Solver for that riddle...

solve.py

import itertools

everybody = ['akira', 'shinya', 'hiroshi', 'chiaki', 'saki', 'yuri', 'mika', 'reina']
akira, shinya, hiroshi, chiaki, saki, yuri, mika, reina = everybody

def conds(p):
  return (
    # no self cites
    p[0] not in (shinya, reina, hiroshi, yuri)
    and p[1] not in (mika, chiaki)
    and p[2] not in (mika, )
    and p[4] not in (chiaki, mika)
    #and p[5] not in (akira, saki) # possible self cite
    and p[6] not in (saki, chiaki)
    and p[7] not in (saki, reina)
    # mouth-no_mouth
    and p[7] not in (mika, )
    # glasses
    and chiaki in (p[2], p[5])
    # girls
    and yuri not in (p[1], p[4], p[6])
    and saki not in (p[1], p[4], p[6])
    # girls only
    and p[3] not in (saki, chiaki)
    # boys
    and (akira in (p[1], p[4], p[6]) or akira not in (p[0], p[3]))
    and (reina in (p[1], p[4], p[6]) or reina not in (p[0], p[3]))
    # backdoor only
    and p[1] not in (shinya, akira)
    # straight
    and saki not in (p[0], p[2])
    # derivates => free factors
    and p[6] == akira # possible affection
    and (p[1] == hiroshi and p[3] == yuri)
    )

for p in itertools.ifilter(conds, itertools.permutations(everybody)):
  print p