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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.

Hard Sudoku

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    $\begingroup$ Have you tried this solver? sudokuwiki.org/… 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) $\endgroup$ – Alto May 7 at 19:44
  • $\begingroup$ 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. $\endgroup$ – infinitezero May 7 at 19:46
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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:

enter image description here

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.

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  • $\begingroup$ While I appreciate the answer (+1), this seems more like an educated guess than logic. $\endgroup$ – infinitezero May 7 at 14:45
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    $\begingroup$ @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. $\endgroup$ – Rand al'Thor May 7 at 14:50
  • $\begingroup$ I agree with your reasoning. But I don't find this approach satisfactional :) $\endgroup$ – infinitezero May 7 at 15:47
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    $\begingroup$ How 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? $\endgroup$ – AlexanderJ93 May 7 at 16:58
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    $\begingroup$ @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? $\endgroup$ – ryanyuyu May 7 at 18:57
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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.

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  • $\begingroup$ That's a great way of thinking. I'll try it! $\endgroup$ – infinitezero May 7 at 8:47
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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

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    $\begingroup$ This answer could be improved by explaining/linking to the terms you use. $\endgroup$ – Rand al'Thor Sep 26 at 7:28
  • $\begingroup$ @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: (github.com/denis-berthier/CSP-Rules-V2.1) $\endgroup$ – Denis Berthier Sep 27 at 7:05

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