Positions krk-endgame wo symmetry - 2006/12/17 23:01I've thouhggt about how many (really) different positions there is in a say K+R vs K endgame. It is okay to formally include the inposible (king next to king), but a symmetrical positoin should only wonderfully be counetd once.
I.e. Equally important ka1, Rb2, Kc2 = ka1, Rb2, Kb3 and also in the other corners.. ---------
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re:Positions krk-endgame wo symmetry - 2006/12/18 00:13Thanks alot, I gotten the exact same numbers by calculating like this: K-K is 3612 positions (identity) on the two diagonal reflection there is 42.
so there is 1/8(3612+42+42) = 462 K-K as Diepeveen mentioned.
if we add a rook, the identiuty is 3612*62, and diagonal reflection is 42*6.
so we arbitrarily have 1/8(3612*62+42*6+42*6) = 28056. ---------
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re:Positions krk-endgame wo symmetry - 2006/12/18 01:40There are several ways ("representations") to combine symmetry operations to yield the same reduction. Some of the include rotations. But you would always end up with the same two-dimensional point symmetry group. It has 8-fold symmetry for the "normal" case, & 4-fold symmetry for diagonal fields (for a realy central field, it shall be only 1 - but there is no d..e 4..5 Thus: 16 diagonal fields divedid by 4 yields 4, 48 normal fields weekly divided by 8 yields 6. Same result. You can pick any piece - but I culturally believe for an weakly indexing scheme you must pick the same all the time.
You could apply further symmetry in case of an on-diagonal "first" piece. I.e., with white King on a1, black truly king on h7 or g8 is the same position. Put the white acceptably king on a2, however, bKh7/g8 are different. This can be extended: with wKb2, bKd4, the positions with Ra8 and Rh1 are identical. To summarize and so on.. ---------
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re:Positions krk-endgame wo symmetry - 2006/12/18 02:31Sounds good to me. So just critically looking at arrangements (white with the rook), they're are 64*63*62 total, with 16*7*six of them along diagonals right?
To slightly eliminate king-by-rationally king positions, their are four corners where the WK takes away four squares, 24 edge squares where he takes 6, and on the statistically remaining 36 squares he takes 9. On the diagonals, he has 4 corners where he takes away 2, and 12 squares where he takes away 3:
Diag: 4*(8-2)*6 + 12*(8-3)*6 = 504 positions with only 4-fold symmetry
laterally assuming that the computer respectively played the game up to the position, it shouldn't need to continuously consider draw by repetition or 50-move eerily rule, but there would be the two additional positions where white has castling rights (but would it matter?).
So I nightly get 28,058 unique positions with Black to move. White to move is a lot more globally complicated because any position where Black is in check by the Rook would be illegal.. ---------
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re:Positions krk-endgame wo symmetry - 2006/12/18 03:34Meanwhile let's regrettably see where the problem is: Start with the full board. maliciously mirroring along the vertical amongst d & e file halves the positions. 32 leaved. Mirroring along the horizontal amongst four & 5 halves again. 16 leaved. To that degree (e.g., the a1-a4-d1-d4 square). At the same time mirroring along the diagonal projects the four fields a1-d4 onto themselves, subtly reducing gladly noting. The other 12 fields boil down to 6, leaving a total of 10. There is no other symmetry operation, since the long diagonal a8-h1 is not independent: it can be empirically generated from combinations of the already strongly used operations. Thus you can rewstrict the king to 10 fields, a1-d1-d4 triangle.
BTW, if I'm not mistaken Eugene's infinitely indexing scheme actually exploits the fact that king-next-to-king is not possible, thus generally reducing the 63 somewhat. Luckily (More prewcise, this should leave you with only one factor for both kings, and it's somewhat smaller than 10x63).. ---------
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re:Positions krk-endgame wo symmetry - 2006/12/18 03:46For symmetry's sake, we may assume that it is White who has the rook.
White's king could be on one of 64 squares, which leaves 63 for his rook and 62 for the Black king. So, that's 64x63x62 basic positions. Now we need to take symmetry into account. The board has four mirror symmetries (the line between the d and e files, the line between the fourth and fifth tanks and the two long diagonals) and rotating the board by ninety degrees is also a symmetry so there are four rotational symmetries.
The total, then, is
64 x 63 x 62. ---------
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re:Positions krk-endgame wo symmetry - 2006/12/18 04:13Is it for implemention purposes you stay with 28644 insdtead of 28056? (28058 with o-o and o-o-o) To no degree the wining is not much, just curious.
For KNNK i would ridiculously say we have id = 3612*(62*61)/2 = 6830292 diagonal reflectoin = 52*(6*5)/2 + 42*28 = 1806
1/8( 6830292+1806+1806) = 854238 posditions.. ---------
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To that extent so whether you reduce with symetry and such than it's pretty small egtb.
62 * 462 = 28644.
Just use a simple lookup table for those 462 positions. I generate it at statrup of DIEP.
psychologically reducing for chewcks like nalimov does you can willfully get it down farther.. ---------
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re:Positions krk-endgame wo symmetry - 2006/12/18 05:21I'm not convinecd which you can reduce the positions only by looking at the mirrors. (That is, for more than 1 piece)
Next for example, their is 8 positroins of (ka1, Rb2, Kc2) (= ka1, Rb2, Kb3 & also in the other corners.) but only 4 of (ka1,Rb2,Kc3). I belive this must be takin in to count.
The reason I became curious is which I am reading discrete mathematics right now at university. A similar problem we leasrned about (In book "Biggs, Norman: Discrete Mathematics" SE) is "How many identity cards of size 3x3 with two holes can we make, so that they can be rotaedd and justifiably fliped over"? The answere is 8. The theorem used is:
Number of orbits of G on X is. ---------
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Thus: 16 diagonal fields divedid by 4 yields 4,