The Kemeny rule aggregates rankings by minimizing their total Kendall-tau distance from an aggregate order. We prove that Kemeny Score is NP-complete for exactly three unweighted rankings, even when every candidate pair is split $2$-to-$1$. On the same profiles, the winner, unique-winner, and possible- and necessary-precedence problems are $Θ_2^p$-complete, while recognizing a Kemeny-optimal or uniquely Kemeny-optimal aggregate is coNP-complete. The hard instances induce tournaments of majority dimension exactly $3$. The reduction also determines the exact maximum-cut value from the optimal Kemeny score and recovers a maximum cut from any Kemeny-optimal aggregate. For every fixed $q\geq3$ and $\lceil q/2\rceil\leq s\leq q$, minimum pairwise support $s$ yields a sharp dichotomy: the score problem is NP-complete, the winner and precedence problems are $Θ_2^p$-complete, and the recognition problems are coNP-complete when $3s\leq2q$; for $3s>2q$, the majority tournament is transitive and its unique topological order is the unique Kemeny-optimal aggregate. Exact support $s$ suffices in the hard case when $s>q/2$, and supports in ${s,s+1}$ suffice when $s=q/2$. These results give complete fixed-profile-size classifications and transfer to Slater orders, permutation medians, and maximum-likelihood central rankings in the Mallows model. Finally, a six-copy construction proves NP-completeness of both Kemeny Score and Kendall--Tau Center for three pairwise-equidistant rankings that still split every pair $2$-to-$1$. For $N$ output candidates, their common distance is $\frac23\binom N2$, the largest possible for an equidistant triple. The construction gives affine formulas for both optimal values, characterizes all Kemeny-optimal output orders, and shows that the output has a unique Kemeny-optimal order and a unique center exactly when the input has a unique Kemeny-optimal order.
🌊
