S2T2

Jim Randell 9:13 am on 16 September 2021 Permalink Reply
Tags: by: Adrian Winder   

• 

  Jim Randell 9:14 am on 16 September 2021 Permalink | Reply

  

  The direct route (h = AB) and the indirect route (z = AB′ = z1 + z2) are the hypotenuses of right-angled triangles (AXB and AXB′ respectively):

  h² = x² + d²
  z² = (146 − x)² + d²

  For a given value of x we can determine values for h, d and z by looking at the divisors of x²:

  h² − d² = x²
  (h − d)(h + d) = x² = a⋅b
  h = (a + b) / 2
  d = (b − a) / 2
  z = √((146 − x)² + d²)

  This Python program runs in 51ms. (Internal run time is 1.3ms).

  Run: [ @replit ]

  from enigma import (irange, divisors_pairs, div, is_square, printf)
  
  for x in irange(1, 72):
    for (a, b) in divisors_pairs(x * x):
      h = div(a + b, 2)
      d = div(b - a, 2)
      if not (h and d): continue
      t = 146 - x
      z = is_square(t * t + d * d)
      if not z: continue
      printf("x={x}; h={h} d={d} z={z}")
  

  Solution: The man walks 169 ft between A and B. The direct distance is 123 ft.

  B is 46 ft from the river (BN = 46). And M and N are 120 ft apart (MN = 120).

  LikeLike

• 

  Frits 11:32 am on 16 September 2021 Permalink | Reply

  This Python program runs in 75ms (with PyPy).

    
  from enigma import SubstitutedExpression, is_square
  
  # h^2 = x^2 + d^2                      d < 2592 ((72^2 - 1) / 2) 
  # z^2 = (146 – x)^2 + d^2  
  
  # the alphametic puzzle
  p = SubstitutedExpression([
      "DE < 26",
      "0 < XY < 73", 
      "is_square(XY**2 + DEFG**2) = HIJK",
      "is_square((146 - XY)**2 + DEFG**2) = ZABC",
      "DEFG > 0",
    ],
    answer="(ZABC, HIJK)",
    distinct="",        # allow symbols with same values
    d2i=dict([(k, "D") for k in range(3, 10)]),
    verbose=256,
  )
   
  # print answers
  for (_, ans) in p.solve(verbose=0):
    print(f"The man walks {ans[0]} ft between A and B. "
          f"The direct distance is {ans[1]} ft.")
  

  LikeLike

• 

  Frits 12:34 pm on 16 September 2021 Permalink | Reply

  This Python program runs in 4ms.

    
  from enigma import pythagorean_triples, is_square
  
  # h^2 = x^2 + d^2                      h < 2593 ((72^2 + 1) / 2) 
  # z^2 = (146 – x)^2 + d^2  
  # x < 73
  
  for (p, q, h) in pythagorean_triples(2592):
    if p > 72: continue
    if q > 72:
      # (x, d) must be (p, q)
      z = is_square((146 - p)**2 + q**2)
    else:
      # (x, d) can be (p, q) or (q, p)
      z = is_square((146 - p)**2 + q**2)
      if not z:
        z = is_square((146 - q)**2 + p**2)
      
    if z:
      print(f"The man walks {z} ft between A and B. "
            f"The direct distance is {h} ft.")
  

  LikeLike

  • 

    Jim Randell 1:58 pm on 16 September 2021 Permalink | Reply

    Here’s a version that only uses the list of Pythagorean triples (and doesn’t use [[ is_square() ]]).

    from enigma import (pythagorean_triples, ordered, printf)
    
    # we can show:
    #
    #   d <= (x^2 - 1) / 2
    #
    # and max x = 72, so:
    #
    #   d < 2592
    #
    # hence:
    #
    #   z^2 <= 145^2 + 2592^2
    #   z < 2597
    
    # find pythagorean triples, map: (a, b) -> c
    triples = dict(((a, b), c) for (a, b, c) in pythagorean_triples(2597))
    
    # consider triples
    for ((a, b), c) in triples.items():
      if a > 72: continue
      # consider x=a, d=b, h=c
      z = triples.get(ordered(b, 146 - a))
      if z:
        printf("x={a}; h={c} d={b} z={z}")
      if b < 73:
        # consider x=b, d=a, h=c
        z = triples.get(ordered(a, 146 - b))
        if z:
          printf("x={b}; h={c} d={a} z={z}")
    

    Although it is not quite as fast as my original program.

    LikeLike

• 

  John Crabtree 2:52 pm on 22 September 2021 Permalink | Reply

  In theory there should be a way to solve brain teasers manually. Today I take this to mean without the use of computers or programable calculators. When this teaser was published in 1964, handheld calculators were some way off.

  Putting h = d + m and z = d + n, and then
  d = (x^2 – m^2)/(2m) = ((146 – x)^2 – n^2)/(2n) with n > m
  where if x is odd, m and n are odd, and if x ix even, m and n are even.
  Effectively, this is a way of finding Pythagorean triples with a common shorter side.

  Then for each value of x, one can look for values of m and n which give the same value of d. I did this with the aid of a calculator. I could have done it by hand with a table of squares, but it would have been tedious. The ratio of n/m is approximately (146 – x)^2/x^2. One can use this to reduce the number of calculations. The results took up two and half sides of paper. I found x = 27, m = 3, n = 49, which gave d = 120. The solution = d + n = 169.

  In summary, this is very challenging but accessible teaser. If Brian Gladman’s table of Pythagorean triples sorted by shorter sides went to 140, it would be very straightforward.

  LikeLike

  • 

    John Crabtree 5:02 pm on 22 September 2021 Permalink | Reply

    The man walks d + n = 169 feet. The direct distance is d + m = 123 feet.

    LikeLike