$\mathit{SB}(3,n)$ has no Hamiltonian cycle when $n$ is even: a sign-of-permutation proof, with extension to all odd $m\equiv 3\pmod 4$

Abstract

We resolve exercise 7.2.2.4--224 of Knuth's Pre-Fascicle 8a (10 April 2026 draft, rated [46]): the digraph $\mathit{SB}(3,n)$ has no Hamiltonian cycle when $n$ is even. The argument is a sign-of-permutation obstruction. Writing the successor map of a candidate Hamiltonian cycle as $f_S = A_b \circ \sigma$, $\operatorname{sgn}(A_b)=+1$ when $m$ is odd, so $\operatorname{sgn}(f_S)=\operatorname{sgn}(\sigma)$ for every choice set $S$. A short dihedral Burnside computation shows $\operatorname{sgn}(\sigma)=-1$ on $\Sigma_3^n$ for even $n$, contradicting the sign $+1$ required of a single $3^n$-cycle. The same argument gives the stronger statement that $\mathit{SB}(m,n)$ has no Hamiltonian cycle whenever $m$ is odd with $m\equiv 3\pmod 4$ and $n$ is even; this restricts the residue classes in which Knuth's hint to Ex.~225 (existence of Hamiltonian cycles in $\mathit{SB}(m,n)$ for all $m>3$ and $n>2$) can hold.