Determining the number of blocks in a graph by the number of components and vertices.

by Idle Math Guy   Last Updated October 09, 2019 17:20 PM

I'm trying to show that the number of blocks in $G$ is equal to $\omega$ + $\sum_{v \in V}(b(v)-1)$. In this statement $\omega$ denotes the number of components of $G$ and $b(v)$ denotes the number of blocks of $G$ containing $v$.

Intuitively, this makes sense but I'm trying to put this together algebraically as a picture does not constitute a proof.

Any help is appreciated.

-IdleMathGuy

Tags : graph-theory


Answers 1


well, the blocks of a component of G are connected to one another via cut vertices. any vertex present in more than one component is a cut vertex otherwise these two components would be one. you can tell how many blocks there are in a component of G by starting with 1 and counting for each cut vertex in it how many additional independent components its removal induces, which is similar to counting in how many blocks it is present and subtracting 1.

Fuseques
Fuseques
October 09, 2019 17:19 PM

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