 Solution

If a player chooses site $x$, the opponent can co-locate and hold its payoff to $k$. Among opposing sites to its right, the lowest payoff is obtained against $x+1$, giving $2x$. Among opposing sites to its left, the lowest is against $x-1$, giving $2(k-x+1)$. Thus the pure-action security level payoff isThe expression remains valid at an endpoint, where the missing-side term is redundant. It reaches its largest possible value $k$ exactly when $k/2\le x\le k/2+1$. Hence the pure maxmin strategies areReflection gives the same answer for the other player.