S2T2

Jim Randell 9:18 am on 30 November 2023 Permalink Reply
Tags: by: Angela Newing ( 37 )   

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  Jim Randell 9:19 am on 30 November 2023 Permalink | Reply

  If we consider the four rows that make up the square, each row sums to the magic constant k, and between all 4 rows each number is included exactly once. So:

  4k = sum(1 .. 16)
  ⇒ k = 34

  We can now use the [[ SubstitutedExpression ]] solver from the enigma.py library to fill out the remaining squares.

  The following run file executes in 96ms. (Internal runtime of the generated program is 853µs).

  Run: [ @replit ]

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  #  A B C D
  #  E F G H
  #  I J K L
  #  M N P Q
  
  # digits 1-16 are used:
  --base="17"
  --digits="1-16"
  
  # rows
  "A + B + C + D == 34"
  "E + F + G + H == 34"
  "I + J + K + L == 34"
  "M + N + P + Q == 34"
  
  # columns
  "A + E + I + M == 34"
  "B + F + J + N == 34"
  "C + G + K + P == 34"
  "D + H + L + Q == 34"
  
  # diagonals
  "A + F + K + Q == 34"
  "D + G + J + M == 34"
  
  # short diagonals
  "B + G + L < 34"
  "C + F + I < 34"
  "E + J + P < 34"
  "H + K + N < 34"
  
  # given values
  "A == 9"
  "J == 13"
  "N == 12"
  "Q == 14"
  
  --template=""
  --verbose="AT"
  

  Solution: Here is the completed square:

  

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  NigelR 6:22 pm on 30 November 2023 Permalink | Reply

  Hi Jim
  I agree your solution, but I’m not sure how you reached it with your short diagonal condition of
  “E + J + P < 34" (which isn't satisfied by the solution). Shouldn't it be E+J+O?

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  • 

    NigelR 6:25 pm on 30 November 2023 Permalink | Reply

    Apologies: Just realised that your notation didn’t include ‘O’!!

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  • 

    Jim Randell 7:55 pm on 30 November 2023 Permalink | Reply

    @NigelR: I often skip O as a variable, as it is easily confused with 0 (zero).

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  GeoffR 7:02 pm on 1 December 2023 Permalink | Reply

  #  A B C D
  #  E F G H
  #  I J K L
  #  M N P Q
  
  from itertools import product, permutations
  from enigma import all_different
  
  # Given grid values
  A, J, N, Q = 9, 13, 12, 14
  
  # Find remaining 12 digits
  digits = set(range(1,17)).difference({9, 13, 12, 14 }) 
  
  # 1st row of grid
  for B, C, D in product(digits, repeat=3):
    if not all_different(A, B, C, D):continue
    if A + B + C + D != 34:continue
    R1 = list(digits.difference({B, C, D}))
    
    # 2nd row of grid
    for E, F, G, H in product(R1, repeat=4):
      if not all_different(E, F, G, H):continue
      if E + F + G + H != 34:continue
      R2 = set(R1).difference({E, F, G, H})
    
      # 3rd row of grid
      for I, K, L in product(R2, repeat=3):
        if not all_different(I, J, K, L):continue
        if I + J + K + L != 34:continue
        if B + G + L >= 34:continue
        if C + F + I >= 34:continue
        
        # only M and P are missing from the 4th row of grid
        R3 = list(set(R2).difference({I, K, L}))
        for M, P in permutations(R3, 2):
          # columns
          if A + E + I + M != 34:continue
          if B + F + J + N != 34:continue
          if C + G + K + P != 34:continue
          if D + H + L + Q != 34:continue
          # diagonals
          if E + J + P >= 34:continue
          if H + K + N >= 34:continue
          if A + F + K + Q != 34:continue
          if D + J + G + M != 34:continue
          
          print('A, B, C, D = ', A,B,C,D)
          print('E, F, G, H = ', E,F,G,H)
          print('I, J, K, L = ', I,J,K,L)
          print('M, N, P, Q = ', M,N,P,Q)
          exit() # only 1 solution needed
  
  # A, B, C, D =  9 6 15 4
  # E, F, G, H =  16 3 10 5
  # I, J, K, L =  2 13 8 11
  # M, N, P, Q =  7 12 1 14
  
  
  
  

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  Hugo 10:01 am on 2 December 2023 Permalink | Reply

  I found a second solution. Did I do something wrong?

  9 2 15 8
  6 7 10 11
  16 13 4 1
  3 12 5 14

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  • 

    Jim Randell 11:03 am on 2 December 2023 Permalink | Reply

    Yes, the shorter diagonals have to sum less than 34.

    You have 16 + 7 + 15 = 38 in your diagram.

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    • 

      Hugo 11:16 am on 2 December 2023 Permalink | Reply

      Aha! I misread the condition as “not equal to that sum”.

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