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If I have an MST, and I add any edge to create a cycle, will removing the heaviest edge from that cycle result in an MST? 7 Do the algorithms of Prim and Krusksal always produce the same minimum spanning tree, given the same tiebreak criterion? Minimum spanning tree containing a given edge after removing edges.

To implement Kruskal's algorithm, we use a priority queue to consider the edges in order by weight, a union-find data structure to identify those that cause cycles, and a queue to collect the MST edges.

This is a part of an exam preparation. Rope pulley for felling trees know it has something to do with max-flow algorithm, but I'd be happy for a hint: Let G= (V,E) an undirected connected graph, and let w:E->R a weight function, e an edge and k > 0.

Jun 04, Adding e to T' creates a cycle, and deleting any edge of that cycle would create another spanning tree. Let's add e and delete the heaviest edge of that cycle. That heaviest edge is definitely not e, because e is not the heaviest edge of any cycle. So we added e to T', then deleted an edge heavier than e. This means that we've reduced the total weight of T': therefore. Removing one edge from the spanning tree will make the graph disconnected, i.e.

the spanning tree is minimally connected.

Using the following remarkable theorem of Aldous and Broder: Start at an arbitrary vertex s and take a random walk until every vertex has been visited choosing an outgoing edge uniformly at random among all incident edges.

Adding one edge to the spanning tree will create a circuit or loop, i.e. the spanning tree is maximally acyclic. Mathematical Properties of Spanning Tree. Spanning tree has n-1 edges, where n is the number of nodes (vertices). Removing an edge from this cycle recreates a spanning tree. Choose the edge to remove as follows: Since one edge of the cycle crosses between S and V-S, another edge, v'w' must cross back Let T' = T + vw - v'w', another spanning tree. Since vw is a minimum weight edge from V-S to S, and v'w' is another W(vw) ≤ W(v'w') and hence W(T') ≤ W(T), but T is an MST so W(T') = W(T), and T' is also an MST.