S2T2

Jim Randell 3:16 pm on 29 July 2022 Permalink Reply
Tags: by: Nick MacKinnon ( 51 )   

• 

  Jim Randell 3:55 pm on 29 July 2022 Permalink | Reply

  The title of this puzzle is presumably an allusion to Sherlock Holmes’ “three pipe problem” in The Red-Headed League.

  Some judicious applications of Pythogoras’ theorem gets us the answer.

  This Python program runs in 54ms. (Internal runtime is 313µs).

  Run: [ @replit ]

  from enigma import (irange, div, is_square, sq, printf)
  
  # suppose the pipes have diameters a, b, c, d, e, f (in cm)
  # we know h = 2a + c < 500; b = (2/3)a
  for a in irange(6, 249, step=3):
    b = div(2 * a, 3)
    # from pipes ABC
    c = a - b
    # calculate the height of the stack (in centimetres)
    h = 2 * a + c
    if not (h < 500): break
    # from pipes AAD: (a + d)^2 = a^2 + (a - d)^2
    d = div(a, 4)
    if d is None: continue
    # from pipes ABC: (a + e)^2 = (a + c - b - e)^2 + (b + c)^2
    e = div(sq(b) + sq(c) - a * (b - c), 2 * a - b + c)
    if e is None: continue
    # from pipes ADF: (d + f)^2 = (d - f)^2 + x^2; (a + f)^2 = (a - f)^2 + y^2; x + y = a
    r = is_square(a * d)
    if r is None: continue
    f = div(sq(a), 4 * (a + d + 2 * r))
    if f is None: continue
    if a > b > c > d > e > f > 0:
      # output solution (in cm)
      printf("height = {h} cm [a={a} b={b} c={c} d={d} e={e} f={f}]")
  

  Solution: The height of the stack is 420 cm.

  Note that the derivation of d requires a to also be divisible by 4 (as well as 3), so we could consider just multiples of 12 for a for a slight increase in speed.

  ---

  Manually:

  If we suppose a = 180k, then we can simplify the expressions used in the program to give the following values:

  a = 180k
  b = 120k
  c = 60k
  d = 45k
  e = 24k
  f = 20k

  And we require all these values to be positive integers, so k takes on a positive integer value.

  The total height of the stack is h = 2a + c = 420k, and we require h < 500.

  So the only permissible value is k = 1, and the answer follows directly.

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• 

  Frits 2:26 pm on 2 August 2022 Permalink | Reply

  Problem is when to stop with your analysis.
  Ultimately each diameter can be expressed in terms of the smallest diameter.

     
  from enigma import SubstitutedExpression
  
  # AGH, BIJ, CK, DL, EM and FN variables are diameters
  # if (x - y)^2 = z^2 then (2x - 2y)^2 = (2z)^2
  # so we can also calculate with diameters in the Pythagorean formulae
  
  # the alphametic puzzle
  p = SubstitutedExpression(
    [
      # type A 50 per cent wider than type B
      # diameter A is equal to diameter C + two times radius B
      # a height of less than 5 metres so 2a + c = 7c < 500
      "1 < CK < 72",
      "3 * CK = AGH",
      "2 * CK = BIJ",
      
      # (a + d)^2 = a^2 + (a - d)^2 so 4ad = a^2
      "div(AGH, 4) = DL",
      
      # (a + e)^2 = (a + c - b - e)^2 + (b + c)^2 using a = 3c and b = 2c
      # 6ce + 9c^2 = 4c^2 - 4ce + 9c^2
      "div(2 * CK, 5) = EM",
      
      # (d + f)^2 = (d - f)^2 + x^2; (a + f)^2 = (a - f)^2 + (a - x)^2;
      # so 4df = x^2 and 4af = (a - x)^2
      "4.0 * AGH * FN == (AGH - 2 * (DL * FN)**.5)**2",
      
      # A to F in decreasing size
      "AGH > BIJ > CK > DL > EM > FN"
    ],
    answer="2 * AGH + CK, (AGH, BIJ, CK, DL, EM, FN)",
    d2i=dict([(k, "CF") for k in range(8, 10)]),
    distinct="",
    verbose=0,    # use 256 to see the generated code
  )
  
  # print answers
  for (_, ans) in p.solve():
    print(f"the final height of the stack in centimetres: {ans[0]}")
  

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