# The Diameter of resulting graph by sum of two unions.

Let's say we have a graph $$T$$, which is also a spanning tree of a graph $$G$$, $$T$$ is connected, now let's say the diameter of $$T$$ is $$x$$.

Now, what would be the diameter of :

$$(T \cup T^c) + (T \cup T^c)$$

where $$T^c$$ is the complement of $$T$$, and we assume that $$T$$, $$T^c$$ are the different copies of $$T$$.

What I know is since $$T$$ and $$T^c$$ are disjoint, the union of them will be a disconnected graph, but I'm not clear how the join operation denoted by $$+$$ sign here change the final resulting graph. Is it still disconnected? (which I hope not)

My idea was that the union of $$T$$ and $$T^c$$ would be a complete graph, but it seems like it's wrong if I plot a sample graph. Hence, I'm looking for help. Thank you for any support in advance.

• This depends a little on the definitions. What do you mean by union specifically, and what do you mean by $+$? Dec 3, 2021 at 4:25
• The union of two graphs G 1 = (V_1 ,E_1 ) and G2 = (V_2 ,E_2 ), where V_1 ∩ V_2 = ∅ , is the graph G1 ∪ G2 = (V_1 ∪ V_2 ,E_1 ∪ E_2 ). And the + signifies the Join operation of two graphs, which follows the standard definition of join operation of two graphs. Dec 3, 2021 at 4:39
• And this is the definition of join: Let G1 = (V_1 ,E_1 ) and G2 = (V_2 ,E_2 ) be two graphs such that V 1 ∩ V 2 = ∅ . The join G1 + G2 is the graph that has the vertices and the edges of the original graphs, in addition to the edges that connect all the vertices of G with all the vertices of H. Dec 3, 2021 at 4:42

The union of $$T$$ and its complement $$T^{c}$$ is a complete graph $$K_n$$, where $$n$$ is the number of nodes in $$T$$. The diameter of a complete graph is ofcourse 1, since all vertices are connected to each other by distance 1. Hence, since the join of graphs $$G+H$$ connects every node of $$G$$ to each one of $$H$$, we obtain that the problem's join graph $$(T\cup T^c) + (T\cup T^c) = K_n + K_n = K_{2n}$$ is also complete, so its diameter is 1.