S2T2

Jim Randell 11:26 am on 10 October 2024 Permalink Reply
Tags: by: Andrew Skidmore ( 88 )   

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  Jim Randell 11:27 am on 10 October 2024 Permalink | Reply

  See also: Enigma 77, Enigma 1769.

  There is a line between each pair of tacks, so the number of lines is:

  L(n) = C(n, 2) = n(n − 1)/2

  Each crossing point is defined by two of the lines (any four different points form the vertices of a quadrilateral, the diagonals of which give the crossing lines), and each line is defined by two tacks.

  So the number of crossings is given by:

  X(n) = C(n, 4) = n(n − 1)(n − 2)(n − 3)/24

  And we are interested in when (for some single digit k):

  X(n) = k L(n)
  k = X(n) / L(n)
  ⇒
  12k = (n − 2)(n − 3)

  So we need to find two consecutive numbers (≥ 3), that give a suitable value for k = 2..9.

  The only candidate is k = 6:

  12 × 6 = 72 = 8 × 9

  Hence n = 11.

  Solution: Callum used 11 tacks.

  L(11) = 55
  X(11) = 330
  330 = 6 × 55

  The next solution would occur at n = 14.

  L(14) = 91
  X(14) = 1001
  1001 = 11 × 91

  but the multiple is not a single digit number.

  ---

  Here is a program that considers increasing n values:

  from enigma import (irange, inf, C, printf)
  
  # consider number of tacks
  for n in irange(7, inf):
    # calculate number of lines and number of crossings
    (L, X) = (C(n, 2), C(n, 4))
    (k, r) = divmod(X, L)
    if k > 9: break
    if r != 0: continue
    # output solution
    printf("n={n}: L={L} X={X}; k={k}")
  

  Alternatively here is a program that solves the equation:

  from enigma import (Polynomial, irange, printf)
  
  # make the polynomial p(n) = (n - 2)(n - 3)
  n = Polynomial('n', var='n')
  p = (n - 2) * (n - 3)
  
  # consider possible single digit multiples k
  for k in irange(2, 9):
    # find integer roots of the polynomial p(n) = 12k
    for r in p.find_roots(12 * k, domain='Z'):
      # look for values > 6
      if r > 6:
        # output solution
        printf("n={r} [k={k}]")
  

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• 

  Ruud 7:29 am on 11 October 2024 Permalink | Reply

  import math
  import itertools
  
  for number_of_tacks in itertools.count(7):
      number_of_crossing_points = math.comb(number_of_tacks, 4)
      number_of_pieces_of_string = math.comb(number_of_tacks, 2)
      if number_of_crossing_points / number_of_pieces_of_string > 9:
          break
      if number_of_crossing_points % number_of_pieces_of_string == 0:
          print(f"{number_of_tacks=}  {number_of_pieces_of_string=}  {number_of_crossing_points=}  {number_of_crossing_points//number_of_pieces_of_string=}")
  

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