Hi Cenoman,
Cenoman wrote:Just to show how eleven was right to chose the DP solution...
- Code: Select all
+------------------------+------------------------+-----------------------+
| 8 E26 1 | D2679 D23679 D67 | 5 23 4 |
| 3 5 2469 | C26 246 1 | 269 7 8 |
| 469 7 2469 | 58 2346 58 | 2369 1 269 |
+------------------------+------------------------+-----------------------+
| 14 ZF24-3 5 | 679 679 67 | 8 234 12 |
| 467 8 Y2467 | 1 5 3 | 247 9 X27 |
| 17 9 zb37 | 4 8 2 | y137 6 5 |
+------------------------+------------------------+-----------------------+
| 5 46 a4678 | 3 A267 9 | x12467 248 W1267 |
| 4679 346 346789 | B25678 1 458 | 24679 248 2679 |
| 2 1 46789 | 678 67 48 | 4679 5 3 |
+------------------------+------------------------+-----------------------+
Kraken row (7)r7c3579
(7)r7c3 - (7=3)r6c3
(7-2)r7c5 = r8c4 - (2=6)r2c4 - r1c456 = (6-2)r1c2 = (2)r4c2
(7-1)r7c7 = (1-3)r6c7 = (3)r6c3
(7)r7c9 - (7=2)r5c9 - r5c3 = (2)r4c2
=> -3 r4c2; ste
You often use the Krakens, which I consider to be networks of several tracks whose interactions are exploited, usually eliminating a candidate who sees all these tracks at the same time. In fact, you mark them with letters as I do with colors.
Here, in your resolution, I could write:
P(7r7c3): 7r7c3->7r6c3
P(7r7c5) : 7r7c5->2r8c4->6r2c4->6r1c2->2r4c2
P(7r7c7): 7r7c7->1r6c7->3r6c3
P(7r7c9) :7r7c9->2r5c9->2r4c2
So we can eliminate the 3r4c2 that sees all four tracks.
My question is this: according to what criteria do you choose 7r7 and what guarantee do you have that if an elimination occurs, it will lead to the solution?
Sincerely
Robert