# Jewel Cave: Long Reach

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!

Text Version

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|   |   |   |   |   |   |   |   |   |   |   |   |   |   |2 6|
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|   |1 9|   |   |1 8|   |   |   |   |   |   |   |   |   |   |
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|   |   |   |   |   |   |   |   |   |   |   |   | 8 |   |   |
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|   |   | 3 |   |   |   |   |   |   |   |   |   |   |   |   |
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|   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |
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|   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |
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|   |   | 4 |   |   |   |   |   |   |   |   |   | 2 |   |   |
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|1 2|   |   | ● |   |   | 5 |   |   |   |   |   |   | 2 |   |
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|   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |
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|   |1 2|   |   |1 0|   |   | 4 |   |   |   |1 2|   |   |   |
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|   |   |   |   |   |   |   |   |   |   |   |   | 3 |   |   |
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|   |   |   |   |   |   |   |   |   |   |   | ● |   | 5 |   |
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|   |   |   |   |   |1 0|   |   | 9 |   |   |   |   |   |   |
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|   |   |1 7|   |   |   |   |   |   |   |   |   |   |   |   |
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|2 3|   |   |   |   |   |   |   |   |   |   |   |   |   |1 6|
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*  *   *   *
*  *  **  **   *    *   ** **  * * **   *  *T*
I  L   Y  N    *  *Z*  *W   F* *U* *P* *X*  *
*  **  *  *  **V  *    *    *           *   *
*


Cave clues alone can give a surprising amount of progress:

And if we remember that all shaded regions are pentominoes, more can be done:

Now it's time for some more in-depth logic:

The 3 near the bottom cannot go to the right -- it would mean that the shape on the bottom couldn't be a pentomino. So the P must be the top, and the N must be on the bottom. This determines a bit more:

Continuing...

If the 9 and the 10 near the bottom can see each other horizontally, then the 10 must extend one more space vertically than the 9. So the cell above the 10 cannot be shaded -- it would make the two cells right of it unshaded, and then we would have a problem.

And now it's finally time to use the connectivity rule!

The P needs another shaded cell next to it -- that means the 10 clue is filled, and then the 9 clue is as well.

To stop the 4 from seeing too far down, one of the two cells under it must be shaded. It can't be the bottom one, because then we'd make another L shape, so it has to be the top one.

We only have five unfinished shaded regions, so two of them must be connected; this can only be the two on the right, which are connected by the I piece.

And finally, there's only one way to place the remaining pieces and satisfy edge connectivity.

• Re "If the 9 and the 10 near the bottom can see each other horizontally" - I don't see anywhere where you explain the other scenario, when they can't see each other... Is your point more that regardless of whether the 9 and 10 can see each other the space above the 10 needs to be unshaded? I see how the logic unfolds but just wondered if this might be an easier explanation for people following along this way? (Feel free to disagree, of course...) Another solid solve anyway! +1
– Stiv
Oct 27, 2020 at 16:10
• Well done! Turns out going with a bigger grid was the secret to getting the edge connectivity rule to be relevant. Hope you enjoyed! Oct 27, 2020 at 16:16
• @Stiv The logic was: If that cell is shaded, then the 9 and 10 see each other, then that cell must be unshaded. So the cell is unshaded no matter what. (Will edit for clarity in a bit!)
– Deusovi
Oct 27, 2020 at 16:21