S2T2

Jim Randell 9:15 pm on 30 October 2019 Permalink Reply
Tags: by: Victor Bryant ( 21 )   

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  Jim Randell 9:17 pm on 30 October 2019 Permalink | Reply

  The following recursive Python 3 program finds two sets of three letters that may be combined to form a “common three-letter” name. But only one of them really works.

  It runs in 433ms.

  Run: [ @repl.it ]

  from enigma import irange, factor, join, update, subsets, cached, printf
  
  factor = cached(factor)
  
  # membership numbers are allocated such that membership numbers share
  # factors exactly when the corresponding members names share letters
  
  # number for the mystery man
  xxx = 84
  
  # check if <name>, <number> can be added to members in dict <d>
  def check(name, number, d):
    ls = set(name)
    fs = set(factor(number))
    return all(bool(ls.intersection(k)) == bool(fs.intersection(factor(v))) for (k, v) in d.items())
  
  # allocate numbers from <numbers> to the names in <names>
  # (the assignment is stored in dictionary <d>)
  def allocate(names, numbers, d):
    # are we done?
    if not names:
      yield d
    else:
      # consider the next name
      for (i, n) in enumerate(numbers):
        # check names share letters only when they share factors
        if check(names[0], n, d):
          yield from allocate(names[1:], numbers[:i] + numbers[i + 1:], update(d, [(names[0], n)]))
  
  # choose a number for Bec (the highest number)
  for b in irange(85, 100):
  
    # remaining allowable numbers
    ns = list(irange(70, b - 1))
    ns.remove(xxx)
  
    # allocate numbers for the remaining names
    for d in allocate(['cal', 'doc', 'ian', 'jon', 'luv', 'rev'], ns, { 'bec': b }):
  
      # output the allocations
      printf("{d}", d=join((join((k, '=', v)) for (k, v) in sorted(d.items(), key=lambda x: x[::-1])), sep=' '))
  
      # find (up to) three letters that form xxx's name
      for ls in subsets(set(join(d.keys())), min_size=1, max_size=3):
        if check(ls, xxx, d):
          ls = join(sorted(ls)).ljust(3, '?')
          names = set(join(s) for s in subsets(ls, size=len, select="mP"))
          printf("letters = {ls} -> names = {names}", names=join(sorted(names), sep=', '))
  

  Solution: (a) Ian’s membership number is 76. (b) The eighth friend is named Vic.

  The solutions found by the program are “civ” and “acv”. The only rearrangement that gives a common three-letter name is “Vic” (which as the setter is Victor Bryant, is probably what is wanted as the answer). Although one could argue that “Cav” or “Vac” are no worse than the names used for other members.

  So the membership numbers are:

  Doc = 75
  Ian = 76
  Rev = 77
  Cal = 78
  Vic = 84
  Luv = 91
  Jon = 95
  Bec = 99

  Shared letters and factors are (each pair only shown once):

  Doc with: Cal (C, 3); Vic (C, 3); Jon (O, 5); Bec (C, 3)
  Ian with: Cal (A, 2); Vic (I, 4); Jon (N, 19)
  Rev with: Vic (V, 7); Luv (V, 7); Bec (E, 11)
  Cal with: Vic (C, 6); Luv (L, 13); Bec (C, 3)
  Vic with: Luv (V, 7); Bec (C, 3)

  However if we were to use “Vac” (or “Cav”) instead of Vic, we would have the following pairings for Vac:

  Vac with: Doc (C, 3); Ian (A, 4); Rev (V, 7); Cal (AC, 6); Luv (V, 7); Bec (C, 3)

  Vac and Cal give us the only pairing that shares more than 1 letter, so if this had been disallowed we would know for certain that the three letters that make up the name of the eighth friend were “civ”.

  ---

  This posting completes my notes for all Sunday Times Teaser puzzles that I have previously posted to repl.it [ @repl.it ].

  I do however have more code for puzzles that I didn’t post, so there may well be more to come.

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