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9200a2e543
* from __future__ import annotations
* fixup! from __future__ import annotations
* fixup! from __future__ import annotations
* fixup! Format Python code with psf/black push
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70 lines
2.0 KiB
Python
70 lines
2.0 KiB
Python
"""
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The number 197 is called a circular prime because all rotations of the digits:
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197, 971, and 719, are themselves prime.
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There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73,
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79, and 97.
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How many circular primes are there below one million?
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To solve this problem in an efficient manner, we will first mark all the primes
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below 1 million using the Seive of Eratosthenes. Then, out of all these primes,
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we will rule out the numbers which contain an even digit. After this we will
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generate each circular combination of the number and check if all are prime.
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"""
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from __future__ import annotations
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seive = [True] * 1000001
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i = 2
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while i * i <= 1000000:
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if seive[i]:
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for j in range(i * i, 1000001, i):
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seive[j] = False
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i += 1
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def is_prime(n: int) -> bool:
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"""
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For 2 <= n <= 1000000, return True if n is prime.
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>>> is_prime(87)
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False
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>>> is_prime(23)
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True
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>>> is_prime(25363)
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False
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"""
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return seive[n]
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def contains_an_even_digit(n: int) -> bool:
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"""
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Return True if n contains an even digit.
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>>> contains_an_even_digit(0)
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True
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>>> contains_an_even_digit(975317933)
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False
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>>> contains_an_even_digit(-245679)
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True
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"""
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return any(digit in "02468" for digit in str(n))
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def find_circular_primes(limit: int = 1000000) -> list[int]:
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"""
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Return circular primes below limit.
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>>> len(find_circular_primes(100))
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13
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>>> len(find_circular_primes(1000000))
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55
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"""
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result = [2] # result already includes the number 2.
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for num in range(3, limit + 1, 2):
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if is_prime(num) and not contains_an_even_digit(num):
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str_num = str(num)
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list_nums = [int(str_num[j:] + str_num[:j]) for j in range(len(str_num))]
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if all(is_prime(i) for i in list_nums):
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result.append(num)
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return result
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if __name__ == "__main__":
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print(f"{len(find_circular_primes()) = }")
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