S2T2

Jim Randell 10:07 am on 26 February 2019 Permalink Reply
Tags: by: R Postill ( 20 ), flawed ( 52 )   

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  Jim Randell 10:07 am on 26 February 2019 Permalink | Reply

  I found two solutions for this puzzle, although with a slight change in the wording we could arrive at a unique solution.

  This Python program runs in 226ms.

  Run: [ @repl.it ]

  from itertools import product, permutations, combinations
  from enigma import icount, unpack, printf
  
  # all the players
  players = "ABCDEFGHIJKLMN"
  
  # players with dual allegiance
  duals = "EFLM"
  
  # the pubs
  pubs = set("PQR")
  
  # do the three sets form a team
  def team(a, b, c):
    for (x, y, z) in permutations(pubs):
      if x in a and y in b and z in c: return True
    return False
  
  # choose the "missing" pub for the duals
  for s1 in product(pubs, repeat=4):
    (E, F, L, M) = (pubs.difference([x]) for x in s1)
  
    # ELM do not form a team
    if team(E, L, M): continue
  
    # choose (single) pubs for K, N
    for s2 in product(pubs, repeat=2):
      (K, N) = (set([x]) for x in s2)
  
      # KLM, FKN, ELN do not form teams
      if team(K, L, M) or team(F, K, N) or team(E, L, N): continue
  
      # CJN do form a team
      for s3 in permutations(pubs.difference(N)):
        (C, J) = (set([x]) for x in s3)
  
        # remaining assignments for given combinations, AGHI
        for s4 in product(pubs, repeat=4):
          (A, G, H, I) = (set([x]) for x in s4)
  
          # check the remaining combinations
          if not (team(C, F, H)) or team(A, F, I) or team(F, G, J): continue
  
          # which leaves B and D
          for s5 in product(pubs, repeat=2):
            (B, D) = (set([x]) for x in s5)
  
            table = (A, B, C, D, E, F, G, H, I, J, K, L, M, N)
  
            # P, Q, R should have teams of 5, 6, 7
            (P, Q, R) = (icount(table, (lambda x: p in x)) for p in 'PQR')
            if (P, Q, R) != (5, 6, 7): continue
  
            # count the total possible teams
            t = icount(combinations(table, 3), unpack(team))
            # there should be 162
            if t != 162: continue
  
            # find how many unused IJ? combinations there are
            ijs = list(p for (p, x) in zip(players, table) if p not in 'IJ' and not team(I, J, x))
            # there should be at least 2
            if len(ijs) < 2: continue
  
            # who plays for R
            Rs = list(p for (p, x) in zip(players, table) if 'R' in x)
  
            printf("Rs = {Rs}")
            printf("  A={A} B={B} C={C} D={D} E={E} F={F} G={G} H={H} I={I} J={J} K={K} L={L} M={M} N={N}")
            printf("  ijs={ijs}")
            printf()
  

  Solution: A, B, D, F, G, I, J are on the Roebuck team.

  This solution assumes that the final team member of the I+J+? team that is picked out of the hat can be any of the other players (i.e. I and J have never played together on a team before).

  The team allegiances in this case are:

  P = C, E, F, L, M [5]
  Q = E, H, K, L, M, N [6]
  R = A, B, D, F, G, I, J [7]

  However, I took the construction of final I+J+? team by picking a name out of a hat to require only that there should be more than one candidate to placed in the hat. If we have the following team allegiances:

  P = E, F, J, L, M [5]
  Q = E, H, K, L, M, N [6]
  R = A, B, C, D, F, G, I [7]

  Then this is also a solution to the puzzle. In this case the remaining possible partners for I+J+? are: A, B, C, D, F, G.

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