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* Add comments and wikipedia link in calculate_prime_numbers * Add improved calculate_prime_numbers * Separate slow_solution and new_solution * Use for loops in solution * Separate while_solution and new solution * Add performance benchmark * Add doctest for calculate_prime_numbers * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * Removed white space --------- Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
163 lines
4.4 KiB
Python
163 lines
4.4 KiB
Python
"""
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Project Euler Problem 187: https://projecteuler.net/problem=187
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A composite is a number containing at least two prime factors.
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For example, 15 = 3 x 5; 9 = 3 x 3; 12 = 2 x 2 x 3.
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There are ten composites below thirty containing precisely two,
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not necessarily distinct, prime factors: 4, 6, 9, 10, 14, 15, 21, 22, 25, 26.
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How many composite integers, n < 10^8, have precisely two,
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not necessarily distinct, prime factors?
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"""
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from math import isqrt
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def slow_calculate_prime_numbers(max_number: int) -> list[int]:
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"""
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Returns prime numbers below max_number.
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See: https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
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>>> slow_calculate_prime_numbers(10)
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[2, 3, 5, 7]
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>>> slow_calculate_prime_numbers(2)
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[]
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"""
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# List containing a bool value for every number below max_number/2
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is_prime = [True] * max_number
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for i in range(2, isqrt(max_number - 1) + 1):
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if is_prime[i]:
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# Mark all multiple of i as not prime
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for j in range(i**2, max_number, i):
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is_prime[j] = False
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return [i for i in range(2, max_number) if is_prime[i]]
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def calculate_prime_numbers(max_number: int) -> list[int]:
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"""
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Returns prime numbers below max_number.
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See: https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
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>>> calculate_prime_numbers(10)
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[2, 3, 5, 7]
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>>> calculate_prime_numbers(2)
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[]
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"""
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if max_number <= 2:
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return []
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# List containing a bool value for every odd number below max_number/2
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is_prime = [True] * (max_number // 2)
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for i in range(3, isqrt(max_number - 1) + 1, 2):
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if is_prime[i // 2]:
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# Mark all multiple of i as not prime using list slicing
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is_prime[i**2 // 2 :: i] = [False] * (
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# Same as: (max_number - (i**2)) // (2 * i) + 1
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# but faster than len(is_prime[i**2 // 2 :: i])
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len(range(i**2 // 2, max_number // 2, i))
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)
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return [2] + [2 * i + 1 for i in range(1, max_number // 2) if is_prime[i]]
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def slow_solution(max_number: int = 10**8) -> int:
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"""
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Returns the number of composite integers below max_number have precisely two,
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not necessarily distinct, prime factors.
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>>> slow_solution(30)
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10
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"""
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prime_numbers = slow_calculate_prime_numbers(max_number // 2)
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semiprimes_count = 0
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left = 0
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right = len(prime_numbers) - 1
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while left <= right:
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while prime_numbers[left] * prime_numbers[right] >= max_number:
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right -= 1
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semiprimes_count += right - left + 1
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left += 1
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return semiprimes_count
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def while_solution(max_number: int = 10**8) -> int:
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"""
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Returns the number of composite integers below max_number have precisely two,
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not necessarily distinct, prime factors.
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>>> while_solution(30)
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10
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"""
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prime_numbers = calculate_prime_numbers(max_number // 2)
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semiprimes_count = 0
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left = 0
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right = len(prime_numbers) - 1
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while left <= right:
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while prime_numbers[left] * prime_numbers[right] >= max_number:
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right -= 1
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semiprimes_count += right - left + 1
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left += 1
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return semiprimes_count
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def solution(max_number: int = 10**8) -> int:
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"""
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Returns the number of composite integers below max_number have precisely two,
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not necessarily distinct, prime factors.
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>>> solution(30)
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10
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"""
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prime_numbers = calculate_prime_numbers(max_number // 2)
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semiprimes_count = 0
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right = len(prime_numbers) - 1
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for left in range(len(prime_numbers)):
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if left > right:
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break
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for r in range(right, left - 2, -1):
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if prime_numbers[left] * prime_numbers[r] < max_number:
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break
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right = r
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semiprimes_count += right - left + 1
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return semiprimes_count
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def benchmark() -> None:
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"""
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Benchmarks
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"""
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# Running performance benchmarks...
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# slow_solution : 108.50874730000032
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# while_sol : 28.09581200000048
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# solution : 25.063097400000515
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from timeit import timeit
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print("Running performance benchmarks...")
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print(f"slow_solution : {timeit('slow_solution()', globals=globals(), number=10)}")
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print(f"while_sol : {timeit('while_solution()', globals=globals(), number=10)}")
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print(f"solution : {timeit('solution()', globals=globals(), number=10)}")
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if __name__ == "__main__":
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print(f"Solution: {solution()}")
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benchmark()
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