I am unable to finish this Sudoku. All numbers are verified by the app. I'm sure there must be a way to solve this without guessing. Even an online tool couldn't tell me a next number.
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2Have you tried this solver? https://www.sudokuwiki.org/sudoku.htm?bd=400203005060010024072406019030004000080000473004300000823040156156832947947561238 I know your board doesn't have any "pencil marks" but the first strategies are quite simple. (edit - strategies ended up not being simple) (btw, "Unique Rectangles" seems to be Glorfindel's strategy, but it's not required and you can uncheck the box) – Alto May 07 '20 at 19:44
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I removed my pencil marks to not bias anyone here. I found many entries to popular strategies but none which lead to something. I tried a different solver but will try your suggestion. – infinitezero May 07 '20 at 19:46
3 Answers
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It's a bit of a hack, but note that the only possibilities for
B4, G4 and G6 are 7 or 9.
That means that
if B6 would be a 7 or a 9, the Sudoku would have two solutions:
- a 7 in B6 and G4, and a 9 in and B4 and G6
- a 9 in B6 and G4, and a 7 in and B4 and G6
Since Sudokus are
required to have a unique solution, B6 cannot be a 7 or 9, but must be a 5 or an 8.
I hope you can take it from there.
Glorfindel
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Glorfindel's method of assuming uniqueness of solutions is OK and valid (for Sudokus), but a puzzle with a unique solution can always logically be solved without resorting to such logic. Here's how I did it.
First, you should look for a cluster of cells which have few possibilities (ideally just two) and are closely related, so that if any one of them is filled all the others go down.
After hunting around for a while, I decided on cells B1 and C1. It is clear that these can only be $3$ and $5$ (in some order), while the $3$ in the top right box must be either in B7 or C7, and the $5$ in the top middle box must be either in B6 or C5. So whatever order of $3$ and $5$ we assume in B1 and C1, stuff can be deduced right away.
Let's assume then that
B1 is $3$ and C1 is $5$. Then the deductions go like this up to a final contradiction:
So it must be instead
B1 is $5$ and C1 is $3$. Then you can fill in the whole top right box pretty quickly, and just keep going from there. I won't solve the whole thing for you since it seems all you need is to get unstuck at this one point.
Rand al'Thor
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While I appreciate the answer (+1), this seems more like an educated guess than logic. – infinitezero May 07 '20 at 14:45
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2@infinitezero This is pure logic: you guess one of two possible options and then reach a contradiction, which proves that option logically impossible and so it MUST be the other option instead. It's a common technique in Sudoku and other grid-deduction puzzles. If you get stuck, try something (especially something unlikely) and see if you get a contradiction, which then enables you to proceed logically from the place where you got stuck. – Rand al'Thor May 07 '20 at 14:50
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I agree with your reasoning. But I don't find this approach satisfactional :) – infinitezero May 07 '20 at 15:47
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@infinitezero Well, it's up to you whether you like this method or not. But it's definitely logically correct, and guarantees an answer even when a uniqueness argument like Glorfindel's doesn't work. – Rand al'Thor May 07 '20 at 15:53
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Sure it does. By I prefer to solve by deduction rather than by trying. In theory I could just try any permutation of numbers and see if that leads to a valid solution or contradiction. – infinitezero May 07 '20 at 16:00
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3How does deduction differ from trying? I think a valid deduction would be something like "if I put a 7 here, then this one must be an 8 or a 9, but if it's 8 then there's no solution over here and if it's 9 then there's no solution over there, therefore this can't be a 7", but that's literally just trying. Maybe the difference is that you didn't write it down? – Alex Jones May 07 '20 at 16:58
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2@infinitezero I see no distinction in "trying" from the "logic". If a column has all the values filled out except the last, how do you fill in the final cell if not by trying 9 possibilities, discovering that 8 of them lead to contradictions with the game rules and then concluding through a process of elimination that the 9th cell must be that value? – ryanyuyu May 07 '20 at 18:57
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This now turned into semantics. I think it's clear enough, what I mean. – infinitezero May 07 '20 at 19:02
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This link "always logically be solved without resorting to such logic" does not give any proof it just states it as given, also the question there is not about sudoku. Is there any indication that this statement hold true for sudoku? – Andrew Savinykh May 07 '20 at 20:11
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@AndrewSavinykh The proof is in the penultimate paragraph of that answer, and applies to any kind of logical deduction puzzle. – Rand al'Thor May 07 '20 at 20:23
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@AlexanderJ93 Regardless of what you call it: In one case you can write down facts like "this is definitely 7, 8 or 9", and then use that fact, and other facts, to directly deduce or narrow down the value of other cells. In the other case you assume (or "guess") something like "this is 7" and if you run into a problem you know it's not 7. It's a direct proof versus a proof by contradiction. I wouldn't try to argue that it's invalid (proofs by contradiction are certainly valid proofs in mathematics and logic), but there is a clear difference and I just don't want that in my puzzles. – NotThatGuy May 08 '20 at 00:33
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It's all about the level of reasoning required, there's no difference other than in difficulty. Do you want to only deal with naked subsets? Pointing pairs and triples? Hidden subsets? X-Wing, Sword and Jellyfish? XY-Wing or XYZ-Wing? The last several are hardly better than just guessing a chain... If you don't want to stick to easy puzzles, the more advanced techniques are required. – Jason Goemaat May 08 '20 at 01:34
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@JasonGoemaat It isn't hard to guess and then backtrack if it doesn't work. In fact, it's probably what one might expect a beginner to do when they get stuck. For the average person, it could in many cases be the long way around when they can't find the shorter but more complex way, so in that sense it could be considered a less advanced technique. It's just time consuming. – NotThatGuy May 08 '20 at 02:55
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Without assuming uniqueness, the puzzle can be solved by elementary techniques:
naked-pairs-in-a-column: c4{r2 r7}{n7 n9} ==> r5c4 ≠ 9, r4c4 ≠ 9, r4c4 ≠ 7
naked-pairs-in-a-column: c1{r2 r3}{n3 n5} ==> r6c1 ≠ 5, r5c1 ≠ 5, r4c1 ≠ 5
whip[1]: b4n5{r5c3 .} ==> r2c3 ≠ 5 (whips are interactions between blocks and rows or columns)
finned-x-wing-in-rows: n8{r3 r4}{c5 c7} ==> r6c7 ≠ 8
finned-x-wing-in-columns: n9{c2 c5}{r1 r6} ==> r6c6 ≠ 9
biv-chain[4]: r4c9{n2 n1} - r4c4{n1 n6} - r5n6{c4 c1} - r5n2{c1 c5} ==> r4c5 ≠ 2
biv-chain[4]: r7c6{n7 n9} - c4n9{r7 r2} - r2c3{n9 n8} - c6n8{r2 r6} ==> r6c6 ≠ 7
whip[1]: b5n7{r6c5 .} ==> r1c5 ≠ 7
hidden-single-in-a-row ==> r1c7 = 7
hidden-single-in-a-block ==> r1c8 = 6
whip[1]: c8n8{r6 .} ==> r4c7 ≠ 8
biv-chain-rc[3]: r5c6{n9 n5} - r6c6{n5 n8} - r6c8{n8 n9} ==> r6c5 ≠ 9
biv-chain[3]: r6c8{n8 n9} - c2n9{r6 r1} - r1c5{n9 n8} ==> r6c5 ≠ 8
biv-chain[3]: r6n7{c1 c5} - c5n2{r6 r5} - r5c1{n2 n6} ==> r6c1 ≠ 6
singles to the end
Denis Berthier
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2This answer could be improved by explaining/linking to the terms you use. – Rand al'Thor Sep 26 '20 at 07:28
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@Rand al'Thor OK; the notations are those in my book "Pattern-Based Constraint Satisfaction" (freely available on research gate ) and in my CSP-Rules solver (freely available on GitHub: (https://github.com/denis-berthier/CSP-Rules-V2.1) – Denis Berthier Sep 27 '20 at 07:05