Ejemplo n.º 1
0
 
""" confusing spiral patterns, read the description on the site

http://projecteuler.net/index.php?section=problems&id=28

"""
# the key to this one is to generate the diagonals, not the whole
# spiral. It's not that hard if you notice each additional layer adds a
# range of numbers, and the diagonals are situated at the 1/4, 1/2, 3/4, 1
# parts of the range.

from lib import spiraldiagonals



diagonals = [1]
last_number = 1

generator = spiraldiagonals.diagonals()

for x in range(3, 1002, 2):
  ds = generator.next()
  diagonals += [d for d in ds]

print sum(diagonals)
Ejemplo n.º 2
0
 
"""
http://projecteuler.net/index.php?section=problems&id=58
"""

from lib import spiraldiagonals
from lib import euler

"""
this is pretty easy - we can generate the diagonals indefinitely from the
formula we came up with in q28 and simply check the prime ratio

it's a bit slow, probably the prime check, but still less than a minute
"""

primes = []
ds = [1]
for i, diagonals in enumerate(spiraldiagonals.diagonals()):
  
  for d in diagonals:
    ds += [d]
    if euler.is_prime(d): primes += [d]
  ratio = float(len(primes)) / len(ds)
  if ratio <= 0.1:
    print (i+1)*2 + 1  # this is the spiral width
    break