Question: In step 3 of Dijkstras algorithm, the least-cost path values are only updated for nodes not yet in T. Is it possible that a lower-cost

In step 3 of Dijkstra’s algorithm, the least-cost path values are only updated for nodes not yet in T. Is it possible that a lower-cost path could be found to a node already in T? If so, demonstrate by example. If not, provide reasoning as to why not.

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