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I have filled all the possible singles. Also identified and shortlisted candidates, pointing pairs. How do I move ahead without guessing?

enter image description here

PS: I applied "Nishio", choosing one of the two possible candidate values on row 1, column 1. 7 will come in top left. Using which I can solve the rest of the puzzle. But it is "guessing". Any other way?

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4 Answers 4

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Others have pointed out the correct start point but not linked it to a method.

I call it Coupling

UPDATE After @justhalf's comment and further review I see that my method is just an extension of pointing pairs, which was already employed on the puzzle. So I'll update the rest of the answer from here to explain what I did in addition. For those unaware of what "pointing pairs" are (like I was, even though I used the method) here's a link to information on that.

Using pointing pairs:

you find couples of pointing pairs that are linked together in a set of 4, which regardless of which number is placed in it can also eliminate candidates.

I was able to do that in this image:

enter image description here You can see here, regardless of whether you put 4 or an 8 in r2c1, r3c5 will be the other, which means (without guessing) you know r3c3 cannot be a 4.

I don't have any other name for it, but the way I see it

these 4 positions are coupled together, forcing an additional elimination.

Using this as a starting point, I was able to logically fill out the rest of the puzzle:

Solved puzzle

Please excuse my poor writing. I think you can get the gist. I did this with one hand on my phone because my 5-month-old refuses to not be held and walked around the house. Lol

This is not a "fix-all" method. There are puzzles at the highest difficulty that cannot be solved without guessing, and indeed some have more than one solution (depending on who made it and how diligently it was designed.)

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  • $\begingroup$ Uh, I think your orange annotation in the first image is incorrect (I assume the orange slash means eliminated) The coupling method only applicable to eliminate those affected by two of those boxes which are confirmed to have different values, right. For example, you can't eliminate 8 from r1c2 in the first image (in fact, your final solution has 8 at this position). Similarly with the 4 and 8s in the bottom left box. $\endgroup$
    – justhalf
    Apr 7 at 13:21
  • $\begingroup$ @justhalf I updated my answer with extensive changes. I realized I missed something from the OP. $\endgroup$
    – Hawkeye
    Apr 8 at 20:30
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You can spot that

the 1, 7 cell cannot be a 4 (and therefore is an 8)

because

setting it to 4 forces an 8 into cell 5,7; which forces a 4 into cell 5,3; which eliminates 4 as a possibility from all of the remaining open cells in the top left block (cells 1,2 and 3,3).

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You could spot that:

In the top left block, the bottom right 4/6 cell must be a 6.

Why?

Because if it is a 4 then in the bottom left block the right side becomes 5/6,5/6. But now the solution isn't unique as setting either of these cells as 5 has no vertical effect on the board (from the bottom right block).

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  • $\begingroup$ I am still unclear. What is the vertical effect in bottom right block please? $\endgroup$
    – jerrymouse
    Dec 10, 2018 at 19:00
  • $\begingroup$ I'm confused. Using (column,row) notation: setting (3,3) to 4 would require (1,7) to be 4, then (2,9) to be 8, (9,9) to be 5, (3,9) would be 6, and (3,8) would be 5. No ambiguity. $\endgroup$ Dec 10, 2018 at 19:31
  • $\begingroup$ @user3294068; well it can't be a 4 according to OP's p.s. $\endgroup$
    – JMP
    Dec 10, 2018 at 19:49
  • $\begingroup$ @jerrymouse; if you follow my logic, we could have both 568:645 or 645:568 equally, which bears no impact on the rest of the grid. $\endgroup$
    – JMP
    Dec 10, 2018 at 19:51
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    $\begingroup$ I might prefer guessing (brute force all options, doesn't seem too difficult here) to this approach, though: if you base any deductions on the solution's assumed uniqueness, you can only reach "an answer" at best. If you want to reach "the answer", you can't assume uniqueness, you have to prove it instead. $\endgroup$
    – Bass
    Dec 10, 2018 at 20:54
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You can notice

r2c1, r7c1, r3c5, r7c5 form a finned X-Wing for 4.

Thus

Consider r2c1. It must be either 4 or 8. If r2c1 is 4 that forces r3c3 to be 6. If r2c1 is 8 that forces r7c1 to be 4 which forces r7c5 to be 8 which forces r3c5 to be 4 which forces r3c3 to be 6. So if r2c1 is either 4 or 8 r3c3 is forced to be 6.

Hopefully you can get it from there.

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  • $\begingroup$ Hi Rocus, welcome to PSE. Please hide your answers using ">!" $\endgroup$ Jan 18, 2019 at 16:45
  • $\begingroup$ Yeah. Just figured out how that works. This was my first answer. $\endgroup$ Jan 18, 2019 at 16:46

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