handshaking lemma and Erdos-Gallai theorem The conditions for a sequence to be the degree sequence of a simple graph are given by the Erdos-Gallai theorem in addition to the handshaking lemma. Is there an example of a degree sequence where the handshaking lemma is satisfied, but the Erdos-Gallai theorem is not satisfied and thus, the sequence is not graphic (or vice versa).
Or, does satisfying one of these conditions ensure the other is always satisfied?
 A: Let $n=3$ and $d_1=d_2=d_3=1$. Then
$$\begin{align*}
&\sum_{i=1}^1d_i=1\le 2=1\cdot0+\sum_{i=2}^3\min\{d_i,1\}\,,\\
&\sum_{i=1}^2d_i=2\le 3=2\cdot1+\sum_{i=3}^3\min\{d_i,1\}\,,\text{ and}\\
&\sum_{i=1}^3d_i=3\le 6=3\cdot2+\sum_{i=4}^3\min\{d_i,1\}\,.
\end{align*}$$
Thus, the sum of the degrees is odd, but the sequence satisfies the inequality in the Erdős-Gallai theorem.
Obviously one cannot satisfy the full hypotheses of the theorem without also satisfying the handshaking requirement that the sum of the degrees be even, since those hypotheses include the requirement that the degrees be even. However, that requirement is made solely to ensure that the handshaking lemma is satisfied. The meat of the theorem is the inequality, and this example shows that it is possible for a sequence to satisfy that without satisfying the handshaking lemma. (Examples that satisfy the handshaking lemma but not the inequalities are not hard to find.)
A: The simplest example would be
$$
(2,0,0)
$$
The handshaking lemma is satisfied, but the Erdős–Gallai inequalities are not, as $2\not\le 1\cdot (1-1)+0+0$.
