0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

well, the blocks of a component of G are connected to one another via cut vertices. any vertex present in more than one block is a cut vertex otherwise these two blocks 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.