I know colouring with one chain (colour) and two chains (colours). Andrew Stuart writes in his solving guide that colouring with three chains (colours) is theoretically possible, but he has no example. My be I have found one accidentally. Please can anyone proof my “discovery” ?
Is it really true that I discovered an example for colouring with four colours?
Does colouring with more than two colours really exist?
This is Ruuds One Trick Pony no. 222 from 27/06/2007
..7..829...4.21...8............5......6....39.9.....4...8.....34..7..5...6..13.8.
617538294954621378823974651741359826586142739392867145178295463439786512265413987
Elementary steps took me to this point.
Code: Select all
.-----------------------.------------------------.----------------------.
| 136 135 7 | 3456 346 8 | 2 9 146 |
| *369 35 4 | *3569 2 1 | 3678 567 678 |
| 8 1235 *1239 | 34569 ~34679 45679 | 1346 156 146 |
:-----------------------+------------------------+----------------------:
| 137 48 13 | +134689 5 +4679 | 1678 1267 12678 |
| 5 48 6 | 1248 47 247 | 178 3 9 |
| 1237 9 123 | 1368 367 67 | 1678 4 5 |
:-----------------------+------------------------+----------------------:
| 129 7 8 | 24569 ~469 24569 | $49 126 3 |
| 4 123 *1239 | 7 8 *269 | 5 126 126 |
| 29 6 5 | 249 1 3 | $479 8 47 |
'-----------------------'------------------------'----------------------'
chain 1 (*): [r2c4]=9=[r2c1]=9=[r3c3]=9=[r8c3]=9=[r8c6]
chain 2 (+): [r4c4]=[r4c6]
chain 3 (~): [r3c5]=9=[r7c5]
Chain 4 ($): [r7c7]=9=[r9c7]
The underlined cells mean for example “true” and the not underlined cells “false”.
One of the cells with ~-sign and +-sign must be true and the other false. One cell of both chains faces r2c4 (underlined) and the other r8c6 (underlined), which is a contradiction.
So the underlined cells of chain 1 (*) can’t contain the nine, the others can be solved with nine.
We found: r3c3<>9, r2c4<>9, r8c6<>9 and r2c1=9, r8c3=9
This results according colouring Type 2.
Now you can also make reductions according colouring Type 1.
One of the cells r7c5 and r7c7 must be false, the other true. That’s why all other nines in row 7 can get eliminated.
Ruuds solver makes a reduction according to the skyscraper, which is hidden in the cells r8c3+r3c35+r7c5 and makes a headstand.
It follows: r7c1<>9, r8c6<>9
If you only look at the skyscraper, you don’t use the full information, which sticks in all the nines. But it’s enough to solve the puzzle and that’s the target.
Gabriele