S2T2

Jim Randell 9:06 am on 5 August 2025 Permalink Reply
Tags: by: Victor Bryant ( 139 )   

• 

  Jim Randell 9:07 am on 5 August 2025 Permalink | Reply

  Here is a solution using the SubstitutedExpression solver from the enigma.py library.

  It runs in 77ms. (Internal runtime of the generated code is 382µs).

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  # [ @a | @b | @c ]
  # [ BR | AI |  N ]
  # [ TE |  A | @d ]
  
  # magic sum is a square number
  --macro="@sum = (3 * AI)"
  "is_square(@sum)"
  
  # missing values are positive integers
  --macro="@a = (@sum - BR - TE)"  # @a + BR + TE == @sum  [col 1]
  --macro="@b = (2 * AI - A)"      # @b + AI + A  == @sum  [col 2]
  --macro="@c = (2 * AI - TE)"     # @c + AI + TE == @sum  [diag 2]
  --macro="@d = (@sum - TE - A)"   # TE +  A + @d == @sum  [row 3]
  "@a > 0" "@b > 0" "@c > 0" "@d > 0"
  
  # remaining rows / cols / diags give the magic sum
  "@a + @b + @c == @sum"  # [row 1]
  "BR + AI + N == @sum"   # [row 2]
  "@c + N + @d == @sum"   # [col 3]
  "@a + AI + @d == @sum"  # [diag 1]
  
  --answer="((@a, @b, @c), (BR, AI, N), (TE, A, @d))"
  --output="lambda p, s, ans: print(ans)"
  

  Solution: The completed square is:

  

  And so the magic constant is 81.

  Which would correspond to: [X, XA, AB], [BR, AI, N], [TE, A, BY] for some unused symbols X and Y.

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• 

  Frits 2:07 pm on 5 August 2025 Permalink | Reply

  Minimising the number of iterations.

  # magic constant M = 3 * AI or AI = 3 * k^2
  for AI in [3 * k * k for k in range(2, 6)]:
    A, I = divmod(AI, 10)
    # A can never be equal to I as AI is not a multiple of 11
    M = 3 * AI
    sum2 = M - AI
    # BR + N = sum2 
    if sum2 > 105: continue
    # BR = A + 2.(c33 - AI) so R and A have the same parity
    # (sum2 - N) has same parity as A so N must have the same parity as A
    for N in set(range(A % 2, 10, 2)) - {A, I}: 
      B, R = divmod(sum2 - N, 10)
      if not (0 < B < 10) or not {A, I, N}.isdisjoint({B, R}): continue
      c33 = ((BR := 10 * B + R) - A + sum2) // 2
      T, E = divmod(M - c33 - A, 10)
      if not (0 < T < 10) or not {A, I, N, B, R}.isdisjoint({T, E}): continue
      c11 = sum2 - c33
      c12 = sum2 - A
      c13 = sum2 - (TE := 10 * T + E)
      
      mat = [(c11, c12, c13), (BR, AI, N), (TE, A, c33)]
      # double check matrix
      if not all(sum(r) == M for r in mat): continue
      if len({sum2, c11 + c33, TE + c13}) > 1: continue
      if not all(sum(r) == M for r in zip(*mat)): continue
      for c1, c2, c3 in mat:
        print(f"|{c1:>2} | {c2:>2} | {c3:>2} |")
  

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• 

  GeoffR 12:22 pm on 6 August 2025 Permalink | Reply

  % A Solution in MiniZinc
  include "globals.mzn";
  
  %   w   x   y
  %   BR  AI  N
  %   TE  A   z
  
  predicate is_sq(var int: y) =
  exists(z in 1..ceil(sqrt(int2float(ub(y))))) (z*z = y );
  
  var 1..9:A; var 1..9:B; var 1..9:T; var 1..9:N; 
  var 0..9:R; var 0..9:I; var 0..9:E;
  var 10..99:BR; var 10..99:AI; var 10..99:TE;
  var 1..99:w; var 1..99:x; var 1..99:y; var 1..99:z;
  var 20..210:m_const;
  
  constraint all_different ([B, R, A, I, N, T, E]);
  constraint BR = 10*B + R /\ AI = 10*A + I /\ TE = 10*T + E;
  
  % All rows are squares
  constraint is_sq(BR + AI + N) /\ is_sq(TE + A + z) /\ is_sq(w + x + y);
  
  constraint m_const  == BR + AI + N;
  
  % Check conditions for a magic square
  constraint m_const == w + x + y /\ m_const == TE + A + z
  /\ m_const == w + BR + TE /\ m_const == x + AI + A /\ m_const == y + N + z
  /\ m_const == w + AI + z /\ m_const == y + AI + TE;
  
  solve satisfy;
  
  output[ show([w, x, y]) ++ "\n" ++ show([BR, AI, N]) ++ "\n" ++ show([TE, A, z]) ];
  
  % [5, 52, 24]
  % [46, 27, 8]
  % [30, 2, 49]
  % ----------
  % ==========
  
  
  

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