S2T2

Jim Randell 10:53 pm on 16 February 2019 Permalink Reply
Tags: by: J L Powell ( 2 )   

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  Jim Randell 10:55 pm on 16 February 2019 Permalink | Reply

  This Python program runs in 370ms.

  Run: [ @repl.it ]

  from collections import namedtuple
  from itertools import permutations, combinations, product
  from enigma import irange, partitions, printf
  
  # represent a die
  Die = namedtuple("Die", "B R W")
  
  # collect possible dice
  dice = set(Die(*s) for ns in partitions(irange(1, 6), 2) for s in permutations(ns))
  
  # can we make the numbers in <ns> from the opposite faces of <fs>
  def make(ns, fs):
    ns = set(ns)
    (f1, f2, f3) = fs
    return any(ns.issubset([f1[i1] + f2[i2] + f3[i3], f1[1 - i1] + f2[1 - i2] + f3[1 - i3]]) for (i1, i2, i3) in product((0, 1), repeat=3))
  
  # change tuple <t> so index <i> is <v>
  def change(t, i, v):
    t = list(t)
    t[i] = v
    return tuple(t)
  
  # possible white, red values for Tommy (white > red)
  ts = set((w, r) for w in irange(6, 10) for r in irange(5, w - 1))
  
  # choose a set of three dice
  for (d1, d2, d3) in combinations(dice, 3):
  
    # "no number appeared in the same colour twice"
    if not all(len(set(x1 + x2 + x3)) == 6 for (x1, x2, x3) in zip(d1, d2, d3)): continue
  
    # "on one of the dice two similary coloured sides add up to 9"
    if not any(sum(x) == 9 for x in d1 + d2 + d3): continue
  
    # can we make a R, W, W combination that sums to 14 and 15
    if not any(make((14, 15), fs) for fs in [(d1.R, d2.W, d3.W), (d1.W, d2.R, d3.W), (d1.W, d2.W, d3.R)]): continue
  
    # Tommy picked one of the dice and swapped the colours of a red number and a white number, suppose he repaints t1
    for (t1, t2, t3) in [(d1, d2, d3), (d2, d1, d3), (d3, d1, d2)]:
      # choose red/white faces of t1 to repaint
      for (r, w) in product((0, 1), repeat=2):
        # the repainted die
        t = Die(B=t1.B, R=change(t1.R, r, t1.W[w]), W=change(t1.W, w, t1.R[r]))
  
        # record possible 3 white/red scores (including the repainted faces)
        ws = set(sum(s) for s in product([t.W[w]], t2.W, t3.W))
        rs = set(sum(s) for s in product([t.R[r]], t2.R, t3.R))
  
        # do these create feasible values?
        vs = ts.intersection(product(ws, rs))
        if vs:
          printf("dice: {t1} {t2} {t3}")
          printf("  red = {r}, white = {w} -> {vs}", r=t.R[r], w=t.W[w])
  

  Solution: (1) Tommy repainted 3 to red, (2) and 4 to white.

  There is only one possible set of dice:

  die 1: blue = 1,5; red = 2,4; white = 3,6
  die 2: blue = 2,6; red = 1,3; white = 4,5
  die 3: blue = 3,4; red = 5,6; white = 1,2

  The throws for Bill, showing 2 white faces and 1 red:

  throw 1: die 1 = white 3; die 2 = white 5; die 3 = red 6; total = 14
  throw 2: die 1 = white 6; die 2 = white 4; die 3 = red 5; total = 15

  Both die 1 and die 2 have white faces that sum to 9.

  Tommy chooses die 1 and repaints 3 from white to red, and 4 from red to white, giving (with the repainted faces indicated in braces):

  die 1: blue = 1,5; red = 2,{3}; white = {4},6
  die 2: blue = 2,6; red = 1,3; white = 4,5
  die 3: blue = 3,4; red = 5,6; white = 1,2

  Tommy then throws 3 whites (totalling 6 to 10), and then 3 reds (totalling 5 to 9), the white total being more than the red total.

  throw 1: die 1 = {white 4}; die 2 = white 4; die 3 = white 2; total = 10
  throw 2: die 1 = {red 3}; die 2 = red 1; die 3 = red 5; total = 9

  So Tommy’s calculation of 8 for the reds and 7 for the whites were both out by 2.

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