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I'd like to decide whether, given a connected graph $G = (V, E)$ and an integer $k$ as input, $G$ admits two vertex-disjoint subgraphs $T_1 = (V_1, E_1)$ and $T_2 = (V_2, E_2)$ such that $T_1$ and $T_2$ are trees, $V_1 \cup V_2 = V$ (so in some way they are spanning), and $|V_1 - V_2| \leq k$ (they aren't too different in size).

Some graphs, such as perfect binary trees or Hamiltonian graphs, do admit such balanced spanning trees ($k=o(1)$), whereas others do not, such as star graphs ($k = n - 2$).

Is this NP-hard? Is it already NP-hard when fixing $k=0$? Also, is the smallest $k$ for which such trees can be found a known parameter, like maybe ``balanced vertex-disjoint spanning tree distance'' or something? I did not find any information on vertex-disjoint spanning trees like this, I always end up with an article on edge-disjoint spanning trees (which each individually are spanning).

EDIT: vertex-arboricity (resp. equitable vertex-arboricity) seems quite close, but here one aims to partition the vertices such that their induced subgraph is a forest (resp. forests of the same size). They don't seem interested in the exact number of partitions (whereas I fix it to two) and I'm not interested in induced subgraphs or forests.

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    $\begingroup$ I wonder if there are any $2$-connected graphs that do not admit balanced spanning trees for $k\leq 1$. These graphs seem to have a lot of freedom in choosing the pair of trees, although I'm not sure if it's enough. If all $2$-connected graphs have a balanced pair of spanning trees, then it seems likely there is a polynomial time algorithm. If not, I think an example may give some insight into why the problem may be hard. $\endgroup$
    – Discrete lizard
    Commented Feb 9, 2023 at 20:38
  • $\begingroup$ @Discretelizard that got me thinking! In the answer's cited paper, a result of Gyori and Lovasz (Theorem 1.1) implies a 2-connected graph always has such a balanced pair of spanning trees. I suppose this implies that it is only hard to decide if 1-connected graphs admit such trees, huh. $\endgroup$
    – J. Schmidt
    Commented Feb 10, 2023 at 18:11

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For $k = 0$, this problem is equivalent to the problem known as Balanced Connected Vertex 2-Partition problem: given a graph $G = (V,E)$, find a partition of the vertex set $V = V_1 \cup V_2$ such that $|V_1| = |V_2| = |V|/2$ and the induced subgraphs $G[V_1]$ and $G[V_2]$ are connected. Note that if an induced subgraph $G[V_i]$ is connected then we can take a spanning tree of it as $T_i$.

This problem is NP-hard even for bipartite graphs [1 Theorem 2.2]. For the approximation version of this problem, see this question on TCS.SE Partition a graph into 2 connected subgraphs.

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  • $\begingroup$ Nice observation about my condition of the subgraphs needing to be trees can actually be relaxed into just needing to be connected. $\endgroup$
    – J. Schmidt
    Commented Feb 10, 2023 at 9:50

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