TITLE: Game on simple finite graphs
QUESTION [14 upvotes]: Consider the following game on graphs (no multiple edges, but graphs can be disconnected).
Players A and B alternate picking a vertex. After picking a vertex, a number is assigned to that vertex such that the number is the smallest non-negative integer that is not already assigned to its neighbours.
The game ends when all vertices have numbers assigned to them. The last player to make a move wins if and only if that vertex was assigned an odd number.
Even for path graphs (that is, $n$ vertices in a line, connected with edges), I have no idea for which $n$ the first player has a winning strategy. 
Some computations suggests that the first player lose if 
$n=1, 2, 4, 5, 6, 7, 9, 11, 13, 15$. I don't find a hit in OEIS that seem to make sense.

REPLY [3 votes]: The reason that, as others have noted, the player who wants the last number to be even wins on large path graphs is that the configuration $0..0$ (and also, less importantly, $1..1$) leads to an even number independent of the order of play, whereas the parity of the last number placed in the configuration $0..1$ depends on the order of play. Thus, when the "even" player sets up $0..0$, the other player has to spend two moves filling in the two dots to prevent these from being the last numbers, whereas there is no corresponding manoeuvre for the "odd" player.
A sketch for a proof that the even player wins sufficiently large path graphs: If the odd player goes first, she has to place a $0$. The even player creates a $0..0$ configuration. If the odd player plays elsewhere, the even player just continues improving his situation by creating more $0..0$ configurations. Likewise, if the odd player plays inside the $0..0$ to destroy it, the even player creates further such configurations (using the sufficiently available space). The best the odd player can do is to create one $0.0$ configuration per move, but for every move she spends doing that, the even player creates another $0..0$.
If the even player goes first, he plays somewhere in the middle. The odd player can then create one $0.0$, or can place a $1$ next to the $0$ in an attempt to destroy opportunities of $0..0$ creation. But the even player still has the other side to create a $0..0$. Then the odd player can create another $0.0$, or shield the $0$ on the other side with a $1$. But neither helps. If the odd player opts for $0.0$s, the even player can win either by creating more $0..0$s or by filling in the $0.0$s (since he's now half a move ahead in the filling race), and if the odd player opts for shielding $1$s, the even player can play another $0$ in a sufficiently distant part of the field and create another $0..0$ there.
Obviously there are some details to fill in to turn this into a rigorous proof, but it seems clear to me that this asymmetry, the ability to build a winning configuration that takes two moves to destroy whereas the other player can only build single winning spots, is the basic reason for the results for large path graphs.