change log for P0000848 (close)
date | time | user | object | table | field | type | old | new |
---|---|---|---|---|---|---|---|---|
2017-09-22 | 7:02:34 | A.Buchanan | Fairies | refgen | add | |||
2017-09-22 | 7:02:14 | A.Buchanan | P0000848 | problems | soltxt | update | Es gibt 112,112 Lösungen, wobei jede spielgerechte Abweichung zählt. Forget about aP for now. Number of paths for bP 7 bK = c(5,4) - collisions. c(5,5) = 10!/5!/5! = 252 Collisions of two types: when bK moves onto bP & vice versa. Can't have both. For each the number is the same: one moves 4 times and other moves 3. Then the critical move when one treads on the other. Finally either can move off first. So that's c(4,3)x2x2 in total = 140 Difference is 112. Now blend in aP moves c(10,4) ways to do this = 1,001, hence result | |
2017-09-22 | 6:52:57 | A.Buchanan | P0000848 | problems | solution | update | Es gibt 112112 Lösungen, wobei jede spielgerechte Abweichung zählt. Eine mögliche Lösung: 1-4.a2 5.-9.Ka1 10.-14.b1L Lg7#. | Eine mögliche Lösung: 1. a5 2. a4 3. a3 4. a2 5. Ke5 6. Kd4 7. Kc3 8. Kb2 9. Ka1 10. b5 11. b4 12. b3 13. b2 14. b1=L Lg7#. |
2013-09-04 | 8:00:06 | SBD | P0000848 | problems | solution | update | Es gibt 112112 Lösungen, wobei jede spielgerechte Abweichung zählt. Eine mögliche Lösung: 1-4.a2 5.-9.Ka1 10.-14.b1L Lg7#. | |
2012-09-05 | 2:56:41 | A.Buchanan | Fairies | refgen | remove | |||
2012-09-04 | 19:34:12 | A.Buchanan | Series mover | refkey | add | |||
2012-09-04 | 19:33:49 | A.Buchanan | h# | refgen | remove | |||
2012-06-30 | 8:51:18 | A.Buchanan | Path enumeration | refkey | keyid | update | S0000069 | S0000797 |
2012-06-30 | 8:50:56 | A.Buchanan | Fairies | refgen | add | |||
2012-06-30 | 8:48:23 | A.Buchanan | P0000848 | problems | stip | update | h#14 Wieviele Lösungen? | ser-h#14 Wieviele Lösungen? |
2012-06-30 | 6:29:53 | A.Buchanan | Mathematics | refgen | add | |||
2012-06-30 | 6:29:53 | A.Buchanan | Retro | refgen | remove |