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POUpdating a Minimum spanning tree when a new edge is inserted
 

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    <p>I've been presented the following problem in University:</p> <p>Let <em>G = (V, E)</em> be an (undirected) graph with costs <em>c<sub>e</sub></em> >= 0 on the edges <em>e</em> &isin; <em>E</em>. Assume you are given a minimum-cost spanning tree <em>T</em> in <em>G</em>. Now assume that a new edge is added to <em>G</em>, connecting two nodes <em>v</em>, <em>t<sub>v</sub></em> &isin; <em>V</em> with cost <em>c</em>.</p> <ol> <li>Give an efficient algorithm to test if <em>T</em> remains the minimum-cost spanning tree with the new edge added to <em>G</em> (but not to the tree <em>T</em>). Make your algorithm run in time O(|E|). Can you do it in O(|V|) time? Please note any assumptions you make about what data structure is used to represent the tree <em>T</em> and the graph <em>G</em>.</li> <li>Suppose <em>T</em> is no longer the minimum-cost spanning tree. Give a linear-time algorithm (time O(|E|)) to update the tree T to the new minimum-cost spanning tree.</li> </ol> <p>This is the solution I found:</p> <pre><code>Let e1=(a,b) the new edge added Find in T the shortest path from a to b (BFS) if e1 is the most expensive edge in the cycle then T remains the MST else T is not the MST </code></pre> <p>It seems to work but I can easily make this run in O(|V|) time, while the problem asks O(|E|) time. Am I missing something?</p> <p>By the way we are authorized to ask for help from anyone so I'm not cheating :D</p>
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    1. COMaybe I'm missing something -- surely G is connected so |V| <= |E| + 1? So your O(|V|) algorithm is automatically O(|E|) also, so where's the problem? Or are you saying, "I'm surprised I was able to solve it even better than the question asked me to"?
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    2. COYes, something like that! I'm surprised it's so simple to do in O(|V|) yet it asks it to be in O(|E|) in both questions. I fear there's something I'm not seeing. Also I have to give a proof of my solution and it's obvious that if e1 is the most expensive edge on a cycle then it can't be in the MST, and that if e2 is more expensive of e1 then e2 can't be in the graph, but can't the possibility of a new path throught e1 create a MST with edges that were in G but not in the previous MST T?
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    3. COUnless you've been told it, I don't think the property that no MST edge can be the longest edge in a cycle is "obvious". So I don't think the problem is "too easy". Re your proof question: suppose such an MST containing e1 existed. Then deleting e1 leaves 2 connected components. Because e1 was originally part of a cycle, there must be some other edge in that cycle that connects these 2 components (proving this part is left as an exercise :-P) and which has weight < w(e1). Add that edge to produce a smaller MST and thus a contradiction. So no MST containing e1 can exist.
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