Is it possible to build this graph? Is it possible to build a graph made from 10 vertices, that has these degrees for each vertex?
1,2,1,2,3,4,3,4,5,6 
it can be directed or un-directed, also it can be connected or not connected. however there is not a self edge (edge that get out of a vertex back to itself)
I have a feeling that its not possible to build such a graph, however I can't find a way to prove this. can anyone please show me the proof? or a graph if thats possible.
Thanks
 A: Writing what Boff said: Suppose you have a graph in which the sum of the degrees of all vertices is n. If you add an edge you will add $1$ to the $2$ vertices connected to that edge, making the sum of the degrees n+2. Using this we can see a graph with $k$ edges has a sum of degrees of $2k$. In other words the sum of the degrees of any graph must always be even. But your list has an odd sum, so the graph can't exist.
A: In a graph, if we add an edge then the degree of the two incident vertices each increases by one. Thus the sum of the degrees increases by two. As a result, the sum of degrees of of vertices in a graph is always an even number. Now calculate that your list of numbers sums up to an odd number.
A: Formulas:
Undirected Graph: $\sum_{v \in V} deg (v)$ = $2$$\mid E \mid$
Directed Graph: $\sum_{v \in V} deg^- (v)$ = $\sum_{v \in V} deg ^+(v)$ = $\mid E \mid$ 
Explanation: 
This problem is an application of the Handshaking Lemma, proven by Euler. The lemma states that every finite, undirected graph (which is one possibility of your graph) has an even number of vertices with an odd degree. The formula is as follows: 
$\sum_{v \in V} deg (v)$ = $2$$\mid E \mid$.
Applying this formula, the sum of all your degrees within the provided set would be 31, which is not divisible by two and hence this graph cannot be built with undirected edges.
I am not sure how to consider this problem with directed edges because with directed edges there is no generalized "degree" term. There are in-degrees and out-degrees for each vertex, however I don't know how many are designated to each node so I couldn't specifically answer your question (unless someone else could elaborate). However, the application is similar to the one for undirected edges. In this situation, you would apply the degree-sum formula for digraphs $-$ which is essentially the Handshaking lemma for digraphs. This lemma states:$\sum_{v \in V} deg^- (v)$ = $\sum_{v \in V} deg ^+(v)$ = $\mid E \mid$ meaning that in order for this graph to exist, the sum of the in-degrees must be equal to the sum of the out-degrees which must be equal to the number of edges. 
A: In general you can use the Havel–Hakimi algorithm (although testing for an even degree sum is a useful and easy first step).
Here your sample degree sequence $1,2,1,2,3,4,3,4,5,6$ fails the even-degree-sum test (handshaking lemma), $\sum d_i = 31$. So for the purpose of Havel-Hakimi testing I'll use the example of $1,2,1,2,3,4,3,5,4,5$ with degree sum $30$ which would normally be written $(5,5,4,4,3,3,2,2,1,1)$.
The process is that you sort the degree sequence descending, as usual, then generate a new sequence by removing a highest-degree node of degree $d$ and subtract one off  each of the following $d$ degree values, to produce a new shorter sequence.
The process can be applied repeatedly until you reach a sequence that is trivial to recognize as possible or impossible. (In principle, for a graphical sequence - one that corresponds to a realizable graph - this process can be continued to finish with all zeroes).
$\begin{array}{c}
 & 5 & 5 & 4 & 4 & 3 & 3 & 2 & 2 & 1 & 1 \\
 &  & 4 & 3 & 3 & 2 & 2 & 2 & 2 & 1 & 1\\
 &  &  & 2 & 2 & 1 & 1 & 2 & 2 & 1 & 1\\
\text{re-sort} &  &  & 2 & 2 & 2 & 2 & 1 & 1 & 1 & 1\\
 &  &  &  & 1 & 1 & 2 & 1 & 1 & 1 & 1\\
\text{re-sort} &  &  &  & 2 & 1 & 1 & 1 & 1 & 1 & 1\\
 &  &  &  &  & 0 & 0 & 1 & 1 & 1 & 1\\
 &  &  &  &  &  &  &  & 0 & 1 & 1\\
 &  &  &  &  &  &  &  &  &  & 0 &\checkmark \\
\end{array}$
So this modified degree sequence can be drawn as a graph.
For directed graphs you would normally specify in-degree and out-degree for each vertex, and of course the sum of the two across all vertices should be equal.
