Jünger, Michael, Gerhard Reinelt, and William R. Pulleyblank. "On partitioning the edges of graphs into connected subgraphs." Journal of graph theory 9.4 (1985): 539-549.

states that for 4-edge-connected graph one can partition its edges into disjoint subsets of size $r$, such that each subset form a connected subgraph.

I wonder if the same kind of statement could be formulated for partition of vertices. For what kind of graphs one find partition of vertices into disjoint subsets of size $\approx r$, such that each subset form a connected subgraph (for each r)? I'm particularly interested in planar graphs, but would be happy with any class.

I can soften some conditions (it will still meet my needs): for what graph classes existence of partition into less than $\frac{\alpha n}{r}$ connected subgraphs of size less than $r$ is guaranteed (for some $\alpha$)?

  • $\begingroup$ Compute planer embedding. Run some kind of sweep line algorithm. Once sweep line hits n/r vertices cut the edges connecting to the rest of the graph. This should work for planer graphs. $\endgroup$ Aug 19, 2014 at 21:40
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    $\begingroup$ Well, you can't do this partition for any planar graph: for example star-graph is planar, but it is impossible to cut it into pieces of size more then 1. $\endgroup$
    – ivmihajlin
    Aug 20, 2014 at 0:57
  • $\begingroup$ I see. This makes the question very interesting. +1 $\endgroup$ Aug 20, 2014 at 1:35

2 Answers 2


Compute a constant degree spanning tree $T$ of your graph, root it, and now greedily find subtrees of roughly size $r$, extract them, and repeat. Naturally, if there is no constant degree spanning tree, then the star example shown above demonstrates that this algorithm can fail.


For a graph $G=(V,E)$, deciding if $V$ can be partitioned into equal sized subsets (say, for a fixed size $r$) where each subset induces a connected subgraph is $\mathsf{NP}$-hard. It remains $\mathsf{NP}$-hard for planar graphs, and also if the number of subsets is fixed instead of the subset size ($|V|/r$ fixed).

However, the problem is polynomial for cycles, and thus for Hamiltonian graphs. It is also polynomial for trees and for series-parallel graphs when the number of subsets is fixed.

You can find these results in:

Edit (bis): I think that I was wrong about Hamiltonian graphs. They may be partitioned even if their Hamiltonian cycle cannot. Thus the problem can be not polynomial. See the first comment.

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    $\begingroup$ Looks like you can do it for Hamiltonian graph. As this graph has a Hamiltonian graph, it has a spanning tree of max degree 2. It is possible to find a spanning tree of max degree 3 in polynomial time and using this tree partition the graph (as stated in other answer). research.microsoft.com/en-us/um/people/mohits/publications/… $\endgroup$
    – ivmihajlin
    Aug 20, 2014 at 21:33

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