Nurikabe minicubes: the Headache, the Panache, the Apache

These are three-dimensional Nurikabe puzzles. In each case, the four squares represent the layers of a $$4\times4\times4$$ cube. The goal is to shade some cells in each layer so that the resulting space satisfies the rules1 of Nurikabe:

• Numbered cells cannot be shaded.
• Unshaded cells are divided into regions, all of which contain exactly one number. The number indicates how many cells there are in that unshaded region.
• Regions of unshaded cells cannot be adjacent to one another, but they may touch at a corner or along an edge.
• Shaded cells must all be orthogonally connected in 3D space.
• There are no groups of shaded cells that form a $$2\times2\times1$$ cuboid in any dimension.

1 Paraphrased from the original rules on Nikoli

In all three explanations, I'll use "LxRyCz" to refer to layer X, row Y, column Z (all counted starting from 1, left-to-right or top-to-bottom). The directions will be "left/right/up/down" within a layer, and "back/forward" between layers -- the first layer is the "front", and the last layer is the "back".

Puzzle 1

The obvious place to start is with the size-1 regions:

Some empty cells were formed because they would make 2x2x1 cuboids.

Next, some empty cells can only be accessed by certain regions:

specifically, the two in layer 3. This forces some more empty cells:

And we've completed more regions, and forced some more unshaded cells:

and the rest resolves with the same techniques.

Puzzle 2

Start with the same techniques as before: finish the size-1 regions, and mark walls in any cells that would connect two regions.

Some regions now have only one way to extend:

the 2 and 3 in the front layer, the 4 in the third layer, and the 3 in the back layer. Each of those regions can be finished off.

Now we've incidentally finished a region by forced empty cells: block it off and mark any newly arising almost-2×2×1s.

Finally, there's one last deduction to finish the puzzle off:

Something has to reach the bottom-left of the front layer. The only region that can do that is the 5, meaning it has to go up through L(2-3)R4C2. That takes up four of its five cells:

and the remaining one has to be used to block a 2x2x1 in the back-bottom-right.

Puzzle 3

Once again, finish off the 1-regions and shade any cell that would connect two rooms.

Some regions have only one way to extend now: do that.

Now, there's a...

cell that can't be reached: the very front-top-left cell. Once that is taken care of, and the front 3 extends upwards into L1R2C2, there are a few more unreachable cells in layer 1, and one in layer 2 at L2R2C3.

The last cell in the previous step forms another almost-2×2×1. This completes a region:

and another unshaded cell is forced, in the top right of the first layer.

The rest of the puzzle resolves just by repeatedly completing regions and checking for new unshaded cells.

• For puzzle 1, doesn't the $7$ region have $6+2 = 8$ cells? Also, the shaded cell in the upper left corner of the third layer doesn't seem connected to any other shaded cell. – Jens Sep 14 '19 at 22:34
• @Jens Whoops, you're right - accidentally marked the top-left of the second layer unshaded instead of shaded. Thanks for pointing it out! – Deusovi Sep 14 '19 at 23:19
• All correct. Nice job! – Jafe Sep 15 '19 at 12:30