S2T2

Jim Randell 10:46 am on 11 May 2023 Permalink Reply
Tags: by: Peter Morris ( 4 )   

• 

  Jim Randell 10:47 am on 11 May 2023 Permalink | Reply

  See also: Teaser 1688, Teaser 2533.

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

  The following run file executes in 66ms. (Internal runtime of the generated program is 80µs).

  Run: [ @replit ]

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  --digits="1-9"
  
  "T + E = S"
  "E + A = T"
  "A + S = U"
  "S + E = D"
  "E + R = Y"
  
  # TEASER is odd
  "R % 2 = 1"
  
  --answer="SUNDAY"
  

  Solution: SUNDAY = 469528.

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• 

  GeoffR 12:09 pm on 11 May 2023 Permalink | Reply

  
  % A Solution in MiniZinc
  include "globals.mzn";
  
  set of int: INT = 1..9;
  var INT: T; var INT: E; var INT: A;
  var INT: S; var INT: R; var INT: U;
  var INT: D; var INT: Y; var INT: N;
  
  constraint all_different([T, E, A, S, R, U, D, Y, N]);
  
  constraint S == T + E /\ T == E + A /\ U == A + S /\ D == S + E
   /\ Y == E + R /\ R mod 2 == 1;
  
  solve satisfy;
  output ["SUNDAY = " ++  "\(S)\(U)\(N)\(D)\(A)\(Y)" ];
  
  % SUNDAY = 469528
  % ----------
  % ==========
  
  

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• 

  GeoffR 1:47 pm on 11 May 2023 Permalink | Reply

  from itertools import permutations
  
  digits = {1, 2, 3, 4, 5, 6, 7, 8, 9}
  
  for p1 in permutations(range(1, 10), 4):
      T, E, A, S = p1
      if S != T + E:continue
      if T != E + A:continue
      
      # find 5 remaining digits
      q1 = digits.difference(p1)
      for p2 in permutations(q1):
         U, D, R, Y, N = p2
         if U != A + S:continue
         if D != S + E:continue
         if R % 2 != 1:continue  # TEASER is odd
         if Y != E + R:continue
         SUNDAY = 100000*S + 10000*U + 1000*N + 100*D + 10*A + Y
         print('SUNDAY = ', SUNDAY)
  
  
  

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