Skip to content

Created problem_35 in project_euler #2309

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
merged 18 commits into from
Aug 16, 2020
Merged
Show file tree
Hide file tree
Changes from all commits
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
1 change: 1 addition & 0 deletions project_euler/problem_35/__init__.py
Original file line number Diff line number Diff line change
@@ -0,0 +1 @@
#
69 changes: 69 additions & 0 deletions project_euler/problem_35/sol1.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,69 @@
"""
The number 197 is called a circular prime because all rotations of the digits:
197, 971, and 719, are themselves prime.
There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73,
79, and 97.
How many circular primes are there below one million?

To solve this problem in an efficient manner, we will first mark all the primes
below 1 million using the Seive of Eratosthenes. Then, out of all these primes,
we will rule out the numbers which contain an even digit. After this we will
generate each circular combination of the number and check if all are prime.
"""
from typing import List

seive = [True] * 1000001
i = 2
while i * i <= 1000000:
if seive[i]:
for j in range(i * i, 1000001, i):
seive[j] = False
i += 1


def is_prime(n: int) -> bool:
"""
For 2 <= n <= 1000000, return True if n is prime.
>>> is_prime(87)
False
>>> is_prime(23)
True
>>> is_prime(25363)
False
"""
return seive[n]


def contains_an_even_digit(n: int) -> bool:
"""
Return True if n contains an even digit.
>>> contains_an_even_digit(0)
True
>>> contains_an_even_digit(975317933)
False
>>> contains_an_even_digit(-245679)
True
"""
return any(digit in "02468" for digit in str(n))


def find_circular_primes(limit: int = 1000000) -> List[int]:
"""
Return circular primes below limit.
>>> len(find_circular_primes(100))
13
>>> len(find_circular_primes(1000000))
55
"""
result = [2] # result already includes the number 2.
for num in range(3, limit + 1, 2):
if is_prime(num) and not contains_an_even_digit(num):
str_num = str(num)
list_nums = [int(str_num[j:] + str_num[:j]) for j in range(len(str_num))]
if all(is_prime(i) for i in list_nums):
result.append(num)
return result


if __name__ == "__main__":
print(f"{len(find_circular_primes()) = }")