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Here’s yet another variant of a chip firing game, now on the (3,7) hyperbolic tiling. The last variant can be found here, and the ones before that can be found in that link.

You have an infinite 3,7-hyperbolic tiling with some (finitely many) chips in some cells. Here are the rules of the game:

  1. You can always add or remove two chips in a cell.
  2. You can fire a chip in a cell (if there are multiple chips, pick one). When you fire a chip, you remove that chip, and add three chips, one in each of the edge-adjacent cells. An example is shown in the figure below.

firing

Your goal is to reach, or prove that you cannot reach, the empty configuration from each of the following configurations in finitely many moves:

final

Image of the hyperbolic tiling taken from Wikipedia, and originally created by Parcly Taxel!

Hint 1:

In the triangular grid variant, the invariants were found to be made of lines of rhombi. This doesn’t immediately extend to the current problem. But there’s a closely related invariant in the triangular grid problem which does extend to this one.

Hint 2:

The A’s are quite similar in both the triangular games.

Hint 3: At this point, A is solved by Hall Livingston, and B and C are solved by franck vivien. So here’s a third hint (in conjunction with the first hint) for D, which is the hardest of all and requires the knowledge of nontrivial invariants (other than the obvious total parity).

The invariants for the game on the triangular grid can be equivalently represented as surfaces with jagged boundaries rather than lines of rhombi. The relation between them is shown in the figure below.
jagged surface This version of the invariant can be extended to the current game on the 3,7-hyperbolic tiling.

Hint 4: Another hint for D.

The green curve is a circular arc that intersects the boundary circle of the Poincaré disk at right angles. In other words, it’s a geodesic. What’s the invariant associated with it? geodesic

Hint 5: Final hint for D.

The shaded (orange) triangles are those that have at least one side completely to the bottom-left side of the green geodesic in Hint 4. invariant

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    $\begingroup$ I am in fact the one who made the hyperbolic tiling you used! $\endgroup$ Commented Jun 7 at 17:22
  • $\begingroup$ @ParclyTaxel. Ah nice, thanks for letting me know! $\endgroup$ Commented Jun 7 at 17:51

3 Answers 3

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Partial answer

First notice that

Rule 1 changes the total number of chips by +2 or -2
Rule 2 adds 2 chips to the board
Hence any initial position with an odd number of chips, will always contain an odd number of chips after any operation

This obviously proves that

starting from B (containing 7 chips, odd number), we will never reach the empty configuration

For problem C,

Firing the 7 "central" cells gives 2*C (configuration C where all chips are doubled). This can then easily be emptied cell by cell by removing the 2 chips in each cell (rule 1)

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Partial answer

Label inner ring as R1:1-7 and 2nd ring as R2:1-7, where R1:n and R2:n share an edge.

  • A. Cells R1:4&5 and R2:2,3,6&7 are occupied.

    Add 2 chips each to cells R1:2&7. Fire one chip in each of those two cells. Fire the chips in cells R1:3&6. Remove all of the double chips. Done.

  • B. Every legal action either adds or removes two chips. B has an odd number of chips, so no combination of legal moves can reduce it to zero chips.

  • C. Fire the seven chips in R1:1-7. Then every cells that originally had one chip will now have two chips. Remove all of the double chips. Done.

  • D.

    1. Recenter two vertices to the left, such that cells R1:1&7 and R2:2,5&6 are occupied.

    2. Add two chips to the cell that has become R1:6 and fire one of them. Then fire the chip just generated in R1:5. Then remove all double chips.

    3. The only chips remaining near the revised center are in cells R1:1&4 and R2:2 (which is adjacent to the original R2:7).

    4. Add two chips to the cell that has become R1:2 and fire one of them. Then fire the chip just generated in R1:3. Then remove all double chips.

    5. Reverting to the original center, the only chips remaining on the left side are R1:6 and R2:6&7.

    6. Repeat steps 1-4 for the right side.

    7. Reverting to the original center, the only chips remaining are R1:3&6 and R2:2,3,6&7.

    8. Firing R1:3&6, then R1:2&7, then removing double chips leaves chips in R1:3-6.

I believe this is unsolvable, but I don't have time now to prove it.

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  • $\begingroup$ @Pranay, Two errors, both of them mine. Do you still need more explanation? I think you understood very well. $\endgroup$ Commented May 29 at 22:15
  • $\begingroup$ +1. Looks good now. I’ll leave more hints for D later. $\endgroup$ Commented May 29 at 23:31
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It's been two weeks since I posted my last hint. A, B, and C have been answered already, so here's the answer for D.

First of all, we can combine rules 1 and 2 into a single rule where we pick a cell and then "toggle" (between "chip" and "no chip") that cell and the three cells adjacent to it, just like in a lights-out game.

Now, such a move doesn't change the parity of the number of chips in all the cells shaded orange in Hint 5. (In fact, for any geodesic that cuts through a "row" of cells, there is an associated invariant with all cells on one side of the geodesic.) But there is exactly one chip in the orange region in D, so it cannot be emptied.

For completeness, here are the figures with answers to all the four parts. The red circles represent the cells about which a toggle is performed.

answers

B is not possible because the parity of the total number of chips is also an invariant, and B has an odd number of them.

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