This is a game came to my mind last month. I have thought a lot and have searched the literature, only to find nothing much related.
Alice and Bob are playing a triangle game on the Euclidean plane $\mathbb{R}^2$. Each turn Alice moves first by placing a white chess on the plane, and then Bob places $N$ black chesses. Each point $(x,y)\in\mathbb{R}^2$ can hold at most one chess. As long as there are three white chesses which form a white equilateral triangle of length 1, Alice wins. Otherwise, the game will last indefinitely. In this case, Bob is said to be the winner. Suppose Alice and Bob are adversaries.
I am curious in the following questions:
- For which $N$ does Bob have a winning strategy?
- What if we tweak the rule a bit so that Alice wins only when there is a white $M$-gon of side length 1? I.e. for which $N$ does Bob have a winning strategy to prevent Alice from constructing such a regular white $M$-gon of length 1?
I came up with these observations (for the first problem):
- If for $N$ Bob has a winning strategy, then for any integer $N'>N$ Bob can win as well.
- For $N\leq3$, Alice has a winning strategy.
- I once turned to the discrete counterpart, in which Alice and Bob play the game on an infinite hexagon grid. I discovered that as long as $N\geq2$, Bob has a winning strategy. If $N=1$, Alice can always win.
Elaborations on my observations:
Quite straightforward.
Alice can always place her chess in the square $[0,0.1]^2$ such that no three of them are collinear, that every circle of radius 1 passing through 2 white chesses is distinct and that there is no white isosceles triangle whose side length is 1. Given this premise, in the $6$-th turn, Alice already has $5$ points, so there are $2\binom{5}2=20$ distinct points which has distance 1 to exactly 2 of the white chesses (which we call the key point henceforth). Say the Alice has points $A,B$ and $C$ is one key point (i.e. $d(A,C)=d(B,C)=1$). Alice now has 4 choices to construct a white equilateral triangle: $C+\omega(A-C),C+\omega^{-1}(A-C),C+\omega(B-C)\text{ and }C+\omega^{-1}(B-C)$, where $\omega=e^{\frac{i\pi}3}$. Since there are 20 key points, at least one key point $C$ and the associated 4 winning points have not yet been covered by Bob so far. Alice now places her $6$-th chess on $C$. However Bob places his 3 black chesses, one winning points for Alice is left uncovered.
The conclusion is trivial if $N\geq3$ or $N\leq1$. Since I think the discrete case barely has anything to do with $\mathbb{R}^2$, I shall omit the winning strategy and its proof for Bob when $N=2$. [Another interesting exercise.]
Here is where I got stuck:
I am afraid that for $N\geq4$ Alice cannot win. But to achieve this, Bob has to prevent Alice from orchestrating any suspicious geometric structure like concyclic points or hidden intersections of multiple unit circles. Can we give the $O(1)$ bound for Bob to do this?
Any help or pointer to related materials is appreciated.
This is very similar to John Conway's "Angel Problem". Maybe It will help to seek a generalized version of that?
Also, what is a chess? As in a chess piece, like a pawn? It is odd wording.