Let $G = (V,E)$ be an undirected graph with $m = |E|$ edges (assume that $m = 3t$ for some $t \in \mathbb{N}$).

Problem: Partition $E$ to $q = \frac{m}{3}$ sets $S_1,S_2,\ldots, S_q \subseteq E$ sets such that for each $i \in \{1,2,\ldots,q\}$ it holds that (1) $|S_i| = 3$ and (2) $S_i$ induces a forest (graph with no cycles) in $G$.

Question: Is the above problem NP-Hard for a general graph $G$?

  • 1
    $\begingroup$ Is your graph simple, or do you allow multiple edges? For simple graphs, you always have such a partition for $m\ge 6$. $\endgroup$
    – domotorp
    Commented Sep 17, 2023 at 4:14
  • $\begingroup$ Even if there exists such a partition, can you find it in polynomial time? additionally, for non-simple graphs, does it become NP-Hard? $\endgroup$
    – John
    Commented Sep 17, 2023 at 6:44
  • $\begingroup$ Please reedit the question to make it more precise. $\endgroup$
    – domotorp
    Commented Sep 17, 2023 at 8:02
  • $\begingroup$ I'm not sure I understand the problem. When you say a set $S_i$ "induces a forest", knowing that $|S_i| = 3$, do you just mean that the three edges of $S_i$ should not form a triangle? $\endgroup$
    – a3nm
    Commented Sep 18, 2023 at 0:21
  • $\begingroup$ Yes, this is what I mean. Assume that there can be repetitions of the edges. $\endgroup$
    – John
    Commented Sep 18, 2023 at 11:35

1 Answer 1


This question boils down to answering to simpler questions.

Can you find a graph with 6 edges where it is impossible to partition the edges?

In a graph of more than 6 edges can you always find a triple of edges that is not a forest?

  • $\begingroup$ Is there a known answer to the question you mentioned? does it imply NP-hardness/polynomial time algorithm for my problem? $\endgroup$
    – John
    Commented Sep 17, 2023 at 6:45

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