This puzzle is another Nurikabe, but with a new twist for me: it's a gigantic Super-Kabe! It's also, in my opinion, quite tricky and has some interesting deductions. I hope you enjoy!

Rules of a Nurikabe (copied from my previous puzzle):

This is a Nurikabe puzzle. The goal is to paint some cells black so that the resulting grid satisfies the rules of Nurikabe:

  • Numbered cells are white. (Think of them as "islands.")
  • White cells are divided into regions, all of which contain exactly one number. The number indicates how many white cells there are in that region.
  • Regions of white cells cannot be adjacent to one another, but they can touch at a corner.
  • Black cells must all be orthogonally connected. (Think of them as "oceans.")
  • There are no groups of black "ocean" cells that form a 2×2 square anywhere in the grid.

Now, the puzzle:

enter image description here

And as always, here is a handy puzz.link solver in case you would like one. The image should also be MS Paint compatible, if you click through to the larger version.

  • 1
    $\begingroup$ Nice puzzle...thanks Mick! $\endgroup$ Commented Jan 18, 2021 at 3:19
  • $\begingroup$ FYI, I found that there was an overreliance on "only this clue can possibly reach this whole 2x2 block". That's visible in the finished solution by there being so many clues entirely surrounded by black, including on the diagonals, and many straight/shortest path lines, especially on the larger clues, most conspicuously the 8, 5, other 5 and 6. But I do really enjoy Nurikabe and this was a satisfyingly straight forward solve. Thanks! $\endgroup$
    – Tirno
    Commented Jan 18, 2021 at 11:26

1 Answer 1



Completed Grid


First we start with basic connectivity and counting deductions, to get to:


Reaching around:

Looking in the lower right corner, the 2 on the right edge must go down to prevent a 2x2. Just above it, the 2x2 R6-7C19-20 must have an unshaded square, but nothing can reach it except the 2 in R5C19. Then the unshaded square in R3C17 can only be reached by the 5 in R7C17, resolving most of the right side.

Moving left, the unshaded square R6C6 can only be reached by the 4 in R6C9. Then the 2x2 block R4-5C4-5 is only reachable by the 3 in R3C5, forcing R4C5 to be unshaded. Now looking at the 5 in R4C10, it can have at most 1 square going to the left. Moreover there is a 2x3 block, R6-R7C10-12 that can only be reached by the 5, and in particular R6C11 must be in its area. The grid thus far:


Moving to the left side:

There is another 2x3 area which cannot be covered except by the 8 in the lower left, namely R9-10C6-8, so we must have R10-C67 unshaded in the 8s area. We cannot isolate any squares underneath this area, so we must in fact have the whole row R10C1-7. Looking at the 3 in R9C10, we see that at least one of its cells must go right, as otherwise R9-10C11-12 would be a shaded 2x2. In particular, this forces the 8 area to continue into R10C8, so that R9-10C8-9 will not be entirely shaded. Now R8-9C1-2 must have an unshaded square, which must include at least R8C2.

Moving up for a moment, we see that the 6 in R2C3 cannot go left, since it would block the 2 in the upper left corner, so the 2 in R5C1 must go up. This forces the 4 in R8C3 to continue up into R7C2, completing the lower left. Finally, the 6 cannot cover the R2-3C1-2 2x2, so the 2 in the upper left corner must come down. The grid thus far:


Lower middle:

The 2x2 R9-10C12-13 can only be covered by the 3 in R9C10, which must extend right to R9C12. This then completes the 4 in R9C16, forcing it to extend to R10C14, and also completes the 5 in R4C10, since R7C11 must be unshaded. Now, the 2x2 R5-613-14 can only be reached by the 6 in R1C13, which must extend down, and easy logic finishes this section. The grid thus far:


Finishing up:

Since the 6 in R2C3 cannot use R2C4, the 2x2 square R1-2C8-9 can only be reached by the 3 in R3C8, and the rest is just fill-in.

  • $\begingroup$ Excellent solve path and perfect explanation, well done! :D $\endgroup$
    – Sciborg
    Commented Jan 18, 2021 at 3:20

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