S2T2

Jim Randell 7:59 am on 21 December 2023 Permalink Reply
Tags: by: Anthony Isaacs   

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  Jim Randell 8:00 am on 21 December 2023 Permalink | Reply

  The following run file executes in 112ms. (Internal runtime of the generated program is 23ms).

  Run: [ @replit ]

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  #  A B C D
  #  E F G H
  #  I J K L
  #  M N P Q
  
  --digits="1-9"
  --distinct=""
  
  # rows read as squares
  "is_square(ABCD)"
  "is_square(EFGH)"
  "is_square(IJKL)"
  "is_square(MNPQ)"
  
  # columns read as squares
  "is_square(AEIM)"
  "is_square(BFJN)"
  "is_square(CGKP)"
  "is_square(DHLQ)"
  
  # [optional] squares must end in 1, 4, 5, 6, 9
  --invalid="2|3|7|8,DHLMNPQ"
  
  --answer="sum([A, B, C, D, E, F, G, H, I, J, K, L, M, N, P, Q])"
  --template=""
  

  Solution: The sum of the 16 digits is 56.

  The completed grid is as follows:

  

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  GeoffR 6:21 pm on 21 December 2023 Permalink | Reply

  
  % A Solution in MiniZinc
  include "globals.mzn";
  
  % A B C D
  % E F G H
  % I J K L
  % M N P Q
  
  var 1..9:A; var 1..9:B; var 1..9:C; var 1..9:D; var 1..9:E; 
  var 1..9:F; var 1..9:G; var 1..9:H; var 1..9:I; var 1..9:J; 
  var 1..9:K; var 1..9:L; var 1..9:M; var 1..9:N; var 1..9:O; 
  var 1..9:P; var 1..9:Q; 
  
  var 16..108:total; % UB = 2*45 + 18
  set of int: sq4 = {n*n | n in 34..96}; 
  % 97,98,99 have zeroes in squares, as do 32 and 33
  
  % ROWS
  var 1156..9216:ABCD == 1000*A + 100*B + 10*C + D;
  var 1156..9216:EFGH == 1000*E + 100*F + 10*G + H;
  var 1156..9216:IJKL == 1000*I + 100*J + 10*K + L;
  var 1156..9216:MNPQ == 1000*M + 100*N + 10*P + Q;
  % COLUMNS
  var 1156..9216:AEIM == 1000*A + 100*E + 10*I + M;
  var 1156..9216:BFJN == 1000*B + 100*F + 10*J + N;
  var 1156..9216:CGKP == 1000*C + 100*G + 10*K + P;
  var 1156..9216:DHLQ == 1000*D + 100*H + 10*L + Q;
   
  % Last digit of squares is 1,4,5,6 or 9.
  set of int: dig_end = {1,4,5,6,9};
  constraint sum([D in dig_end, H in dig_end, L in dig_end, Q in dig_end]) == 4;
  constraint sum([M in dig_end, N in dig_end, P in dig_end]) == 3;
  
  constraint sum([ABCD in sq4, EFGH in sq4, IJKL in sq4, MNPQ in sq4,
  AEIM in sq4, BFJN in sq4, CGKP in sq4, DHLQ in sq4]) = 8; 
  
  constraint total = A+B+C+D+E+F+G+H+I+J+K+L+M+N+P+Q;
  solve satisfy;
  
  output ["Total of all digits = " ++ show(total) 
  ++ "\n" ++ show(ABCD) ++ "\n" ++ show(EFGH)
  ++ "\n" ++ show(IJKL) ++ "\n" ++ show(MNPQ)];
  
  % Total of all digits = 56
  % 2116
  % 1225
  % 1296
  % 6561
  % ----------
  % ==========
  % Finished in 442msec.
  
  

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• 

  Frits 12:58 pm on 22 December 2023 Permalink | Reply

  Using permutations is a lot slower (1 second with PyPy).

     
  from itertools import permutations
  
  sqs =  {x2 for x in range(32, 100) if '0' not in (x2 := str(x * x))}
  
  for hor in permutations(sqs, 4):
    # check columns, transpose the permutation to get the columns
    if any(''.join(x) not in sqs for x in (zip(*hor))): continue
    for h in hor: print(h)
    print("\nanswer:", sum(int(d) for h in hor for d in h))
  

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