If G is tree with 18 vertices, with max degree of vertex 6 and with no vertices of degree 2, prove that number of leaves $l$ satisfies $12\le l\le 14$. From handshaking lemma I get that $l\ge 10$, because I used that $34\ge l+3(18-l)$. So I am not sure if I'm mistaking because lower bound should be $12$. Also not sure how to get upper bound.
Asked
Active
Viewed 443 times
2
-
Somewhat related question you might find helpful: Fewer degree-3+ nodes than leaf nodes in a tree - especially for the numerical relation. "No degree-2 nodes" is key here. Upper bound relies on the max degree 6. – Joffan Apr 27 '21 at 14:25
1 Answers
2
Hope you are doing well.
So, you know that the sum of degrees over the $18$ vertices of $G$ is 34. What degree does each leaf contribute to this and what degree do the other vertices contribute? Well, the leaves add $1$ and the other vertices add $3$-$6$ (since the max is 6 and the min is 3). Let $\ell$ denote the number of leaves. From the above statements we have the following inequality, $$\ell + \color{#0B2}6 + 3((18-1) - \ell) \leq 34 \leq \ell + \color{#0B2}6 + 6((18-1) - \ell)$$ (note that it is $18-1$ since we treat the known vertex of degree $\color{#0B2}6$ separately)
this inequality implies $$-2\ell + 57 \leq 34 \leq -5\ell + 108 $$ which gives us that $12 \leq\ell \leq 14$ (do you think you can work out how to get from the last inequality to this one?; if not, please comment).
I hope this helps in your understanding of Graph Theory.
-
OK, I get it. So the point is that we need to use the fact that there is node of degree 6, even we are bounding this? – Trevor Apr 27 '21 at 14:38
-
1@Trevor Yes, the fact that there is at least one node of degree 6 is the key to this problem. – R509 Apr 27 '21 at 14:39
-
Nice work, I tweaked your answer a little for further clarity. The opening and closing paragraphs are not really MSE style. – Joffan Apr 27 '21 at 15:46
-
@Joffan Solid, looks great! Thank you for the tweaks and wish you the best! – R509 Apr 27 '21 at 15:47