Is there an algorithm to cover a partite set of a bipartite graph with disjoint 6-cycles? Does an algorithm exist that finds a vertex-disjoint set of 6-cycles on a bipartite graph that covers one of the partite sets or determines that one does not exist? 
I've found some papers on related subjects, specifically Disjoint Small Cycles in Graphs and Bipartite Graphs (Ma, Gao 2013), which gives conditions for being able to decompose bipartite graphs into disjoint 6-cycles and 6-paths or quadrilaterals, but I haven't found a useful algorithm for checking if a disjoint 6-cycle cover exists. Does anyone know if such an algorithm exists?
Thank you.
 A: Your problem is, I think, NP-complete. Here is a tentative proof of the hardness (feel free to polish it). The reduction is from Partition into Triangles.
Given a graph G=(V,E) let Subd(G) be obtained from G by subdividing each edge exactly once. Note that Subd(G) is bipartite, with its two partite sets corresponding to V and E, respectively. 
Assume G can be partitioned into vertex-disjoint triangles. To any such partition we can associate a set of vertex-disjoint cycles of length six in Subd(G) that cover the partite set V (simply add the vertices that correspond to the edges in these triangles). 
For the converse direction, assume that G has maximum degree at least three (otherwise, G is a disjoint union of cycles and paths, and it is easy to check whether G can be partitioned into vertex-disjoint triangles)
1) Let e = uv be an edge of G and let xe be the corresponding vertex in Subd(G). Then, in any cycle of length six in Subd(G) that contains xe, the two neighbours of xe must be u and v. In particular, suppose a vertex v of G is incident to three edges e1, e2, e3. We cannot have two vertex-disjoint cycles of length six in Subd(G) that contain xe1, xe2, xe3. 
2) The above implies that if there exists a collection of vertex-disjoint cycles of length six that cover one partite set of Subd(G), this partite set must be V (i.e., it cannot be E). Then, it is easy to see that we can associate to such a collection a partition of G into triangles.
Since Partition into Triangles is NP-complete, so is your problem.
