S2T2

Jim Randell 10:26 am on 21 March 2024 Permalink Reply
Tags: by: Pete Bradley ( 7 )   

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  Jim Randell 10:27 am on 21 March 2024 Permalink | Reply

  Using the [[ SubstitutedExpression ]] solver from the enigma.py library.

  The following run file executes in 76ms. (Internal runtime of the of the generated program is 85µs).

  Run: [ @replit ]

  #! python3 -m enigma -rr
  
  SubstitutedExpression.split_sum
  
  # the alphametic sum
  "DEAR + READER + HAPPY = EASTER"
  
  # the total is odd
  --extra="R % 2 = 1"
  
  --answer="EGGS"
  

  Solution: EGGS = 4882.

  The sum is: 1403 + 340143 + 60997 = 402543.

  Which assigns 9 of the ten digits, leaving G = 8.

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  GeoffR 7:03 pm on 21 March 2024 Permalink | Reply

  from itertools import permutations
  digits = set('0123456789')
  
  for R in (1,3,5,7,9):
      r = str(R)
      q1 = digits - {r}
      for p1 in permutations(q1, 3):
          d, e, a = p1
          dear = int(d + e + a + r)
          reader = int(r + e + a + d + e + r)
          q2 = q1 - set(p1)
          for p2 in permutations(q2, 3):
              h, p, y = p2
              happy = int(h + a+ p + p + y)
              EASTER = dear + reader + happy
              easter = str(EASTER)
              if easter[0] == e and easter[4] == e:
                  if easter[1] == a and easter[-1] == r:
                     s, t = easter[2], easter[3]
                     used = set((e, a, s, t, r, d, h, p, y))
                     if len(used) == 9:
                         g = digits - used
                         G = int(g.pop())
                         E, S = int(e), int(s)
                         EGGS = 1000*E + 110*G +  S
                         print(f"EGGS = {EGGS}")
                         print(f"Sum: {dear} + {reader} + {happy} = {EASTER}")
  # EGGS = 4882           
  # Sum: 1403 + 340143 + 60997 = 402543
  

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• 

  Frits 11:11 am on 22 March 2024 Permalink | Reply

  from functools import reduce
  
  # convert digits sequence to number
  d2n = lambda s: reduce(lambda x,  y: 10 * x + y, s)
  
  dgts = set(range(10))
  
  # R is odd as total is odd but not 5 (Y becomes 5) or 9 (E becomes 10)
  for R in [1, 3, 7]:
    if len(q1 := dgts - {R, E := R + 1, Y := 10 - R}) != 7: continue
    for A in q1:
      # carry + E + H = 10 + A --> A < carry + E
      if not (A < 2 + E): continue
     
      if len(q2 := q1 - {A, P := 9 - A}) != 5: continue
      APPY = d2n([A, P, P, Y])
      for D in q2:
        if D == 0: continue
        DEAR = d2n([D, E, A, R])
        READER = d2n([R, E, A, D, E, R])
     
        # sum last 4 columns
        carry, tot = divmod(DEAR + READER % 10000 + APPY, 10000)
        # carry + E + H = 10 + A
        H = 10 + A - carry - E
        if H in {0, D} or H not in q2: continue
       
        # tot must be equal to STER
        S, T = divmod(tot // 100, 10)
     
        if len(q3 := q2 - {D, H, S, T}) != 1: continue
       
        HAPPY = 10000 * H + APPY
        EASTER = d2n([E, A, S, T, E, R])
        if DEAR + READER + HAPPY != EASTER: continue
  
        print(f"answer: EGGS = {d2n([E, (G := q3.pop()), G, S])}")
        '''
        print()
        print(f"{DEAR:>6}")
        print(f"{READER:>6}")
        print(f"{HAPPY:>6}")
        print("------")
        print(f"{EASTER:>6}")
        '''
  

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• 

  Ruud 6:42 am on 21 April 2024 Permalink | Reply

  Brute force, extremely simple, solution. Runs within 30 seconds …

  import itertools
  from istr import istr
  
  for d, e, a, r, h, p, y, s, t,g in istr(itertools.permutations("0123456789", 10)):
      if r.is_odd() and (d | e | a | r) + (r | e | a | d | e | r) + (h | a | p | p | y) == (e | a | s | t | e | r):
          print('eggs = ', e|g|g|s)
  

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  • 

    Frits 10:55 am on 22 April 2024 Permalink | Reply

    Hi Ruud, with CPython your version runs for 140seconds on my PC. A similar program without the istr package runs for 3 seconds with CPython (PyPy is faster).

    from itertools import permutations
     
    for d, e, a, r, h, p, y, s, t, g in permutations("0123456789"):
      if (r in "13579" and "0" not in (d + r + h + e) and
          int(d+e+a+r) + int(r+e+a+d+e+r) + int(h+a+p+p+y) == int(e+a+s+t+e+r)):
        print('eggs = ', e+g+g+s)     
    

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• 

  GeoffR 11:10 am on 21 April 2024 Permalink | Reply

  Using primary school addition method of columns and carry digits.

  
  % A Solution in MiniZinc
  include "globals.mzn";
  
  var 1..9:D; var 1..9:E; var 0..9:A; var 1..9:R;
  var 0..9:S; var 0..9:T; var 0..9:G; var 1..9:H;
  var 0..9:P; var 0..9:Y;
  var 1000..9999:EGGS = 1000*E + 110*G + S;
  
  constraint R in {1,3,5,7,9};
  
  % Column carry digits from right hand end
  var 0..2:c1; var 0..2:c2; var 0..2:c3; var 0..2:c4; var 0..2:c5; 
  
  constraint all_different ([D, E, A, R, S, T, G, H, P, Y]);
  
  % Adding columns from the right hand end
  constraint (R + R + Y) mod 10 == R /\ c1 == (R + R + Y) div 10;
  constraint (c1 + A + E + P) mod 10 == E /\ c2 == (c1 + A + E + P) div 10;
  constraint (c2 + E + D + P) mod 10 == T /\ c3 ==  (c2 + E + D + P) div 10;
  constraint (c3 + D + A + A) mod 10 == S /\ c4 ==  (c3 + D + A + A) div 10;
  constraint (c4 + E + H) mod 10 == A /\ c5 == (c4 + E + H) div 10;
  constraint E == R + c5;
  
  solve satisfy;
  output ["EGGS = " ++ show(EGGS) ++ "\n" 
  ++ "([D, E, A, R, S, T, G, H, P, Y] = "  ++ show([D, E, A, R, S, T, G, H, P, Y]  )];
   
  % EGGS = 4882
  % ([D, E, A, R, S, T, G, H, P, Y] = 
  %  [1, 4, 0, 3, 2, 5, 8, 6, 9, 7]
  % ----------
  % ==========
  
  

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