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  1. #1
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    Discrete question.

    Graph G = (V,E) has n vertices, m edges and p connected components.
    If m = n - p, prove that G is a forest.
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  2. #2
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    In other words, you want to show that each component of that graph is a tree.
    None of the components can contain a cycle.
    Consider the fact |E|=|V|-p. Is a cycle possible?
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  3. #3
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    Quote Originally Posted by Plato View Post
    In other words, you want to show that each component of that graph is a tree.
    None of the components can contain a cycle.
    Consider the fact |E|=|V|-p. Is a cycle possible?
    I can't prove that a cycle doesn't exist in every tree.
    What can I refer to when I suppose that there is a connected component contains a cycle?
    Last edited by le_su14; November 5th 2007 at 05:42 PM.
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  4. #4
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    This is again one of those situations in which I donít for sure what you have to use. In general a tree has one less edge than vertices. If a component (btw. saying connect is redundant components are connected) has a cycle then it has at least as many edges as vertices. That would contradict the given that |E|=|V|-p.
    If every component is a tree then  \left| E \right| = \sum\limits_{k = 1}^p {\left| {E_k } \right|}  = \sum\limits_{k = 1}^p {\left( {\left| {V_k } \right| - 1} \right)}  = \left| V \right| - p.
    So a cycle any where changes that equation.
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  5. #5
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    Thumbs up

    Quote Originally Posted by Plato View Post
    This is again one of those situations in which I don’t for sure what you have to use. In general a tree has one less edge than vertices. If a component (btw. saying connect is redundant components are connected) has a cycle then it has at least as many edges as vertices. That would contradict the given that |E|=|V|-p.
    If every component is a tree then  \left| E \right| = \sum\limits_{k = 1}^p {\left| {E_k } \right|}  = \sum\limits_{k = 1}^p {\left( {\left| {V_k } \right| - 1} \right)}  = \left| V \right| - p.
    So a cycle any where changes that equation.
    Thank you very much! Now I can solve it.
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