change log for P1308318 (close)
date | time | user | object | table | field | type | old | new |
---|---|---|---|---|---|---|---|---|
2015-08-26 | 14:07:28 | Alfred Pfeiffer | P1308318 | problems | solution | update | a) 1.Ka6 d8=T 2.Ka5 Td6= 1.Ka6 d8=D 2.Ka7 Dc8= b) 1.Ka6 a8=S 2.Ka5 Sc7= 1.Ka6 a8=L 2.Ka5 Lb7= | a1) 1. Ka6 d8=T 2. Ka5 Td6= a2) 1. Ka6 d8=D 2. Ka7 Dc8= b1) 1. Ka6 a8=S 2. Ka5 Sc7= b2) 1. Ka6 a8=L 2. Ka5 Lb7= |
2015-08-26 | 14:06:09 | Alfred Pfeiffer | wKc5 wBb3a7 sKa5 | position | add | |||
2015-08-26 | 14:06:09 | Alfred Pfeiffer | wKc5 wBb3d7 sKa5 | position | squares | update | ||
2015-08-26 | 14:06:09 | Alfred Pfeiffer | wKc5 wBb3d7 sKa5 | position | posnr | update | a | |
2015-08-26 | 14:06:09 | Alfred Pfeiffer | wKc5 wBb3d7 sKa5 | position | position | update | B k K B | B k K B |
2015-08-26 | 14:05:40 | Alfred Pfeiffer | P1308318 | problems | stip | update | h=2 2.1...; a)Diagramm, b)wBd7->a7; | h=2 2.1... b) wBd7->a7 |
2015-08-26 | 12:10:58 | Erich Bartel | Kindergarten-Problem | refkey | add | |||
2015-08-26 | 12:10:17 | Erich Bartel | Allumwandlung | refkey | add | |||
2015-08-26 | 12:10:17 | Erich Bartel | Tivadar Kardos | refaut | add | |||
2015-08-26 | 12:09:57 | Erich Bartel | Fairies | refgen | add | |||
2015-08-26 | 12:09:57 | Erich Bartel | wKc5 wBb3d7 sKa5 | position | add | |||
2015-08-26 | 12:09:57 | Erich Bartel | 736 Feladvanykedvelök Lapja 01/972 | refsrc | add | |||
2015-08-26 | 12:09:57 | Erich Bartel | P1308318 | problems | cpluscomm | update | %C+ PY 3.21 am 4.3.1995 | |
2015-08-26 | 12:09:57 | Erich Bartel | P1308318 | problems | fcooked | update | F | |
2015-08-26 | 12:09:57 | Erich Bartel | P1308318 | problems | cplus | update | T | |
2015-08-26 | 12:09:57 | Erich Bartel | P1308318 | problems | comment | update | Die Quellenangabe stimmt nicht. Nr.736 FL ist eine ganz andera Aufgabe. Ich vermute, daß diese Aufgabe die Korrektur der Nr/55.4 ist. \eb 17.3.1995 --> P1308287 | |
2015-08-26 | 12:09:57 | Erich Bartel | P1308318 | problems | solution | update | a) 1.Ka6 d8=T 2.Ka5 Td6= 1.Ka6 d8=D 2.Ka7 Dc8= b) 1.Ka6 a8=S 2.Ka5 Sc7= 1.Ka6 a8=L 2.Ka5 Lb7= | |
2015-08-26 | 12:09:57 | Erich Bartel | P1308318 | problems | stip | update | h=2 2.1...; a)Diagramm, b)wBd7->a7; | |
2015-08-26 | 12:09:57 | Erich Bartel | P1308318 | problems | add |