Graph theory partitioning game

I believe that the first player cannot always guarantee a tie, even when $N=1$. Some observations:

• Given a graph $G$ and a vertex subset $X \subset V(G)$, we can always modify $G$ so that the players are forced to choose a vertex in $X$: add a huge independent set $I$ to $G$ and make all the vertices in $I$ adjacent only to the vertices in $X$, so that if one player plays outside $X$ and the other plays inside it, then the player inside $X$ grabs all vertices in $I$ and wins even if the other player gets everything else.

• This trick also lets us assign nonnegative integer weights to the edges, and calculate distances according to the weights: subdivide each edge an appropriate number of times, and choose our subset $X$ so that the internal vertices of the subdivided edges aren't feasible to play in. (Edit: As originally written this doesn't quite work, because the players in the modified graph would score points for the subdivided edges, but I think it can be made to work if you "blow up" the real vertices into large cliques or large independent sets, so that the value of getting one more "real vertex" far exceeds the value of the new vertices inside all the subdivided edges. This transformation doesn't quite preserve the rules of the game: in the transformed graph, Bob is essentially allowed to pick a vertex already chosen by Alice, so the transformed graph is friendlier to Bob than the original graph, but if the point is to come up with a graph where Bob can guarantee a win, then this is not a problem for us.)

Once you've done these tricks, I think you can create a rock-paper-scissors scenario using the complete bipartite graph $K_{3,3}$. Say that the three vertices one partite set are $R,P,S$ and that the three vertices of the other set are $X,Y,Z$. The vertex $R$ has weights $1,2,3$ to $X,Y,Z$ respectively; $P$ has weights $3,1,2$, and $S$ has weights $2,3,1$. The players are only allowed to play in $\{R,P,S\}$, via the above tricks.

Now, if Alice picks $R$, then Bob picks $P$ and gets $Y,Z$ versus Alice's $X$. ($S$ is equidistant to both $R$ and $P$). If Alice picks $P$, Bob picks $S$ and gets $X,Z$ versus Alice's $Y$. If Alice picks $S$, Bob picks $R$ and gets $X,Y$ versus Alice's $Z$.