1) C & MI cross
2) MI returns
3) C & C cross (now all cannibals are across, and all MI at square 1)
4) C returns
5) MI & MI cross
6) C & MI return
7) MI & MI cross (now all MI are across, and all but 1 C at square 1)
8) The 1 cannibal who is across will make 2 ferry trips to bring his or her companions across one at a time.
My closing comment in my first solution attempt at least was correct.