For integers $k\geq 1$ and $n\geq 2k+1$, the Schrijver graph $S(n,k)$ has as vertices all $k$-element subsets of $[n]:=\{1,2,\ldots,n\}$ that contain no two cyclically adjacent elements, and an edge between any two disjoint sets. More generally, for integers $k\geq 1$, $s\geq 2$, and $n \geq sk+1$, the $s$-stable Kneser graph $S(n,k,s)$ has as vertices all $k$-element subsets of $[n]$ in which any two elements are in cyclical distance at least $s$. We prove that all the graphs $S(n,k,s)$, in particular Schrijver graphs $S(n,k)=S(n,k,2)$, admit a Hamilton cycle that can be computed in time $\mathcal{O}(n)$ per generated vertex.

翻译：对于整数 $k\geq 1$ 和 $n\geq 2k+1$，Schrijver图 $S(n,k)$ 的顶点集为 $[n]:=\{1,2,\ldots,n\}$ 中所有不含任意两个循环相邻元素的 $k$ 元子集，且任意两个不相交集合之间连边。更一般地，对于整数 $k\geq 1$、$s\geq 2$ 和 $n \geq sk+1$，$s$-稳定Kneser图 $S(n,k,s)$ 的顶点集为 $[n]$ 中任意两个元素之间的循环距离至少为 $s$ 的所有 $k$ 元子集。我们证明所有图 $S(n,k,s)$（特别是 Schrijver图 $S(n,k)=S(n,k,2)$）均存在哈密顿圈，且该圈可在每个生成顶点的时间复杂度 $\mathcal{O}(n)$ 内计算得到。