Alternating count of common-center graphs
= Alternating count of common-center graphs
{title2=$\sum_G(-1)^{|E(G)|}\mathrm{STAR}_n(G)=(n-1)(n-2)/2$}
For $n\geq3$, count the empty <graph> once, all one-edge <graphs> once, and each larger accepted <graph> under its unique center. The latter weighted contribution is $n(n-2)$, giving $1-\binom n2+n(n-2)$. Its nonzero value proves evasiveness by the <alternating-sum criterion for decision-tree evasiveness>.