Affidavit of Assumption 4

 20 August 01:26   

    For a blueprint $G$, the afterward are equivalent:

    : A blueprint is a timberline if and alone if for every brace of audible vertices $u,v$, there is absolutely one $u,v$-path.

    Suppose $G$ is a tree, and let $uv$ be an bend not in $G$. By , there is absolutely one $u,v$-path. If this aisle is $ux\_1ldots\; x\_kv$, then abacus the bend $uv$ will make the aeon $ux\_1ldots\; x\_kvu$. As a timberline is to be a affiliated backwoods and a backwoods is to be acyclic. A timberline is accordingly a acute acyclic graph.

    Now accept $G$ is a tree, and let $uv$ be an bend of $G$. By , there is absolutely one $u,v$-path. This aisle is acutely just the bend $uv$. Thus, in the blueprint $G-uv$, there is no $u,v$-path, and appropriately $G-uv$ is disconnected. Back a timberline is to be a affiliated forest, a timberline is accordingly a basal affiliated graph.

    Now accept $G$ is a acute acyclic graph, but is not a tree. Back an acyclic blueprint is a forest, and a timberline is to be a affiliated forest, $G$ haveto be disconnected. Appropriately there exists vertices $u,v$ such that there is no $u,v$-path in $G$. In accurate $uv$ is not an bend of $G$. As $G$ is a acute acyclic graph, $G+uv$ haveto accept a cycle. Back $G$ has no cycle, this aeon haveto cover the bend $uv$. Let this aeon be $ux\_1ldots\; x\_kvu$. Then $ux\_1ldots\; x\_kv$ is a $u,v$-path in $G$. This bucking shows that acute acyclic graphs are trees.

    Finally, accept $G$ is a basal affiliated graph, but is not a tree. As a timberline is to be a affiliated forest, it follows that $G$ is not a forest, and appropriately (by ), $G$ haveto accept a cycle. Let $u,v$ be adjoining vertices on the cycle, and let the aeon be $ux\_1ldots\; x\_kvu$. As $G$ is a basal affiliated graph, $G-uv$ haveto be disconnected. Accede any acme $x$. In $G$, there was a aisle from $x$ to $v$. If the aisle went through the bend $uv$, then in $G-uv$, there is acutely an $x,u$-path. Otherwise, there is acutely an $x,v$-path in $G-uv$. Either way, the affiliated basic of $G-uv$ contains either $u$ or $v$. Back $ux\_1ldots\; x\_kv$ is a $u,v$-path in $G-uv$, the affiliated basic absolute $u$ aswell contains $v$. Appropriately every acme in $G-uv$ is independent in the aforementioned affiliated component, acceptation that $G-uv$ is connected. This bucking shows that basal affiliated graphs are trees.