S2T2

Jim Randell 11:47 am on 10 December 2019 Permalink Reply
Tags: by: G M L Tuchmann   

• 

  Jim Randell 11:47 am on 10 December 2019 Permalink | Reply

  I did this one manually before I wrote a program.

  This Python program looks at the girls in ascending age order, and the possible assignment of forms, such that the setter would be unable to deduce any celebrants, and only deduce one non-celebrant. From the information given we can then assign to the girls and verify the remaining conditions. It runs in 80ms.

  Run: [ @repl.it ]

  from collections import defaultdict
  from itertools import product
  from enigma import subsets, irange, tuples, intersect, diff, join, printf
  
  # update a sequence with (<index>, <delta>) pairs
  def update(s, *ps):
    s = list(s)
    for (i, v) in ps:
      s[i] += v
    return tuple(s)
  
  # is a sequence ordered?
  def is_ordered(s):
    return not any(x > y for (x, y) in tuples(s, 2))
  
  # indices for the girls
  girls = list(irange(0, 4))
  
  # record the celebrants by their forms (on Wednesday)
  r = defaultdict(list)
  
  # choose the forms that the girls are in on Wednesday
  for fs in subsets([1, 2, 3, 4], size=5, select="R"):
    # at most 2 girls can be promoted to form 4
    if fs.count(4) > 2: continue
  
    # choose the girls with birthdays on Tuesday and Wednesday
    for (Tu, We) in subsets(girls, size=2, select="P"):
      # the situation on Tuesday
      fsT = update(fs, (We, -1))
      if 0 in fsT or not is_ordered(fsT): continue
      # the situation on Monday
      fsM = update(fsT, (Tu, -1))
      if 0 in fsM or 4 in fsM or not is_ordered(fsM): continue
      r[fs].append((Tu, We))
  
  # for each key determine girls that definately are or aren't celebrants
  for (fs, vs) in r.items():
    # we can't positively identify any celebrant
    pos = intersect(vs)
    if pos: continue
    # we can only negatively identify one non-celebrant
    neg = set(girls).difference(*vs)
    if len(neg) != 1: continue
    # that non-celebrant is A
    A = neg.pop()
    # I and E are either side of A
    for (I, E) in product(girls[:A], girls[A + 1:]):
      # and U and O are the other two (in that order)
      (U, O) = diff(girls, (A, E, I))
  
      # we are then told Wednesday's celebrant is older than Tuesday's
      for (Tu, We) in vs:
        if not(We > Tu): continue
        # and U and O are in the same form on Monday
        fsM = update(fs, (We, -1), (Tu, -1))
        if not(fsM[U] == fsM[O]): continue
        # output solution
        d = dict(zip((A, E, I, O, U), "AEIOU"))
        s = join((join((d[i], '=', x)) for (i, x) in enumerate(fsM)), sep=", ")
        printf("Tu={Tu} We={We}; Mon=[{s}]", Tu=d[Tu], We=d[We])
  

  Solution: Isabel’s birthday is on Tuesday. Olive’s birthday is on Wednesday.

  The order of the girls (youngest to oldest) is: I, U, A, O, E.

  Mon: I=1, U=2, A=2, O=2, E=3
  Tue: I=2, U=2, A=2, O=2, E=3 [I moves up to Form 2]
  Wed: I=2, U=2, A=2, O=3, E=3 [O moves up to Form 3]

  LikeLike