Sunday Times Teaser 3106 – Underground Umbra Conundrum

by BRG on April 2, 2022

by Stephen Hogg

Published Sunday April 03 2022 (link)

An underground car park has rectangular walls, each covered with 300 abutting square tiles arranged in three repeats of red, orange, yellow, green and blue columns twenty tiles high. A sharp tapering shadow from a duct played across one wall from top to bottom (with straight lines similar to that shown).

Each shadow edge crossed every colour and, curiously, hit just two points where four tile corners met. The upper and lower pairs of these were on levels the same distance up and down from the mid-level line. I also noticed that no column was completely free of shadow, but just one column was completely in shadow.

Give the shadow’s area relative to the wall area as a fraction in simplest form.

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Brian Gladman permalink

This solution runs in 160 milliseconds with profile timing and finds two solutions with the same area. Its performance can be improved to 55 milliseconds by using the GMPY2 rational implementation in place of the standard Python fractions module.


from itertools import combinations
from math import ceil, floor
try:
  from gmpy2 import mpq as RF
except ImportError: 
  from fractions import Fraction as RF

class Line(object):

  # define a line through two points (a, b)
  def __init__(self, a, b):
    self.x0, self.y0 = a[0], a[1]
    self.t, self.b = b[0] - a[0], b[1] - a[1]

  # a function giving x as a function of y on the line
  def x(self, y):
    return RF(self.x0 * self.b + (y - self.y0) * self.t, self.b)

# consider lines above and below the horizontal centre line at
# y = v and -v where the shadow edges cross four tile corners
for v in range(1, 10):
  
  # consider shadow edge positions on the lines above (tl, tr)
  # and below (bl, br) the midline
  for bl, tl, br, tr in combinations(range(17), 4):
    l = Line((bl, -v), (tl, v))
    r = Line((br, -v), (tr, v))
    
    # the left/right shadow edges must include the first/last 
    # columns respectively; each must span more than 4 columns
    
    if not (0 < l.x(-10) < 1 and l.x(10) > 4 and
           14 < r.x(10) < 15 and r.x(-10) < 11):
      continue
          
    # one column is totally in the shadow
    if floor(r.x(-10)) - ceil(l.x(10)) != 1:
      continue

    # enforce two four tile corners on each shadow edge    
    if (sum(l.x(y).denominator == 1 for y in range(-10, 10)) != 2 or
        sum(r.x(y).denominator == 1 for y in range(-10, 10)) != 2):
      continue
    
    # check for a positive line slope difference and find shadow area
    if tr - tl < br - bl:
      s = RF((tr + br) - (tl + bl), 2)
      print(f"Area = {s}/15 [v = {v}, l = {(bl, tl)}, r = {(br, tr)}]")

from itertools import combinations

from math import ceil, floor

try:

from gmpy2 import mpq as RF

except ImportError:

from fractions import Fraction as RF

class Line(object):

# define a line through two points (a, b)

def __init__(self, a, b):

self.x0, self.y0 = a[0], a[1]

self.t, self.b = b[0] - a[0], b[1] - a[1]

# a function giving x as a function of y on the line

def x(self, y):

return RF(self.x0 * self.b + (y - self.y0) * self.t, self.b)

# consider lines above and below the horizontal centre line at

# y = v and -v where the shadow edges cross four tile corners

for v in range(1, 10):

# consider shadow edge positions on the lines above (tl, tr)

# and below (bl, br) the midline

for bl, tl, br, tr in combinations(range(17), 4):

l = Line((bl, -v), (tl, v))

r = Line((br, -v), (tr, v))

# the left/right shadow edges must include the first/last

# columns respectively; each must span more than 4 columns

if not (0 < l.x(-10) < 1 and l.x(10) > 4 and

14 < r.x(10) < 15 and r.x(-10) < 11):

continue

# one column is totally in the shadow

if floor(r.x(-10)) - ceil(l.x(10)) != 1:

continue

# enforce two four tile corners on each shadow edge

if (sum(l.x(y).denominator == 1 for y in range(-10, 10)) != 2 or

sum(r.x(y).denominator == 1 for y in range(-10, 10)) != 2):

continue

# check for a positive line slope difference and find shadow area

if tr - tl < br - bl:

s = RF((tr + br) - (tl + bl), 2)

print(f"Area = {s}/15 [v = {v}, l = {(bl, tl)}, r = {(br, tr)}]")

Sunday Times Teaser 3106 – Underground Umbra Conundrum

by Stephen Hogg

Published Sunday April 03 2022 (link)

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