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bcfca67faa
* [mypy] fix type annotations for problem003/sol1 and problem003/sol3 * [mypy] fix type annotations for project euler problem007/sol2 * [mypy] fix type annotations for project euler problem008/sol2 * [mypy] fix type annotations for project euler problem009/sol1 * [mypy] fix type annotations for project euler problem014/sol1 * [mypy] fix type annotations for project euler problem 025/sol2 * [mypy] fix type annotations for project euler problem026/sol1.py * [mypy] fix type annotations for project euler problem037/sol1 * [mypy] fix type annotations for project euler problem044/sol1 * [mypy] fix type annotations for project euler problem046/sol1 * [mypy] fix type annotations for project euler problem051/sol1 * [mypy] fix type annotations for project euler problem074/sol2 * [mypy] fix type annotations for project euler problem080/sol1 * [mypy] fix type annotations for project euler problem099/sol1 * [mypy] fix type annotations for project euler problem101/sol1 * [mypy] fix type annotations for project euler problem188/sol1 * [mypy] fix type annotations for project euler problem191/sol1 * [mypy] fix type annotations for project euler problem207/sol1 * [mypy] fix type annotations for project euler problem551/sol1
114 lines
3.0 KiB
Python
114 lines
3.0 KiB
Python
"""
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https://projecteuler.net/problem=51
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Prime digit replacements
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Problem 51
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By replacing the 1st digit of the 2-digit number *3, it turns out that six of
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the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime.
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By replacing the 3rd and 4th digits of 56**3 with the same digit, this 5-digit
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number is the first example having seven primes among the ten generated numbers,
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yielding the family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993.
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Consequently 56003, being the first member of this family, is the smallest prime
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with this property.
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Find the smallest prime which, by replacing part of the number (not necessarily
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adjacent digits) with the same digit, is part of an eight prime value family.
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"""
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from __future__ import annotations
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from collections import Counter
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def prime_sieve(n: int) -> list[int]:
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"""
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Sieve of Erotosthenes
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Function to return all the prime numbers up to a certain number
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https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
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>>> prime_sieve(3)
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[2]
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>>> prime_sieve(50)
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[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
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"""
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is_prime = [True] * n
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is_prime[0] = False
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is_prime[1] = False
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is_prime[2] = True
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for i in range(3, int(n ** 0.5 + 1), 2):
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index = i * 2
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while index < n:
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is_prime[index] = False
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index = index + i
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primes = [2]
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for i in range(3, n, 2):
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if is_prime[i]:
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primes.append(i)
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return primes
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def digit_replacements(number: int) -> list[list[int]]:
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"""
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Returns all the possible families of digit replacements in a number which
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contains at least one repeating digit
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>>> digit_replacements(544)
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[[500, 511, 522, 533, 544, 555, 566, 577, 588, 599]]
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>>> digit_replacements(3112)
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[[3002, 3112, 3222, 3332, 3442, 3552, 3662, 3772, 3882, 3992]]
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"""
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number_str = str(number)
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replacements = []
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digits = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
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for duplicate in Counter(number_str) - Counter(set(number_str)):
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family = [int(number_str.replace(duplicate, digit)) for digit in digits]
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replacements.append(family)
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return replacements
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def solution(family_length: int = 8) -> int:
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"""
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Returns the solution of the problem
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>>> solution(2)
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229399
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>>> solution(3)
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221311
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"""
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numbers_checked = set()
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# Filter primes with less than 3 replaceable digits
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primes = {
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x for x in set(prime_sieve(1_000_000)) if len(str(x)) - len(set(str(x))) >= 3
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}
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for prime in primes:
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if prime in numbers_checked:
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continue
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replacements = digit_replacements(prime)
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for family in replacements:
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numbers_checked.update(family)
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primes_in_family = primes.intersection(family)
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if len(primes_in_family) != family_length:
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continue
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return min(primes_in_family)
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return -1
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if __name__ == "__main__":
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print(solution())
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