Author: Dieter Buerssner
Date: 14:09:27 02/28/04
Go up one level in this thread
On February 28, 2004 at 16:36:32, David H. McClain wrote: >For the 5 piece knnkn.nbb.emd I guess what follows is not only statistics for knnkn.nbb.emd, but rather statistics for both, knnkn.nbb.emd (the later part) and knnkn.nbw.emd (the former part of the table). >wtm: Lost in 0: 56 >wtm: Draws: 44082657 >wtm: Mate in 7: 38 >wtm: Mate in 6: 757 >wtm: Mate in 5: 3887 >wtm: Mate in 4: 5955 >wtm: Mate in 3: 9382 >wtm: Mate in 2: 10535 >wtm: Mate in 1: 4973 >btm: Lost in 0: 964 >btm: Lost in 1: 1007 >btm: Lost in 2: 1404 >btm: Lost in 3: 726 >btm: Lost in 4: 796 >btm: Lost in 5: 177 >btm: Lost in 6: 18 >btm: Draws: 48090675 >btm: Mate in 1: 293 > >Could someone interpret these figures to a layman? wtm means white to move, btm black to move. However, these are perhaps not the best abbrevetions. Actually wtm means "left side" to move (left side in knnkn, so knn side) and btm means the side with the one knight and the king has the right to move. For the side with the 2 knights, it turns out, that there are 56 + 44082657 + 38 + 757 + 3887 + 5955 + 9382 + 10535 + 4973 "different" possible posisions. Obviously, most of those are draws. Very few are "lost in 0". This just means, that the NN side is mated already in those positions. Mate in 1 means just, can force mate in 1. Similar to the others. "btm: Lost in 6: ..." means, after my best move, opponent can force a mate in 6. Some elobaration on "different". There are no pawns in this endgame. So we can do many symmetry operations on the board, without changing the prinicpal position. Basically, you can force by mirroring operations on the board (for example) the king with the 2 knights to be always inside the triangle a1-d1-d4-a1. There are even few more restrictions (for example, when the K is on the long diagonal a1-h8, you can apply more syymetry operations, and force the opponent K to not be above that diagonal). So, only positions, that are not transferrable to another already counted position by any symmetry operation are counted. (Most probably some positions are counted more than once - namely those, where both kings are on the same long diagonal.) You see, that both TBs (nbb and nbw) have something like 50 million "different" positions. We can try to estimate this by some math. Without symmetry considerations, and considerations about illegal positions (because the side not to move would be in check) we could estimate, that there are 64*63*62*61*60/2 different positions. The first piece we place on the board has 64 possible squares, the second has 63 left, ... The divisor of 2, is because the 2 knights of the same color are indistinguishable. First knight on a1 and second knight on a2 is the same as first knight on a1 and second knight on a2. If we force the "white" (=side of the two knights) K to the above triangle, this will be 10*63*62*61*60/2 We can estimate a little bit better, by enumerating all legal KK positions with the restrictions noted above. It turns out, that there are not 10*63 such positions, but rather only 462 such positions. So the last estimate will be 462*62*61*60/2 = 52418520 A bit over 50 millions. The ones, that are missing in the above table above are the ones, that are because side not to move is in check by an opponent knight (and therfore illegal). Regards, Dieter
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