theoretical-computer-science / Computational social choice
Is computing a Kemeny-optimal aggregate ranking NP-hard when the input consists of exactly three complete rankings? Hardness was known for every even $n \ge 4$; three voters was the minimal open case, and $n = 2$ is polynomial-time solvable.
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Is computing a Kemeny-optimal aggregate ranking NP-hard when the input consists of exactly three complete rankings? Hardness was known for every even $n \ge 4$; three voters was the minimal open case, and $n = 2$ is polynomial-time solvable.
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Is computing a Kemeny-optimal aggregate ranking NP-hard when the input consists of exactly three complete rankings? Hardness was known for every even $n \ge 4$; three voters was the minimal open case, and $n = 2$ is polynomial-time solvable.
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