Timeline for What proportion of chess positions that one can set up on the board, using a legal collection of pieces, can actually arise in a legal chess game?
Current License: CC BY-SA 3.0
18 events
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| Jan 30, 2014 at 8:44 | comment | added | Peter Taylor | Yes, I also get 17932. I had the correct order for the notation, but I was reading "unopposed" as "passed" and trying to account for cases where a capture by a pawn didn't pass the pawn. | |
| Jan 29, 2014 at 23:49 | comment | added | Douglas Zare | @Peter Taylor: The order in my notation was (White opposed pawns, White unopposed pawns, White pieces, Black opposed pawns, Black unopposed pawns, Black pieces). For example, if a Black piece takes a White opposed pawn, Black loses an opposed pawn and gains an unopposed pawn, so the change would be $(-1,0,0,-1,+1,0)$. | |
| Jan 29, 2014 at 23:44 | comment | added | Douglas Zare | @Peter Taylor: Yes, I made a fence-posting error and left out the $9^2$ term. I also should have said that the projections were contained in the set rather than that they are equal to it. So, did you confirm the value $17932$? | |
| Jan 29, 2014 at 23:37 | comment | added | Peter Taylor | The notation is copied from the answer, except that I've skipped the $+$ for brevity. The key point, which the use of a 6-tuple obscures, is that the definition of "opposed" means that both players must have the same number of opposed pawns. Thanks for your comment, which highlighted this for me. For the benefit of anyone else trying to reproduce this, it's actually 285 possible projections to pawn type triples, because with $0$ opposed pawns there are a further $9^2$ cases. | |
| Jan 29, 2014 at 15:44 | comment | added | Douglas Zare | @Peter Taylor: If you project to the possible triples of pawn types (opposed, white unopposed, black unopposed) there are $1+4+9+...+64=204$ possible projections. If you project to the number of White pieces there are $17$ possibilities, and the same for the number of Black pieces. So, the total number of possible vectors is at most $204\times 17^2 = 58956$ which is lower than your figure. Please check your work. I did a BFS with a particular set of move vectors, eliminating duplicates. I could say the move vectors, but your notation is different from the one I used in the answer. | |
| Jan 28, 2014 at 23:46 | comment | added | Douglas Zare | @Peter Taylor: I'll take a look again when I have time, perhaps in a few days. | |
| Jan 28, 2014 at 23:35 | comment | added | Peter Taylor | I'm trying to reproduce the first stage of your calculations and failing. With vectors $(-2,2,0,-2,1,0), (-1,1,0,0,0,-1), (0,0,0,-1,0,0), (0,0,0,0,-1,0), (0,0,0,0,0,-1)$ (representing respectively pawn takes pawn, pawn takes piece, piece takes piece, and piece takes pawn in two forms) and the symmetric vectors for the other colour, I already get 105464 different vectors - and I think there are other deltas required in some cases. Would you mind expanding your description of that phase? | |
| Aug 14, 2013 at 17:28 | comment | added | The Masked Avenger | Thank you for the distribution. To me it would seem artistic to leave an angled notch in each display of numbers. | |
| Aug 14, 2013 at 9:28 | history | edited | Douglas Zare | CC BY-SA 3.0 |
Added filtered counts.
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| Aug 13, 2013 at 20:16 | comment | added | The Masked Avenger | Also, I have an OPN chat room, but am willing to chat there now about this, if you wish. | |
| Aug 13, 2013 at 20:10 | comment | added | Douglas Zare | @Joel David Hamkins: Once you have the distribution of multisets of pieces, you can sample from it, and sample random positions, and see how often there are double checks. I haven't done this. | |
| Aug 13, 2013 at 20:08 | comment | added | Joel David Hamkins | Douglas, thank you very much for your outstanding answer! I appreciate it very much. If most positions have a lot of pieces, then it seems to me that double checks on one side or both will seriously cut into the proportion of legal positions. | |
| Aug 13, 2013 at 20:03 | comment | added | The Masked Avenger | Also, my first idea was to label the pawns, and note that legal labeled placements were fewer than .006% of all labeled placements. Granted it is a different problem, and ignores pawns passing through each other, but is that idea useful, and how poor is that estimate? | |
| Aug 13, 2013 at 19:55 | comment | added | The Masked Avenger | I am attempting estimates by hand, and am interested in verification, but do not have a good enough handle on the number and shape of legal multisets yet. Also, I believe bishop placement will turn your 14% into 4%, and likely smaller if games with more bishops are more numerous. | |
| Aug 13, 2013 at 19:46 | comment | added | Douglas Zare | I didn't keep track of the number of distinct multisets. I did keep track of the numbers of pieces for each side, so I could add up the values for 27 or more total pieces, but why not execute a similar calculation yourself? I didn't distinguish the bishop colors. I doubt trapped bishops or bishop color violations are significant. As I said, there is a sense in which the weighted average of the number of pawns is about 7 compared with 16 for the initial position, so there is a lot of capturing and promotion in a typical position. You can underpromote to a second bishop on a color. | |
| Aug 13, 2013 at 19:29 | comment | added | The Masked Avenger | Also, it was conjectured that 28-piece games would be a significant contributor to the ratio, and I conjectured that leaving out all arrangements with 26 pieces or fewer would not seriously change the ratio by a factor alpha where (1 - alpha) has size at most .001. Does your work shed light on those conjectures? | |
| Aug 13, 2013 at 19:24 | comment | added | The Masked Avenger | How many distinct multisets (collections of legal chessmen) did you find? And was respecting bishop placement (but perhaps ignoring trapped bishops) observed? | |
| Aug 13, 2013 at 18:09 | history | answered | Douglas Zare | CC BY-SA 3.0 |