FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in          Games  Calendar Log in to check your private messages    is there a rule for this?          Sudoku Programmers Forum Index -> Solving sudoku View previous topic :: View next topic   Author Message arsoncupid Joined: 22 Nov 2005 Posts: 10 : Items Posted: Tue Nov 22, 2005 8:07 pm    Post subject: is there a rule for this? I am trying to hand solve a wicked puzzle I found here. This puzzle has no business be solved by hand! Code: ..1 .8. 6.4 .37 6.. ... 5.. ... ... ... .5. ... ..6 .1. 8.. ... 4.. ... ... ... ..3 ... ..7 52. 8.2 .9. 7.. Through basic logic I reach this state in the middle "row of groups": Code: 12379 124789 349 | 2789 2367 5 | 12349 134679 12679 2379 2479 6 | 279 1 29 | 8 34579 2579 12379 12789 5 | 4 2367 2689 | 1239 13679 12679 Notice in the middle row, the 3's and 4's. There are a pair of each, and the pairs intersect at the 8'th column. Late one night I assumed this meant I could exclude 5,7,9 from the 8th column because either a 3 or a 4 must go in that square. But now I see no way of saying that without excluding the other options. I did prove that this must be the case, in fact the eighth column's number must be a 3. But I had to do it by assuming a 5,7,9 in that possition, then proving it leads to a contradiction (just found out this is called Nishio). This is a rather large branch of logic and my work on this puzzle, still largely unfinished, is 16 pages! ... Another sample, one I've not proved but is the same sort of generalization, is with this puzzle: Code: .96 ... .1. .5. 6.. 7.. ..1 8.. ... 5.. .94 .68 ..6 ... .4. 97. 16. ..5 ... ..1 3.. ..5 ..2 .7. .1. ... 59. Through basic steps, the bottom right group (3x3) is: Code: 3 28 246 468 7 1 5 9 24 The same reduction applied here would drop the 4 from 4,6,8. Is there a logical way to say the following? In any row, column, or group, where two pairs intersect, you can exclude the other numbers from the intersecting square. Back to top Lummox JR Joined: 07 Sep 2005 Posts: 202 : Items Posted: Tue Nov 22, 2005 9:58 pm    Post subject: I'm not sure what you mean when you say there's a pair each of 3's and 4's in that middle section. There are definitely many more 3's and 4's there than that. Nor do the 3 and 4 appear "naked" anywhere (without other candidates). However it looks like a clue in your first puzzle is misplaced or missing, because it has multiple solutions and is therefore invalid. It looks like the 5 in r4c5 may belong in r4c6, or else perhaps the 4 in r6c4 may belong in r6c5. Which is it? Back to top arsoncupid Joined: 22 Nov 2005 Posts: 10 : Items Posted: Tue Nov 22, 2005 10:27 pm    Post subject: Lummox JR wrote: I'm not sure what you mean when you say there's a pair each of 3's and 4's in that middle section. There are definitely many more 3's and 4's there than that. Nor do the 3 and 4 appear "naked" anywhere (without other candidates). However it looks like a clue in your first puzzle is misplaced or missing, because it has multiple solutions and is therefore invalid. It looks like the 5 in r4c5 may belong in r4c6, or else perhaps the 4 in r6c4 may belong in r6c5. Which is it? I mean the 5th row. Sorry about the lack of clarity. I got the puzzle from the thread What is the hardest sudoku puzzle? -- It is the second post, availabel on the first page. Are you certain it does not have a unique solution? Sudoku Solver verifies it has only one solution. BTW, the second puzzle I listed above. I have verified with the Nishio method, the 4 can be dropped from the 4,6,8 square. Back to top Lummox JR Joined: 07 Sep 2005 Posts: 202 : Items Posted: Wed Nov 23, 2005 4:49 am    Post subject: You did misplace a clue from that puzzle. The 5 at r4c5, as I suspected, belongs in r4c6. Back to top rkral Joined: 21 Oct 2005 Posts: 233 : Items Posted: Wed Nov 23, 2005 11:20 am    Post subject: Lummox JR wrote: I'm not sure what you mean when you say there's a pair each of 3's and 4's in that middle section. I think arsoncupid is trying to dedce eliminations from the two conjugate pairs ... the 3s at r5c1 and r5c8 and the 4s at r5c2 and r5c8 ... which "intersect" (share the cell r5c8). My answer to the same question on another board was ... "if r5c1<>3, then r5c8=3, r8c3<>4, and r5c2=4" was the most that could be inferred. Of course, implications could start with r5c2<>4 as well. 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