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Here is a sudoku, but with additional special rule.

The highlighted cells(which are gray) should have distinct number, from 1 to 9.

Sudoku

Could be quite hard… ;)

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2 Answers 2

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Using only singles and the fact that the shaded cells must be all distinct gets us this far – all 5s and 6s placed: step 1 There must be a 2 in r6c12. This points into box 6 and shows r4c9 = 2. The same reasoning shows that 4 is in r56c6 and hence in r2c4, which eventually allows putting 9 in r2c8 and from there all 9s and 7s: step 2 step 3

r5c7 can only be 3 or 4. If it is 3, r5c9 is 4, r6c8 8 and r8c3 8. If 4, r8c3 can only be 8. We thus fill in r8c3 = 8, and no matter how the last two shaded cells are filled, r6c8 = 8. We thus solve all 8s. step 4

The rest is easy once you realise that 4 cannot go into r8c7. solved

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  • $\begingroup$ With your first picture, surely the cell at r4c9 can be easily filled in by noting that the digit in question cannot be placed anywhere else in r4; similarly the digit at r2c4 cannot be placed anywhere else in c4? $\endgroup$ Commented Apr 15 at 8:44
  • $\begingroup$ @Neil One does not absolutely have to put each number by the simplest line of reasoning. $\endgroup$ Commented Apr 15 at 12:21
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Here’s my approach:

I think my first steps are similar to Parcly Taxel’s answer. I figured out all the 5’s, 6’s, 7’s, and 9’s and a bunch of other cells and got the following grid

special sudoku 1

At this point, there are two options for 3 in the gray cells: bottom right or middle right. I tried the bottom right first and it soon led to inconsistency: there’s no way to place a 2 in the bottom right 3×3 square.

special sudoku 2

So that means there’s a 3 in the middle right gray cell. The rest was again straightforward.

special sudoku 3

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    $\begingroup$ Nice deduction and logic(+1). Just for the improvement of your logic, from your first result, rot13(E8P7 unf 2, fvapr E8P3, E8P8, E5P7 unf qvfgvapg ahzoref, barf bs 3, 4, 8.) Maybe you can then solve it without guessing. $\endgroup$ Commented Apr 12 at 15:08
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    $\begingroup$ @RDK. Ah yes, that’s wonderful. $\endgroup$ Commented Apr 12 at 15:11
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    $\begingroup$ Thanks(, and that was why I decided to post this :) ). $\endgroup$ Commented Apr 12 at 15:14

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