S2T2

Jim Randell 9:01 am on 11 July 2023 Permalink Reply
Tags: by: Danny Roth ( 86 )   

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  Jim Randell 9:01 am on 11 July 2023 Permalink | Reply

  The house numbers in positions k = 1 .. 99 are defined as:

  a[1] = n
  a[k + 1] = (a[k] + n) mod 100

  or:

  a[k] = (k.n) mod 100

  So numberers with the same age mod 100 will give the same sequence. We can suppose that the numberers age lies in the range 16 – 115, so we can identify the ages by the residue mod 100.

  We are looking for a 2-digit house number XY that appears in the “correct” position, i.e.:

  a[XY] = XY

  And has a neighbouring house whose number is YX:

  a[XY ± 1] = YX

  I am assuming that 1-digit numbers are represented with a single digit.

  This Python program runs in 54ms. (Internal runtime is 1.4ms).

  Run: [ @replit ]

  from enigma import (irange, nreverse, printf)
  
  # find solutions for age <n>
  def solve(n):
    # map: position -> house number
    (d, seen) = (dict(), set())
    (p, k) = (1, n)
    while p < 100:
      if k == 0 or k in seen: break
      d[p] = k
      seen.add(k)
      k = (k + n) % 100
      p += 1
    # did we allocate all the houses?
    if p == 100:
      # look for a house with a 2-digit number the same as position
      for (p, k) in d.items():
        if not (k > 9 and k == p): continue
        # look for neighbouring houses with the reverse of the number
        r = nreverse(k)
        if not (r > 9): continue
        if d.get(p - 1, 0) == r: yield (n, (p, k), (p - 1, r))
        if d.get(p + 1, 0) == r: yield (n, (p, k), (p + 1, r))
  
  # consider possible ages (mod 100)
  for n in irange(0, 99):
    for (n, (p0, k0), (p1, k1)) in solve(n):
      printf("age = {n} -> pos {p0} = {k0}, pos {p1} = {k1}")
  

  Solution: George and Martha’s house number is 25. The numberer’s age was 73.

  We have:

  a[25] = (25×73) mod 100 = 1825 mod 100 = 25
  a[24] = (24×73) mod 100 = 1752 mod 100 = 52

  ---

  The complete mapping is:

  (0 → 0), 1 → 73, 2 → 46, 3 → 19, 4 → 92, 5 → 65, 6 → 38, 7 → 11, 8 → 84, 9 → 57, 10 → 30,
  11 → 3, 12 → 76, 13 → 49, 14 → 22, 15 → 95, 16 → 68, 17 → 41, 18 → 14, 19 → 87, 20 → 60,
  21 → 33, 22 → 6, 23 → 79, 24 → 52, 25 → 25, 26 → 98, 27 → 71, 28 → 44, 29 → 17, 30 → 90,
  31 → 63, 32 → 36, 33 → 9, 34 → 82, 35 → 55, 36 → 28, 37 → 1, 38 → 74, 39 → 47, 40 → 20,
  41 → 93, 42 → 66, 43 → 39, 44 → 12, 45 → 85, 46 → 58, 47 → 31, 48 → 4, 49 → 77, 50 → 50,
  51 → 23, 52 → 96, 53 → 69, 54 → 42, 55 → 15, 56 → 88, 57 → 61, 58 → 34, 59 → 7, 60 → 80,
  61 → 53, 62 → 26, 63 → 99, 64 → 72, 65 → 45, 66 → 18, 67 → 91, 68 → 64, 69 → 37, 70 → 10,
  71 → 83, 72 → 56, 73 → 29, 74 → 2, 75 → 75, 76 → 48, 77 → 21, 78 → 94, 79 → 67, 80 → 40,
  81 → 13, 82 → 86, 83 → 59, 84 → 32, 85 → 5, 86 → 78, 87 → 51, 88 → 24, 89 → 97, 90 → 70,
  91 → 43, 92 → 16, 93 → 89, 94 → 62, 95 → 35, 96 → 8, 97 → 81, 98 → 54, 99 → 27

  So 25 is in the correct position, and 24 has the number that is the reverse of 25.

  50 and 75 are also in the correct position, but are not neighbours of 5 or 57 (respectively).

  62 is given the number 26, the reverse of its position.

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  Jim Randell 10:22 am on 11 July 2023 Permalink | Reply

  Without checking that each house is assigned a different number there is only one candidate solution (but we can provide an additional check to ensure the answer is viable).

  This run file executes in 62ms. (Internal runtime of the generated program is 1.3ms).

  Run: [ @replit ]

  #! python3 -m enigma -rr
  
  # AB = age of the numberer
  # XY = George and Martha's house number
  
  SubstitutedExpression
  
  --distinct="XY"
  --code="num = lambda n, k: (n * k) % 100"
  
  # G&M's house number is the same as its position
  "num(AB, XY) == XY"
  
  # a neighbouring house is numbered YX
  "YX in { num(AB, XY - 1), num(AB, XY + 1) }"
  
  # answer is: G&M's number, numberers age
  --answer="(XY, AB)"
  
  # check each house is assigned different number
  --reorder=0
  "seq_all_different(num(AB, k) for k in irange(1, 99))"
  

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