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Here's a problem in computational social choice which is not known to be in P, and may or may not be NP-complete.

Agenda control for balanced single-elimination tournaments:

Given: tournament graph $T$ on $n=2^k$ nodes, node $a$

Question: does there exist a permutation of the nodes (a bracket) so that a is the winner of the induced single-elimination tournament?

Given a permutation $P_k$ on $2^k$ nodes of $V$ and a tournament graph $T$ on $V$, one can obtain a permutation $P_{k-1}$ on $2^{k-1}$ nodes as follows. For every $i>0$, consider $P_k[2i-1]$ and $P_k[2i]$ and the arc $e$ between them in $T$; let $P_{k-1}[i]=P_k[2i-1]$ if $e=(P_k[2i-1],P_k[2i])$ and $P_{k-1}[i]=P_k[2i]$ otherwise. That is, we match up pairs of nodes according to $P_k$ and use $T$ to decide which nodes (winners) move on to the next round $P_{k-1}$. Hence given a permutation on $2^k$ one can actually define $k$ rounds $P_{k-1},\ldots,P_0$ inductively as above, until the last permutation contains only one node. This defines a (balanced) single-elimination tournament on $2^k$ nodes. The node which remains after all the rounds is the winner of the tournament.

Agenda control for balanced single-elimination tournaments (graph formulation):

Given: tournament graph $T$ on $n=2^k$ nodes, node $a$

Question: does $T$ contain a (spanning) binomial arborescence on $2^k$ nodes rooted at $a$?

A binomial arborescence on $2^k$ nodes rooted at a node $x$ is defined recursively as $a$ binomial arborescence on $2^{k-1}$ nodes rooted at $x$ and a binomial arborescence on $2^{k-1}$ nodes rooted at a different node $y$ and an arc from $x$ to $y$. (If $k=0$, a binomial arborescence is just the root.) The spanning binomial arborescences in a tournament graph capture exactly the single-elimination tournaments which can be played, given the match outcome information in the tournament graph.

Some references: