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Jim Randell 11:17 am on 1 November 2020 Permalink Reply
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  Jim Randell 11:18 am on 1 November 2020 Permalink | Reply

  I assume the missing condition means that it is not possible for an 8th member of the family to have the same “day of week” and “day of month” birthday, but not share a birthday month with one of the group of 7.

  The current calendar repeats itself every 400 years, so this program looks for sets of dates in a 400 years span that share the same “day of week” and “day of month” values, but that only involve 7 different months. (So that any further additions to the family could not be born of the same day of the week and day of the month, but a month that has not yet been used. (The condition that was missing when the puzzle was originally published)).

  It then looks for dates that satisfy the required differences, and checks the all use different months.

  It runs in 199ms.

  Run: [ @repl.it ]

  import datetime
  from enigma import group, catch, printf
  
  # generate dates between years a and b
  def dates(a, b):
    d = datetime.date(a, 1, 1)
    i = datetime.timedelta(days=1)
    while d.year < b:
      yield d
      d += i
  
  # the date n months earlier than d
  def earlier(d, n):
    (yy, mm, dd) = (d.year, d.month, d.day)
    (y, m) = divmod(mm - n - 1, 12)
    return catch(datetime.date, yy + y, m + 1, dd)
  
  # group 400 years of dates by (<day of week>, <day of month>)
  d = group(dates(1850, 2250), by=(lambda d: (d.weekday(), d.day)))
  
  # now look for keys that involve exactly 7 different months
  for ((dow, dom), vs) in d.items():
    ms = set(d.month for d in vs)
    if len(ms) != 7: continue
  
    # collect possible birthdates for the granddaughter
    (gda, gdb) = (datetime.date(1945, 5, 15), datetime.date(1946, 5, 14))
    for gdd in vs:
      if gdd < gda: continue
      if not(gdd < gdb): break
  
      # find birthdates for the grandsons (earlier than gd)
      for gsd2 in vs:
        if not(gsd2 < gdd): break
        gsd1 = earlier(gsd2, 9)
        if gsd1 is None or gsd1 not in vs: continue
  
        # find birthdates for the son and daughter (earlier than gs1 - 15 years)
        dx = gsd1 - datetime.timedelta(days=5479)
        for dd in vs:
          if not(dd < dx): break
          sd = earlier(dd, 18)
          if sd is None or sd not in vs: continue
  
          # find birthdates for the husband and wife (earlier than sd - 15 years)
          wx = sd - datetime.timedelta(days=5479)
          for wd in vs:
            if not(wd < wx): break
            hd = earlier(wd, 41)
            if hd is None or hd not in vs: continue
  
            # check the months are all different
            if len(set(d.month for d in (gdd, gsd2, gsd1, dd, sd, wd, hd))) != 7: continue
  
            printf("{dow} {dom}", dow=["Mon", "Tue", "Wed", "Thu", "Fri", "Sat", "Sun"][dow])
            printf("-> husband = {hd}, wife = {wd}")
            printf("-> son = {sd}, daughter = {dd}")
            printf("-> grandsons = {gsd1}, {gsd2}")
            printf("-> granddaughter = {gdd}")
            printf()
  

  Solution: The birthdates are all Mondays. The full list is (by generation):

  husband = 31st October 1898; wife = 31st March 1902
  son = 31st January 1921; daughter = 31st July 1922
  grandson1 = 31st August 1942; grandson2 = 31st May 1943; granddaughter = 31st December 1945

  The only “day of month” that allows exactly 7 months to be used is the 31st of the month, as there are only 7 months that have 31 days.

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