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Jim Randell 8:28 am on 29 November 2022 Permalink Reply
Tags: by: Angela Newing ( 23 )   

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  Jim Randell 8:28 am on 29 November 2022 Permalink | Reply

  I used the [[ SubstitutedExpression ]] solver from the enigma.py library to assign the fence values (1-5) to the 3 groups of jockeys, horses and colours.

  The following run file executes in 66ms. (Internal runtime of the generated program is 94µs).

  Run: [ @replit ]

  #! python3 -m enigma -r
  
  # assign fence numbers (1-5) to the following groups:
  #
  #  jockeys = A (Wally), B (Bob), C (Chris), D (Dave), E (Fred)
  #  horses  = F (Egg Nog), G (Big Gun), H (Long Gone), I (Nig Nag), J (Dragon)
  #  colours = K (mauve), L (yellow), M (green), N (blue), P (red)
  
  SubstitutedExpression
  
  --base="6"
  --digits="1-5"
  --distinct="ABCDE,FGHIJ,KLMNP"
  --template="(A B C D E) (F G H I J) (K L M N P)"
  --solution=""
  
  # "Dave (D) riding Egg Nog (F) lasted longer than Bill (B) who fell at the 2nd fence"
  "D = F"
  "D > B"
  "2 = B"
  
  # "Big Gun (G) fell at the 3rd"
  "3 = G"
  
  # "mauve (K) lasted the longest"
  "5 = K"
  
  # "Long Gone (H) lasted longer than yellow (L)
  "H > L"
  
  # "Chris's (C's) horse fell one fence later than the horse ridden by green (M)"
  "M + 1 = C"
  
  # "Fred (E) and his friend blue (N) did not fall at adjacent fences"
  "abs(E - N) > 1"
  
  # "Nig Nag (I) was ridden by Wally (A)"
  "I = A"
  
  # "Dragon's (J's) jockey wore red"
  "J = P"
  

  We can wrap this in Python code to format the solution in a more friendly form:

  Run: [ @replit ]

  from enigma import (SubstitutedExpression, group, printf)
  
  # load the run file
  p = SubstitutedExpression.from_file("teaser2685.run")
  
  # map symbols to jockey, horse, colour names
  name = dict(
    A="Wally", B="Bob", C="Chris", D="Dave", E="Fred",
    F="Egg Nog", G="Big Gun", H="Long Gone", I="Nig Nag", J="Dragon",
    K="mauve", L="yellow", M="green", N="blue", P="red",
  )
  
  # solve the puzzle
  for s in p.solve(verbose=0):
    # group symbols by value
    d = group(s.keys(), by=s.get)
    # output solution for each fence
    for k in sorted(d.keys()):
      # extract jockey, horse, colour
      (j, h, c) = (name[s] for s in sorted(d[k]))
      printf("{k}: {j} ({c}) on {h}")
    printf()
  

  Solution: Fred was in yellow, riding Big Gun.

  The full solution is:

  1st fence: Wally (blue) on Nig Nag
  2nd fence: Bob (red) on Dragon
  3rd fence: Fred (yellow) on Big Gun
  4th fence: Dave (green) on Egg Nog
  5th fence: Chris (mauve) on Long Gone

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