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For $n\ge 3$, 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 $1-\left(\genfrac{}{}{0ex}{}{n}{2}\right)+n(n-2)$. Its nonzero value proves evasiveness by the alternating-sum criterion for decision-tree evasiveness.

A Boolean function is evasive when every deterministic decision tree computing it has a worst-case path that queries all input bits. The alternating-sum criterion for decision-tree evasiveness is one sufficient method for proving this lower bound.

A decision tree queries individual input bits, chooses subsequent queries from previous answers and labels each leaf with an output. Its decision-tree depth is the largest number of queries on any root-to-leaf path; $D(f)$ is the least such depth over trees computing $f$. An evasive Boolean function on $n$ bits has $D(f)=n$.

Remove repeated queries along any path, since their answers are already known. A leaf at depth $r\le d<n$ fixes $r$ bits and leaves at least one bit free. The inputs reaching it form a subcube on which $f$ is constant, say $b$. Its contribution to the alternating sum isHere $\mathrm{wt}(x)$ is the Hamming weight. The leaf subcubes partition the input cube, so adding their contributions provesThe contrapositive is the alternating-sum criterion for decision-tree evasiveness: a nonzero alternating sum forces all $n$ bits to be necessary in the worst case.