We present an efficient labeling scheme for answering connectivity queries in graphs subject to a specified number of vertex failures. Our first result is a randomized construction of a labeling function that assigns vertices $O(f^3\log^5 n)$-bit labels, such that given the labels of $F\cup \{s,t\}$ where $|F|\leq f$, we can correctly report, with probability $1-1/\mathrm{poly}(n)$, whether $s$ and $t$ are connected in $G-F$. However, it is possible that over all $n^{O(f)}$ distinct queries, some are answered incorrectly. Our second result is a deterministic labeling function that produces $O(f^7 \log^{13} n)$-bit labels such that all connectivity queries are answered correctly. Both upper bounds are polynomially off from an $\Omega(f)$-bit lower bound. Our labeling schemes are based on a new low degree decomposition that improves the Duan-Pettie decomposition, and facilitates its distributed representation. We make heavy use of randomization to construct hitting sets, fault-tolerant graph sparsifiers, and in constructing linear sketches. Our derandomized labeling scheme combines a variety of techniques: the method of conditional expectations, hit-miss hash families, and $\epsilon$-nets for axis-aligned rectangles. The prior labeling scheme of Parter and Petruschka shows that $f=1$ and $f=2$ vertex faults can be handled with $O(\log n)$- and $O(\log^3 n)$-bit labels, respectively, and for $f>2$ vertex faults, $\tilde{O}(n^{1-1/2^{f-2}})$-bit labels suffice.

翻译：我们提出了一种高效的标记方案，用于回答图中在指定数量顶点故障情况下的连通性查询。第一个结果是随机化构建的标记函数，该函数为顶点分配$O(f^3\log^5 n)$比特的标签，使得给定$F\cup \{s,t\}$的标签（其中$|F|\leq f$），我们能够以$1-1/\mathrm{poly}(n)$的概率正确报告$s$和$t$在$G-F$中是否连通。然而，在所有$n^{O(f)}$个不同查询中，部分查询可能得到错误回答。第二个结果是确定性的标记函数，它生成$O(f^7 \log^{13} n)$比特的标签，使得所有连通性查询均被正确回答。这两个上界与$\Omega(f)$比特的下界相差多项式因子。我们的标记方案基于一种新的低度分解，该分解改进了Duan-Pettie分解，并便于其分布式表示。我们大量使用随机化技术来构造命中集、容错图稀疏化器以及线性草图。去随机化标记方案结合了多种技术：条件期望法、命中-未命中哈希族以及轴对齐矩形的$\epsilon$-网。Parter和Petruschka的先前标记方案表明，$f=1$和$f=2$的顶点故障可分别用$O(\log n)$和$O(\log^3 n)$比特的标签处理，而对于$f>2$的顶点故障，$\tilde{O}(n^{1-1/2^{f-2}})$比特的标签即足够。