Tapa on a Penrose Tiling

Question

^{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.

Example

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)

Answer 1

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

Next,

connectivity forces the wedge piece left of the [1111] to be shaded. Applying the vertex rule, this lets the top get resolved.

More connectivity deductions on the right: the rightmost shaded region can't escape by going downwards.

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].

I've highlighted one cell in red: can the top group connect through there? If it did, we'd get this:

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:

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.

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...

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:

And with that, the puzzle is solved!