13
$\begingroup$

The goal of this puzzle, a hybrid of Statue Park and Cave, is to place the given shapes into the grid, subject to the following rules. As in both original puzzles, all unshaded squares must form a single orthogonally (on a side) connected region. As in Statue Park, two shapes cannot be orthogonally adjacent, and all squares with darkened circles must be in a shape. As in Cave, all squares with numbers must have that number of unshaded squares directly connected to it horizontally and vertically, including itself. A final rule inspired by Cave is that all shaded squares must be connected, orthogonally or diagonally (necessary at shape boundaries), by other shaded squares to the edge of the grid. The example below is a legal Jewel Cave grid where the set of shapes is the tetromino set.

Sample

The puzzle below uses the standard pentomino set. I hope you enjoy!

Grid

Text Version

-------------------------------------------------
|   |   | 7 |   | 8 |   |   |1 0|   |   |   |   |
-------------------------------------------------
|   |   |   | 3 |   |   |   | 4 |   | 8 |   |   |
-------------------------------------------------
| ● |   | 4 | ● |   |   |   |   |   |   |   |   |
-------------------------------------------------
|   |   |   |   |   |   |   | 4 |   |   |   |   |
-------------------------------------------------
|   |   |   |   |   |   |   |   |   |   | ● |   |
-------------------------------------------------
|   |   |   |   |   |   |   |   |   |   |   |   |
-------------------------------------------------
|   |   |   |   |   |   | 4 |   |   |   |   |   |
-------------------------------------------------
|   |   |   |   | 3 |   |   |   |   |   |   |   |
-------------------------------------------------
|   |   |   |   |   | 5 |   | 2 |   |   |   |   |
-------------------------------------------------
| 5 |   |   |   |   |   | ● |   |   |   |   | 9 |
-------------------------------------------------
|   |   | 6 |   |   |   |   |   |   |   |   |   |
-------------------------------------------------
|   | ● |   |   |   |1 0|   | 6 |   | 8 |   |   |
-------------------------------------------------

 *  *   *   *
 *  *  **  **   *    *   ** **  * * **   *  *T*
 I  L   Y  N    *  *Z*  *W   F* *U* *P* *X*  *
 *  **  *  *  **V  *    *    *           *   *
 *

Poster Notes

Hybridizing the Statue Park and Cave puzzles is natural in the sense that both puzzles target a fairly large unshaded area that must be contiguous. But the challenge is that the targets of the two puzzles are not identical; generally a Statue Park shading will not be a legal Cave shading and vice versa. When designing the puzzle, I figured that the clues would generally be more Cave-like, so it made sense to make the shading more Statue Park-like, which is why the Cave shading rule needed to compromise. As far as the name, all of the combinations of Cave with Statue or Park sounded dumb, but Cave Park made me think of Jewel Cave National Park in the United States (OK, it's really a National Monument, but work with me here); there are more famous cave parks, but I've actually been in Jewel Cave. As a final note, I am really happy with how this came out, so ironically I'm specifically interested in any criticism, since I'll probably make some more of these (spoiler: already have two more).

$\endgroup$
4
  • $\begingroup$ This looks awesome! Unfortunately for me, my lunch break is over and it'll doubtless go in the next few hours. Looking forward to having a go later anyway :) $\endgroup$
    – Stiv
    Oct 20 '20 at 13:18
  • $\begingroup$ I thought I could wait until Friday...in reality, I almost didn't make it to this morning. I'm really excited about this one! Hope you enjoy whenever you're able to tackle it...and there are more. $\endgroup$ Oct 20 '20 at 13:22
  • $\begingroup$ This was a very fun puzzle! I don't think the connectivity of the shapes to the edge was really used though? Makes it seem more like a mix of Statue Park and Kurodoko (which has Cave-style clues, but "no two adjacent shaded cells"-style shading) - maybe that expression would be more natural. $\endgroup$
    – Deusovi
    Oct 20 '20 at 15:02
  • $\begingroup$ @Deusovi Thank you for the feedback. I agree that the rooting of the shapes to the edge was not used here, but at least for now I think that's a deficiency of the setter, not necessarily the puzzle type. I tend to be much more combinatorial than topological in my thinking, so I need to get myself out of my comfort zone. $\endgroup$ Oct 20 '20 at 15:39
10
$\begingroup$

First, I will fill in all unshaded squares as yellow and all shaded squares as dark grey.

SP_Cave_1

Note that we can fill in the dark grey next to '2' at the bottom since it cannot touch the '5'.

A few deductions at the top. We try satisfying the '7' by extending it to '4', but it has to touch the '8' next to it anyway. We perform a similar operation for the '10' in the top row and realise it has to touch the '8' as well. This gives us

SP_Cave_2

Now, we realise the '7' cannot touch the '4' since it already has 6 squares. The top left square must also be shaded black. The resulting formation gives us 'T' at the top and we can carry out some further deductions.

SP_cave_4

Looking at the '8' at top right, we realise some of its squares must extend downwards. Also, the dark grey cells will block the yellow cells near the '4', so to ensure connectivity, it has to go towards the '8' side. Additionally, the two cells near '4' can only be connected to the dark grey cells below '3'. Note that 'U' is also complete now. This means the pentomino above '8' must be a 'L'. This gives us

SP_Cave_5

The '8' at the bottom cannot extend far enough for 8 cells, so it must be connected to the '6'. Similar logic can be applied for the '10'. Because of the '6', the '8' can only have 6 cells in the bottom row, so 2 cells must be extended in the vertical direction. This also means the bottom right corner is a grey square, since the 8 cannot reach it. Since a 'L' has already been used, the cell below '9' must be shaded. The cell to the left of '9' must be shaded as well for 5 dark grey cells. This gives us

SP_Cave_5

The '10' can have only 6 cells in the bottom row, so it must extend 4 cells upwards. This completes the '5'. Then, there is only one way to complete the '4' in the middle, and this actually leads to a chain deduction where we fill in the 'W', 'F' and 'P' pentomino.

SP_Cave_6

Filling in a few additional squares gives us

SP_cave_7

Now, the cell in last row, column 4 cannot be connected to the cell near the '3' and '5'. Thinking about it further, the only shape that can fit in there is the 'X'. There can only be a 'V' or 'y' near the bottom left. This information allows us to complete a few more squares.

SP_cave_8

The pentomino near the '4' can only be a 'Z' and we need a 'V' in the right orientation to restrict the '5' at the bottom left. This allows us to complete the puzzle.

SP_Cave_9

$\endgroup$
1
  • $\begingroup$ Nailed it! Great job, and great explanation! Hope you enjoyed... $\endgroup$ Oct 20 '20 at 15:39
9
$\begingroup$

Cave clues by themselves can get us this far:

enter image description here

Since the shaded cells need to be in pentominoes,

we can't cut off the bottom right corner, so the 9 cannot go left. Also, the single cell at the top trapped in the 7-3-8 clues cannot be shaded. This gives some more Cave progress:
enter image description here

Time to do some actual Statue Park thinking!

The region in R2C7-8 must be an N -- if it doesn't extend in that one cell that would need to be added, it would either block off a cell or make a shaded region too big.

The bottom-right region must be a P, because at least one of the bottom two cells must be shaded. We can also make progress in the top right.
enter image description here

Now the path to the end of the puzzle is very smooth:

The 4 in the middle can't extend two more to the right, so it must extend only one more to the right. This pretty quickly gives us two more pentominoes.
enter image description here
Since we've used up our L and N already, all of ⓐ,ⓑ,ⓒ,ⓓ must be distinct. ⓑ can only be Z, then ⓒ can only be X (because if it was Y it would make another pentomino). And finally, to block off the 5 from seeing all of the left column, we need ⓐ to be V, and ⓓ to be Y.

And then we have the solution!

enter image description here

$\endgroup$
1
  • $\begingroup$ +1. I completely missed the fact that you could extend the '10's vertically in the top and bottom row from the very beginning. Very concise answer! $\endgroup$
    – Alaiko
    Oct 20 '20 at 15:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.