S2T2

Jim Randell 8:25 am on 28 February 2021 Permalink Reply
Tags: by: Herbert Wright   

• 

  Jim Randell 8:26 am on 28 February 2021 Permalink | Reply

  The house number must be 5 or 6 letters, so there is only a limited number of possibilities.

  We can use some handy routines from the enigma.py library to solve this puzzle. The [[ int2words() ]] function converts a number into English, and then we can use the [[ SubstitutedSum() ]] solver to find solutions to the corresponding alphametic sums.

  The following Python program runs in 127ms.

  from enigma import (irange, int2words, translate, SubstitutedSum, printf)
  
  # suppose the house number is n (not 7)
  for n in irange(1, 100):
    if n == 7: continue
    word = translate(int2words(n).upper(), (lambda x: x if x.isalpha() else ''))
    if not (4 < len(word) < 7): continue
  
    # construct the alphametic sum
    for terms in [('ROGER', 'GRAY'), ('ROGER', 'GREY')]:
      p = SubstitutedSum(terms, word)
      for s in p.solve(verbose=0):
        # one of the letters has to be 8
        if not (8 in s.values()): continue
        # output solution
        printf("{p.text} / {t}", t=p.substitute(s, p.text))
  

  Solution: Roger Gray lives at number 12. The sum is: 94729 + 7953 = 102682.

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• 

  Frits 10:43 pm on 1 March 2021 Permalink | Reply

  Only calling SubstitutedExpression once, looking for trouble.

  In order to get a decent run time (1 second with PyPy) I had to use some analysis (like skipping all the words ending on “Y”). Only 4 words (THREE, EIGHT, ELEVEN and TWELVE) remain to be checked.

  from enigma import SubstitutedExpression, int2words, translate
  
  def check(num, txt):        # txt has format (_.....)
    # first letter may not be zero
    if len(str(num)) < len(txt) - 3:
      return False
    # number of different digit must match number of letters
    return len(set(str(num))) == len(set(txt)) - 3
    
  words = [] 
  # suppose the house number is n
  for n in range(1, 100):
    if n == 7: continue
    word = translate(int2words(n).upper(), (lambda x: x if x.isalpha() else ''))
    if not(4 < len(word) < 7): continue
    # units: R + Y can't be equal to Y (else R would have to be zero)
    if word[-1] == "Y": continue
    words.append(word)
  
  lower = "".join([x for x in "abcdefghijklmnopqrstuvwxyz" 
                   if x not in "if else or and not sum check str"])
  bits = lower[:len(words)]
  exprs = ["sum([" + ','.join(bits) + "]) == 1"]
  
  for i, word in enumerate(words):
    if len(word) == 6:
      exprs.append(bits[i] + " == 0 or R == 9")
      exprs.append(bits[i] + " == 0 or " + word[0] + " == 1")
      exprs.append(bits[i] + " == 0 or Y - 1 == " + word[-1])
    else:
      exprs.append(bits[i] + " == 0 or " + word[0] + " - 1 == R")
   
    exprs.append(bits[i] + " == 0 or " + word + " - ROGER in {GRAY, GREY}")
    exprs.append(bits[i] + " == 0 or '8' in str(" + word + ")")
    exprs.append(bits[i] + " == 0 or check(" + word + ", '" + word + "')")
    
  # collect letters not used in words, they can't have value 8  
  outside = "".join([x for x in "ROGEYA" if x not in "".join(words)])
  symbols = "".join(set("ROGEYA" + "".join(words))) + bits
  dist = ["ROGEYA" + c for c in [x for x in set(''.join(words)) 
          if x not in "ROGEYA"]]
  
  # puzzle
  p = SubstitutedExpression(
    exprs,
    symbols=symbols,
    distinct=dist,
    answer="(ROGER + GRAY if '8' in str(ROGER + GRAY) else ROGER + GREY), " + 
           " + ".join([x + ' * ' + str(i) for i, x in enumerate(bits)]),
    env=dict(check=check),
    verbose=0,
    d2i=dict([(0, "RG")] +  
             [(k, bits) for k in range(2, 8)] +
             [(8, outside + bits)] +
             [(9, bits)])
  )
  
  # print answer
  sol = set()
  for (_, ans) in p.solve():
    sol.add(words[ans[1]] + " = " + str(ans[0]))  
  
  print(*sol)
  

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