S2T2

Jim Randell 8:40 am on 12 September 2025 Permalink Reply
Tags: by: Graham Smithers ( 82 )   

• 

  Jim Randell 8:41 am on 12 September 2025 Permalink | Reply

  This is another puzzle that can be solved directly with the [[ grouping ]] solver from the enigma.py library.

  The following Python program runs in 66ms. (Internal runtime is 1.4ms).

  from enigma import grouping
  
  names = ["Bubba", "Chris", "Derek", "Emiren", "Finsan", "Gilson"]
  foods = ["apple", "carrot", "pear", "pinto bean", "raspberry", "watermelon"]
  months = [
    "January", "February", "March", "April", "May", "June", "July",
    "August", "September", "October", "November", "December",
  ]
  
  grouping.solve([names, foods, months], grouping.share_letters(1))
  

  Solution: The favourite foods are: watermelon, pear, pinto bean, carrot, apple, raspberry.

  The complete solution is:

  Bubba, watermelon, July
  Chris, pear, August
  Derek, pinto bean, March
  Emiren, carrot, May
  Finsan, apple, November
  Gilson, raspberry, June

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• 

  Frits 2:30 pm on 12 September 2025 Permalink | Reply

  names = ["Bubba", "Chris", "Derek", "Emiren", "Finsan", "Gilson"]
  ns = [x.lower() for x in names] 
  foods = ["apple", "carrot", "pear", "pinto bean", "raspberry", "watermelon"]
  fs = [x.strip() for x in foods]
  months = [
    "January", "February", "March", "April", "May", "June", "July",
    "August", "September", "October", "November", "December",
  ]
  ms = [x.lower() for x in months] 
   
  # does sequence <x> share exactly one item with all sequences in <g>
  share1 = lambda x, g: all(len(set(x).intersection(z)) == 1 for z in g)
  
  # pick one value from each entry of a <k>-dimensional dictionary <dct>
  # so that all elements in the <k> values are different
  def pickOneFromEach(k, dct, seen=set(), ss=[]):
    if k == 0:
       yield ss
    else:
      for vs in dct[k - 1]:
        # values in vs may not have been used before
        if seen.isdisjoint(vs):
          yield from pickOneFromEach(k - 1, dct, seen | set(vs), ss + [vs])
  
  # dictionary of food, month per name index
  d = {i: [(fo, mo) for fo in fs if share1(fo, [na]) for mo in ms
           if share1(mo, [na, fo])] for i, na in enumerate(ns)}
  
  for s in pickOneFromEach(len(names), d):
    for i, (f, m) in enumerate(s[::-1]):
      print(f"{names[i]}: {foods[fs.index(f)]}, {months[ms.index(m)]}")
    print()  
  

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