5
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Solve this Sudoku, with rules :

  • Every blue square means, it is the left-top of a 3x3 block with each numbers [1 to 9] appearing only once there.
  • There is only one unique solution.
  • Check the example.

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Example
enter image description here

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    $\begingroup$ Just to check whether I've understood correctly: if there's a blue square then the 3x3 block whose northwest corner is that blue square has exactly one of each digit in it? $\endgroup$ – Gareth McCaughan Nov 23 '17 at 11:42
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    $\begingroup$ (So e.g. the top left blue square is unnecessary because the 3x3 block it's the NW corner of is already known, by ordinary Sudoku rules, to have exactly one of each digit.) $\endgroup$ – Gareth McCaughan Nov 23 '17 at 11:42
  • $\begingroup$ Really fun. I enjoyed solving it. :) $\endgroup$ – Maria Deleva Nov 23 '17 at 12:43
5
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I'm pretty sure this is the solution:

527 891 346
346 527 891
891 346 527

752 918 634
634 752 918
918 634 752

275 189 463
463 275 189
189 463 275

The way I solved it:

The positioning of the blue squares gives a very particular pattern in the numbers. Looking at the first 3 columns, the 1st, 4th and 7th row have the numbers 2,5,7 in different orders. 2nd, 5th and 8th have 3,4 and 6; and the 3rd, 6th and 9th row have 8,9,1. The same pattern applies per 3 rows. With the given numbers, this allows solving quite easily.
Also, though I only see it now, groups of 3 numbers appear in the same horizontal order in every 3x3 block, and go down diagonally within every group of 3 rows. See e.g. 527: top row in the upper left block, middle row in the upper middle block, lower row in the upper right block. The same doesn't seem to hold true for vertical ordering.

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5
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The solution is

5 2 7 8 9 1 3 4 6
3 4 6 5 2 7 8 9 1
8 9 1 3 4 6 5 2 7
7 5 2 9 1 8 6 3 4
6 3 4 7 5 2 9 1 8
9 1 8 6 3 4 7 5 2
2 7 5 1 8 9 4 6 3
4 6 3 2 7 5 1 8 9
1 8 9 4 6 3 2 7 5

and a key observation is that

when you have two adjacent blue squares, it tells you that two "parallel" blocks of three squares contain the same set of three numbers, which is extremely constraining.

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    $\begingroup$ Note: Lolgast posted his solution about five seconds before I posted mine. So, Jamal, unless you find mine markedly better in some way, Lolgast should get the coveted green checkmark. $\endgroup$ – Gareth McCaughan Nov 23 '17 at 12:21

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