S2T2

Jim Randell 9:45 am on 13 December 2020 Permalink Reply
Tags: by: RBJT Allenby ( 4 )   

• 

  Jim Randell 9:45 am on 13 December 2020 Permalink | Reply

  Evidently we need to find some selection from the three types of whatsit, such that from the value of their combined weight we can determine the weights of the individual types.

  If the scale did not have such a low maximum weight we could weigh 100 type A whatsits, 10 type B whatsits, and 1 type C whatsit, to get a reading of ABC, which would tell us immediately the weights of each type. (For example, if the total weight was 768g, we would know A = 7g, B = 6g, C = 8g).

  We first note that if we can determine the weights for two of the types, then the third types weight must be the remaining value. (Equivalently if the selection (a, b, c) determines the weights, so does selection (0, b − a, c − a) where a is the smallest count in the selection).

  So we need to find a pair of small numbers (as 6 or more of the lightest whatsits will give an error on the scales), that give a different outcome for each possible choice of whatsits.

  This Python program runs in 45ms.

  Run: [ @repl.it ]

  from enigma import group, subsets, printf
  
  # display for a weight of x
  def weigh(x):
    return ("??" if x > 30 else str(x))
  
  # check the qunatities <ns> of weights <ws> are viable
  def check(ns, ws):
    # collect total weight
    d = group(subsets(ws, size=len, select="P"), by=(lambda ws: weigh(sum(n * w for (n, w) in zip(ns, ws)))))
    # is this a viable set?
    return (d if all(len(v) < 2 for v in d.values()) else None)
  
  # consider (a, b) pairs, where a + b < 6
  for (a, b) in subsets((1, 2, 3, 4), size=2):
    if a + b > 5: continue
    ns = (0, a, b)
    d = check(ns, (6, 7, 8))
    if d:
      # output solution
      printf("ns = {ns}")
      for k in sorted(d.keys()):
        printf("  {k:2s} -> {v}", v=d[k])
      printf()
  

  Solution: You should choose one whatsit from one of the boxes, and three whatsits from another box.

  The combined weight indicates what the weight of each type of whatsit is:

  25g → (0 = 8g, 1 = 7g, 3 = 6g)
  26g → (0 = 7g, 1 = 8g, 3 = 6g)
  27g → (0 = 8g, 1 = 6g, 3 = 7g)
  29g → (0 = 6g, 1 = 8g, 3 = 7g)
  30g → (0 = 7g, 1 = 6g, 3 = 8g)
  <error> → (0 = 6g, 1 = 7g, 3 = 8g) (actual weight = 31g)

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• 

  Frits 12:01 pm on 13 December 2020 Permalink | Reply

  from enigma import SubstitutedExpression, group
  
  # the alphametic puzzle
  p = SubstitutedExpression(
    [# pick not more than 5 whatsits from one or two boxes
     "min(X, Y, Z) == 0 and sum([X, Y, Z]) < 6",
     
     # don't allow all whatits to be picked from one box only
     "max(X, Y, Z) != sum([X, Y, Z])"
    ],
    answer="(X, Y, Z), X*6 + Y*7 + Z*8",
    d2i=dict([(k, "XYZ") for k in range(5,10)]),
    distinct="",
    verbose=0)   
  
  # collect weighing results
  res = [y for _, y in p.solve()]
  
  # group items based on X,Y,Z combinations    
  grp = group(res, by=(lambda a: "".join(str(x) for x in sorted(a[0]))))
  
  for g, vs in grp.items(): 
    # there should be 6 different weighing results in order to determine 
    # which whatsit is which 
    if (len(vs)) != 6: continue
    
    # group data by sum of weights
    grp2 = group(vs, by=(lambda a: a[1]))
    
    morethan30 = [1 for x in grp2.items() if x[0] > 30]
    if (len(morethan30) > 1): continue 
    
    ambiguous = [1 for x in grp2.items() if len(x[1]) > 1]
    if len(ambiguous) > 0: continue
    
    print("Solution", ", ".join(g), "\nweighings: ", vs)
  

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  • 

    Frits 12:17 pm on 13 December 2020 Permalink | Reply

    I could have used “[X, Y, Z].count(0) == 1” but somehow count() is below my radar (probably because I have read that count() is/was quite expensive).

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