vidarino wrote:The state of the grid at the point of the first guess is:
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49 49 5 | 8 3 1267 | 17 126 267
236 1 267 | 2457 456 2567 | 357 9 8
26 2367 8 | 9 156 12567 | 157 4 2367
----------------------+-----------------------+----------------------
124 2345 9 | 2357 8 257 | 6 123 234
7 2356 126 | 235 569 4 | 189 1238 239
2346 8 246 | 1 69 2369 | 349 7 5
----------------------+-----------------------+----------------------
5 4679 1467 | 34 149 189 | 2 368 34679
124689 2469 3 | 45 7 1589 | 489 68 469
489 479 47 | 6 2 389 | 34789 5 1
Vidar
After this my first step was noticing that there are no patterns for the digit 2 involving (4,1) or (4,2) or (5,2). (But this is not yet needed for the next step.) Next
(7,9)4 > (9,7)7 > (9,6)3 > (7,4)4 > (7,9)!4 shows that (7,9) is not 4. Next
(8,4)4 > (8,6)5 > (8,1)1 > (4,1)4 > (4,9)!4 > (8,9)4 > (8,4)!4 shows that (8,4) is 5.
Afterwards things get messy. I have several solutions, but none is very short.
One of the reasons is that there is an almost-solution to your puzzle:
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945837126
312456798
678912543
439785612
751264839
286193475
596341287
123578964
847629351
where all rows and all boxes and 7 columns are ok. This is forced by the assumption (7,4)3, so shows after many easy steps that (7,4)4. Alternatively, a possible next step is showing that (7,3) is not 7. If it is, then (7,3)7, (9,3)4, (9,2)9, (9,1)8, (9,6)3, (6,6)!3, (6,3)!4, and there is a triple 269 in r6c356 so that (6,1) is not 2,6. There is also a pair 26 in r27c3 so that (5,3)1, (4,1)4 so that (6,1) is not 4. There is also a pair 26 in box 1, so that (2,1)3 and (6,1) is not 3. That does not leave any value for (6,1) so the assumption was wrong and (7,3)!7.