this is the exercise:

If possible draw a Tree with $15$ vertices having
3 vertices with degree $4$;
4 vertices with degree $3$;
6 vertices with degree $2$;
0 vertices with degree greater than the ones of the above.

This is what I have done:
considering the definition of a tree: with $d_i \ge 1, \,\, \forall \,i \, \, \,1 \le i \le n$

$$\sum_{i=1}^n d_i = 2n-2$$

in the exercise is given the degree of only $13$ vertices, and not $15$ so, $(4,4,4,3,3,3,3,2,2,2,2,2,2,x,y) \\ 36 + x + y = 2(15) - 2 \\ 36 +x+y = 28$
$x+y = -8$
i.e. I must add two vertices and (they must have a degree $<4$ as said above) the sum of their degree must result $-8$.
But a tree can't have a vertex with a negative degree by definition.
So it is impossible to draw a tree with the data given in the exercise.

What do you think? Please, can you help me? Thanks!

  • $\begingroup$ Your argument looks good to me. $\endgroup$ – quasi Feb 14 '17 at 23:05
  • $\begingroup$ Followup question: With the same degree conditions, what number of total vertices would allow there to be a solution? The answer will suggest that the number $15$ was possibly a typo. $\endgroup$ – quasi Feb 14 '17 at 23:09
  • $\begingroup$ If you follow the further discussion, you'll see that if the number of vertices was specified as $25$ rather than $15$, the condition on the sum of the degrees wouldn't imply that it was impossible. So maybe $15$ was a typo? $\endgroup$ – quasi Feb 15 '17 at 0:23
  • $\begingroup$ @quasi Probably not. $\endgroup$ – JB-Franco Feb 15 '17 at 0:45
  • 1
    $\begingroup$ @JB-Franco I asked a question inspired by this (which led to another question) about the number of trees that might have a given degree sequence... turns out that was a hard question. :-) $\endgroup$ – Joffan Feb 20 '17 at 19:27

It's well-known that a tree has one fewer edges than the number of nodes, hence your summation. Since there are already at least $(3\cdot 4 + 4\cdot3 + 6\cdot2)/2 = 18$ edges indicated by the degree values already, definitely no such tree on fewer than $19$ vertices is possible, and the negative degree you calculated is an indicator of that.

In fact, since we specify a tree (implicitly a connected graph) and not a forest, any additional nodes would need to have degree no less than $1$ so we'd need an extra $5\cdot2 = 10$ nodes of degree $1$ (plus the two mystery nodes which bump the original base edge count to $19$) to be able to draw a tree with the specified higher-degree nodes:

enter image description here

  • $\begingroup$ Wouldn't it require $12$ leaf nodes to make it work? $\endgroup$ – quasi Feb 14 '17 at 23:31
  • $\begingroup$ So, provided there are exactly 12 leaf nodes, the condition on the sum of the degrees is satisfied. Does that automatically imply that such a tree exists? $\endgroup$ – quasi Feb 15 '17 at 0:19
  • $\begingroup$ @ quasi Yes, it does. Trees are too simple to require any more elaborate analysis like Havel-Hakimi $\endgroup$ – Joffan Feb 15 '17 at 0:27
  • $\begingroup$ @JB-Franco: Given the above comments, aren't you glad that the problem specified $15$ vertices rather than $25$? With $15$, you prove it's impossible, and you're done. With $25$, you would need to figure out how to actually draw a tree.satisfying the given conditions. $\endgroup$ – quasi Feb 15 '17 at 0:30
  • $\begingroup$ @quasi not that difficult really... added an example. :-) $\endgroup$ – Joffan Feb 15 '17 at 0:57

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