S2T2

Jim Randell 12:36 pm on 7 January 2024 Permalink Reply
Tags: by: Graham Smithers ( 82 )   

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  Jim Randell 12:37 pm on 7 January 2024 Permalink | Reply

  Consider a cubiod with dimensions a × b × c. Then we can calculate the number of cubelets that are painted on the required number of faces:

  3 faces = 8
  2 faces = 4(a + b + c − 6)
  1 face = 2((a − 2)(b − 2) + (a − 2)(c − 2) + (b − 2)(c − 2))
  0 faces = (a − 2)(b − 2)(c − 2)

  providing the smallest dimension is greater than 1.

  And as some of the cubelets are unpainted the cuboids we are interested in must have smallest dimension of at least 3.

  This Python program runs in 56ms. (Internal runtime is 335µs).

  Run: [ @replit ]

  from enigma import (defaultdict, irange, inf, decompose, map2str, printf)
  
  # total number of faces is indicated by key 7
  total = 7
  
  # for a cuboid of dimensions a x b x c
  # return number of cubelets by faces painted, and the total number of cubelets
  def cubelets(a, b, c):
    assert min(a, b, c) > 1
    r = dict()
    r[3] = 8
    r[2] = 4 * (a + b + c - 6)
    r[0] = (a - 2) * (b - 2) * (c - 2)
    t = a * b * c
    r[1] = t - sum(r.values())
    r[total] = t
    return r
  
  def solve():
    # collect cubes by total
    g = defaultdict(list)
    # consider increasing a + b + c dimensions
    for s in irange(3, inf):
      # consider cuboids for M
      for (a, b, c) in decompose(s, 3, increasing=-1, sep=0, min_v=3):
        # calculate number of cubelets
        r = cubelets(a, b, c)
        # we are interested in those cubes where r[0] = r[2]
        if not (r[2] == r[0] > 0): continue
  
        # consider smaller totals (for O)
        for (k, vs) in g.items():
          # difference between totals
          d = r[total] - k
          if d < 1: continue
          for (r_, (a_, b_, c_)) in vs:
            if r_[0] == d:
              # return (M, O) as (cubelets, dimensions)
              yield ((r, (a, b, c)), (r_, (a_, b_, c_)))
        # record this cuboid
        g[r[total]].append((r, (a, b, c)))
  
  # find possible (M, O) values
  for ((r, (a, b, c)), (r_, (a_, b_, c_))) in solve():
    fmt = lambda r: map2str(r, arr='->', enc="")
    printf("M = {a} x {b} x {c} [{r}]; O = {a_} x {b_} x {c_} [{r_}]", r=fmt(r), r_=fmt(r_))
    break  # only need the first answer
  

  Solution: Megan has 112 cubelets painted on just one face.

  Megan has a cuboid with dimensions: 12 × 5 × 4.

  12 × 5 × 4 = 240 cubelets
  8 painted on 3 faces
  60 painted on 2 faces
  112 painted on 1 face
  60 painted on 0 faces

  Oliver has a cuboid with dimensions: 8 × 6 × 4.

  8 × 6 × 4 = 192 cubelets
  8 painted on 3 faces
  48 painted on 2 faces
  88 painted on 1 face
  48 painted on 0 faces

  And:

  240 − 192 = 48

  ---

  If cuboids with a dimensions less than 3 were allowed, then we can find further solutions using cuboids that give 0 unpainted cubelets, such as:

  Oliver = 8 × 6 × 4 (= 192 cubelets, 48 unpainted, 48 painted on 2 faces)
  Megan = 120 × 2 × 1 (= 240 cubelets, 0 unpainted, 0 painted on 2 faces)

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