S2T2

Jim Randell 3:59 pm on 7 May 2020 Permalink Reply
Tags: by: Howard Williams ( 9 )   

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  Jim Randell 4:33 pm on 7 May 2020 Permalink | Reply

  Assuming an x by y rectangle, with x > y.

  If there are k repeats of the (red, grey yellow) bands, then y = 16k – 1.

  And the first band of red stones requires 2x + 2y – 4 = 402 stones, so x + y = 203.

  Here is a constructive program that counts the number of stones in each band. It runs in 78ms.

  from enigma import irange, inf, printf
  
  # calculate the number of stones with <k> repeats
  # and the initial band being <x> by <y>
  def stones(k, y, x):
    d = dict(R=0, G=0, Y=0)
    for i in ("RRRRGGGY" * k):
      if y < 2:
        d[i] += x * y
        break
      d[i] += 2 * (x + y - 2)
      x -= 2
      y -= 2
    return (d['R'], d['G'], d['Y'])
  
  # consider the number of repeats of the bands
  for k in irange(1, inf):
    y = 16 * k - 1
    x = 203 - y
    if not(x > y): break
   
    # calculate the number of stones needed
    (R, G, Y) = stones(k, y, x)
   
    # output solution
    if 5 * Y == R:
      printf("k={k}, y={y} x={x}; R={R} G={G} Y={Y}")
  

  Solution: James requires 5520 red stones.

  The patio consists of 6 red/grey/yellow layers, finishing with a single row of 14 yellow stones:

  

  The number of stones required is: red = 5520, grey = 3636, yellow = 1104; total = 10260.

  The outer band measures 108×95 stones.

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  Jim Randell 9:33 am on 10 May 2020 Permalink | Reply

  Each band requires 8 fewer stones than the previous band, so we can just start with the outermost band (402 stones) and work our way in. When we get to the number blocks for the yellow band we can adjust the number to account for a single line to see if we have reached the middle of the design, and if not carry on. We don’t need to do any analysis to work out the dimensions of the patio, or consider the number of repeats.

  This gives us a simpler, and slightly shorter, program.

  Run: [ @repl.it ]

  from enigma import irange, chunk, printf
  
  # number of blocks in outer band
  n = 402
  
  # record Red, Grey, Yellow blocks required
  R = G = Y = 0
  
  # work inwards in chunks of 8 bands
  # each band has 8 less blocks than the previous band
  for s in chunk(irange(n, 1, step=-8), 8):
    if len(s) < 8: break
    (r1, r2, r3, r4, g1, g2, g3, y) = s
    R += r1 + r2 + r3 + r4
    G += g1 + g2 + g3
    # the middle is a single row
    m = 1 + y // 2
    if 5 * (Y + m) == R:
      printf("R={R} G={G} Y={Y}", Y=Y + m)
    # otherwise, carry on
    Y += y
  

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