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arxiv: 1504.04169 · v1 · pith:JXVW6S7Snew · submitted 2015-04-16 · 💻 cs.DS

Fault Tolerant BFS Structures: A Reinforcement-Backup Tradeoff

classification 💻 cs.DS
keywords setminusbackupedgesepsilonfault-pronenamelycomponentsdist
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This paper initiates the study of fault resilient network structures that mix two orthogonal protection mechanisms: (a) {\em backup}, namely, augmenting the structure with many (redundant) low-cost but fault-prone components, and (b) {\em reinforcement}, namely, acquiring high-cost but fault-resistant components. To study the trade-off between these two mechanisms in a concrete setting, we address the problem of designing a $(b,r)$ {\em fault-tolerant} BFS (or $(b,r)$ FT-BFS for short) structure, namely, a subgraph $H$ of the network $G$ consisting of two types of edges: a set $E' \subseteq E$ of $r(n)$ fault-resistant {\em reinforcement} edges, which are assumed to never fail, and a (larger) set $E(H) \setminus E'$ of $b(n)$ fault-prone {\em backup} edges, such that subsequent to the failure of a single fault-prone backup edge $e \in E \setminus E'$, the surviving part of $H$ still contains an BFS spanning tree for (the surviving part of) $G$, satisfying $dist(s,v,H\setminus \{e\}) \leq dist(s,v,G\setminus \{e\})$ for every $v \in V$ and $e \in E \setminus E'$. We establish the following tradeoff between $b(n)$ and $r(n)$: For every real $\epsilon \in (0,1]$, if $r(n) = {\tilde\Theta}(n^{1-\epsilon})$, then $b(n) = {\tilde\Theta}(n^{1+\epsilon})$ is necessary and sufficient.

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