Rules of Nurikabe: (copied from here)

  • Numbered cells are unshaded.
  • Unshaded cells are divided into regions, all of which contain exactly one number. The number indicates how many unshaded cells there are in that region.
  • Regions of unshaded cells cannot be (orthogonally) adjacent to one another, but they may touch at a corner.
  • All shaded cells must be connected.
  • There are no groups of shaded cells that form a 2 × 2 square anywhere in the grid.

Special rule:

  • Every row and every column must have exactly 5 cells shaded (and the remaining 4 cells unshaded).

Now, solve the following puzzle. Unlike the last one (where you had to do at least some amount of case bashing), I confirmed there's a clean solving path this time.



2 Answers 2


Note: "islands" are unshaded regions, "oceans" are shaded cells. Also, sorry for the length!

Step 1:

step 1
Quick initial deductions: islands can't touch, the 1s must be surrounded by ocean, and C2 has all unshaded cells already identified.

Step 2:

step 2
To prevent a 2x2 of ocean cells at R23C12, the 4 must cover R2C1. No other island can reach that area. The 6 must also extend right in order to fit.

Step 3:

step 3
The 2 on the right must have another island cell in R1, so that accounts for all of R1's unshaded cells. So the 4 can't have any more R1 cells, which lets us place it fully on the left.

Step 4:

step 4
Only one more C1 cell can be unshaded, so R6C1 must be shaded. The 3 must therefore extend right a bit.

Step 5:

step 5
Over on the left side, islands can't touch. Only one more cell in R3 can be island. If the 10 has that cell, then the 5 is forced to cover four cells in R2 - but that's no allowed, since there is already an unshaded cell in R2. Therefore the 10 cannot have another unshaded cell in R3

Step 6:

step 6
There is no way for the 5 to reach R3C7 without using any other R3 cells (and remember, only one more can be used). Therefore it is ocean.

Step 7:

step 7
To prevent a 2x2 of ocean in R23C67, the 5 must reach to R2C6. There is only one way to do this while using a single C3 cell and preventing 2x2s, so we can place the 5.

Step 8:

step 8
C3 has almost all its cells spoken for - R8C3 must be unshaded to complete it. This forces the ocean in the bottom left to escape up. Now the 3 must extend left (to complete R1), which allows us to place the 3. I also shaded R2C9, because no island can reach it.

Step 9:

step 9
Both R6 and C4 have all of their ocean cells placed, so the rest must be unshaded.

Step 10:

step 10
There is now only one way to complete the 6 without having it touch the 10. Ocean can be placed around the 6 for connectivity.

Step 11:

step 11
For connectivity of top and bottom oceans, R7C8 must be shaded. This forces R7C7 to be island to complete the row.

Step 12:

step 12
The middle-bottom ocean must connect to the rest, which forces the placement of the 4.

Step 13:

step 13
For ocean connectivity R5C8 must be shaded. Also, C6 has all of its shaded cells spoken for so the rest must be unshaded.

Step 14:

step 14
The rest of the cells in R5 must be shaded to complete the row. This forces several cells to be unshaded to prevent 2x2s and allow the 10 to be connected to itself.

Step 15 (also solution):

The 10 now has all its unshaded cells. The rest of the cells are forced by the column/row rules.



I don't know if this is for sure the answer, but i think it is.

  • 3
    $\begingroup$ Yes, that is correct. Just note that answers to logical-deduction (and especially grid-deduction) puzzles are expected to include the steps taken to solve the puzzle. $\endgroup$
    – Bubbler
    Commented Nov 11, 2020 at 2:07
  • $\begingroup$ Ok, Ill type up my method $\endgroup$
    – Benja
    Commented Nov 11, 2020 at 2:16
  • $\begingroup$ So first I filled in the boxes i knew should be filled in, such as the ones in between numbered boxes, the boxes directly adjacent to a box with a 1 in it ,all of the empty boxes in the second column, and the box above the box in the lower left corner. I also figured out how many blank boxes that the end result would have (36), and from that, i found out the number of shaded boxes there would be(45). Then i did a bit of trial and error using the boxes that i knew from the bottom three rows and found a few solutions that worked, and all of them required the box with the 10 to extend into the... $\endgroup$
    – Benja
    Commented Nov 11, 2020 at 2:26
  • $\begingroup$ bottom of the puzzle, compared to taking up space in the top. Using THAT information i found that the box in the 6th column, 2nd row, had to be empty, but not part of the region that the 10 box would cover, which means that the region with the 5 in it had to extend completely to make sure that the box could be empty, and i was able to outline the area around that with shaded boxes. At that point it was basically just simple math and making sure that i didn't break any of the rules in order to complete the puzzle. $\endgroup$
    – Benja
    Commented Nov 11, 2020 at 2:33
  • 4
    $\begingroup$ You should put that information as part of your answer, not as a comment. Also, next time for future reference, it would be good to show the grid at intermediate steps so readers can visualise what you are saying. $\endgroup$
    – Alaiko
    Commented Nov 11, 2020 at 2:52

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