TITLE: Determine if a graph exists knowing the degree of its vertices
QUESTION [6 upvotes]: Problem: There is a graph on ten vertices whose degrees are $9,8,8,8,6,5,4,4,2,2$

Answer: False.
My attempt: Since we know that $2 | E | = \sum _ { v \in V } d ( v )$, $$2\cdot 28 = 9+8\cdot3+6+5+4\cdot2+2\cdot2$$
Hence $| E | = 28$. Since we don't now more about the graph (it may not be a tree), how can I determine that such graph does not exist?

REPLY [9 votes]: The answer is FALSE.
We proceed with the Havel–Hakimi Theorem, which determines whether a degree sequence can represent a simple graph.
Applying this theorem, we note that the original degree sequence is graphical if and only if the degree sequence given by $\{7, 7, 7, 5, 4, 3, 3, 2, 1\}$ is graphical. We can keep on reiterating to get the following degree lists:
$$\{6, 6, 4, 3, 2, 2, 1, 0\} \rightarrow \{5, 3, 2, 1, 1, 0, 0\} \rightarrow \{2, 1, 0, 0, 0, -1\}$$
But, the last degree list is clearly not graphical: it has an edge with a negative degree! So, the answer is false.