Sunday Times Teaser 3241 – Users Pay

by BRG on November 3, 2024

by Howard Williams

Published Sunday November 03 2024 (link)

I live in a cul-de-sac which had planning permission for building on ten identical plots along the whole length of one side of the road. The developer started building, and numbered the properties, from the closed end. The plots aren’t all finished, and the cost of surfacing the whole road has to be paid by the owners of the completed properties. It was decided that the owners would only contribute to the cost of the road leading to their property, so the owner of number 1 pays for the road section in front of their property, the cost of the section outside number 2 is shared equally between numbers 1 and 2, and so on.

My contribution is £1,000, while that of another homeowner is £3,800.

What is my house number and how many houses have been built?

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BRG permalink

I didn’t understand how this one worked until Jim Randell explained it here.


from itertools import combinations
from fractions import Fraction as RF

# consider the number of houses built
for built in range(1, 11):
  # collect the road costs for the built houses
  share = [0] * (built + 1)
  
  # allocate a proportion of the cost of
  # each road section to the built houses
  for s in range(1, 11):
    if s <= built:
      # built houses share only the cost
      # of the road sections they use  
      for i in range(1, s + 1):
        share[i] += RF(1, s)
    else:
      # built houses share the the cost of
      # the remaining road sections equally
      for i in range(1, built + 1):
        share[i] += RF(1, built)
  # look for two costs (me : other) in ratio (10 : 38)
  for other, me in combinations(range(1, built), 2):
    if 10 * share[other] == 38 * share[me]:
      print(f"House number {me} of {built} [other house {other}].")

from itertools import combinations

from fractions import Fraction as RF

# consider the number of houses built

for built in range(1, 11):

# collect the road costs for the built houses

share = [0] * (built + 1)

# allocate a proportion of the cost of

# each road section to the built houses

for s in range(1, 11):

if s <= built:

# built houses share only the cost

# of the road sections they use

for i in range(1, s + 1):

share[i] += RF(1, s)

else:

# built houses share the the cost of

# the remaining road sections equally

for i in range(1, built + 1):

share[i] += RF(1, built)

# look for two costs (me : other) in ratio (10 : 38)

for other, me in combinations(range(1, built), 2):

if 10 * share[other] == 38 * share[me]:

print(f"House number {me} of {built} [other house {other}].")

Frits permalink


from fractions import Fraction as RF

# unit fractions
uf = [RF(1, f) for f in range(1, 11)]
   
# number of houses built
for n in range(2, 10):
  # caculate total cost for each built house
  # credit: B. Gladman
  costs = [RF(10 - n, n) + sum(uf[h:n]) for h in range(n)]
  
  # look for one cost which is 3.8x smaller than another
  smaller38 = set()
  for i, c in enumerate(costs, start=1):
    if c in smaller38:
      print(f"answer: my house number is {i}, number of houses built: {n}")
    smaller38.add(c * 5 / 19)

from fractions import Fraction as RF

# unit fractions

uf = [RF(1, f) for f in range(1, 11)]

# number of houses built

for n in range(2, 10):

# caculate total cost for each built house

# credit: B. Gladman

costs = [RF(10 - n, n) + sum(uf[h:n]) for h in range(n)]

# look for one cost which is 3.8x smaller than another

smaller38 = set()

for i, c in enumerate(costs, start=1):

if c in smaller38:

print(f"answer: my house number is {i}, number of houses built: {n}")

smaller38.add(c * 5 / 19)

John Z permalink


# arbitrarily set unit cost of road in front of a house at lcm(1, 2, 3 ... 10)
uc = 9*8*5*7

# road cost fractions
rc = tuple(uc // i for i in range(1, 11))


for n in range(2, 11): # number of houses

  # cost of road with no houses built is divided equally among all houses
  ec = (10 - n) * uc // n

  # assign cost of road with houses built to each house as below (plus ec)
  cph = tuple(sum(rc[i: n]) + ec for i in range(n))

  for i in range(n - 1):
    for j in range(i + 1, n):
      if cph[i] * 1000 == cph[j] * 3800: # cost ratio found?
        print(n, "houses have been built;", "mine is number", j + 1)

# arbitrarily set unit cost of road in front of a house at lcm(1, 2, 3 ... 10)

uc = 9*8*5*7

# road cost fractions

rc = tuple(uc // i for i in range(1, 11))

for n in range(2, 11): # number of houses

# cost of road with no houses built is divided equally among all houses

ec = (10 - n) * uc // n

# assign cost of road with houses built to each house as below (plus ec)

cph = tuple(sum(rc[i: n]) + ec for i in range(n))

for i in range(n - 1):

for j in range(i + 1, n):

if cph[i] * 1000 == cph[j] * 3800: # cost ratio found?

print(n, "houses have been built;", "mine is number", j + 1)

John Z permalink

Interestingly, in my code above, replacing the integer division (double slash) operations with floating point (single slash) operations and setting the unit cost to 1 gives the same result. I realize, of course, that rounding errors can creep in with floating point that prevent getting a (correct) result.

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Sunday Times Teaser 3241 – Users Pay

by Howard Williams

Published Sunday November 03 2024 (link)

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