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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.

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1 Paraphrased from the original rules on Nikoli

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Answers
Puzzle 1

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Puzzle 2

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Puzzle 3

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Quick notes on notation

This is a blank square. We have not deduced anything yet.
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This is an unshaded square.
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This is a shaded square.
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I make use of both an absolute coordinate system... RxCyLz, standing for Row x, Column y, Layer z.
To refer to multiple cells (usually to mark some 2x2's) I parenthesize squares to group them together. So R(2-3)C3L(3-4) refers to the four squares R2C3L3, R2C3L4, R3C3L3, R3C3L4.
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I also make use of a relative coordinate system. North, south, east and west always refer to staying in the same layer. Going up is going towards layer 1; going down is going towards layer 4.
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Puzzle 1 full solution

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Puzzle start. First, the standard Nurikabe deduction of having two diagonally adjacent clues still applies. But in the 3D version, some of these are harder to spot!
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The 9 must go north, and then up.
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Then north again. The bottom layer 4-clue must go north.
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The shaded cells in the top right needs to escape north, and then up.
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R4C4L3 must escape to R4C4L2. R3C4L2 must also be shaded to separate R3C4L1 and R3C4L3.
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Top layer 4-clue must resolve as follows. It can't go down because it's blocked by shaded squares, and at R4C3L1, going down would result in touching the 3rd layer 4-clue.
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Now we surround the top 4-clue.
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The 3rd layer 4-clue must continue west to R4C2L3, and the 9-clue must continue up to R2C4L2.
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Also, R3C3L2 must be unshaded to avoid creating a 2x2x1.
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However, R2C3L2 must be shaded, because otherwise it would make the 2-clue be at least size 3.
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We get two more extensions in layer 2 as a result.
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R3C2L4 is unshaded or else we make a 2x2x1 from R3C(2-3)L(3-4). Tricky to see, eh?
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Extend the isolated shaded group in R1C4L3 to R1C3L3.
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Now we can complete the 4th layer 4.
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And fill in adjacent squares.
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Now, the 9-clue is now 6 long. Its only squares of freedom starts in R1C4L2, where it can go either up or west. Up only has room for 1 square, so it must go west at least twice.
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This makes R2C4L1 shaded pretty isolated and it must escape that way.
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This finishes up the 2-clue. Also the R3C2L1 must escape west.
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Finish the 9 clue...
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At this point the two unshaded squares in R3C2-3L2 must go to the 6 clue, in only one way.
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And the 4 connects to R3C2L4 like that.
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Puzzle 2 full solution

1 clues are great and give us lots of information.
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Also some squares clues are separated apart by a block.
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Already, layer 3 needs R2C2 to be unshaded to avoid the 2x2.
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That completes the 2-clue.
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4-clue only has 1 room for expansion going south; so it must go east and down.
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R2C4L1 gets shaded to separate. Also, R2C1L4's 3 clue must go up as its only expansion.
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Now, to employ some tricky logic. One of R3C1-2L1 must be unshaded, or else we make a 2x2 (R3C(1-2)L(1-2)). Either the 3 extends south, or the 6 gets there, but the 6 would exhaust all of its length doing so. (west, west, up, up, north, say.) That would leave R(2-3)C4L3 unattended (The 4-clue requires 5 to get there; other clues are more hopeless). So the 6 must care for that (we can't immediately mark R3C4L3 as unshaded, because we could conceivably satisfy it by first going down, north, north, up). But we can declare that 3 must go south, south.
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Fill adjacents, then note that R2C3L1 must be unshaded to avoid a 2x2. That completes the 4-clue...
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Fill those adjacents, note that R1C1L3 must also be unshaded to avoid 2x2's, which completes the R2C1L4 3 clue...
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And fill yet more adjacents. Whew.
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Now we should look at layer 2's R4C(1-2). One of these must be unshaded, and it must be satisfied by the 6-clue. It must hit R4C2L2 first, so let's path that.
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Actually the 3-clue in R3C4L1 is pretty trapped and has only one way to resolve so let's do that.
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Fill adjacents. R2C3L4 is too far from the 6-clue now, as it's already size 4. So it completes the size 3 region as follows:
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Nothing can reach R1C4L4. Let's shade it.
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That means R2C4L3 must be unshaded to avoid another 2x2 (R(1-2)C4L(3-4)) and that completes the 6-clue. We're done!
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Puzzle 3 full solution

Let's do the spacing deductions again, and fill out all of the 1's.
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We use another 2D nurikabe deduction that often appears with 2's - if a 2-clue only has two places to expand, then the diagonally adjacent square that is adjacent to both of these places is shaded. Otherwise the 2-clue is trapped to either being a 1 or 3+ clue. This shades R2C3L4, since the 2-clue only has 2 places to go now.
enter image description here R1C1L1's 5 cannot go down at all! This finishes the 5-clue.
enter image description here And fill adjacents.
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Both the 9 clue and above it R1C4L1 must go south to escape. R4C1L1 must go east to escape. This also shades R2C3L2.
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Now R3C2L2 must be unshaded to avoid the 2x2 in R(3-4)C2L(1-2). That completes that 2-clue, and fill adjacents.
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The 5-clue in R3C3L1 must go down at some point. The only accessible place is R4C4L2.
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After shading in R4C3L2, R4C4L3 (both blocked by the 2-clue in R4C3L3), and R3L4C2 (blocked by the 9-clue), this completes the 5-clue.
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9-clue must go down. This allows us to fill R2C4L4 (blocked by the 2-clue in R1C4L4)...
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...which resolves the 2-clue that way.
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The 9-clue must keep going down, however. Only place to do that is via R3C4L4.
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Thus R4C4L3 must escape down, west.
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9-clue continues westward.
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Shade R3C3L3; that resolves the last 2-clue.
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R4C3L4 must keep escaping west, west.
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R2C3L3 belongs to the 9-clue or else R(2-3)C(2-3)L3 is a 2x2 shaded
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R2C3L4 escapes west...
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Then the 9-clue escapes west, completing it. Fill adjacents...
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Everything else goes to the 4-clue. We're done!
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  • 1
    $\begingroup$ Everything's correct, nice work! I really like the detailed explanations. You actually found a different way to get #2 than what I had planned -- my idea was to start with the fact that rot13(gurer ner gjb pbearef gung ner bayl ernpunoyr ol gur 6, v.r. obggbz yrsg ba yri 1-2 naq gbc evtug ba yri 3-4, naq gurer vf bayl bar jnl sbe gung 6 gb ernpu obgu). $\endgroup$ – jafe Mar 27 at 9:17
  • 2
    $\begingroup$ This is a thorough, slick, nicely illustrated and (somehow) speedy answer! It's hard enough to think in 3D yet alone explain it to somebody else - well done! $\endgroup$ – Stiv Mar 27 at 9:23

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