Denise’s birthday problem solution (in Python)
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# -*- coding: utf-8 -*- | |
from collections import defaultdict | |
# (dd, mmm, yyyy) | |
dates = [ | |
(17, "Feb", 2001), | |
(13, "Mar", 2001), | |
(13, "Apr", 2001), | |
(15, "May", 2001), | |
(17, "Jun", 2001), | |
(16, "Mar", 2002), | |
(15, "Apr", 2002), | |
(14, "May", 2002), | |
(12, "Jun", 2002), | |
(16, "Aug", 2002), | |
(13, "Jan", 2003), | |
(16, "Feb", 2003), | |
(14, "Mar", 2003), | |
(11, "Apr", 2003), | |
(16, "Jul", 2003), | |
(19, "Jan", 2004), | |
(18, "Feb", 2004), | |
(19, "May", 2004), | |
(14, "Jul", 2004), | |
(18, "Aug", 2004), | |
] | |
# Albert = mmm | |
# Bernard = dd | |
# Cheryl = yyyy | |
dd = 0 | |
mmm = 1 | |
yyyy = 2 | |
def my_filter(dates, what_i_know, what_other_knows): | |
# count repetitions | |
other_counter = defaultdict(int) | |
for d in dates: | |
other_counter[d[what_other_knows]] += 1 | |
# group | |
others_by_me = defaultdict(list) | |
for d in dates: | |
others_by_me[d[what_i_know]].append(d[what_other_knows]) | |
possible = [] | |
for k, v in others_by_me.iteritems(): | |
if len(filter(lambda x: other_counter[x] > 1, v)) == len(set(v)): | |
possible.append(k) | |
return filter(lambda x: x[what_i_know] in possible, dates) | |
# Albert: I don’t know when Denise’s birthday is, but I know that Bernard does not know. | |
dates = my_filter(dates, mmm, dd) | |
# Bernard: I still don’t know when Denise’s birthday is, but I know that Cheryl still does not know. | |
dates = my_filter(dates, dd, yyyy) | |
# Cheryl: I still don’t know when Denise’s birthday is, but I know that Albert still does not know. | |
dates = my_filter(dates, yyyy, mmm) | |
print dates |
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