|
| 1 | +""" |
| 2 | +https://projecteuler.net/problem=51 |
| 3 | +Prime digit replacements |
| 4 | +Problem 51 |
| 5 | +
|
| 6 | +By replacing the 1st digit of the 2-digit number *3, it turns out that six of |
| 7 | +the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime. |
| 8 | +
|
| 9 | +By replacing the 3rd and 4th digits of 56**3 with the same digit, this 5-digit |
| 10 | +number is the first example having seven primes among the ten generated numbers, |
| 11 | +yielding the family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993. |
| 12 | +Consequently 56003, being the first member of this family, is the smallest prime |
| 13 | +with this property. |
| 14 | +
|
| 15 | +Find the smallest prime which, by replacing part of the number (not necessarily |
| 16 | +adjacent digits) with the same digit, is part of an eight prime value family. |
| 17 | +""" |
| 18 | + |
| 19 | +from collections import Counter |
| 20 | +from typing import List |
| 21 | + |
| 22 | + |
| 23 | +def prime_sieve(n: int) -> List[int]: |
| 24 | + """ |
| 25 | + Sieve of Erotosthenes |
| 26 | + Function to return all the prime numbers up to a certain number |
| 27 | + https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes |
| 28 | +
|
| 29 | + >>> prime_sieve(3) |
| 30 | + [2] |
| 31 | +
|
| 32 | + >>> prime_sieve(50) |
| 33 | + [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47] |
| 34 | + """ |
| 35 | + is_prime = [True] * n |
| 36 | + is_prime[0] = False |
| 37 | + is_prime[1] = False |
| 38 | + is_prime[2] = True |
| 39 | + |
| 40 | + for i in range(3, int(n ** 0.5 + 1), 2): |
| 41 | + index = i * 2 |
| 42 | + while index < n: |
| 43 | + is_prime[index] = False |
| 44 | + index = index + i |
| 45 | + |
| 46 | + primes = [2] |
| 47 | + |
| 48 | + for i in range(3, n, 2): |
| 49 | + if is_prime[i]: |
| 50 | + primes.append(i) |
| 51 | + |
| 52 | + return primes |
| 53 | + |
| 54 | + |
| 55 | +def digit_replacements(number: int) -> List[List[int]]: |
| 56 | + """ |
| 57 | + Returns all the possible families of digit replacements in a number which |
| 58 | + contains at least one repeating digit |
| 59 | +
|
| 60 | + >>> digit_replacements(544) |
| 61 | + [[500, 511, 522, 533, 544, 555, 566, 577, 588, 599]] |
| 62 | +
|
| 63 | + >>> digit_replacements(3112) |
| 64 | + [[3002, 3112, 3222, 3332, 3442, 3552, 3662, 3772, 3882, 3992]] |
| 65 | + """ |
| 66 | + number = str(number) |
| 67 | + replacements = [] |
| 68 | + digits = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"] |
| 69 | + |
| 70 | + for duplicate in Counter(number) - Counter(set(number)): |
| 71 | + family = [int(number.replace(duplicate, digit)) for digit in digits] |
| 72 | + replacements.append(family) |
| 73 | + |
| 74 | + return replacements |
| 75 | + |
| 76 | + |
| 77 | +def solution(family_length: int = 8) -> int: |
| 78 | + """ |
| 79 | + Returns the solution of the problem |
| 80 | +
|
| 81 | + >>> solution(2) |
| 82 | + 229399 |
| 83 | +
|
| 84 | + >>> solution(3) |
| 85 | + 221311 |
| 86 | + """ |
| 87 | + numbers_checked = set() |
| 88 | + |
| 89 | + # Filter primes with less than 3 replaceable digits |
| 90 | + primes = { |
| 91 | + x for x in set(prime_sieve(1_000_000)) if len(str(x)) - len(set(str(x))) >= 3 |
| 92 | + } |
| 93 | + |
| 94 | + for prime in primes: |
| 95 | + if prime in numbers_checked: |
| 96 | + continue |
| 97 | + |
| 98 | + replacements = digit_replacements(prime) |
| 99 | + |
| 100 | + for family in replacements: |
| 101 | + numbers_checked.update(family) |
| 102 | + primes_in_family = primes.intersection(family) |
| 103 | + |
| 104 | + if len(primes_in_family) != family_length: |
| 105 | + continue |
| 106 | + |
| 107 | + return min(primes_in_family) |
| 108 | + |
| 109 | + |
| 110 | +if __name__ == "__main__": |
| 111 | + print(solution()) |
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