The undirected version of a rooted tree is a free tree
Dependencies:
The undirected version of a rooted tree is a free tree.
Proof that the undirected version of a rooted tree is a free tree
Let $T$ be a rooted tree. Since $T$ has a path from the root to every vertex, the undirected version of $T$ is connected.
Assume that the undirected version of $T$ has a cycle $C$ of length $n$. ($C$ may not be a cycle in the directed version of $T$).
No vertex can have in-degree in $C$ greater than 1, otherwise there will be multiple paths from the root to that vertex. Since $C$ has $n$ edges, sum of in-degree of vertices is $n$ and sum of out-degree of vertices is $n$. Therefore, every vertex has in-degree 1 and out-degree 1. This means $C$ is a directed cycle, which contradicts the fact that there is a unique simple path from the root to every vertex.
Therefore, the undirected version of $T$ cannot have a cycle. Therefore, the undirected vertion of $T$ is connected and acyclic, which means it is a free tree.
Dependency for: None
Info:
- Depth: 3
- Number of transitive dependencies: 8
Transitive dependencies:
- /sets-and-relations/equivalence-classes
- /sets-and-relations/relation-composition-is-associative
- /sets-and-relations/equivalence-relation
- Graph
- Degree sum and number of edges
- Path in Graph
- Tree
- Rooted tree