A tournament on $n$ vertices: an irreflexive, complete, asymmetric directed-edge relation on $\mathrm{Fin}\, n$, equipped with a decidability instance.
- decEdge : DecidableRel self.edge
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Number of Hamilton paths of $T$: permutations $\sigma$ of $\mathrm{Fin}\, n$ such that every consecutive edge $\sigma(i) \to \sigma(i+1)$ lies in $T$.
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Number of Hamilton cycles of $T$: permutations $\sigma$ of $\mathrm{Fin}(n+1)$ such that every cyclic edge $\sigma(i) \to \sigma(i+1)$ lies in $T$; zero if $n = 0$.
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Trivial bound: the number of Hamilton paths is at most $n!$, the total number of permutations of $\mathrm{Fin}\, n$.
Adjacency matrix of a tournament as a real $0/1$-matrix: entry $(i, j)$ is $1$ if $T$ has the edge $i \to j$, else $0$.
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Every entry of the tournament's adjacency matrix is $0$ or $1$.
Successor permutation derived from $\sigma$: the conjugate $\sigma \circ (+1) \circ \sigma^{-1}$, which sends $\sigma(j)$ to $\sigma(j + 1)$.
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Successor relation: $\sigma(j + 1) = (\mathrm{toSuccPerm}\,\sigma)(\sigma(j))$.
If $\sigma$ encodes a Hamilton cycle of $T$, then for every vertex $v$ the directed edge $v \to (\mathrm{toSuccPerm}\,\sigma)(v)$ lies in $T$.
Two permutations with the same successor permutation and the same value at $0$ must be equal: induct on the index using the successor relation.
Permanent lower bound: if $S$ is a set of permutations $\sigma$ with $A_{\sigma(i), i} = 1$ for all $i$, then $\mathrm{perm}(A) \ge |S|$.
Adjacency-matrix entry is $1$ iff the corresponding tournament edge exists.
Key counting bound: the number of Hamilton cycles is at most $(n + 1) \cdot \mathrm{perm}(A_T)$, where $A_T$ is the adjacency matrix of $T$.
Permanent optimization bound for tournament adjacency matrices: combining Brégman's inequality with smoothing/Stirling, $(n + 1) \cdot \mathrm{perm}(A_T) \le \sqrt{\pi/2 + 1} \cdot \sqrt{n+1} \cdot (n+1)! / 2^{n+1}$.
Alon's Hamilton-cycle bound: there exists $C > 0$ such that every $n$-vertex tournament has at most $C \sqrt{n} \cdot n! / 2^n$ Hamilton cycles.
Averaging step: for $n \ge 2$ there exists a one-vertex extension $T'$ of $T$ such that $\mathrm{numHamiltonPaths}(T) \le 4 \cdot \mathrm{numHamiltonCycles}(T')$.
Theorem 10.2.4 (Alon 1990, Hamilton paths). There exists $C > 0$ such that every $n$-vertex tournament has at most $C \cdot n \sqrt{n} \cdot n! / 2^n$ Hamilton paths.