To block to the south, player 2 MUST play AZ, to which player 1 responds with CO, to which player 2 MUST respond NM, player 1 plays OK, player 2 must block with TX, and player 1 finally plays at an unblockable location, AR.

To block to the north, player 2 MUST play ID, player 1 responds with WY, and then player 2 MUST play MT. Player 1 then plays at the unblockable location, SD.

No matter how player 2 plays, player 1 can win north-to-south with an initial play of UT.

Let G be a graph (planar or not!), and let N, E, S, and W be 4 consecutive paths with overlapping endpoints nw, ne, se, and sw, where the notation, hopefully, is self-explanatory. The system (G,N,E,S,W) is “hex-agreeable” if for every black/white coloring of the vertices of G, there is at least one path that’s either all black or all white running either from N to S or from E to W. The problem is to decide whether a given system is hex-agreeable.

Note that by momentarily adding one more vertex attached to each of the vertices in N, E, S, and W and then drawing the augmented graph on a surface of some possibly high genus, one can interpret N,E,S,W as constituting the border states of a country on some topologically weird planet. So one can think of hex-agreeability as saying that electoral hex for this country cannot end in a draw.

It’s fairly easy to see that the Yes/No decision problem of hex-agreeability is in co-NP: If the answer is No, you can demonstrate it by giving a black/white coloring that lacks any transcontinental path. (Verifying the lack of a path is not completely trivial, but it’s clear you can do it in polynomial time.) However, it’s not clear (to me, at least) that hex-agreeability is in NP, and far from clear (again, to me) if it’s in P. Maybe some other bit-player commentator can make it clear.