A and B play a game on a $(2n+1)$-agon where $n\ge 1, n\in\mathbb{N}$
They draw diagonals such that every diagonal drawn
i) Cannot be previously drawn
ii) only intersect an EVEN number of drawn diagonals (if two diagonals share the same vertex it doesn't count in version 1, counts in version 2)
The player who has no legal move loses. Who has the winning strategy?
For $n=2$ A wins in version 1 and 2. For $n=3$ it's quite complicated to bash. My prediction is that A wants to prevent B from copying them, so A plays $v_1$ to $v_n$. Then the case bash is quite annoying