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This is a follow up to my previous two questions, here and here, but this time the game is played on a triangular grid.

You have an infinite triangular grid 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 as shown in the figure below. Clarification: This rule applies to both kinds of triangles, not just the one shown below.

firing

Your goal is to reach, or prove that you can’t reach, an empty configuration starting from each of the following configurations.

configs

Bonus: (not necessary for acceptance) In cases where you can reach the empty configuration, minimize the number of chips fired. And in cases where you can’t, minimize the number of left over chips.

Hint 1: SmarthBansal already answered A, C, and D, so this is a hint for B.

In this answer to my first question, Tim Seifert identified a lot of invariants (one for each row and each column). What is the analog of that here?

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  • $\begingroup$ While the obvious next step involved hexagons, I am hoping for and looking forward to versions played on hyperbolic tilings, there are quite a few interesting ones out there, starting with the expected {5,5} $\endgroup$ Commented May 21 at 10:47
  • $\begingroup$ @htmlcoderexe. Thanks for the suggestion! What kind of games are you looking for on hyperbolic lattice? I am more interested in finding the invariants of the game, and I know a few games on hyperbolic lattice that give some interesting invariants, but I am wondering what you are looking for in these games. $\endgroup$ Commented May 21 at 15:23
  • $\begingroup$ Honestly, I am mostly really curious how the same thing as this one or the square lattice one would work on such a plane, how would the equivalent moves look like and indeed what invariants would apply. Perhaps {3, 7}, which is this one but 7 triangles meet at the vertices? Would it still have those lines of rhombuses? {4, 5} is interesting, too - would it do unexpected things? {4, 6}, maybe, if my feeling that this requires even numbers of polygons meeting at each vertex? $\endgroup$ Commented May 22 at 7:21
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    $\begingroup$ @htmlcoderexe. There are indeed games with such invariants on the tilings you mention. Also, you don’t need even number of polygons to meet at a vertex, as can be seen in the honeycomb (hexagonal lattice). I’ll try to come up with an interesting game when I have time. $\endgroup$ Commented May 22 at 8:41
  • $\begingroup$ Completely forgot about hexagons meeting at 3, haha. I'm looking forward to what you come up with! $\endgroup$ Commented May 22 at 8:50

3 Answers 3

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As with the previous variants, the game can be simplified by always removing any chip pairs so that a cell is either empty or has one chip. You can fire an empty cell by adding two chips first (and removing any pairs afterwards). This reduces the game to a lights-out variant, where each move is simply to toggle 4 cells that make up a triangle (add a chip to any empty cells, remove a chip from any non-empty cell).

My first idea for an invariant was to use a four-colouring of the grid, so that every move changes one cell of each colour. This already proves many patterns impossible, simply because the number of filled cells of each colour must have the same parity at all times (i.e. all odd or all even).

four colouring

Unfortunately that is not powerful enough. Pattern B passes this test as it has 6 of one colour and 0 of the other colours, i.e. all even.

It turns out there is a much more powerful set of invariants:

Consider a line of rhombuses as shown in these pictures: Invariants
Every firing move affects zero or exactly two of the cells of such a line. Therefore every such line must have an even number of chips inside it.

Clearly pattern B does not pass this test,

as there is for example a horizontal line of rhombuses that contains three filled cells, the top half of the hexagonal pattern.

For completeness, here are solutions to the solvable patterns (the cells that need to be fired are shown in red), and proof that B is impossible.

Solutions

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    $\begingroup$ +1. The powerful invariants are exactly what I had in mind. In fact, these invariant completely characterize the configurations that are related by the moves. $\endgroup$ Commented May 20 at 12:48
  • $\begingroup$ Just to finish the bonus question, the minimal number of chips obtained from B is 2, obtained by removing A. Indeed, B doesn’t satisfy six of the invariant-rhombus-lines (one of which you show), whereas a single chip doesn’t satisfy only three of them. $\endgroup$ Commented May 20 at 13:43
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EDIT: Updated the game so that you can start initial state then start the game. and you can play it with A to add two chips, S to fire, D to delete two chips by hovering on the triangles.

This is partial solution with a playable html file attached, but to help make the game more accessible and playable for anyone curious, I’ve created a small interactive version of the Triangular Chip Game as described in this puzzle.

You can experiment with the rules directly in your browser using the link below:

👉 Play the Triangular Chip Game

All the mechanics are implemented as per the original rules. It’s a simple HTML interface so no installation required. Just click and play.

Happy puzzling and let me know if something is not working as it is intended.

I solved A as below:

enter image description here

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    $\begingroup$ A is correct but there’s a much simpler way to solve A. $\endgroup$ Commented May 19 at 12:54
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    $\begingroup$ @Pranay maybe as a new puzzle or by editing puzzle with the least move? ;) $\endgroup$ Commented May 19 at 12:55
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    $\begingroup$ The goal is to start with the given positions and clear all chips. $\endgroup$ Commented May 19 at 14:14
  • $\begingroup$ You’re right, @DanielMathias I had it the other way around. But this might still be reversible, as long as you approach it through the shape itself rather than defining the game based on initial conditions? $\endgroup$ Commented May 19 at 16:02
  • $\begingroup$ Your solution is still valid. If you can reach a given state from an empty grid, then you can double all the chips, then simply remove them. $\endgroup$ Commented May 19 at 21:00
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Two observations:

We have following operations available: Add 2 chips (+2), Remove 2 chips (-2), Blast a chip (+2).

1. All three operations don't change the parity of the board.
2. All operations are reversible you can go back and forth.

Therefore, a board that can reach an empty configuration must have even number of chips and we need to bring the chips together, so they pair up into twos.

Thanks to @Oray for making the interactive game. The screenshots are from game.

A:

Add 4 pairs of chips as shown below. And then blast one chip at each of the four locations. You would be left with two chips at all places, which we can remove.
Step 1 A ----------> Step 2 A

B:

A is a subset of B. Therefore, we can already remove the 4 side chips in the same way. The remaining two chips can't collapse into an empty configuration because we can't bring them together enough. (The closest they can come is occupy 1 chip each of a hexagon.) Don't have a formal proof though, can see it intuitively.

C:

We can start collapsing the chips from the outside and work inwards. It becomes same as A after few steps.
c

D:

This one is similar, start with dealing with the extremes and work inwards.
d

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  • $\begingroup$ +1. A, C, D are correct but a proof for B is missing. $\endgroup$ Commented May 19 at 17:21
  • $\begingroup$ you mentioned blast a chip is reversible. How to reverse it? $\endgroup$ Commented May 19 at 21:22
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    $\begingroup$ ah, place 2 then fire and remove doubles, got it. $\endgroup$ Commented May 19 at 22:07

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