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Research PaperResearchia:202607.31026

The Complexity of Kemeny Aggregation with Three Rankings

Péter Madarasi

Abstract

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 ...

Submitted: July 31, 2026Subjects: Mathematics; Mathematics

Description / Details

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 22-to-11. On the same profiles, the winner, unique-winner, and possible- and necessary-precedence problems are Θ2pΘ_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 33. 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 q3q\geq3 and q/2sq\lceil q/2\rceil\leq s\leq q, minimum pairwise support ss yields a sharp dichotomy: the score problem is NP-complete, the winner and precedence problems are Θ2pΘ_2^p-complete, and the recognition problems are coNP-complete when 3s2q3s\leq2q; for 3s>2q3s>2q, the majority tournament is transitive and its unique topological order is the unique Kemeny-optimal aggregate. Exact support ss suffices in the hard case when s>q/2s>q/2, and supports in s,s+1{s,s+1} suffice when s=q/2s=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 22-to-11. For NN output candidates, their common distance is 23(N2)\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.


Source: arXiv:2607.28588v1 - http://arxiv.org/abs/2607.28588v1 PDF: https://arxiv.org/pdf/2607.28588v1 Original Link: http://arxiv.org/abs/2607.28588v1

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Date:
Jul 31, 2026
Topic:
Mathematics
Area:
Mathematics
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