number of coupons, the difference between each coupon and the next is a certain value d (d=1 in the example given).

so, first take the simplest stack: a stack of two coupons, one has value 2d, and the next has value d (we’re making the floor equal to d just for simplicity), so if you go first, you take value 2d, and they take value d, resulting in a margin of 2d-d = d, which is half the value of the largest coupon on the stack, we’ll see that that’s not just a fluke.

Next up we’ll look at four coupons: 4d, 3d, 2d, and d. A takes 4d, B takes 3d, A takes 2d, and B takes d, the A has 4d+2d = 6d, B has 3d+d = 4d, meaning A wins by 6d-4d = 2d once again half the size of the largest coupon.

Now, let’s show this in a more general case, where the stack is made of Nd, (N-1)d, (N-2)d, (N-3)d, …, 3d, 2d, d, where N is an even number. We’re gonna skip the part where we go over who gets what and the total, but instead get the margin immediately through Nd - (N-1)d + (N-2)d - (N-3)d + … - 3d + 2d - d. We can group these together like [Nd - (N-1)d] + [(N-2)d - (N-3)d]+…+[4d - 3d] + [2d - d], one thing you’ll notice is that each group is equal to d, making the resulting amount d+d+…d+d, where there are half as many d’s as there were coupons on the stack, resulting in a marginal value of (N/2)d.

so if we have a largest coupon of any stack that goes to 0, we’ll call this value V, where V=Nd, the amount A gets for going first is V/2 = (N/2)d.

There’s a slightly different calculation if the smallest coupon is not d (I originally said it didn’t matter, but the truth is that it’s about the size of the stack and not the largest coupon in that case)

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Or a time-out, or in the case of a tournament game, a disqualification.

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@mekliff
Wow that turned out unexpectedly neat.

I tried something similar with coupons that have half the value of previous one. If the smallest coupon has value S (which stands for small or scale), the coupons in descending order are: S * 2^(n - 1), S * 2^(n - 2), …, S * 2^1, S * 2^0

I didn’t feel too inventive with algebra so I just calc’d some margins with recursive formula and searched in OEIS: (4^n - 1)/3

I got: 4/3 * V^2 - 1/3

I’m buffled by “- 1/3” term as it doesn’t depend on V, becoming negligible for large V.


I wonder if there’s some kind of pattern for the way the value of moves decreases in a perfect game. (Would the height of stairsteps be always decreasing or periodic?) If there is, even an approximation of it might help guess the existence of a better move.

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