# Sierpinski triangle game for 3 players

The players are red, green and blue. The game is played on a n-deep Sierpinski triangle.

Each player colors a (black) triangle, starting at one of the main vertices. They then take turns to color an adjacent (black) triangle. There is only a loser, which is the first player not to be able to make a move.

On 1-deep, player 2 loses

On 2-deep, player 1 loses.

On 3-deep, player 2 can't lose (if they play well).

Is there a general strategy for n-deep? Can player 2 always avoid defeat for n>1 (or similarly, does player 1 always lose for n>N)?

• Does "color an adjacent triangle" mean "color an uncolored triangle that is adjacent to a triangle of your color"? – mjqxxxx Apr 1 '15 at 19:50
• I've edited my post to make this (hopefully) clearer, adjacency in this case is defined by common vertices on the black triangles. – JonMark Perry Apr 1 '15 at 19:54
• I would be a little surprised if players $3$ and $1$ couldn't gang up to force player $2$ to lose. Think of a red-blue vs. green game, where red-blue colors a red and a blue triangle each turn. It seems like a big advantage. – mjqxxxx Apr 1 '15 at 20:26
• the bigger they are... - red-blue has 2 weaknesses - either red or blue can lose. – JonMark Perry Apr 1 '15 at 20:33

I'm making a lot of assumptions because of the ambiguity of the question. Firstly, I'll assume each player moves triangle to trianlge. Secondly, I'll assume the players start at specific locations, play their best, and have a specific order in which they make moves. Thirdly, I'll assume its a equilateral seirpinski (doesn't make any difference). Red starts at bottom left, Green at top left, and Blue at bottom right. (This will be more of a soft analysis, and you'll have to try some of this out yourself, so this might (annoy you) not answer all your questions)

First of all, although this is a fractal, you're controlling how much complexity is in the thing. So instead of thinking of this as a fractal, let's think of it as a graph with edges that expire, (vertices are triangles and edges are connections between them). If you make a matrix of the edges between vertices. You'll see that the vertices with the highest degree, or highest number of edges, are the most mobile spots on the triangle. However, something becomes very apparent. Its the person who goes first who loses. Here's why, They move first, but then all the other players are free to move to places that close themselves off. However, because of the ordering, the first person to go is the one who loses and the other two players lose in the order they begin in.

The first color to go loses, the second color comes in second, the third color comes in third. By symmetry of the sierpinski triangle, it doesn't matter which color goes first or on which edge they start; the result is the same.

For all $n>1$, the Sierpinski triangle has $120^\circ$ rotational symmetry and no center triangle. So players $2$ and $3$ can always make the same move as player $1$ just did, rotated by $1/3$ and $2/3$ of a full turn respectively, and force player $1$ to lose.

For $n=1$, player $2$ loses.

For $n=2$, player $1$ loses.

For $n>2$, players $2$ and $3$ can force player $1$ to lose.

The more interesting specific question, I think, is which other coalitions can force a win.

A. For which $n$, if any, can players $1$ and $3$ force player $2$ to lose?

B. For which $n$, if any, can players $1$ and $2$ force player $3$ to lose?

Players $1$ and $3$ can always force player $2$ to lose (case A). For their initial moves, they make a beeline for player $2$'s corner. This ensures that player $2$ can't penetrate either of their territories (i.e., thirds of the board), and at least one opponent will penetrate player $2$'s territory. Once this happens, player $2$ has at least one fewer reachable triangle than either opponent. They can then fill in their own territories, and player $2$ will be the first to run out of moves.

Similarly, players $1$ and $2$ can force player $3$ to lose (case B), again by smothering player $3$ in his own territory.