S2T2

Jim Randell 9:19 am on 27 September 2024 Permalink Reply
Tags: by: Victor Bryant ( 139 )   

• 

  Jim Randell 9:20 am on 27 September 2024 Permalink | Reply

  I used the [[ SubstitutedExpression ]] solver from the enigma.py library to assign positions 1-5 in the queue to the forenames and surnames, such that the required conditions are met.

  We can then use the [[ filter_unique() ]] function to select solutions from the viable assignments, such that knowing Alan’s surname gives a single solution for all names.

  The following Python program runs in 89ms. (Internal runtime is 5.8ms).

  from enigma import (SubstitutedExpression, trim, filter_unique, update, peek, join, printf)
  
  # generate possible queues
  def queues():
    # allocate positions 1-5 in the queue to:
    #
    #  Alan = A; Brian = B; Colin = C; Dave = D; Ed = E
    #  Smith = S; Jones = J; Rogers = R; Mason = M; Hall = H
    #
    p = SubstitutedExpression(
      [
        # the surnames are not given in the correct order
        "A != S or B != J or C != R or D != M or E != H",
        # B is ahead of S, who is ahead of E
        "B < S", "S < E",
        # J is ahead of C, who is ahead of D, who is ahead of H
        "J < C", "C < D", "D < H",
        # M's neighbours (on different sides) are A and R
        "abs(A - R) = 2", "abs(A - M) = 1", "abs(R - M) = 1",
      ],
      base=6,
      digits=[1, 2, 3, 4, 5],
      distinct=["ABCDE", "SJRMH"],
    )
  
    # solve the puzzle
    for s in p.solve(verbose=0):
      # assemble the queue
      q = ["??"] * 6
      for k in "ABCDE": q[s[k]] = update(q[s[k]], [0], k)
      for k in "SJRMH": q[s[k]] = update(q[s[k]], [1], k)
      # return the queue
      yield trim(q, head=1)
  
  # knowing A's surname gives a unique solution
  f = (lambda q: peek(x for x in q if x[0] == 'A'))
  for q in filter_unique(queues(), f).unique:
    printf("{q}", q=join(q, sep=" "))
  

  Solution: Alan Smith is second in the queue.

  The queue is as follows:

  1st = Brian Jones
  2nd = Alan Smith
  3rd = Colin Mason
  4th = Dave Rogers
  5th = Ed Hall

  There are 5 candidate solutions:

  BJ, AS, CM, DR, EH
  BJ, CS, ER, DM, AH
  BJ, CS, DR, EM, AH
  AJ, BM, CR, DS, EH
  AJ, CM, BR, DS, EH

  The first of these is unique for AS. The next two both have AH, and the final two have AJ.

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  • 

    ruudvanderham 5:39 pm on 27 September 2024 Permalink | Reply

    import itertools
    
    for firstnames, lastnames in zip(itertools.repeat("Alan Brian Colin Dave Ed".split()), itertools.permutations("Smith Jones Rogers Mason Hall".split(), r=5)):
        for positions in itertools.permutations(range(1, 6)):
            position = {firstname: position for firstname, position in zip(firstnames, positions)}
            position.update({lastname: position for lastname, position in zip(lastnames, positions)})
            if position["Brian"] < position["Smith"] < position["Ed"]:
                if position["Jones"] < position["Colin"] < position["Dave"] < position["Hall"]:
                    if {position["Mason"] - position["Alan"], position["Mason"] - position["Rogers"]} == {-1, 1}:
                        if list(lastnames) != "Smith Jones Rogers Mason Hall".split():
                            for firstname, lastname in zip(firstnames, lastnames):
                                print(f"{firstname+ " "+lastname:12} {position[firstname]}  ", end="")
                            print()
    

    This prints:

    Alan Smith   2  Brian Jones  1  Colin Mason  3  Dave Rogers  4  Ed Hall      5  
    Alan Jones   1  Brian Rogers 3  Colin Mason  2  Dave Smith   4  Ed Hall      5
    Alan Jones   1  Brian Mason  2  Colin Rogers 3  Dave Smith   4  Ed Hall      5
    Alan Hall    5  Brian Jones  1  Colin Smith  2  Dave Rogers  3  Ed Mason     4  
    Alan Hall    5  Brian Jones  1  Colin Smith  2  Dave Mason   4  Ed Rogers    3
    

    So, it has to be Alan Smith (the others are not unique) and he is in second position.

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