S2T2

Jim Randell 9:39 am on 27 May 2025 Permalink Reply
Tags: by: Angela Newing ( 37 )   

• 

  Jim Randell 9:40 am on 27 May 2025 Permalink | Reply

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

  The following run file executes in 77ms. (Internal runtime of the generated program is 170µs).

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  # the number is: 0ABCDEFGHI0
  --digits="1-9"  # for A-I
  
  # a keypad is laid out as:
  #
  #   1 2 3
  #   4 5 6
  #   7 8 9
  #     0
  #
  # map digits to (<row>, <col>):
  #              0  1  2  3  4  5  6  7  8  9
  --code="row = [4, 1, 1, 1, 2, 2, 2, 3, 3, 3]"
  --code="col = [2, 1, 2, 3, 1, 2, 3, 1, 2, 3]"
  
  # no digits share row or col with an adjacent digit
  --code="check = lambda x, y: col[x] != col[y] and row[x] != row[y]"
  "check(0, A)"
  "check(A, B)"
  "check(B, C)"
  "check(C, D)"
  "check(D, E)"
  "check(E, F)"
  "check(F, G)"
  "check(G, H)"
  "check(H, I)"
  "check(I, 0)"
  
  # one of 5th or 6th digits is double the other
  "(D == 2 * E) or (E == 2 * D)"
  
  # 2nd and 3rd digits differ by 2
  "abs(A - B) = 2"
  
  # 7th digit is lower than 6th
  "F < E"
  
  # 10th digit is odd, and one more than 8th
  "I % 2 = 1"
  "G + 1 = I"
  
  --template="0 A B C D E F G H I 0"
  --solution=""
  

  Solution: The number is: 03594816270.

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• 

  Ruud 3:02 pm on 27 May 2025 Permalink | Reply

  Very brute force, but still vey fast:

  row = {0: 0, 1: 1, 2: 1, 3: 1, 4: 4, 5: 4, 6: 4, 7: 7, 8: 7, 9: 7}
  column = {0: 2, 1: 1, 2: 2, 3: 3, 4: 1, 5: 2, 6: 3, 7: 1, 8: 2, 9: 3}
  
  for n in itertools.permutations(range(1, 10)):
      if (
          ((number:=(None,0)+ n + (0,))[5] == 2 * number[6] or number[6] == 2 * number[5])
          and abs(number[2] - number[3]) == 2
          and number[7] < number[6]
          and number[10] % 2 == 1
          and number[10] == number[8] + 1
          and all(row[number[i]] != row[number[i + 1]] and column[number[i]] != column[number[i + 1]] for i in range(1, 11))
      ):
          print("".join(str(number[i]) for i in range(1, 12)))
  

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• 

  Frits 6:29 pm on 27 May 2025 Permalink | Reply

  #! python3 -m enigma -rr
   
  SubstitutedExpression
  
  # the number is: 0ABCDEFGHI0
  --digits="1-9"  # for A-I
   
  # a keypad is laid out as:
  #
  #   1 2 3      
  #   4 5 6      
  #   7 8 9      
  #     0       
  #
  # map digits to (<row>, <col>):
  #              0  1  2  3  4  5  6  7  8  9
  --code="row = [4, 1, 1, 1, 2, 2, 2, 3, 3, 3]"
  --code="col = [2, 1, 2, 3, 1, 2, 3, 1, 2, 3]"
  
  --invalid="1|3|5|7|9,G"     # G is even
  --invalid="1|3|5|6|7|9,DE"  # {D, E} is either {2, 4} or {4, 8}
  --invalid="8|9,F"           # F < E and E is 2, 4 or 8
  --invalid="4,ABCFGH"        # either D = 4 or E = 4
  --invalid="2|4|6|8,AB"      # D, E and G are even, so A and B can't be even
  --invalid="1|9,AB"          # A and B not both on same row
  --invalid="5,CDEFGH"        # {3, 5, 7} remains for A and B
  --invalid="2|5|8,A"         # not adjacent to "0" button
  
  --assign="B,5"              # either A = 5 or B = 5 but also A != 5
  --invalid="2|4|6|8,AC"      # not on same row/col as B(5)
   
  # no digits share row or col with an adjacent digit
  --code="check = lambda x, y: col[x] != col[y] and row[x] != row[y]"
  "check(A, B)"
  "check(B, C)"
  "check(C, D)"
  "check(D, E)"
  "check(E, F)"
  "check(F, G)"
  "check(G, H)"
  "check(H, I)"
  
  # parity of F and H is unknown
  "45 - (A + B + C + D + E + F + G + I) = H"
   
  # one of 5th or 6th digits is double the other
  "(D == 2 * E) or (E == 2 * D)"
   
  # 2nd and 3rd digits differ by 2
  "abs(A - B) = 2"
   
  # 7th digit is lower than 6th
  "F < E"
   
  # 10th digit is odd, and one more than 8th
  "G + 1 = I"
   
  --template="0 A B C D E F G H I 0"
  --distinct="ABCDEFGI"
  --solution=""
  

  or

  N = 3
  r = lambda i: (i - 1) // N
  c = lambda i: (i - 1) % N if i else 1
  
  forbidden = {i: set(j for j in range(10) if j != i and
                  (r(i) == r(j) or c(i) == c(j))) for i in range(10)}
  
  def solve(k=9, ns=set(range(1, 10)), ss=[0]):
    if k == 0:
      # check 10th number
      if 0 not in forbidden[ss[-1]] and ss[-1] == ss[-3] + 1:
        yield ss + [0]
    else:
      match(10 - k):
       case(3): 
         if abs(ss[-1] - ss[-2]) != 2: return
       case(6): 
         if not (ss[-1] == 2 * ss[-2] or ss[-2] == 2 * ss[-1]): return
       case(7):   
         if ss[-1] > ss[-2]: return
       case(8):    
         if ss[-1] % 2: return
  
      for n in ns:
        if n not in forbidden[ss[-1]]:
          yield from solve(k - 1, ns - {n}, ss + [n])
  
  for s in solve():
    print("answer:", ''.join(str(x) for x in s))
  

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