S2T2

Jim Randell 8:28 am on 27 June 2025 Permalink Reply
Tags: by: C Higgins ( 35 ), by: H Bradley ( 35 )   

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  Jim Randell 8:29 am on 27 June 2025 Permalink | Reply

  We assume the cuts themselves make a negligible difference to the length of the pieces.

  Here is a constructive program that counts the number of dissections of an n-foot rod into 3 dissimilar whole inch lengths.

  I decided that as the puzzle was set in 2010, viable years are in the range [1940, 1985].

  The program runs in 71ms. (Internal runtime is 6.1ms).

  from enigma import (irange, inf, decompose, icount, printf)
  
  # generate number of dissections
  def generate():
    # consider number of feet
    for n in irange(1, inf):
      # divide into different sets of inches
      k = icount(decompose(12 * n, 3, increasing=1, sep=1))
      printf("[{n} ft -> {k} sets]")
      yield (n, k)
  
  # find lengths that give up to 1985 sets
  ss = list()
  for (n, k) in generate():
    ss.append((n, k))
    if k > 1985: break
  
  # output solution (between 1940 and 1985)
  for (_, y) in ss:
    if 1939 < y < 1986:
      printf("year = {y}")
  

  Solution: Pat’s father was born in 1951.

  1951 is the number of dissections of a 13-foot rod.

  ---

  We can use the values accumulated by the program to determine a polynomial equation to generate the sequence:

  % python3 -i teaser/teaser2471.py
  [1 ft -> 7 sets]
  [2 ft -> 37 sets]
  [3 ft -> 91 sets]
  [4 ft -> 169 sets]
  [5 ft -> 271 sets]
  [6 ft -> 397 sets]
  [7 ft -> 547 sets]
  [8 ft -> 721 sets]
  [9 ft -> 919 sets]
  [10 ft -> 1141 sets]
  [11 ft -> 1387 sets]
  [12 ft -> 1657 sets]
  [13 ft -> 1951 sets]
  [14 ft -> 2269 sets]
  year = 1951
  
  >>> from enigma import Polynomial
  >>> p = Polynomial.interpolate(ss)
  >>> print(p)
  Polynomial[(+12)x^2 (-6)x (+1)]
  >>> print(p(13))
  1951
  >>> print(p(5280))
  334509121
  

  So, for an n-foot rod (integer n ≥ 1), the number of dissections D(n) is given by:

  D[n] = 12n^2 − 6n + 1

  This corresponds to OEIS A154105 [@oeis.org].

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    Frits 3:32 pm on 27 June 2025 Permalink | Reply

    It is not too difficult to find this formula for the sum 1 + 2+4 + 5+7 + 8+10 + 11+13 + 14+16 + ….

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    Frits 4:08 pm on 27 June 2025 Permalink | Reply

    Another formula for D(n) is:

    sum(6 * n – (k + 3) // 2 – k – 1 for k in range(4 * n – 1))

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  Jim Randell 8:10 am on 28 June 2025 Permalink | Reply

  See also: BrainTwister #25.

  Where we had:

  t[n] = intr(sq(n) / 12)

  Then D[n] appears as a subsequence of t[n] taking every 12th element (as there are 12 inches in 1 foot).

  Specifically:

  D[n] = t[12n − 3]

  And this gives rise to the quadratic equation given above.

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