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?


You can spot that

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


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).


You could spot that:

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


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).

  • $\begingroup$ I am still unclear. What is the vertical effect in bottom right block please? $\endgroup$
    – jerrymouse
    Dec 10 '18 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 '18 at 19:31
  • $\begingroup$ @user3294068; well it can't be a 4 according to OP's p.s. $\endgroup$
    – JMP
    Dec 10 '18 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 '18 at 19:51
  • 1
    $\begingroup$ 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 '18 at 20:54

You can notice

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


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.

  • $\begingroup$ Hi Rocus, welcome to PSE. Please hide your answers using ">!" $\endgroup$ Jan 18 '19 at 16:45
  • $\begingroup$ Yeah. Just figured out how that works. This was my first answer. $\endgroup$ Jan 18 '19 at 16:46
  • $\begingroup$ No problem. Happy puzzling! $\endgroup$ Jan 18 '19 at 16:47

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