Sunday Times Teaser 3100 – Crate Expectations

by BRG on February 19, 2022

by Nick MacKinnon

Published Sunday February 20 2022 (link)

Pip and Estella know each other to be honest and clever. At the reading of Miss Havisham’s will they see a large crate on which are placed two envelopes addressed V+S+E and V, and a lipstick. The solicitor gives Estella the V+S+E envelope and Pip the V envelope and says, “Inside this crate Miss Havisham has left a cuboidal gold bar whose dimensions in mm are different whole numbers greater than 1, and whose volume, surface area and total edge length are V mm³, S mm² and E mm respectively. Inside your envelope, which you should now open, is the secret value of your address. Every fifteen minutes a bell will ring, and if at that point you know the other’s value, write both values on the crate with lipstick and the gold is yours.” At the first bell Pip and Estella sat still, but when the bell rang again Estella lipsticked 3177 and Pip’s value on the crate.

What was Pip’s value?

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from number_theory import divisor_pairs

# for the cuboid with sides (a, b, c):
# 
#  V = a.b.c ; S = 2.(a.b + b.c + c.a) ; E = 4.(a + b + c)
#
#  V + S + E = (a + 2).(b + 2).(c + 2) - 8

# generate the ordered triples of the divisors of <n>
def triples(abc):
  for a, bc in divisor_pairs(abc):
    for b, c in divisor_pairs(bc):
      if a < b < c:
        yield a, b, c

# return V + S + E for the cuboid <t>
vse_sum = lambda t: (t[0] + 2) * (t[1] + 2) * (t[2] + 2)

# hence find possible cuboids from the known V + S + E value (3177) 
for t in triples(3177 + 8):
  if t[0] < 4:
    continue
  # the possible cuboid dimensions
  a, b, c = (i - 2 for i in t)
  # Estella's address (V + S + E) cannot be deduced from Pip's (V)
  if len(set(vse_sum(t) for t in triples(a * b * c) if 1 < t[0])) > 1:
    print(f"Pip's value (V) is {a * b * c} [cuboid = ({a}, {b}, {c})]")

from number_theory import divisor_pairs

# for the cuboid with sides (a, b, c):

# V = a.b.c ; S = 2.(a.b + b.c + c.a) ; E = 4.(a + b + c)

# V + S + E = (a + 2).(b + 2).(c + 2) - 8

# generate the ordered triples of the divisors of <n>

def triples(abc):

for a, bc in divisor_pairs(abc):

for b, c in divisor_pairs(bc):

if a < b < c:

yield a, b, c

# return V + S + E for the cuboid <t>

vse_sum = lambda t: (t[0] + 2) * (t[1] + 2) * (t[2] + 2)

# hence find possible cuboids from the known V + S + E value (3177)

for t in triples(3177 + 8):

if t[0] < 4:

continue

# the possible cuboid dimensions

a, b, c = (i - 2 for i in t)

# Estella's address (V + S + E) cannot be deduced from Pip's (V)

if len(set(vse_sum(t) for t in triples(a * b * c) if 1 < t[0])) > 1:

print(f"Pip's value (V) is {a * b * c} [cuboid = ({a}, {b}, {c})]")

Doing it Estelle’s way using partition unique (the hard way!).


from partition_unique import partition_unique

maxss = 100
# consider cuboids (a, b, c) with 9 <= a + b + c < maxss
sol = []
for a in range(2, maxss // 3):
  for b in range(a + 1, maxss - a):
      for c in range(b + 1, maxss - a - b):
        sol.append((a, b, c))

# given a solution (a, b, c) return V + S + E [where some
# algebra gives v + s + e = (a + 2).(b + 2).(c + 2) - 8]
s2svse = lambda a, b, c: (a + 2) * (b + 2) * (c + 2) - 8

# given a solution return V
s2v = lambda a, b, c: a * b * c

# solutions for which Pip's known V does not identify Estelle's V + S + E
pip_nu = partition_unique(sol, s2v, s2svse)[1]
# solutions for which Estelle's known V + S + E does not identify Pip's V
est_nu = partition_unique(sol, s2svse, s2v)[1]

# now find solutions for which Estelle can identify Pip's address
for a, b, c  in set(est_nu).intersection(pip_nu):
  if s2svse(a, b, c) == 3177:
    v, s, e = a * b * c, 2 * (a * b + b * c + c * a), 4 * (a + b + c)
    print(f"Pip's value (V) is {v} [S = {s}, E = {e}, cuboid = {(a, b, c)}]")

from partition_unique import partition_unique

maxss = 100

# consider cuboids (a, b, c) with 9 <= a + b + c < maxss

sol = []

for a in range(2, maxss // 3):

for b in range(a + 1, maxss - a):

for c in range(b + 1, maxss - a - b):

sol.append((a, b, c))

# given a solution (a, b, c) return V + S + E [where some

# algebra gives v + s + e = (a + 2).(b + 2).(c + 2) - 8]

s2svse = lambda a, b, c: (a + 2) * (b + 2) * (c + 2) - 8

# given a solution return V

s2v = lambda a, b, c: a * b * c

# solutions for which Pip's known V does not identify Estelle's V + S + E

pip_nu = partition_unique(sol, s2v, s2svse)[1]

# solutions for which Estelle's known V + S + E does not identify Pip's V

est_nu = partition_unique(sol, s2svse, s2v)[1]

# now find solutions for which Estelle can identify Pip's address

for a, b, c in set(est_nu).intersection(pip_nu):

if s2svse(a, b, c) == 3177:

v, s, e = a * b * c, 2 * (a * b + b * c + c * a), 4 * (a + b + c)

print(f"Pip's value (V) is {v} [S = {s}, E = {e}, cuboid = {(a, b, c)}]")

Sunday Times Teaser 3100 – Crate Expectations

by Nick MacKinnon

Published Sunday February 20 2022 (link)

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