Steiner tree problem in graphs of diameter 3

Question

I have an unweighted undirected graph $G(V, E)$ of diameter 3 and a subset $T\subseteq V$ of these vertices. I want to find the minimum tree $(V', E')$ that contains all vertices in $T$, minimizing the number of nodes $|V'|$.

Is it true that $|V'|\le 2|T|$? If $G$ is of diameter 2 it is easy to prove that this is true. Also for diameter 4, it is not true always since we can take a path graph of 5 vertices with $T$ as two leaves.

Answer 1

This is an interesting question.

It is not true that $|V'|$ is always less than or equal to $2|T|$. Here is the smallest example of $G$ and $T$ where $|V'|$ must be greater than $2|T|$.

Graph $G$ as depicted above has 9 vertices $A,A_1,A_2,B,B_1,B_2,C,C_1,C_2$ and 12 edges $AA_1, A_1A_2, A_2A,$ $BB_1,B_1B_2,B_2B,$ $CC_1, C_1C_2, C_2C,$ $A_1B_2, B_1C_2, C_2A_1$. It is easy to check the diameter of $G$ is 3.

Let $T=\{A,B,C\}$. Suppose $T'=(V',E')$ is a minimal tree in $G$ that spanns $T$.

• If $V'$ contains all 9 vertices, then $|V'|=9$.
• Otherwise, WLOG, assume $V'$ does not contain $A_1$. If anyone of $A_2$, $C_1$, $C_2$ and $B_1$ is not in $T'$, then $A$ and $B$ will not be connected in $T'$. Since $A$ and $B$ are connected in $T'$, all of those 4 vertices must be in $T'$. So $|V'|\ge3+4=7$.

So $|T'|\gt 6= 2|T|.$

Exercise 1. Verify that the following two graphs are also examples where $|V'|$ must be greater than $2|T|$ with $T=\{A, B, C\}$

Exercise 2. (less than 1 minute) If $|T|=3$, then $|V'|\le7$.

Exercise 3. Let $n\ge3$. Construct $G$ and $T$ such that $G$ is connected of diameter 3, $|T|=n$ and $|V'|\gt 2|T|$. (Hint, imitate the simpler example in exercise 2.)