I know that simple graph has no parellel edges and loops. My question is that I have to draw the graph on six vertices with degree sequence $(3,3,5,5,5,5)$. I draw the the graph with the given degree sequence and everytime I got the graph which is not simple. Does there exist a simple graph with these degree sequence? If no why?
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$\begingroup$ Do you know the Havel-Hakimi Theorem? $\endgroup$– EuYuSep 18, 2012 at 18:14
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$\begingroup$ No! Can you please state this theorem! $\endgroup$– KnsSep 18, 2012 at 18:17
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2 Answers
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By the Havel-Hakimi Theorem, a descending degree sequence $\{a_1, \cdots ,\ a_k\}$ is graphical if and only if $a_1 \le k-1$ and $\{a_2 - 1, \cdots, a_{a_1+1}-1, a_{a_1 +2}, \cdots, a_k\}$ is graphical.
Applying the theorem to your case, we have $$\{5,5,5,5,3,3\} \iff \{4,4,4,2,2\}\iff\{3,3,1,1\}\iff\{2,0,0\}$$ The latter graph is evidently impossible.
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$\begingroup$ Evidently not. $\{1,1,1,1\}$ is the degree sequence of two disjoint $K_2$s. However, it is impossible for the sequence as originally stated. $\endgroup$ Sep 18, 2012 at 19:46
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$\begingroup$ @RickDecker Sorry about that, I must've mistyped the sequence in the middle and started using it. It is fixed now. $\endgroup$– EuYuSep 18, 2012 at 20:28
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$\begingroup$ A minor point, but if you edit a post of yours it's a good idea to indicate the edit as part of your post. Otherwise someone new to this thread might be confused by the existing comments (as here, for example). $\endgroup$ Sep 18, 2012 at 23:38
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No fancy theoretical machinery is required.
Call the vertices $v_1,v_2,v_3,v_4,v_5$, and $v_6$. Let $v_1,v_2,v_3$, and $v_4$ be the vertices of degree $5$. There are only six vertices, so each of these vertices must be connected by an edge to every other vertex in the graph. For instance, $v_2$ must be connected to $v_1,v_3,v_4,v_5$, and $v_6$. This means that $v_5$ and $v_6$ are both connected to each of the vertices $v_1,v_2,v_3$, and $v_4$ and must therefore have degree at least $4$: the degree sequence $(3,3,5,5,5,5)$ is impossible for a simple graph.
If you’d like to explore further, this kind of reasoning leads to the easy half of the Erdős-Gallai theorem, which gives a simple computational criterion for whether a sequence is the degree sequence of some simple graph. This kind of reasoning shows that a sequence that fails the criterion cannot possibly be a degree sequence; proving that every sequence that satisfies it really is the degree sequence of some simple graph is harder. See also the comments here.
answered Sep 18, 2012 at 18:22