Thread: Rhombus side length 'n'. Who has the winning strategy?

Chris and Michael play a game on a board which is a rhombus of side length n (a
positive integer) consisting of two equilateral triangles, each of which has been divided into equilateral triangles of side length 1. Each has a single token, initially on the leftmost and rightmost squares of the board, called the “home” squares.
A move consists of moving your token to an adjacent triangle (two triangles are adjacent only if they share a side). To win the game, you must either capture your opponent’s token (by moving to the triangle it occupies), or move on to your opponent’s home square. Supposing that Chris moves first, which, if any, player has a winning strategy?

Wouldnt it effectively come down to probability? I don't quite know how to approach this bizarre question. I realise beginning with n=1 would be a good idea, and proving that he will indeed have an advantage, as he will with n=2. Help as to what to do would be great.

Why did you bring in "squares"? Isn't it ALL triangles?

Its a colloquial term, when i say squares i do indeed mean triangles.

Yes i do indeed mean 'triangle' when i said square, its just a rhombus of side length 'n', cut up into equilateral triangles.
<l> is an example of side length 1. Any help?

You have two classes of triangles in the rhombus, such that triangles in each class are translations of each other. Call their members red and blue triangles respectively. WLOG let Chris's home-triangle be red. In each turn what will be the colour of the triangles that each token occupies?

Well it depends on Side length 'n' but generally Chris would have an advantage because he would reach the centre first - how exactly do i prove this? I need it done by 2moz.

It does not depend on the side-length. If Chris's home triangle is red then Michael's home triangle will be blue. Therefore Michael will never be able to capture Chris's piece, while Chris will be able to capture Michael's if they are in adjacent triangles and it's Chris's turn to move. So Chris will take a shortest path to Michael's home triangle and win the game.