Nurikabe: Build it yourself

Question

This is a Nurikabe puzzle. First form a 16 × 16 grid using the provided squares, then solve the grid by shading some cells so that the resulting grid satisfies the rules¹ of Nurikabe:

• 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 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.

_{¹ Paraphrased from the original rules on Nikoli.}

Answer 1

Solution

First, let's make some observations for assembly:

Red quadrant cannot be on the right, or else the 5-1-5 in the top right corner would lead to a shaded cell not being able to escape. Likewise, Yellow quadrant cannot be on the left due to the 2-2 bottom-left. In fact, neither Yellow nor White can be to the right of Red, since the top row prohibits this (since two clues can never be adjacent to each other since they must be in different islands). So: Blue must be to the right of Red, and Yellow must be to the left of White.

And now, Blue cannot be on top of White, due to their first column. So we can finish our grid as follows:

Now, time to solve the puzzle:

Initial deductions:

Extend shaded squares:

We can more or less complete the yellow quadrant like this:

Bottom right 6 cannot escape left, and must go up at least a few:

Likewise R15C12 5 needs to go up at least one.

Let's focus on the red quadrant now. The bottom left corner 2x2 is only reachable by the 3, so the 3 must go there.

Same goes for the R(10-11)C(1-2); it must be addressed by the 5.

In fact R(12-13)C(3-4) must be addressed by the 2, and the R(12-13)C(4-5) must be addressed by the 7. This uses up all of the 7's length, so the 7 must only go left and a single up. By column 7 it must have already gone up, so it looks like this now.

This finishes the 2, and also the 5 in R16C9 only has 5 squares to go, so it fills that.

So 7 goes up at the very beginning and thus it is also completed.

Then 5 and 6 resolve:

Let's focus our efforts on the 8. It seems as if it has lots of space, but this is not really true. For one, it cannot touch the right border, or else the bottom right shaded cells cannot escape. Also since R11C14 is unshaded, R11C15 is shaded to allow for that mass to escape. So this means the 8 really only has 8 squares it can occupy.

7 must escape like this:

The 2's must resolve like this. We use the shape of the 7 blocking several shaded cells, plus the fact that if a 2 has only 2 places to go then something blocking both is shaded.

That resolves the central 5, also in order to escape the bottom right mass must escape between the 7 and the 5. This resolves the rightmost 5 as well.

The R4C7 5 must address R6C5 and goes down. So it can't go up, so the R1C6 2 needs to address R(2-3)C(5-6) so it must go down. Furthermore this resolves the R5C9 2.

That finishes up the 7.

The 6 cannot touch the top border or else the right shaded mass gets isolated. So it resolves like that.

So R1C9 3 resolves like this to avoid 2x2. The R6C5 which we've identified as going to the 5 must go right and never down, so R7C6 is shaded.

R9C4 3 must resolve like this, or else it goes flat, which makes a 2x2. Since R8C9 is unshaded R7C9 is shaded so that mass can escape.

The 5's must finish resolving like this to connect the shaded cells, complete!