Fast algorithms for finding nearest common ancestors
SIAM Journal on Computing
Fibonacci heaps and their uses in improved network optimization algorithms
Journal of the ACM (JACM)
Dynamic Maintenance Versus Swapping: An Experimental Study on Shortest Paths Trees
WAE '00 Proceedings of the 4th International Workshop on Algorithm Engineering
Point-of-Failure Shortest-Path Rerouting: Computing the Optimal Swap Edges Distributively
IEICE - Transactions on Information and Systems
Single backup table schemes for shortest-path routing
Theoretical Computer Science - Foundations of software science and computation structures
Swapping a failing edge of a shortest paths tree by minimizing the average stretch factor
Theoretical Computer Science
The Swap Edges of a Multiple-Sources Routing Tree
Algorithmica
Computing all the best swap edges distributively
Journal of Parallel and Distributed Computing
A distributed algorithm for finding all best swap edges of a minimum diameter spanning tree
DISC'07 Proceedings of the 21st international conference on Distributed Computing
Finding best swap edges minimizing the routing cost of a spanning tree
MFCS'10 Proceedings of the 35th international conference on Mathematical foundations of computer science
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In network communication systems, frequently messages are routed along a minimum diameter spanning tree (MDST) of the network, to minimize the maximum travel time of messages. When a transientfailure disables an edge of the MDST, the network is disconnected, and a temporary replacement edge must be chosen, which should ideally minimize the diameter of the new spanning tree. Preparing for the failure of any edge of the MDST, the all-best-swaps (ABS) problem asks for finding the best swap for every edge of the MDST. Given a 2-edge-connected weighted graph G= (V,E), where |V| = nand |E| = m, we solve the ABS problem in $ O\left( m\log n \right) $ time and O(m) space, thus considerably improving upon the decade-old previously best solution, which requires $O(n\sqrt{m})$ time and O(m) space, for $m=o\left(n^2/ \log^2 n\right)$.