S2T2

Jim Randell 5:10 pm on 28 February 2020 Permalink Reply
Tags: by: Stephen Hogg ( 36 )   

• 

  Jim Randell 5:31 pm on 28 February 2020 Permalink | Reply

  This Python program runs in 239ms.

  Run: [ @repl.it ]

  from collections import defaultdict
  from enigma import irange, subsets, divisors, printf
  
  # check x and y share a single divisor (not 1, x, y)
  def divisor(x, y):
    ds = set(divisors(x)).intersection(divisors(y)).difference([1, x, y])
    if len(ds) == 1:
      return ds.pop()
  
  # collect sets of 4 numbers by sum
  ns = defaultdict(list)
  
  # consider possible values for the units digits
  for (a, b, c, d) in subsets(irange(0, 9), size=4, select="M"):
    # a cannot be 0
    if a == 0: continue
    # only one of them is odd
    if sum(x % 2 == 1 for x in (a, b, c, d)) != 1: continue
    # construct the numbers
    b += 10
    c += 20
    d += 30
    # each consecutive pair shares a single divisor
    s = list(divisor(*xy) for xy in ((a, b), (b, c), (c, d), (d, a)))
    if None in s: continue
    if len(set(s[:3])) != 3: continue
  
    ns[a + b + c + d].append((a, b, c, d))
  
  # consider possible sums
  for (t, vs) in ns.items():
    # choose two lines
    for (p, q) in subsets(vs, size=2):
      if len(set(p + q)) != 8: continue
      # the largest number is on the top line
      if max(q) > max(p): (p, q) = (q, p)
      printf("{p} / {q}")
  

  Solution: The numbers on the bottom line are: 6, 15, 20, 34.

  The numbers on the top line are: 4, 14, 21, 36.

  These are the only two candidate lines that have the same sum (they both sum to 75).

  There are three other candidate lines, but they all have different sums.

  The full list of candidate lines is:

  sum=69: (4, 14, 21, 30)
  sum=73: (8, 14, 21, 30)
  sum=75: (4, 14, 21, 36) (6, 15, 20, 34)
  sum=79: (6, 15, 20, 38)

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• 

  Frits 12:26 pm on 24 December 2020 Permalink | Reply

  from enigma import SubstitutedExpression
  
  # check if four combinations of 2 digits all have one common factor
  def one_common_factor(s):
    # build list where nth and (n+1)th entry of <s> have only one common factor
    li = [a[0] for a in list(factors(x, y) for x, y in zip(s, s[1:] + s[:1])) 
          if len(a) == 1]
    # first three numbers in a list of four must be different
    return len(li) == 4 and len(set(li[:3])) == 3
    
  # return common factors of <x> and <y> except 1 and numbers it self  
  factors = lambda x, y: [i for i in range(2, min(x, y) // 2 + 1) 
                          if x % i == 0 and y % i == 0]
   
  # the alphametic puzzle
  p = SubstitutedExpression(
    [
    # top line      A, 10 + D, 20 + F, 30 + H
    # bottom line   I, 10 + K, 20 + M, 30 + O
     
    # the sum of the four numbers equals that sum for the other line
    "A + D + F + H == I + K + M + O",
    
    # just one number per line is odd
    "sum(x % 2 == 1 for x in (A, D, F, H)) == 1",
    "sum(x % 2 == 1 for x in (I, K, M, O)) == 1",
    
    # my top line includes the largest of the eight numbers
    "H > O",
    
    # the 1st and 2nd numbers share just one factor as do the 2nd and 3rd 
    # (a different factor) and the 3rd and 4th (another different factor) 
    # and, finally, the 4th and 1st.
    "one_common_factor([A, 10 + D, 20 + F, 30 + H])",
    "one_common_factor([I, 10 + K, 20 + M, 30 + O])",
    
    # eight different numbers
    "A != I",
    "D != K",
    "F != M",
    "H != O",
    ],
    answer="(A, 10 + D, 20 + F, 30 + H), (I, 10 + K, 20 + M, 30 + O)",
    # exclude single digit prime numbers
    d2i=dict([(k, "AI") for k in range(0, 4)] + [(5, "AI")] + [(7, "AI")]),
    env=dict(one_common_factor=one_common_factor),
    distinct="",
    verbose=0)   
  
  # print answers 
  for (_, ans) in p.solve():
    print(f"{ans}")
    
  # ((4, 14, 21, 36), (6, 15, 20, 34))  
  

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