Given a graph $G$, if we can partition the edges into pairwise disjoint subsets of $G$, such that the union of all the subsets is equal to the edgeset of G, then this is a decomposition. If such a partition can be formed from only spanning k-regular graphs, where all the graphs are disjoint, then each of these graphs is a k-factor, and the graph is k-factorizable. If a graph is 2-factorizable, then what conditions are nessasary to make the connected components of each 2-factor the same length?

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spanning2-regular subgraph. So if the whole graph has $n$ vertices, a 2-factor has $n$ vertices and $n$ edges. Perhaps you are just trying to decompose the edge set into cycles? $\endgroup$ – Brendan McKay Mar 5 '16 at 10:08