Given a connected graph G, is there a way to find the spanning tree which covers as many as degree 1 vertex as possible, and to minimise the leaves of the spanning tree which are not degree 1 in G?
2026-04-02 04:34:42.1775104482
spanning tree which covers as many as degree 1
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A partial answer.
We can construct a required spanning tree $T$ as follows. Let $E_0$ be the set of pendant edges of the graph $G$, that is edges of $G$ which are incident to a leaf and $G_1$ be the graph constructed by removing from $G$ all its leaves. Note that the graph $T_0$ spanned by $E_0$ belongs to each spanning tree of $G$ (so each such tree which covers all leaves of $G$) and the graph $G_1$ is empty or connected. So it remains to find a spanning tree $T_1$ for the graph $G_1$ containing as few leaves not incident to edges of $E_0$ as possible and put as $T$ the union of graphs $T_0$ and $T_1$.