1. Construction

Hi,

I need some help with the following question.

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

2. Re: Construction

This follows from the handshaking lemma.

Questions from graph theory should be posted to Discrete Math subforum.

3. Re: Construction

Originally Posted by emakarov
This follows from the handshaking lemma.

Questions from graph theory should be posted to Discrete Math subforum.
If I suppose that all $\displaystyle v$ vertices are of degree greater then or equal to 4 then the sum is at least $\displaystyle \sum_{v \in v} \mbox{deg}(v) = 4v$, but $\displaystyle 2|E|=2(2v-3) =4v-6$, 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?