Here is an e-mail I received from Dave Stigant before adding the last two clues.
I think there's another solution with 2 cycles instead of 1: Bruce loves Mary loves Bruce (By my reading this fulfills the requirement that Bruce loves the woman (Mary) who loves the man (Bruce) who loves Mary) Arthur loves Hazel loves Chad loves Ellen loves David loves Gloria
Incidently, assuming everybody is heterosexual there are no other 2 cycle solutions and no 3 cycle solutions. Arthur cannot love Ellen: either Bruce loves Mary (in which case you must have Hazel loves x loves Gloria) or not (in which case either H loves x loves G or G loves B). Chad and David must be in the same cycle, so the only way to have 3 cycles is A loves E and B loves M and the only way to have 2 cycles is if either A loves E or B loves M.
If you allow for homosexuality, we know that Chad, Arthur and Bruce must be straight. We know that Hazel is straight. We know that at least 2 women are straight (A loves girl, Gx, who loves the man who loves E... and B loves the girl, Gy, who loves the man who loves M. Gx and Gy must be straight but not the same since each is loved by a different man) Also, the number of gay men must be the same as the number of gay women (the number of straight men + the number of gay women = the number of women = number of gay men + the number of straight women = the number of men = 4). So if anybody is gay it must be David and one woman other than Hazel.
This is not an exhaustive list, but here are a few possible solutions:
This problem appears in Games for the Superintelligent by James F. Fixx (page 54).
Michael Shackleford, ASA - November 29, 2000
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