S2T2

Jim Randell 8:46 am on 21 April 2022 Permalink Reply
Tags: by: Danny Roth ( 84 )   

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  Jim Randell 8:46 am on 21 April 2022 Permalink | Reply

  This Python program collects possible (house-number, wrong-number, percentage-reduction) values, and then uses the [[ filter_unique() ]] function from the enigma.py library to find the required values for the first and second house numbers.

  It runs in 55ms. (Internal runtime is 333µs).

  Run: [ @replit ]

  from enigma import (irange, div, filter_unique, unpack, printf)
  
  # record (n, w, p) results
  rs = list()
  
  # consider two digit house numbers
  for n in irange(20, 99):
    # consider different numbers with the first digit less than the
    # correct digit, but the second digit is correct
    for w in irange(10 + (n % 10), n - 1, step=10):
      # calculate the percentage w / n
      p = div(100 * w, n)
      if p is None: continue
      rs.append((n, w, p))
  
  # if you knew the percentage you would know the correct house number
  fn_n = unpack(lambda n, w, p: n)
  fn_p = unpack(lambda n, w, p: p)
  for (n1, w1, p1) in filter_unique(rs, fn_p, fn_n).unique:
    printf("p1={p1}% -> n1={n1} w1={w1}")
  
    # look for results with a different house number
    rs1 = filter(unpack(lambda n2, w2, p2: n2 != n1), rs)
    for (n2, w2, p2) in filter_unique(rs1, fn_p, fn_n).unique:
      printf("  p2={p2}% -> n2={n2} w2={w2}")
    printf()
  

  Solution: The first house number was 50. The second house number was 75.

  There are two possible scenarios:

  Daughter’s first house was 50, but she wrote 20 (= 40% of 50).
  Daughter’s second house was 75, but she wrote 15 (= 20% of 75).

  Daughter’s first house was 50, but she wrote 40 (= 80% of 50).
  Daughter’s second house was 75, but she wrote 15 (= 20% of 75).

  But in both cases the first house number is 50 and the second house number is 75.

  ---

  Here are the 15 possible (house number, wrong number) pairs, grouped by percentage:

  20%: (50 → 10) (75 → 15)
  25%: (40 → 10) (80 → 20)
  40%: (50 → 20)
  50%: (20 → 10) (40 → 20) (60 → 30) (80 → 40)
  60%: (25 → 15) (50 → 30) (75 → 45)
  75%: (40 → 30) (80 → 60)
  80%: (50 → 40)

  From which we see the only values with a unique percentage are:

  40%: (50 → 20)
  80%: (50 → 40)

  Both cases give a first house number of 50.

  If we remove pairs with a first house number of 50 from those listed above we get:

  20%: (75 → 15)
  25%: (40 → 10) (80 → 20)
  50%: (20 → 10) (40 → 20) (60 → 30) (80 → 40)
  60%: (25 → 15) (75 → 45)
  75%: (40 → 30) (80 → 60)

  And we see there is only one value with a unique percentage:

  20%: (75 → 15)

  And so the second house number is 75.

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