An unweighted, undirected graph $G$ on $n$ nodes is said to have \emph{bandwidth} at most $k$ if its nodes can be labelled from $0$ to $n - 1$ such that no two adjacent nodes have labels that differ by more than $k$. It is known that one can decide whether the bandwidth of $G$ is at most $k$ in $O(n^k)$ time and $O(n^k)$ space using dynamic programming techniques. For small $k$ close to $0$, this approach is effectively polynomial, but as $k$ scales with $n$, it becomes superexponential, requiring up to $O(n^{n - 1})$ time (where $n - 1$ is the maximum possible bandwidth). In this paper, we reformulate the problem in terms of bipartite matching for sufficiently large $k \ge \lfloor (n - 1)/2 \rfloor$, allowing us to use Hall's marriage theorem to develop an algorithm that runs in $O(n^{n - k + 1})$ time and $O(n)$ auxiliary space (beyond storage of the input graph). This yields polynomial complexity for large $k$ close to $n - 1$, demonstrating that the bandwidth recognition problem is solvable in polynomial time whenever either $k$ or $n - k$ remains small.

翻译：对于一个具有 $n$ 个节点的无权无向图 $G$，若其节点可以被标记为从 $0$ 到 $n - 1$，且任意两个相邻节点的标记差不超过 $k$，则称该图的 \emph{带宽} 至多为 $k$。已知通过动态规划技术，可以在 $O(n^k)$ 时间和 $O(n^k)$ 空间内判定图 $G$ 的带宽是否至多为 $k$。对于接近 $0$ 的小 $k$ 值，此方法实际上是多项式时间的，但随着 $k$ 随 $n$ 增长，其复杂度会变得超指数级，最多可能需要 $O(n^{n - 1})$ 时间（其中 $n - 1$ 是最大可能带宽）。在本文中，我们针对足够大的 $k \ge \lfloor (n - 1)/2 \rfloor$，将问题重新表述为二分图匹配问题，从而可以利用霍尔婚姻定理，开发出一种运行时间为 $O(n^{n - k + 1})$、辅助空间为 $O(n)$（不计输入图的存储空间）的算法。这为接近 $n - 1$ 的大 $k$ 值带来了多项式复杂度，证明了只要 $k$ 或 $n - k$ 保持较小，带宽识别问题即可在多项式时间内求解。