Author: Eugene Nalimov
Date: 09:31:13 09/24/99
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Yes, I prefer to ignore extra symmetry when both kings are on the same diagonal. IMHO saving is not worth extra work. Also, for the cases that are more complicated than KXK, I am sure my indexing schema is doing the better job. Eugene On September 24, 1999 at 08:55:43, Guido wrote: >Before all, I would thank Mr. Nalimov for the program tbgen, that I found very >fast and efficient. > >I have developed (and continue to debug) a similar program for generating >tablebases. But the statistics I found are slightly different from the >correspondent statistics in tbgen. > >To explain the problem, let us consider the simple case of KQK. I rewrite here >the file KQK.TBS with at right the values obtained by my program: > > tbgen my program > >wtm: Mate in 10: 1 1 >wtm: Mate in 9: 375 335 >wtm: Mate in 8: 1936 1877 >wtm: Mate in 7: 4102 4016 >wtm: Mate in 6: 4101 4010 >wtm: Mate in 5: 3313 3273 >wtm: Mate in 4: 2546 2499 >wtm: Mate in 3: 1157 1135 >wtm: Mate in 2: 649 629 >wtm: Mate in 1: 312 306 >wtm: Broken positions: 7137 > my index algorithm is not so good :-) >btm: Lost in 0: 46 46 >btm: Lost in 1: 175 169 >btm: Lost in 2: 397 372 >btm: Lost in 3: 949 936 >btm: Lost in 4: 1823 1773 >btm: Lost in 5: 3230 3190 >btm: Lost in 6: 5077 4997 >btm: Lost in 7: 6883 6769 >btm: Lost in 8: 5603 5483 >btm: Lost in 9: 1500 1417 >btm: Lost in 10: 8 8 >btm: Draws: 2953 2896 > >As KQK.NBW (white moves) and KQK.NBB (black moves) occupy respectively 25,629 >bytes and 28,644 bytes, we have that the correct number of different positions >of the two endings are: > >KQK.NBW 25,629 - 7,137 = 18,492 my program gives a total of 18,081 >KQK.NBB 28,644 - 0 = 28,644 my program gives a total of 28,056 > >Now, while computation of correct KQK.NBW positions is not easy because we have >to eliminate checking situations, for KQK.NBB this problem doesn't arise and I >try here to compute the positions. > >K(white) has 10 different positions. > >K(white) + K(black) has 462 different positions as I report here from a past >thread: > > 1 x 4 = 1 x 33 = 33 white king in a1 > 3 x 6 = 3 x 58 = 174 white king in b1, c1, d1 > 3 x 9 = 3 x 55 = 165 white king in c2, d2, d3 > 3 x 9 = 3 x 30 = 90 white king in b2, c3, d4 > ---------- >Total 462 positions > >Now we have to add the white queen. If the queen can be put in all the remaining >positions not occupied by kings, we obtain: > >462 * 62 = 28,644 exactly the number given by tbgen. > >But this is true only if the the two kings are not both on the diagonal a1-h8. >In these cases the squares available for the queen would be only 34, because the >other squares generate positions symmetrically equivalent. How many are these >situations? > >For white king in a1 6 (black king in c3, d4, e5, f6, g7, h8) >For white king in b2 5 (black king in d4, e5, f6, g7, h8) >For white king in c3 5 (black king in a1, e5, f6, g7, h8) >For white king in d4 5 (black king in a1, b2, f6, g7, h8) > ----- > Total 21 > >Therefore the correct computation for KQK.NBB should be as follows: > >(462 - 21) * 62 = 27,342 > 21 * 34 = 714 > ------- > Total 28,056 > >IMHO the conclusion should be that in this case statistics of tbgen consider >28,644 - 28,056 = 588 duplicated positions. > >Obviously this fact doesn't not affect the correctness of the tablebases, nor >the exceptional speed of the program and related algorithms (I am very far from >such speed). > >I found analogous difference for all the other cases (3 & some 4-man endings for >now), while only KPK and KPPK give identical results. > >Best regards >Guido
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