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So I have been doing research on the minimal amount of numbers needed to create an 8x8, and found this paper which went over the minimum amount of clues needed for a sudoku to have a unique solution for different sized Sudokus.

This is the 14-clue case for the 8x8 sudoku:

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which I have been trying to solve, although I'm not really able to for some reason. Here is my attempt at solving this:

  1. We can put a 1 in R2C1 and a 4 in R3C8:

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  1. Then, we can put a 2 and a 3 in R4C1 and R4C3 respectively:

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  1. Then, placing another 2 and 3 we get:

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which is where I am currently stuck. No matter what I do, I can't fill out any other numbers nor am I able to fill out any rows/columns.

So my question is, how do I continue solving the minimally clued 8x8 Sudoku?

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2 Answers 2

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You were on the right track with your previous step:

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Consider where the 2s can go in the bottom two right boxes. They form a pair in the blue cells, meaning the 2 in the top right box must go in the red cell.

If you then continue to look at the 2s and 3s, you'll quickly see where a 3 must go at the top, and this then lets you place a lot more 2s and 3s on the right

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  • $\begingroup$ oh my that's a lot of numbers I just placed. Thank you! $\endgroup$
    – CrSb0001
    Commented Jan 11 at 18:41
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    $\begingroup$ You don't need the blue squares. Just look at the fifth column, and the red square is the only possible location for the 2. $\endgroup$
    – fljx
    Commented Jan 12 at 8:51
  • $\begingroup$ @fljx good point, looks like I slightly overcomplicated things $\endgroup$
    – Beastly Gerbil
    Commented Jan 12 at 10:38
  • $\begingroup$ @BeastlyGerbil Maybe in this case, but it is an insightful attack. I wrote a generic solver (6x6, 8x8, 9x9, 25x25, with and without diagonals), but I can't even figure out whether there is an exhaustive list of attacks. I eventually wrote a brute-force guessing rule with back-tracking. Then I figured I had a Sudoku generator -- just start from empty and make ransom guesses until it becomes deterministic. $\endgroup$
    – Paul_Pedant
    Commented Jan 12 at 12:03
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This was a fun little challenge! I'll post my completed puzzle below if you're still stuck (spoilers), but the initial answer by Beastly was correct. To add on, once you're able to get more of the 2s and 3s, you'll surely find yourself stuck again as you'll have a lot of spaces for the 6, 7, and 8s. How I overcame that issue was by writing out possible numbers for where the numbers could be and using those potential numbers in that row or column to eliminate their placement in other parts of the puzzle.

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Looking at the blue columns, if we assume any could be 2 or 3, we can then eliminate them as a possibility for their horizontal rows. The if_then and if_and_then logical processes are your friends here.

Below is my completed puzzle if you're interested (SPOILERS)

Completed Puzzle

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