S2T2

Jim Randell 12:11 pm on 2 August 2020 Permalink Reply
Tags: by: J. D. Robson   

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  Jim Randell 12:12 pm on 2 August 2020 Permalink | Reply

  We can use the [[ SubstitutedExpression() ]] solver from the enigma.py library to solve this puzzle.

  The following Python program runs in 55ms.

  Run: [ @repl.it ]

  from enigma import (SubstitutedExpression, substitute, map2str, printf)
  
  # the 5 expressions
  exprs = [
    "{8} + {7} = {62}",
    "{59} + {4} = {50}",
    "{5} + {3} = {5}",
    "{11} * {1} = {55}",
    "{12} + {8} = {23}"
  ]
  
  # and the final expression
  expr = "{751} * {7}"
  
  # solve the expressions using the alphametic solver
  p = SubstitutedExpression(exprs, answer=expr, verbose=0)
  for (s, ans) in p.solve():
    # and convert the answer back into symbols
    r = dict((str(v), int(k)) for (k, v) in s.items())
    x = substitute(r, str(ans))
    # output solution
    printf("{expr} -> {texpr} = {ans} <- {{{x}}} / {s}", texpr=p.substitute(s, expr), s=map2str(s, sep=" ", enc=""))
  

  Solution: Tom’s written answer would be 0622.

  The map from Tom’s digits to standard digits is:

  0→7
  1→3
  2→4
  3→0
  4→2 (or 5)
  5→9
  6→1
  7→8
  8→6
  9→5 (or 2)

  Tom uses 4 and 9 for the standard digits 2 and 5, but we can’t determine which way round.

  To work out the answer, Tom would interpret the sum: 751 × 7, as: 893 × 8, which gives a result of 7144, but Tom would write this down as: 0622.

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• 

  GeoffR 8:38 pm on 2 August 2020 Permalink | Reply

  
  # Equations
  # use a + b = cd 
  # for 8 + 7 = 62
  
  # use ef + g = eh
  # for 59 + 4 = 50
  
  # use e + j = e
  # for 5 + 3 = 5
  
  # use ii * i = ee 
  # for 11 * 1 = 55
  
  # use id + a = dj
  # for 12 + 8 = 23
  
  # use bei * b = 751 * 7 for Tom's result
  
  from itertools import permutations
  
  digits = set(range(10))
  
  # 1st stage permutation
  for p1 in permutations (digits, 4):                       
    a, b, c, d = p1
    if a + b != 10*c + d: continue
  
    # 2nd stage permutation
    qs = digits.difference(p1)
    for p2 in permutations(qs, 4):
      e, f, g, h = p2
      if e == 0: continue
      if 10*e + f + g != 10*e + h: continue
  
      # 3rd stage permutation
      qs2 = qs.difference(p2)
      for p3 in permutations(qs2):
        i, j = p3
        
        if e + j != e: continue
        if 11 * i * i != 11 * e: continue
        if 10*i + d + a == 10*d + j:
            # Tom's answer
            res =(100*b + 10*e + i) * b
            print(f"Tom's answer = {res}")
            
            print(a,b,c,d,e,f,g,h,i,j)
            #     6 8 1 4 9 5 2 7 3 0
  
            # Tom's answer is 7144 = hcdd in solution,
            # which is 0622 with variables used in original equations
       
            print(f"Equation 1: {a} + {b} = {10*c+d}")
            print(f"Equation 2: {10*e+f} + {g} = {10*e+h}")
            print(f"Equation 3: {e} + {j} = {e}")
            print(f"Equation 4: {11*i} * {i} = {11*e}")
            print(f"Equation 5: {10*i+d} + {a} = {10*d+j}")
            print()
            
  # Actual Equations
  # Equation 1: 6 + 8 = 14
  # Equation 2: 92 + 5 = 97 or 95 + 2 = 97
  # Equation 3: 9 + 0 = 9
  # Equation 4: 33 * 3 = 99
  # Equation 5: 34 + 6 = 40
  
  
  
  
  

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