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Rules of Nurikabe: Shade some cells in the grid. Each area of unshaded cells must contain exactly one number, which indicates the size of that area. Two unshaded areas may not share an edge, but they can touch at a corner. Shaded cells must form one orthogonally connected area, and no 2x2 square may be fully shaded.

Empty Nurikabe grid
Solve on Penpa+

No partial solutions, please. In particular, a correctly filled grid is not enough without an explanation of how to arrive at that solution and why no other solution works.

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    $\begingroup$ I got the answer almost immediately by guessing, lol. The interesting part would be how to prove it, hmm... $\endgroup$
    – justhalf
    Jan 27 at 16:31

1 Answer 1

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Solved grid:

enter image description here

The key idea here is different from the usual Nurikabe. Usually, we start with finding numbers separated by one square (shaded square in between), or from squares numbered 1 (shades all around).

But here,

we first calculate the total area of unshaded squares, which is 9+8+12+20 = 49. The grid is 8x8, so total shaded area is 64-49 = 15.

Now, consider the following green region:

enter image description here
At least one of the green squares are shaded, otherwise, 9 will connect to 8. Similarly, there is at least one shaded square on the right edge (red), bottom edge (yellow), and left edge (blue).
enter image description here

Based on the rules of Nurikabe, all shaded squares are connected, and we also observe:

to connect the shaded square(s) in top edge to the bottom edge, at least one square of each row (excluding the edges) needs to be shaded. This requires at least 6 shaded squares (excluding the edges). Similarly, to connect the shaded square(s) in left edge to the right edge, at least one square of each column (excluding the edges) needs to be shaded. This requires at least 5 more shaded squares, since we already have at least one column covered by the first step.
Including the at least 4 squares at the edges, we already covered the only 15 shaded squares that we have.
This means we need to connect the edge squares in more or less straight lines, with no leftover budget to spare. This will divide the 8x8 grid into 4 rectangular unshaded regions.
This means the unshaded region containing the 9 must be 3x3 (1x9 is impossible), and the unshaded region containing the 20 must be 4x5 (2x10 and 1x20 is impossible), and the unshaded region containing the 9 must be 2x4 (1x8 is impossible since it'll hit 9 or 20). These unshaded regions should cover the corners, so the region containing 9 is fixed, and since there can be no 2x2 shaded squares, the orientation of the unshaded region containing 8 is also fixed (can only be yellow+red in the image below:)
enter image description here
And with 4 shaded squares left, there is only one possible way to make the rectangular region containing the 12, as per the solution.

As for the title, "Back to Basic",

I think it's about the simple multiplications for the areas, as in they are basic maths? Or it can also refer to the fact that we rely quite strongly on one of the basic rules of Nurikabe, the connectivity of shaded squares?

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  • $\begingroup$ Not really proud of the low level of rigor at the last part, but I suppose it's a good start. $\endgroup$
    – justhalf
    Jan 27 at 16:59
  • $\begingroup$ Another way to formulate the explanation: The minimal shaded area to separate the 4 areas is two straight lines (like a cross) - and this already uses up the complete budget of shaded cells we can have. And from this position the only neutral changes (not increasing the shaded area) are moving the lines of this cross, without creating any corners. After that your arguments for the 20 and 9 follow and leave only one solution. $\endgroup$
    – Falco
    Jan 29 at 9:51
  • $\begingroup$ @Falco that's the idea, yea. The explanation you provided was on my list of consideration, but neither feels "right" yet to me. haha $\endgroup$
    – justhalf
    Jan 29 at 15:17

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