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I'm an absolute beginner and was stuck at the first puzzle of Sven's Sudoku Pad. I reached the following board state, where I noted for every cell which values were possible:

the state of the sudoku to solve

After a while I simply tried to brute force a solution. I set the value of the cell R7C4 to 5, which quickly resulted in a contradiction: I could not place a 7 in the bottom middle 3x3 block. After that I was able to solve the puzzle quite easily.

My question is which technique (instead of pure brute force) I should have applied here. I tried reading up on Y wings and similar techniques, but spotting and applying them are quite hard indeed.

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  • $\begingroup$ Out of curiosity, why do some cells have duplicate numbers? $\endgroup$ Commented Mar 21, 2023 at 15:41
  • $\begingroup$ @2012rcampion The little marks represent the Snyder notation which I initially conducted. For this I only jot down for values that have exactly 2 positions in a 3x3 grid. The center marks just note all the possible values that can go in a cell which helped me brute force it. $\endgroup$ Commented Mar 21, 2023 at 16:21

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You can deduce that R7C1=5.

Consider columns 7 and 9, and look at where the 7s can go in those columns. Either R1C7 + R8C9 (call that case 1), or R3C9 + R9C7 (call that case 2).

In case 1, R1C7=7 restricts 7 to R2C1 or R3C1 (column 1) in box 1, so R7C1=5.

In case 2, R9C7=7 forces 7 into R7C4 (row 7) in box 8, so again R7C1=5.

There are no more cases, so R7C1=5.

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  • $\begingroup$ Thanks for the answer and the edit which made it clearer! Although this works out nicely, how would I go looking for this? And is there a name for this technique? $\endgroup$ Commented Mar 21, 2023 at 15:17
  • $\begingroup$ @FelixSchön You're welcome. I don't know if there's a name for it, but Cracking the Cryptic sometimes features these. I don't find them easy to spot. In this case, it was a nice surprise to find that the 7s only had 2 configurations in C7 & C9, and both configurations led to the same number in another cell. Watching the linked solve gave me the idea to look for this kind of thing, though the technique in the video is somewhat different - more complex but also more elegant. $\endgroup$
    – Lawrence
    Commented Mar 21, 2023 at 15:28

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