An entry in Fortnightly Topic Challenge #51: Non-rectangular grids.

This is a standard Tapa puzzle, on a non-standard Tapa grid...the "grid" is a Penrose tiling! Since the grid is not rectangular, we need to interpret the standard Tapa rules in this scenario. Your goal is to shade some of the tiles so that:

  • The shaded tiles form a single orthogonally connected region.
  • The given clues describe the pattern of shaded squares around each shape. There are a lot more possibilities than in a normal Tapa, but the only difference is that at each vertex of a tile there may be more than one tile diagonally connected to it.
  • No interior vertex (vertex: place where two or more tiles meet; interior: not on the border of the figure) can have all of the tiles it adjoins be shaded. This is the analog of the "no shaded 2x2" rule for normal Tapa, and notice that this formulation also works for Tapa on a rectangular grid.
  • Tiles containing clues cannot be shaded.



The picture above shows a "2 3" clue. The reddish tiles are shaded (different shades used to clearly distinguish the tiles), while the blueish tiles are unshaded. Empty tiles are not adjacent to the clue. As you walk around the boundary of the clued tile, you pass through a group of 3 shaded tiles, and then another group of 2 shaded tiles, and these groups are separated by at least one unshaded tile on either side.

I hope you enjoy!


(h/t https://github.com/sarahmarshy/PenroseGenerator for the code to generate the basic grid structure)

  • 2
    $\begingroup$ In a normal Tapa puzzle, I believe there is a rule that cells with numbers in can't be shaded. You haven't stated any such rule. Should we take it to apply to your puzzle, or not? $\endgroup$
    – Gareth McCaughan
    Mar 18, 2021 at 13:00
  • $\begingroup$ @GarethMcCaughan Ah shoot...yes, cells with clues cannot be shaded. Will add to description. $\endgroup$ Mar 18, 2021 at 13:47
  • $\begingroup$ Any chance of getting an explanation on what the clues mean? Or at least a link to the "standard Tapa rules". $\endgroup$
    – Bass
    Mar 18, 2021 at 14:05
  • $\begingroup$ @Bass Like Tapa, the clues are the pattern of shaded tiles around the clued tile. So for example, "2 3" implies that in the ring of tiles that surround the clue (orthogonally and diagonally), there is a group of 2 consecutive shaded tiles, and another group of 3 consecutive shaded tiles. These must be separated by an unshaded tile on both sides (otherwise the clue would be "5"), and all other tiles are unshaded. $\endgroup$ Mar 18, 2021 at 14:08

1 Answer 1


This was a surprisingly difficult puzzle! To start, some basic single-clue deductions:

enter image description here


connectivity forces the wedge piece left of the [1111] to be shaded. Applying the vertex rule, this lets the top get resolved.
enter image description here
More connectivity deductions on the right: the rightmost shaded region can't escape by going downwards.
enter image description here

Next, some deductions involving:

the left side. If the cell just ↘wards of the leftmost [1,2] is unshaded, that forces a 3-long segment near the other [1,2].
enter image description here
I've highlighted one cell in red: can the top group connect through there? If it did, we'd get this:
enter image description here
and then that would break the big [1,1,1,2] clue, because there's not enough space.
So it must connect through the top instead, and that lets us resolve the [1,1,1,2].

Some steps flow nicely from that:

enter image description here
enter image description here

Next, check

the [2,2] near the bottom. If the left shaded cell went downwards, that would block it off. So it must take the cell above it.
enter image description here

And now I had to use an advanced connectivity deduction:

Assume the bottom-right region does not sneak under the [2,3] clue. That means it must connect upwards...
enter image description here
then the cell marked with an × is unshaded. (The tail of the arrow would need to be used because otherwise, the two cells by the rightmost red bar would be both shaded, breaking the [2,4].) But then this breaks the top bar. So, to connect the bottom to the left, we need to go under the [2,3].

And that breaks in to the rest of the puzzle:

enter image description here
enter image description here
enter image description here

And with that, the puzzle is solved!

  • $\begingroup$ That is 100% correct! Got all of the deductions perfectly...nice job! $\endgroup$ Mar 18, 2021 at 14:39
  • $\begingroup$ @JeremyDover My immediate thought was that the given puzzle grid was the central sample from an implied infinite penrose tiling. First steps of this answer seem to rely on the puzzle space being finite. Is it [intended to be] solvable, at least within the finite region shown, if the tiling is assumed infinite? $\endgroup$
    – Steve
    Mar 18, 2021 at 14:46
  • 1
    $\begingroup$ @Steve I didn't really think of that...the puzzle was just meant to be the finite segment displayed, as (not well) indicated by the distinction between interior and exterior vertices. But that said, in the sense you mean, this does not have a unique solution...there are places along the given boundary where one could wander off infinitely with a shaded path, as long as you never circle back around to the clued region. However, this begs the question of whether the type of question you posit COULD be designed! $\endgroup$ Mar 18, 2021 at 15:01
  • $\begingroup$ @JeremyDover Sure, I think you could design something like that by just putting a bunch of 0s around the grid here to restrict this puzzle. $\endgroup$
    – Deusovi
    Mar 18, 2021 at 15:10

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