Rules of Tapa:

  1. Shade some cells so that all shaded cells form one orthogonally connected area.
  2. Clues cannot be shaded and represent the lengths of the shaded cell surrounding that clue.
  3. If there are two or more numbers in a clue, then each group of shaded cells must be separated by at least one unshaded cell.
  4. No 2x2 region may be entirely shaded.

The standard Tapa rules apply.

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1 Answer 1


Vanilla? Perhaps. Easy? No.

The resolved grid should look like this:

Resolved grid

The logical break-in I found was the following:

Consider the 'arrowhead' arrangement of cells between the three 7's and the 6 in the top-left quadrant:

Arrowhead cells

It is impossible for the whole arrowhead in a single colour to be fully shaded - otherwise it forces both neighbouring arrowheads to have the middle cell of their adjacent wall unshaded, which then makes it impossible for the fourth arrowhead to be satisfied, since it will need to have non-adjacent cells of both of its two arrowhead walls unshaded, which is impossible for a 6 or 7 clue.

As a result, we know that the cells around each of the 7's not covered by their arrowhead shapes must be shaded. This enables us to mark some adjacent spaces (shaded green) as needing to contain an unshaded space to prevent the formation of a 2x2 block:

Logic from the arrowheads

This now enables us to shade some more spaces around the other two 7's:

Shading around the other two 7's

And now the 5 in the bottom-left corner can be fully resolved, and some of the shading around the 2-4 clue in the bottom-right:

More spaces resolved around the 5 and 2-4 clues

Now, we just need to ask ourselves which one of the two remaining spaces to the left of the 2-4 clue needs to be shaded.

If we shade the topmost of these two squares (in dark green in the next image), many of the other numbered clues can be fully resolved fairly speedily until we reach a point where the 6 cannot be, since both centre cells of its arrowhead walls need to be unshaded, which is impossible:

Cannot shade R8C7

Instead, we shade the bottom one and simple logic unfolds until the grid is completed!

The resolved grid, again

  • $\begingroup$ Very well done, and the break-in was spot-on! :D For the last spoiler block (in case you want to avoid the branching), you can apply the same principle as the break-in (especially R4C6 and R6C4). $\endgroup$
    – athin
    Sep 15, 2021 at 1:19

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