Rules of 3D Tapa:

  1. The grid is a 3-dimensional cubes, represented here by layers of cubes from top to bottom.
  2. Shade some cubes such that all shaded cubes form a single orthogonally (in 3-axis) connected region.
  3. In addition, no 1x2x2 or 2x1x2 or 2x2x1 cubes are completely shaded.
  4. Some cubes have clues and these cubes cannot be shaded.
  5. Observing only up to their 26 neighboring cube, the clues represent the size of the blocks of orthogonally (in 3-axis) connected shaded cubes. (This is similar to Tapa rules, but in 3-dimensional.)

The "standard" 3D Tapa rules apply.

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  • $\begingroup$ If I understand it - it just happened to be that the layers are connected in such a way that the first one has only diagonal connection, the second to two ... and so on? Or just for convenience - you are not relating to the depth of the cube and describe only the 2D relation (the Tapa rule is a 2D with ALL 2D relation - are you limiting this to 2D relation for the 3D structure?) $\endgroup$
    – Moti
    Jun 25, 2021 at 17:18
  • $\begingroup$ @Moti The full puzzle consists of 5x5x5 unit cubes, the 3D connection is linked whenever two unit cubes share a side. $\endgroup$
    – athin
    Jun 28, 2021 at 3:07

1 Answer 1


Here is the solved puzzle


I didn't capture my step by step process, but once you figure out the first few steps it isn't too difficult:

Starting in the first layer, you can mark quite a few cubes as empty, such as the upper-left corner - if it was filled in, then one of the adjacent cubes would need to be filled in as well in order to satisfy rule #2, but that would violate the nearby clue of only being 1 contiguous block. Similar reasoning can be applied for quite a few cubes in that first layer.

When you take into account the second layer, you can mark even more cubes in the first layer as empty. For example, a filled cube in the upper right corner would not be able to get past the clue with 3 2's.

From there:

I pretty much followed the cubes that I had marked as filled in the first layer. The need for all the filled cubes to form a single connected region drives the solution.

  • $\begingroup$ This is correct! $\endgroup$
    – athin
    Jun 28, 2021 at 3:07

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