Walkthrough Of Puzzle 26 (Nurikabe)

Here is a walkthrough of one of my puzzles.

This puzzle starts in the lower-right corner. The seven in R8C8 can not extend into the red cell because it would surround the shaded cell in R8C9; this would make the shaded cells unable to connect. Therefore, the seven must be placed in the remaining cells (shaded blue).

There are a few easier steps next.

The next place to look is in the upper left corner. You can place four shaded cells in that corner because they can not be reached by any of the numbers in the grid. This also gives two unshaded cells in R2C2 and R2C4 because of the no 2x2 square logic. (The cell in R2C2 can only be reached by the seven at R1C7, so I have drawn a line connecting them to show that they are in the same island. Similarly, I have drawn a line connecting 

R2C4 and R7C5.)

The next step is in the upper right corner. If the seven at R7C9 did not continue into the red cell, that cell would be shaded, as well as the two above it. This creates a 2x2 square, which is not allowed. Therefore, the seven must continue up into the red cell.

There are some more easy steps next.

Next, look at the red cells. The sevens in R1C7 and in R7C5 have just enough cells to connect to R2C2 and R4C2. They are not large enough, however, to enter the red cells and still reach R2C2 and R4C2. Therefore, the red cells must be shaded.

Now, look at the red and blue cells. The red cell must be unshaded, because of the 2x2 square rule. Also, the red cell must be part of the 7 at R7C5. You can therefore mark the blue cells as unshaded and complete the polyomino.

After that, you can use the same logic to prove the position of the 7 at R7C7.

The last step of the puzzle uses the red cells shown. The only way R2C2 can be a part of the seven at R1C7 is if it extends right, through the red cells.

The puzzle is solved!