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Math Help - Construction

  1. #1
    MHF Contributor Siron's Avatar
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    Construction

    Hi,

    I need some help with the following question.

    Suppose you have a Laman construction F=(E,V) (with E the edges and V the vertices) that means E=2V-3 and for every subset E' we have e' \leq 2v'-3. I need to prove there's at least one vertex of degree less then or equal to 3 (a vertex of degree 1 for example is a point that belongs to 1 edge, ...)

    Can someone help me with this proof? I tried by saying 'Suppose all of the vertices have degree greater then or equal to 4' and to come to a contradiction, but this didn't work.

    Thanks in advance!
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  2. #2
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    Re: Construction

    This follows from the handshaking lemma.

    Questions from graph theory should be posted to Discrete Math subforum.
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Construction

    Quote Originally Posted by emakarov View Post
    This follows from the handshaking lemma.

    Questions from graph theory should be posted to Discrete Math subforum.
    If I suppose that all v vertices are of degree greater then or equal to 4 then the sum is at least \sum_{v \in v} \mbox{deg}(v) = 4v, but 2|E|=2(2v-3) =4v-6, thus they're not equal, hence a contradiction.

    Is there a way to prove it without the handshaking lemma and just by using the two conditions for Laman constructions I gave?
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