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The August 26 edition of Puzzle Page Sudoku

I eventually used a trial and error method to solve it:

  • Assume B1 = 2
  • C2 and F1 = 4
  • C3 = 2
  • F3 = 6
  • There are no candidates left for A3

This let me set B1 = 4 and hence solve the puzzle. But is there a generic method I could have applied here to solve this?

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closed as off-topic by Glorfindel, gabbo1092, Brandon_J, Rubio Aug 27 at 1:33

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  • $\begingroup$ There's only two 4's in 3 blocks: TL, CL, CR, so trial and error on 4 forces one of only two options into the remaining TR block. $\endgroup$ – JMP Aug 26 at 20:25
  • $\begingroup$ Have you got a link to the site this came from? $\endgroup$ – JMP Aug 26 at 20:26
  • $\begingroup$ It's from an app called Puzzle Page. $\endgroup$ – Somebody Aug 27 at 12:15
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Just look at the remaining 4s. If B4 is a 4, then so is C2 and then so is F1. But F1 and B4 block both 4 options in the right-center section, so the left-center 4 must be C3. The rest of the solution thereafter is trivial.

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  • $\begingroup$ You can do this starting from F3=4 as well, which blocks the left-centre block. Or start from C2=4. $\endgroup$ – JMP Aug 26 at 16:14
  • $\begingroup$ Yep! Since there are only four 4s remaining, that's where I'd start. (I play this game too so that is indeed where I started when I got to this point!) $\endgroup$ – Somebody Aug 26 at 16:17
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You can prove:

D4=4, otherwise D4,E4,D5,E5 isn't unique.

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    $\begingroup$ That's only usuable if you assume each sudoku is unique. Depending on which "game" you're playing this might not be guaranteed... On the one hand I fully agree and use this technique. On the other hand I don't like resorting to this kind of reasoning ... still +1 for being the only non trial and error trick left. $\endgroup$ – Imus Aug 26 at 14:31
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This is another trial and error attempt I did:

If you choose F1=4, this will reduce the options of the '4' in rows 2 and 3 to be at column 'C'. C2=C3=4. Then, the '4' of the first block can't be at C2 and must be at B1.

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I would generally start looking at forced chains of cells with 2 options left.

In this particular case my first idea would be to follow the chain

A3 -> A6 -> B6 -> B1 -> C2 -> C3

This means:

either A3 is a 2 or if it's a 6 (following the forced chain) C3 is a 2.
Thus B4 and F3 cannot be 2.

And next F needs a 2 in top block so D2 can't be a 2

Similar chaing reasoning can be used to see if a cell with 3 options is part of such a chain as well to reduce it to only 2 options instead:

In this case (after previous steps) I would look at F2 starting the chain from F3:

F3 = 4 -> F2 = 4
F3 = 6 -> A3 = 2 -> C3 = 4 -> C2 = 2 -> F2 = 6
=> F2 cannot be 2
Which then leads us to F1 = 2 and solving the rest trivially.


I do want to point out though that the above steps took longer than your trial and error solution.

Which one is the best way to solve it depends largely on what you want.
I personally don't like guessing without reasoning so will only fall back on trial and error if all other options are exhausted ... or at least untill it takes too long to spot any of those other options. Nothing wrong with at least trying something instead of staring at a really hard puzzle without progress.

Final note: sometimes when just trying something randomly makes it easier to spot other forced hints even if your initial guess didn't directly lead to reducing any options.

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F1 must be 2

Why?

enter image description here

Let's say we put in a 4 at the orange marked cell F1. Then B1 must be a 2; C2 a 4; C3 a 2; A3 a 6; and B4 a 4.
However, we now have a 4 at F1 and a 4 at B4, eliminating the 4 from the entire middle-right block (F3/D4), which isn't correct. Because of that, we can conclude that F1 cannot be a 4 and must be a 2. After that it should be trivial to continue.

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  • $\begingroup$ Thanks for your answer, but how is this different from the trial and error process I posted in my original question? $\endgroup$ – Kidburla Aug 26 at 18:58

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