S2T2

Jim Randell 5:07 pm on 13 November 2020 Permalink Reply
Tags: by: Graham Smithers ( 50 )   

• 

  Jim Randell 5:44 pm on 13 November 2020 Permalink | Reply

  We can solve this puzzle using applications of Pythagoras’ Theorem.

  This Python program runs in 47ms.

  Run: [ @repl.it ]

  from enigma import powers, div, is_square, printf
  
  # difference of 2 squares
  diff_sq = lambda x, y: (x + y) * (x - y)
  
  # length of the larger bridge (= B)
  B = 3969
  
  # consider the length of the road to the larger reservoir (= R)
  for R in powers(1, 22, 2):
    # calculate distance OY (= t)
    t2 = is_square(diff_sq(B + 2 * R, B))
    if t2 is None: continue
    t = div(t2, 2)
    if t is None: continue
  
    # consider length of the road to the smaller reservior (= r)
    for r in powers(1, 22, 2):
      # calculate the length of the bridge over the smaller reservoir (= b)
      b = div(diff_sq(t, r), r)
      if b is None or not(r < b < B and is_square(b)): continue
  
      # output solution
      printf("B={B} R={R}, t={t}, r={r} b={b}")
  

  Solution: The smaller bridge is 2304 m long.

  The layout looks like this:

  

  And the puzzle can be solved by considering the right-angled triangles OYX’ and OYX.

  The calculated distances are:

  OY = 1040 m
  OA = 400 m (20²)
  AB = 2304 m (48²)
  OA’ = 256 m (16²)
  A’B’ = 3969 m (63²)

  LikeLike

  • 

    Jim Randell 2:29 pm on 14 November 2020 Permalink | Reply

    Or, with a bit more analysis:

    from enigma import irange, inf, is_square, div, printf
    
    # length of the larger bridge (= B)
    B = 3969 # = 63^2
    
    # consider increasing squares (larger than B)
    for z in irange(64, inf):
      # calculate length of the road to the larger reservoir (= R)
      R = z * z - B
      if not(R < 500): break
      # calculate the length of the track OY (= t)
      t = is_square(R)
      if t is None: continue
      t *= z
    
      # consider length of the road to the smaller reservoir (= r)
      for j in irange(1, 22):
        r = j * j 
        # calculate the length of the bridge over the smaller reservoir (= b)
        b = div((t + r) * (t - r), r)
        if b is None or not(r < b < B and is_square(b)): continue
    
        # output solution
        printf("B={B} R={R}, t={t}; r={r} b={b}")
    

    LikeLike

• 

  Tony Brooke-Taylor 11:34 am on 17 November 2020 Permalink | Reply

  I first did this assuming that (using your notation) R+B also had to be a square. If I interpret your code correctly, I think that is also what you have done. I got the same value for R as your code produces. Following this through you get a unique solution for r. However, on reflection I don’t think the puzzle applies that constraint. In the puzzle’s notation, that would require OB and its equivalent to be squares, which I cannot see in the puzzle. If I relax the constraint I get three solutions. I am guessing that either the puzzle requires the addition of that constraint or I cannot read, but the more interesting question for me is whether or not it is more than a coincidence that the unique solution has that additional property.

  LikeLike

  • 

    Jim Randell 11:56 am on 17 November 2020 Permalink | Reply

    @Tony: Thanks for your comment.

    We don’t need to assume that (R + B) is a perfect square (and the puzzle doesn’t tell us that), but it follows from considering the right-angled triangle OYX’ (where X’ is the centre point of the larger reservoir).

    We have (where integer t is the length of the track OY):

    t² + (B/2)² = (R + B/2)²
    ⇒ t² = R(R + B)

    Now, we are told that R is a square number, and obviously t² is, so it follows that (R + B) must also be a perfect square.

    LikeLike

• 

  Tony Brooke-Taylor 6:31 pm on 17 November 2020 Permalink | Reply

  Thanks Jim. The constraint I had failed to apply was that t must be an integer.

  LikeLike

• 

  Frits 9:20 pm on 17 November 2020 Permalink | Reply

  from enigma import SubstitutedExpression
  
  # the alphametic puzzle
  p = SubstitutedExpression(
    [
     # PQRS = track OY, AE = sqrt(road OA), BX = sqrt(AB), FGH = road OC 
     # X is center of smaller reservoir
     
     # roads are shorter than 500m
     "AE > 0 and AE < 23",
     
     # roads are shorter than their associated bridges 
     "AE < BX",
     
     # one diameter is larger
     "BX > 0 and BX**2 < 3969",
   
     "PQRS > 0",
     
     # Pythagoras: OY^2 + YX^2 = (OA + YX)^2
     # so OY^2 = OA^2 + OA*AB
     "PQRS**2 == AE**4 + AE**2 * BX**2",
     
     "FGH < 500",
     # one road is longer
     "FGH < AE**2",
        
     # Pythagoras: OY^2 + (3969/2)^2 = ((3969/2) + OC)^2
     "PQRS**2 + 3938240.25 == (1984.5 + FGH)**2", 
    ],
    answer="BX**2, AE**2, PQRS, FGH",
    d2i={},
    distinct="",
    reorder=0,
    verbose=256)   
  
  # Print answers 
  print("Large bridge  small bridge  long road  track  short road")
  for (_, ans) in p.solve():
    print(f"    3969          {ans[0]}         {ans[1]}       {ans[2]}"
          f"     {ans[3]}")
  

  LikeLike