Calculating number of edges from a degree sequence Problem: Suppose a graph $G$ has degree sequence $25, 18, 18, 5, 2, 2$.
How many edges are there in $G$?
I am assuming since the definition of degrees of a vertex that each one of the 6 vertex's (since there are $6$ nums in the degree sequence) has $n$ edges(paths) ($n$ being one of the six numbers in the degree sequence). So would the total number of edges in this case be $25+18+18+5+2+2$? 
I don't think this graph exists as a simple graph, but would have to be drawn with the three vertices having degrees of $18$, $18$, and $25$ have loops or multiple edges between them. Do those count in the total number of edges? 
 A: Hint: For a vertex $v$, the degree of $v$ is the number of edges incident to $v$.
Counting the total degree, you will actually count each edge $\{u,v\}$ twice -- once when you count the degree of $u$, and once when you count the degree of $v$.
So, how should the total degree and the total number of edges be related?
If each edge $\{u,v\}$ gets counted for both $u$ and for $v$, then it has been counted twice -- and so the total degree is just twice the number of edges!
(This works if we allow multiple edges... it works with loops as well, with the convention that a loop on vertex $v$ contributes two to its total degree.)
So, if you notice that your total degree is 70, it must be the case that there are 35 edges.
A: it would be sum of degrees divided by two.
A: I believe the answer is the sum of the sequence divided by two. Each edge is incident on two vertices. I have this same problem due tomorrow too. This link supports my idea http://mathworld.wolfram.com/DegreeSequence.html
A: We can find the number of edges from the following observations:


*

*The degree of a vertex $v$ counts the number of times $v$ appears as the endpoint of an edge.

*If we sum the degrees, we get the total number of times any vertex appears as the endpoint of an edge.

*Each edge has two endpoints.
These observations will lead you to the result known as the Handshaking Lemma.
This particular degree sequence obviously cannot be realized by a simple graph: the vertex of degree $25$ would have $25$ distinct neighbours, but there are only $6$ vertices.
It can be realized as a multigraph, however, as the following picture indicates:

In this figure, numbers on edges indicate their multiplicity.
