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From this point is it necessary to guess or is it possible to find a logic solution?

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The short answer is

Yes there is a logical solution

However, it is a very complicated one

To solve this we need to use the 'BUG' principal.

A 'BUG' in Sudoku stands for 'Bi-value Universal Grave', which sounds scary. However all it means is that, for any given Sudoku, if there are two candidates in every remaining cell then there is not a unique solution. There is either no solutions, or 2 solutions.

As we can see in the above Sudoku, all cells contain two candidates, except one. This cell has 3 candidates and is in D1. This is an example of a 'BUG+1' where the grid is a BUG with one cell having three candidates. BUG+1s can be solved.

Please note:

BUGs work on the assumption that there is a unique solution. For the problem, I checked by putting the sudoku seed into a solver which calculated the number of solutions as 1, so I knew this could be solved using BUGs

To solve:

We need to look at the effect of removing one of the three candidates. One, and only one, of the candidates when removed will result in the rest of the grid having two candidates.

So:

If we remove 5 from the candidates, we get:

- 2 of each candidate in row D
- 2 of each candidate in column 1
- 2 of each candidate in box 4

So we don't have to look at the other candidates. To remove the BUG, D1 MUST be a 5.

And from there, there is a nice logical solution to the final answer, starting with removing 5 from the candidates in box 4, leaving two remaining possibilities. I can include it if needed.

Very tricky and advanced Sudoku this one!

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    $\begingroup$ Thanks a lot, but isn't this move a guess itself? $\endgroup$ – Yordan Apr 13 at 14:48
  • $\begingroup$ Oh, never heard about the <rot13>OHT+1 cevapvcyr</rot13>, that's pretty interesting. I was checking if one of these techniques would work, but apart from the very last one (which is basically guessing) I couldn't come up with something. It's relatively easy to come to the correct solution by just guessing a single cell and go from there till it's either solved or it contains an error, but I couldn't find a logical reason for filling in a number somewhere. $\endgroup$ – Kevin Cruijssen Apr 13 at 14:48
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    $\begingroup$ @Yordan I wouldn't say so. We don't actually try filling in any of the grid itself, and by doing this we KNOW that D1 must be above value. So I would say this is logical and not a guess, as we know for certainty one of the values $\endgroup$ – Beastly Gerbil Apr 13 at 14:52
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    $\begingroup$ @Yordan - Well, technically, most of the techniques are guesses, in the sense that their solutions are based on enumerating possible moves and finding that only one kind of move works; others lead to a dead end either immediately or in a few more moves. The question is - how many moves "look-ahead" do you consider to be "logical reasoning", and how many - "guessing and trying". $\endgroup$ – Vilx- Apr 13 at 22:39
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    $\begingroup$ @Vilx-: Using the BUG like this is a “guess” in a slightly stronger sense than most moves: it is only justified by the knowledge that the solution is unique. Most moves (however much look-ahead they have) don’t rely on uniqueness: they would still be valid for a Sudoku with a non-unique solution, where as this reasoning with the BUG could lead you astray there. $\endgroup$ – Peter LeFanu Lumsdaine Apr 14 at 16:15
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35pair

If the $3,5$ pair (shown connected by the green line) is set with $3$ at the bottom (light blue), then the $3$ triggers the other end of the blue line to be a $5$, and the green $5$ triggers the other end of the orange line to be a $3$. This cancels $r8c2$.

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The currently highest voted solution is good, however BUGs require you to assume that the puzzle has only one solution, which is usually reasonable to assume, but this has drawn criticism from some. So, if that's you for some reason, I'll offer another non-BUG solution (with the help of my favorite Sudoku Solver).

So, what you're missing is this:

There is an XY-chain on row 7.

The basic logic of it goes like this:

Column 1, Row 6 has only 2 possibilities.

First, it can be a 9. Looking at Box 5 (that is, the center box), its lower-left square can't be a 9.

Second, it could be a 5. If it is, the square immediately below it is an 8, which places a 2 in the same row, which places a 5 also in that row. This, in turn, removes the 5 as a possibility from the square immediately below it as well. Having only 2 possibilities itself, it must be a 9.

Now, notice what happens to the middle box when we do this. It removes 9 as a possibility from the same exact square as before.

Thus, if both of our original square's only two options eliminate 9 as a possibility from the same square, than we can safely eliminate 9 from that square, since it will be eliminated regardless.

This, in turn, places a 3 in the center box, and it all just caves in from there.

If you don't like that, you can apparently use a 3D Medusa on it as well (at least, according to the tool I'm using) and solve the whole thing almost instantly (since most of the remaining squares only have 2 options). Though this is quite complex.

If you want use an even more complex strategies, there is also an Alternating Inference Chain, a couple different types of Forcing Chains (which seem to be based on Alternating Inference Chains), Almost Locked Sets, and even a Death Blossom (which seems to be based off of locked sets.


All of these strategies mentioned (including the BUG strategy) are labeled "Diabolical Strategies" or "Extreme Strategies" by my tool, meaning this is quite a hard puzzle. Don't feel too bad if you had trouble with it. Some people (like myself) have trouble finding the "Tough Strategies" (like a swordfish), which are probably considerably easier than all of the above strategies.

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    $\begingroup$ True about the BUGs, I used a solver to find how many solutions before I wrote my answer to make sure there was a unique solutions :) $\endgroup$ – Beastly Gerbil Apr 13 at 19:21
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    $\begingroup$ Fair enough. That certainly was a good idea :) I wasn't trying to say this puzzle had two solutions, but rather the technique in general requires you to assume this. $\endgroup$ – Chipster Apr 13 at 19:30
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    $\begingroup$ oh Dw I know, I think there’s multiple ways you can solve this. BUGs was just the one I know about and noticed first, your answer is also totally valid! :) $\endgroup$ – Beastly Gerbil Apr 13 at 19:31
  • $\begingroup$ I think there's an error in your logic/explanation. If C1R6 is 5, then both C4R6 and C4R8 are both forced to be 9, which is obviously invalid. $\endgroup$ – Herb Wolfe Apr 13 at 21:28
  • $\begingroup$ @HerbWolfe I think that's less of an error and more so another way of looking at it. If you continue the logic in R6, then yes, you end up putting a 9 in two places. Of course, if you eliminate the 9 in C4R6 by placing a 5 in C1R6, then place the 3 in C9R6, you could eliminate 3 as a possibility in C4R6, eliminating all it's possibilities. It's just another way of looking at it. $\endgroup$ – Chipster Apr 13 at 22:22
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If the cell in row 6 column 4 is a 9, then this forces both the cells in row 6 column 1 and row 8 column 4 to be a 5. However this makes it impossible to place a 5 in the bottom left square. The cell in row 6 column 4 is therefore a 3, and the rest of the puzzle is readily solved.

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  • $\begingroup$ Simple and elegant. $\endgroup$ – Lanny Strack Apr 15 at 10:35

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