change log for P1243431 (close)
| date | time | user | object | table | field | type | old | new |
|---|---|---|---|---|---|---|---|---|
| 2014-12-31 | 18:19:13 | A.Buchanan | Loss of tempo | refkey | add | |||
| 2014-12-31 | 18:18:55 | A.Buchanan | P1243431 | problems | soltxt | update | Well obviously it must be 60 solutions. How does this happen? Black has taken at least 6 moves, yet White is checking him, so must have taken at least 7, although White appears to have only made 6 moves. The only way for White to lose 1 tempo is with Bc1-a3-b2. So let's aim for PG in 6.5. Black's moves are deterministic now, but White's consist of 3 independent chains: b3, Ba3, Bb2; Nf3, Nh4; Nc3. These must all be complete before the final check. 3,2 & 1 moves can be ordered in exactly 6!/(3!2!) = 60 ways. | |
| 2014-12-31 | 18:09:11 | A.Buchanan | Path enumeration | refkey | add | |||
| 2014-05-21 | 5:11:32 | A.Buchanan | Non-Unique Proof Game | refkey | add | |||
| 2012-06-30 | 17:57:47 | A.Buchanan | P1243431 | problems | comment | update | Problem 9 original: N.D.Elkies, New Directions in Enumerative Chess Problems, The Electronic Journal of Combinatorics, vol. 11(2), 2004. | |
| 2012-06-29 | 20:52:17 | A.Buchanan | Mathematics | refgen | add | |||
| 2012-06-29 | 20:52:17 | A.Buchanan | Retro | refgen | add | |||
| 2012-06-29 | 20:52:17 | A.Buchanan | wKe1 wDd1 wTh1a1 wLf1b2 wSh4d5 wBh2g2f2e2d2c2a2b3 sKf6 sDg5 sTh8a8 sLg4f8 sSg8b8 sBe6d6h7g7f7c7b7a7 | position | add | |||
| 2012-06-29 | 20:52:17 | A.Buchanan | Noam D. Elkies | refaut | add | |||
| 2012-06-29 | 20:52:17 | A.Buchanan | P1243431 | problems | fcooked | update | F | |
| 2012-06-29 | 20:52:17 | A.Buchanan | P1243431 | problems | cplus | update | F | |
| 2012-06-29 | 20:52:17 | A.Buchanan | P1243431 | problems | dedication | update | for Richard Stanley on his 60th birthday | |
| 2012-06-29 | 20:52:17 | A.Buchanan | P1243431 | problems | stip | update | How many shortest proof games? | |
| 2012-06-29 | 20:52:16 | A.Buchanan | P1243431 | problems | add |