Sunday Times Teaser 2612 – Star Signs

by Graham Smithers

Published: 14 October 2012 (link)

Five friends, Abel, Bob, Jedd, Tim and Will, were investigating birthdays. Abel commented that there was just one letter of the alphabet that occurred in both his star sign and in his month of birth. After further investigation he found that for any pair of his star sign, his month of birth, his name, and the day of the week on which he was born, just one letter of the alphabet occurred in both. Remarkably, the same turned out to be true for each of the friends.

Their names are listed alphabetically. What, respectively, are their star signs?

4 Comments Leave one →

Brian Gladman permalink

Here is my solution:


from __future__ import print_function
from collections import defaultdict as df_dict

days = ( 'monday', 'tuesday', 'wednesday', 'thursday', 'friday',
         'saturday', 'sunday' )

months = ( 'january', 'february', 'march', 'april', 'may', 'june',
           'july', 'august', 'september', 'october', 'november',
           'december' )

signs = ( 'aquarius', 'pisces', 'aries', 'taurus', 'gemini',
          'cancer', 'leo', 'virgo', 'libra', 'scorpio',
          'sagittarius', 'capricorn' )

names = ( 'abel', 'bob', 'jedd', 'tim', 'will' )

# for each name create diictionaries of the days, months and
# star signs that share only one letter with the name - sets
# of letters are stored (except for the star sign because it
# is needed later)

day, star_month = df_dict(list), df_dict(list)
for name in names:
  nset = set(name)

  # names and days having only one letter in common
  for d in days:
    if len(nset & set(d)) == 1:
      day[name] += [set(d)]

  # names and star signs having only one letter in common
  for i, s in enumerate(signs):
    if len(nset & set(s)) == 1:

      # in the above lists the month for each star sign is
      # either in the same position or is one above it
      for m in (set(months[i]), set(months[(i + 1) % 12])):

        # the month shares one letter with the name and one
        # letter with the star sign
        if len(nset & m) == len(set(s) & m) == 1:
          star_month[name] += [(s, m)]

print('Answer: ', end='')
# for each name
for name in names:
  sgns = []
  # for the possible days
  for d in day[name]:
    # and the possible star signs and months
    for s, m in star_month[name]:
      # find those that share only one letter
      if len(d & set(s)) == len(d & m) == 1:
        # save any star signs meeting the criteria
        sgns += [s]

  # exit if the star sign for any name is not unique
  if len(sgns) != 1:
    print('no unique solutions')
    break
  # otherwise print the star sign
  print(sgns[0].capitalize(), end=' ')

from __future__ import print_function

from collections import defaultdict as df_dict

days = ( 'monday', 'tuesday', 'wednesday', 'thursday', 'friday',

'saturday', 'sunday' )

months = ( 'january', 'february', 'march', 'april', 'may', 'june',

'july', 'august', 'september', 'october', 'november',

'december' )

signs = ( 'aquarius', 'pisces', 'aries', 'taurus', 'gemini',

'cancer', 'leo', 'virgo', 'libra', 'scorpio',

'sagittarius', 'capricorn' )

names = ( 'abel', 'bob', 'jedd', 'tim', 'will' )

# for each name create diictionaries of the days, months and

# star signs that share only one letter with the name - sets

# of letters are stored (except for the star sign because it

# is needed later)

day, star_month = df_dict(list), df_dict(list)

for name in names:

nset = set(name)

# names and days having only one letter in common

for d in days:

if len(nset & set(d)) == 1:

day[name] += [set(d)]

# names and star signs having only one letter in common

for i, s in enumerate(signs):

if len(nset & set(s)) == 1:

# in the above lists the month for each star sign is

# either in the same position or is one above it

for m in (set(months[i]), set(months[(i + 1) % 12])):

# the month shares one letter with the name and one

# letter with the star sign

if len(nset & m) == len(set(s) & m) == 1:

star_month[name] += [(s, m)]

print('Answer: ', end='')

# for each name

for name in names:

sgns = []

# for the possible days

for d in day[name]:

# and the possible star signs and months

for s, m in star_month[name]:

# find those that share only one letter

if len(d & set(s)) == len(d & m) == 1:

# save any star signs meeting the criteria

sgns += [s]

# exit if the star sign for any name is not unique

if len(sgns) != 1:

print('no unique solutions')

break

# otherwise print the star sign

print(sgns[0].capitalize(), end=' ')

arthurvause permalink

I went out on the fells yesterday, so a bit late answering, but here is my contribution:


signs =  [ 'aries', 'taurus', 'gemini', 'cancer', 'leo',
           'virgo', 'libra', 'scorpio', 'sagittarius',
           'capricorn', 'aquarius', 'pisces' ]

months = [ 'january', 'february', 'march', 'april', 'may',
           'june', 'july', 'august', 'september', 'october',
           'november', 'december' ]

days =   [ 'monday', 'tuesday', 'wednesday', 'thursday',
           'friday', 'saturday', 'sunday' ]

names =  [ 'abel', 'bob', 'jedd', 'tim', 'will' ]

common_sign_month = [ (signs[i], months[(i + offset) % 12])
    for i in range(12) for offset in [2, 3]
      if len(set(signs[i]) & set(months[(i + offset) % 12])) == 1]

common_sign_month_day_name = [ (n, s, m, d)
    for d in days for n in names for (s, m) in common_sign_month
      if len(set(s) & set(n)) == 1 and len(set(m) & set(n)) == 1 and
         len(set(d) & set(n)) == 1 and len(set(s) & set(d)) == 1 and
         len(set(m) & set(d)) == 1 ]

common_sign_month_day_name.sort()

for x in common_sign_month_day_name:
  print(x)

signs = [ 'aries', 'taurus', 'gemini', 'cancer', 'leo',

'virgo', 'libra', 'scorpio', 'sagittarius',

'capricorn', 'aquarius', 'pisces' ]

months = [ 'january', 'february', 'march', 'april', 'may',

'june', 'july', 'august', 'september', 'october',

'november', 'december' ]

days = [ 'monday', 'tuesday', 'wednesday', 'thursday',

'friday', 'saturday', 'sunday' ]

names = [ 'abel', 'bob', 'jedd', 'tim', 'will' ]

common_sign_month = [ (signs[i], months[(i + offset) % 12])

for i in range(12) for offset in [2, 3]

if len(set(signs[i]) & set(months[(i + offset) % 12])) == 1]

common_sign_month_day_name = [ (n, s, m, d)

for d in days for n in names for (s, m) in common_sign_month

if len(set(s) & set(n)) == 1 and len(set(m) & set(n)) == 1 and

len(set(d) & set(n)) == 1 and len(set(s) & set(d)) == 1 and

len(set(m) & set(d)) == 1 ]

common_sign_month_day_name.sort()

for x in common_sign_month_day_name:

print(x)

Brian Gladman permalink

HI Arthur,

Something was wrong with the layout of your code above so I have taken the liberty of reformatting it. Do let me know if you are unhappy with the revised version..

Reply
- arthurvause permalink
  
  Thanks Brian – looks much better now.
  
  Reply

Sunday Times Teaser 2612 – Star Signs

by Graham Smithers

Published: 14 October 2012 (link)

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