1

As per Wikipeida, adding one more edge anywhere to any spanning tree would create a cycle. What's the proof?

I saw this property used in a proof about a unique spanning tree for graphs with edges with unique weights. In a class a teacher once said you can use properties/theorems given earlier on in the textbook to where the question is being asked. When you're trying to prove something when can you use a property or theorem like this one?

Community
  • 1
Celeritas
  • 2,839
  • 7
  • 37
  • 62

2 Answers2

7

Proof by contradiction:

Assume that adding an edge to a spanning tree does not create a cycle. Suppose we add edge $(u,v)$ between two vertices $u$ and $v$ to $T$, a spanning tree of a graph $G$. Since we assumed adding an edge to $T$ does not create a cycle, this implies that prior to the insertion of $(u,v)$, there was no path from $u$ to $v$ in $T$. But $T$ is a spanning tree, so such a path must have existed and we have arrived at a contradiction.

Therefore, adding an edge to a spanning tree creates a cycle.

SplitInfinity
  • 952
  • 11
    There's really no need to phrase this as a proof by contradiction: you can just say "since $T$ is connected, there's a $u,v$-path in $T$; adding the new edge $uv$ to this path yields the desired cycle". – Gregory J. Puleo Feb 27 '16 at 04:17
  • 1
    @GregoryJ.Puleo How would you prove it introduces exactly one cycle –  Nov 05 '18 at 11:44
  • 1
    @Anush consider it creates two distinct cycles $u,v, x_1, ..., u$ and $u, v, y_1, ..., u$. Then remove edge $u,v$ and we have two paths from $v$ to $u$ in original tree- a contradiction. – Vimal Patel Dec 28 '19 at 08:28
0

You could also use the degree argument. Let G be the graph where T is the spanning tree, n is the number of vertices. We already know T is connected, so adding an edge would still be connected. Suppose it doesn't have a cycle, which would make the new graph also a tree. T has n-1 degrees by a property of trees. Adding an edge would make it not a tree since it would then have n edges. The new graph can't be a tree, contradiction.

I believe you also need to show that it doesn't create 2 or more cycles. AFSOC that it does. WLOG, say the extra edge {u,v} is in two cycles. The first one is u-i1-i2-...-ik-v, the second is u-j1-j2-...-jm-v. Note that even without the edge {u,v}, i1-...-ik-jm-...-j1 is a cycle in the original spanning tree T. This is a contradiction.

Sol He
  • 33
  • 4