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Basic Endgame Theory/ Discussion
Sub-page of BasicEndgameTheory
I have always had a problem with this page. To start off with 'gote = half sente' assumes a hell of a lot really: certainly it presumes both some idea of miai counting, and the sophistication to understand why miai counting might be useful, even though it frequently gives wrong answers in practical play. Anyway, I think 'basic' is a long way off base. Possibly true. What might be more useful is to highlight the importance of sente over gote in yose. Also Mutual Damage is an important principle for beginners to learn in the endgame. The page does seem a little too packed with confusing buzzwords and phrases though. It all depends on what level you are pitching to of course. Are we describing basic as sub 20kyu, sub 10kyu or sub dan? I suppose you would want to sneak in the importance of good shape as well ;-). IanDavis Naughty Ian. Charles
Well what level do you want to pitch toward? For real beginners you could be highlighting
If you want an introduction to endgame theory maybe this is totally different.
Charles It doesn't much matter what conventions you use, in considering small examples; one can verify the results by hand. So in 'theorising' one is trying to counter the problem of scale, really. That's one point; I thought I'd post some examples this morning, but one indeed does need a clear head. It seems that pages with 'basic' in the title generally have problems. OK, this is the sort of thing I had in mind. I'm trying to connect with the kind of models I regard as truly basic. Apologies if I have misplaced a factor of two, yet again. The wacky terminology traditional here has a lot to answer for, In my view. The basic question, as Bill knows, is to know when to play gyaku yose?. That is, the decision is of the type, play reverse sente, or play the largest gote, in a given position (i.e. whole board position). Let's look at a concrete example. Suppose there is an 4-point sente play for White, and it's Black's turn, in a situation where all the other plays are gote. Yes, let's use deiri values for those: say they have values 8, 6, 5, 3, 2, 1. If we take as baseline the mean position, then for example the 8 point play yields +4 for Black if Black plays there, or –4 if White plays there Black has essentially two ways to play here. Black can play reverse sente, White will take the 8 point gote, Black the 6 point gote and so on. Black's result will be then (- 8 + 6 - 5 + 3 - 2 + 1)/2 or –2.5 relative to the mean. On the other hand Black can take the 8 point gote, White will play out the 4 point sente and then take the 6 point gote, etc. Black's result is (8 - 6 + 5 - 3 + 2 - 1)/2 – 4 = -1.5. Therefore this is a case where the reverse sente is not correct play. It is easy to see the structure of the algebra here: we should compare -(a – b + c – d + e - …)/2 with (a – b + c – d + e - …)/2 – s; or equally (a – b + c – d + e - …) with s. Now in cases where there are few endgame plays we may be able to see what the sum is. If not, we can guess that it comes out like a/2. In which case we are comparing a/2 with s; or a with 2s. That is the primary meaning, I say, of regarding gote plays as counted as half of sente plays: it takes us directly here. It doesn’t however, recognise that the modelling assumption is quite likely to break down, if for example a is as large as 20. There is likely to be a sparse set of gote plays of value 19, 18, …, down to about a dozen. This would cause a bias in a real game.
Jonathan Cano: underlying this discussion is an unspecified set of objectives. Two somewhat competing goals to consider are:
Perhaps we might bifurcate the discussion along these lines.
Charles A theory worth the name ought to be concerned with the truth of things; which is not to say that heuristics and rules-of-thumb have no value - most of us play go that way. Bill: Traditional go endgame theory is almost all concerned with estimation and heuristics. Tedomari, which is mainly concerned with correctness (as I think you mean it), gets only a nod in the textbooks. HolIgor: This is what I think. You have to take the score of all gote moves. If it is zero then the gote part is reduced to miai and you may play reverse sente. Otherwise you just lose the value of the sente for nothing. Let there are gote plays with miai values of 5, 5, 1 and a reverse sente play with the value of 2. You cannot play 5 gote, because the opponent plays 2, then without losing tempo, 5, and you are left with 1. You have to play 2 point reverse sente and then let the opponent win 1. Change 5 to 20 here, if the remaining plays are 20, 20 and 4 then the value of the reverse sente has to be compared with 4. The above rule of reverse sente about a half of the largest gote is true for the case when about the half of the largest gote is the balance of the gote game, which is quite often not true, because all miai gote plays have to be eliminated first. I am not able to make the balance of gote part in my games for sure. But that does not reduce the correctness of the approach. Don't unbalance the gote part. And in real games one can safely assume that the gote part is balanced in about half the cases. Or, actually, this is quite a simple math problem. Suppose you have 20 random numbers in the range from 1 to 20. What is the balance? You have to generate the numbers, sort them in decreasing order and then take plus, minus, plus, minus to the end. I am not sure the even statistically the balance would be half of the largest number. There is a lot of maths pros in the library. Is the solution known? Otherwise I am going to run a test. Charles I believe the obvious mathematical result to support this is true, and not hard to prove. By removing miai, we can formulate the question like this: consider random sums formed, for example, with numbers 1, 2, ..., N , where each number is present or absent with probability 0.5. Form in each case the sum S = N - a + b - c + ... where we assume the largest value N is definitely present, and a > b > c ... represents the ordering by size of the other numbers. Then the expected value of S is N/2. Actually, I wrote about this at stacks of coins; before getting fed up of the whole issue there because of the comment that it was 'trivial', page title was misleading - the whole picky bit that has infected SL from time to time. ilanpi: Here is a proof that the expected value of a - b + c - ..., is N/2, where a > b > c > ... are taken with probability 1/2 among the integers 1,..., N. Since each sum is equally probable, the expected value is (S1 + S2 + ... )/K, where S1, S2,..., are all the various sums, and K is the total number of sums (= 2^N). The result then states that S1 + S2 + ... = N * K/2. This is proved as follows: For each sum S which does not include N, one associates the sum constructed by including N as an extra term in S. This new sum will have value N - S since including a new largest term N to the sequence in S will change the signs in the alternating sum coming from S. Adding S and N - S gives N, so the total sum of sums is N * K/2, since there are K/2 such pairs, the correspondence I constructed being a proof of this last fact. This argument also generalizes to alternating sums taken from a sequence A > B > C > ..., showing that the expected value is A/2. It also follows that the expected value in the problem given by Charles above is (N+1)/2 = N - (N-1)/2, since you are subtracting from N the problem with summands N-1,...,1. <Charles Yes, I see that there is a slight glitch in my formulation. When one says 'a play of value N is definitely present', there might be 2 (or any even number). Which means that a could be N, rather than < N. Bill: It's easy to show with only two numbers. What is the expected value of the difference, S = n - m where n and m are positive numbers and m is selected with equal probability among the numbers between 0 and n. The expected value of m is obviously n/2, so E(S) = n - n/2 = n/2 It is also easy to show asymptotically. S = n - m + p - q + ... As the number of numbers approaches infinity, the expected value of m approaches n, and the expected value of S approaches E(S) -> n - (n - a + b - ...) so E(S) -> n - E(S) and E(S) -> n/2 Actually, it was always n/2 when the number of numbers was even. HolIgor: Great, now we know that the number is really a half of the value of the laargest move statistically. Of course, it can be either larger or smaller than that. At least, and this can be easily evaluated, we have to exclude miai at the highest level in the case when there is a gap in the distribution. This is a copy of the living page "Basic Endgame Theory/ Discussion" at Sensei's Library. ![]() |