S2T2

Jim Randell 9:49 am on 25 February 2025 Permalink Reply
Tags: by: Bob Walker ( 10 )   

• 

  Jim Randell 9:50 am on 25 February 2025 Permalink | Reply

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

  It runs in 80ms. (And the internal runtime of the generated program is just 53µs).

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  #  +---+---+---+---+---+---+
  #  | A |   | B |   | C |!=0|
  #  +---+---+---+---+---+---+
  #  | 1 | D | >2| E | <2| F |
  #  +---+---+---+---+---+---+
  #  | G |   | H |!=1| I |   |
  #  +---+---+---+---+---+---+
  #  |   | J |   | K | >1| L |
  #  +---+---+---+---+---+---+
  #  | M | >2| N |!=3| P |   |
  #  +---+---+---+---+---+---+
  #  |!=2| Q |   | R | =1| S |
  #  +---+---+---+---+---+---+
  
  --distinct=""
  --digits="0,1"
  
  # the clues
  "C + F != 0"
  "A + D + G = 1"
  "B + D + E + H > 2"
  "C + E + F + I < 2"
  "E + H + I + K != 1"
  "I + K + L + P > 1"
  "J + M + N + Q > 2"
  "K + N + P + R != 3"
  "M + Q != 2"
  "P + R + S = 1"
  
  --answer="sum([A, B, C, D, E, F, G, H, I, J, K, L, M, N, P, Q, R, S])"
  --template=""
  

  Solution: 10 of the black cards have a cross on them.

  There are 4 possible arrangements, that can be summarised as follows:

  

  8 of the crosses are marked X, and the remaining 2 crosses appear in squares marked a and b.

  LikeLike

• 

  ruudvanderham 12:26 pm on 1 March 2025 Permalink | Reply

  import itertools
  
  
  def s(*args):
      return sum(a[arg] for arg in args)
  
  
  for a in itertools.product((0, 1), repeat=18):
      if (
          s(2, 5) != 0
          and s(0, 3, 6) == 1
          and s(1, 3, 4, 7) > 2
          and s(2, 4, 5, 8) < 2
          and s(4, 7, 8, 10) != 1
          and s(8, 10, 11, 14) > 1
          and s(9, 12, 13, 15) > 2
          and s(10, 13, 14, 16) != 3
          and s(12, 15) != 2
          and s(14, 16, 17) == 1
      ):
          print(a, sum(a))
  

  LikeLike