Timeline for Can someone solve this sudoku without trial and error?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 24, 2020 at 9:48 | comment | added | Jaap Scherphuis | According to your link, an XY-wing needs one cell to "intersect" the other two. How does R6C5 intersect with either R7C3 or R7C6? | |
| Jul 24, 2020 at 9:31 | comment | added | dtc348 | @JaapScherphuis I have updated my solution with correct method. | |
| Jul 24, 2020 at 9:29 | history | edited | dtc348 | CC BY-SA 4.0 |
exact answer
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| Jul 23, 2020 at 13:39 | comment | added | Jaap Scherphuis | For that you would need 3 cells with only 234 (or a subset), but here one has 23489. Note that a hidden tuplet allows you to remove canidates only from the tuplets own cells, and a naked tuplet allows you to remove candidates only from the other cells in the house. You cannot somehow mix the two to remove candidates from both sets. | |
| Jul 23, 2020 at 13:03 | history | edited | dtc348 | CC BY-SA 4.0 |
deleted 2 characters in body
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| Jul 23, 2020 at 12:45 | comment | added | dtc348 | Through this link, might be naked triple will fit. | |
| Jul 23, 2020 at 12:39 | comment | added | Jaap Scherphuis | I don't think that row is anything. There are only 5 open cells. A naked triple would correspond to a hidden pair on the other unused digits in the row, and a hidden triple would correspond to a naked pair on the other unused digits. A priori it is perfectly possible for R5C4=4, as that leaves two naked pairs 2&3 in R5C2+R5C8 and 7&9 in R5C7+R5C9. Nothing in that row rules that out, although the rest of the grid may well make that impossible. | |
| Jul 23, 2020 at 12:32 | comment | added | dtc348 | @JaapScherphuis I think, it is naked triple. Am I right? | |
| Jul 23, 2020 at 12:03 | comment | added | Jaap Scherphuis | But how can it be a hidden triple if the 4 pencilmark occurs more than 3 times in that row? If that was allowed you might as well say the 8 and 9 form a hidden pair in that row. | |
| Jul 23, 2020 at 11:55 | comment | added | dtc348 | @JaapScherphuis I have corrected my printing mistake. $2$, $3$, and $4$ of R5C2, R5C7 and R5C8 will follow the hidden triple technique. That why 2, 3 and 4 will be in either R5C2 or R5C7 or R5C8. Therefore we will remove 4 from R5C4, 8 and 9 from R5C7. | |
| Jul 23, 2020 at 11:52 | history | edited | dtc348 | CC BY-SA 4.0 |
Printing mistake
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| Jul 23, 2020 at 11:40 | comment | added | Jaap Scherphuis | I don't see how that hidden triple works. There should be a 4 pencilmark at R5C4 but it somehow went missing from the hidden triple picture. | |
| Jul 23, 2020 at 11:19 | history | answered | dtc348 | CC BY-SA 4.0 |