10
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As in a previous puzzle, the goal of this Jewel Cave is to place the given shapes (in this case, the standard pentomino set) into the grid, subject to the following rules:

  1. All unshaded squares must form a single orthogonally (on a side) connected region.
  2. Two shapes cannot be orthogonally adjacent.
  3. All squares with darkened circles must be in a shape.
  4. All squares with numbers are unshaded, and must have that number of unshaded squares directly connected to it horizontally and vertically, including itself.
  5. All shaded squares must be connected, orthogonally or diagonally (necessary at shape boundaries), by other shaded squares to the edge of the grid.

I hope you enjoy!

enter image description here

Text Version

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|   |   |   |   |   |   |1 2|   |   |   |   |   |   |
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|   | 3 |   |   | 2 |   |   |   |   |   |   | 3 |   |
-----------------------------------------------------
|   | ● |   | 2 |   |   |   |   |   |   |   |   |   |
-----------------------------------------------------
|   |   | 8 |   |   |   |   | 5 |   |   |   |   |   |
-----------------------------------------------------
|1 5|   |   |   |   |   |   |   | 7 |   |   |   |   |
-----------------------------------------------------
|   |   |   | ● |   |   |   |   |   |   |   |   |   |
-----------------------------------------------------
|   |   |   |   |   | 4 |   |   |   |   |   | 2 |   |
-----------------------------------------------------
|   |   |   |   | 9 |   | 5 |   |   |   |   |   |1 0|
-----------------------------------------------------
|   | 7 |   |   |   |   |   |   |   |   | 3 |   |   |
-----------------------------------------------------
|   |   |   |   |   |   |   | 3 |   | 7 |   |   |   |
-----------------------------------------------------
|   | 4 |   | 2 |   |   |   |   | 3 |   |   |   |   |
-----------------------------------------------------
|   |   |   |   | 2 |   | 2 |   |   |   |   |   |   |
-----------------------------------------------------
|   |   |1 3|   |   |   |   |   |   |   |   |   |   |
-----------------------------------------------------

 *  *   *   *
 *  *  **  **   *    *   ** **  * * **   *  *T*
 I  L   Y  N    *  *Z*  *W   F* *U* *P* *X*  *
 *  **  *  *  **V  *    *    *           *   *
 *
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8
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Number clue logic alone gets us this far:

enter image description here

And if we include "black cells must be in groups of exactly 5", this gets a bit more:

enter image description here (Some of the unshaded cells here, such as the ones by the 10 at the right and the 2 at the top, come from interactions of two clues.)

Now it's time to make heavier use of the piece bank!

First, we've already used our L, so the 4-in-a-row on the left must become an I. (We can also fill in a bit more on the right; we've already used our W, so that must be the N (though we can't quite place it yet).

enter image description here And since the white cell near the lower-right must be unshaded, we can actually place that N.

We can make a lot more progress considering a particular cell:

no matter whether the tetromino at the bottom makes an X or an F, the cell two above its center must be unshaded. This completes the 4 clue above, determining that it is indeed an X.
enter image description here

Now the 12 clue can't be satisfied by just going right and down. So the nearby tetromino must be the F, and the shape just under it must be the Y.
enter image description here

And finally, there is only one way to pack the remaining pieces into the grid:

The bottommost two groups of shaded cells must join to form a T; then the part above it must be a V, and the single shaded cell to its left must be a U. This completes the puzzle!
enter image description here

| improve this answer | |
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  • $\begingroup$ Nice! I'm working on a version that uses the connection to the edge condition more strongly...hopefully coming soon. But I hope this version was still enjoyable. $\endgroup$ – Jeremy Dover Oct 23 at 18:57

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