TU Wien:Diskrete Mathematik für Informatik UE (Gittenberger)/Übungen WS13/Beispiel 5
- Every two nodes of T are connected by exactly one path.
- T is connected and α0(T) = α1(T) + 1.
- T is a minimal, connected graph, i.e., deleting an edge destroys connectivity
- T is a maximal, acyclic graph, i.e., adding an edge generates a cycle.
Theory[Bearbeiten | Quelltext bearbeiten]
Solution[Bearbeiten | Quelltext bearbeiten]
5.1.
Proof by contradiction
1. Assume: , where there are two or more paths between v and w
2. This implies there is a cycle. This contradicts with the definition of a tree
5.2.
Proof by induction
T=(V={v},E={})
1. Add a vertex and connect it with an two existing, vertices with the edges .
2. Now either:
2.a. |V|=1 and since there are no two vertces, then it is not possible to add two edges
2.b. In a tree there is always a path between two distinct vertices (as per definition it's connected). The edges introduce a second path. Use the same arguments as in 5.1
5.3
1. In a tree there is always a path between two distinct vertices . If there is an edge , it is that path, and removing it contradicts with the definition.
5.4
Use the same arguments as in 5.2.2.b.
NOTE:" These are proves for " is a tree". To show equivalence () " is a tree " must also be proved