S2T2

Jim Randell 11:44 am on 22 March 2022 Permalink Reply
Tags: by: Andrew Skidmore ( 85 )   

• 

  Jim Randell 11:45 am on 22 March 2022 Permalink | Reply

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

  The following run file executes in 114ms.

  Run: [ @replit ]

  #! python3 -m enigma -rr
  
  #  # A # #
  #  B C D E
  #  # F # G
  #  H J # K
  
  SubstitutedExpression
  
  # across
  "is_square(BCDE)"
  "is_square(HJ)"
  
  # down
  "is_square(ACFJ)"
  "is_square(EGK)"
  
  # answer
  --answer="ordered(BCDE, HJ, ACFJ, EGK)"
  

  Solution: The squares are: 49, 576, 1089, 3025.

  (49, 576, 1089, 3025) = (7², 24², 33², 55²).

  

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  GeoffR 2:33 pm on 22 March 2022 Permalink | Reply

  % A Solution in MiniZinc
  include "globals.mzn";
  
  %    # A # #
  %    B C D E
  %    # F # G
  %    H J # K
  
  var 1..9:A; var 1..9:B; var 0..9:C; var 0..9:D;
  var 1..9:E; var 0..9:F; var 0..9:G; var 1..9:H;
  var 0..9:J; var 0..9:K;
  
  constraint all_different([A,B,C,D,E,F,G,H,J,K]);
  
  % sets of 2,3 and 4-digit squares
  set of int: sq2 = {n*n | n in 4..9};
  set of int: sq3 = {n*n | n in 10..31};  
  set of int: sq4 = {n*n | n in 32..99};  
  
  constraint (10*H + J) in sq2;
  constraint (100*E +10*G + K) in sq3;
  constraint (1000*A + 100*C + 10*F + J) in sq4 
  /\ (1000*B + 100*C + 10*D + E) in sq4;
  
  solve satisfy;
  
  output [ "[A,B,C,D,E,F,G,H,J,K] = " ++ show([A,B,C,D,E,F,G,H,J,K])];
  % [1, 3, 0, 2, 5, 8, 7, 4, 9, 6]
  % [A, B, C, D, E, F, G, H, J, K]
  % ----------
  % ==========
  % Squares : HJ = 49, EGK = 576, ACFJ = 1089, BCDE = 3025
  
  
  

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• 

  GeoffR 8:10 am on 23 March 2022 Permalink | Reply

  A three stage permutation solution.

  from itertools import permutations
  from enigma import is_square
  
  digits = set('1234567890')
  
  # two digit square HJ
  for p1 in permutations(digits, 2):
    h, j = p1
    if h == '0':continue
    hj = int(h + j)   
    if not is_square(hj):continue
    q1 = digits.difference({h, j})
      
    # three digit square EGK
    for p2 in permutations(q1, 3):
      e, g, k = p2
      if e == '0':continue
      egk = int(e + g + k)
      if not is_square(egk):continue
          
      # two four digit squares BCDE and ACFJ
      q2 = q1.difference({e, g, k})
      for p3 in permutations(q2):
        a, b, c, d, f = p3
        if '0' in (a, b): continue
        bcde = int(b + c + d + e)
        if is_square(bcde):
          acfj = int(a + c + f + j)
          if is_square(acfj):
            print(f"HJ={hj}, EGK={egk}, ACFJ={acfj}, BCDE={bcde}")
                    
  # HJ=49, EGK=576, ACFJ=1089, BCDE=3025
  
  
  

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