Improved project structure.

This commit is contained in:
2017-06-01 14:50:23 +02:00
parent 5e06b03248
commit 89f0d81598
25 changed files with 2 additions and 0 deletions

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python/__init__.py Normal file
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python/e023.py Normal file
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import math
def get_proper_divisors(n):
proper_divisors = set([1])
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
proper_divisors.add(i)
proper_divisors.add(n / i)
return proper_divisors
def is_abundant(n):
return sum(get_proper_divisors(n)) > n
def get_abundant_numbers_smaller(n):
ret = []
for i in range(1, n):
if is_abundant(i):
ret.append(i)
return ret
def is_sum_of_two_abundant(n, abundant_numbers):
abundant_numbers_set = set(abundant_numbers)
for a1 in abundant_numbers:
if a1 > n:
return False
elif (n - a1) in abundant_numbers_set:
return True
if __name__ == "__main__":
abundant_numbers = get_abundant_numbers_smaller(30000)
cannot_be_written_as_sum_of_abundant = []
for i in range(28129):
if not is_sum_of_two_abundant(i, abundant_numbers):
cannot_be_written_as_sum_of_abundant.append(i)
print(sum(cannot_be_written_as_sum_of_abundant))

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from itertools import permutations
print("".join(list(permutations("0123456789"))[1000000-1]))

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python/e025.py Normal file
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from copy import deepcopy
from itertools import islice
def primes(n):
""" Nice way to calculate primes. Should be fast. """
l = range(2, n + 1)
_l = []
while True:
p = l[0]
if p * p > n:
return _l + l
l = [i for i in l if i % p != 0]
_l.append(p)
def calculate_decimal_places(numerator, denominator):
numerator = (numerator - (numerator / denominator) * denominator) * 10
digits = []
while True:
digit = numerator / denominator
numerator = (numerator - digit * denominator) * 10
digits.append(digit)
if digits[-3:] == [0, 0, 0]:
raise StopIteration
yield digit
def has_cycle(decimal_places, n):
d = decimal_places
return list(islice(d, 0, n)) == list(islice(d, 0, n)) and \
list(islice(d, 0, n)) == list(islice(d, 0, n))
def f_025_1():
""" Second try. I realized that only primes must be
checked. Therefore, my brute force approach worked. """
l = []
for d in primes(1000):
for i in range(5, 10000):
decimal_places = calculate_decimal_places(1, d)
if has_cycle(decimal_places, i):
l.append((i, d))
break
print(max(l))
def calculate_cycle(numerator, denominator):
numerator = (numerator - (numerator / denominator) * denominator) * 10
remainders = set([])
while True:
digit = numerator / denominator
remainder = (numerator - digit * denominator)
if remainder in remainders:
raise StopIteration
remainders.add(remainder)
numerator = remainder * 10
yield digit
def f_025_2():
""" Understood trick with remembering remainder... """
s = [(len(list(calculate_cycle(1, d))), d)
for d in range(1, 1001)]
print(max(s))
def f_025_3():
""" Only testing primes... """
s = [(len(list(calculate_cycle(1, d))), d)
for d in primes(10000)]
print(max(s))
f_025_3()
#print([(find_cycle_count(calculate_decimal_places(1, d)), d)
# for d in range(1, 100)])
#print(find_cycle_count(calculate_decimal_places(22, 7)))
#l = []
#for d in range(1, 1000):
# for i in range(5, 1000):
# decimal_places = calculate_decimal_places(1, d)
# if has_cycle_one_off(decimal_places, i):
# l.append((i, d))
# break
# decimal_places = calculate_decimal_places(1, d)
# if has_cycle(decimal_places, i):
# l.append((i, d))
# break
#def find_cycle_count(decimal_places):
# cycles = []
# for digit in decimal_places:
# new_cycles = []
# for cycle in cycles:
# digits, length = cycle
# if digits[0] == digit:
# if len(digits[1:]) == 0:
# return length
# new_cycles.append((digits[1:], length))
# new_cycles.append((digits + [digit], length + 1))
# new_cycles.append(([digit], 1))
# cycles = new_cycles

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def primes(n):
""" Nice way to calculate primes. Should be fast. """
l = range(2, n + 1)
_l = []
while True:
p = l[0]
if p * p > n:
return _l + l
l = [i for i in l if i % p != 0]
_l.append(p)
def produce_prime(a, b, n, primes):
x = n*n + a*n + b
return x in primes
def f_027():
""" n^2 + a*n + b
1) b must be prime
"""
p6 = set(primes(1000000))
p3 = primes(1000)
options = [(a, b)
for a in range(1, 1000, 2)
for b in p3]
print(len(options))
for n in range(100):
options = [(a, b)
for a, b in options
if produce_prime(a, b, n, p6)]
print(options)
print(len(options))
f_027()