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2104fa7aeb
* fixes #5434 * fixes broken solution * removes assert * removes assert * Apply suggestions from code review Co-authored-by: John Law <johnlaw.po@gmail.com> * Update project_euler/problem_003/sol1.py Co-authored-by: John Law <johnlaw.po@gmail.com>
154 lines
3.9 KiB
Python
154 lines
3.9 KiB
Python
"""
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Prime permutations
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Problem 49
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The arithmetic sequence, 1487, 4817, 8147, in which each of
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the terms increases by 3330, is unusual in two ways:
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(i) each of the three terms are prime,
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(ii) each of the 4-digit numbers are permutations of one another.
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There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes,
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exhibiting this property, but there is one other 4-digit increasing sequence.
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What 12-digit number do you form by concatenating the three terms in this sequence?
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Solution:
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First, we need to generate all 4 digits prime numbers. Then greedy
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all of them and use permutation to form new numbers. Use binary search
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to check if the permutated numbers is in our prime list and include
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them in a candidate list.
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After that, bruteforce all passed candidates sequences using
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3 nested loops since we know the answer will be 12 digits.
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The bruteforce of this solution will be about 1 sec.
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"""
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import math
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from itertools import permutations
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def is_prime(number: int) -> bool:
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"""Checks to see if a number is a prime in O(sqrt(n)).
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A number is prime if it has exactly two factors: 1 and itself.
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>>> is_prime(0)
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False
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>>> is_prime(1)
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False
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>>> is_prime(2)
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True
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>>> is_prime(3)
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True
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>>> is_prime(27)
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False
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>>> is_prime(87)
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False
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>>> is_prime(563)
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True
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>>> is_prime(2999)
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True
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>>> is_prime(67483)
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False
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"""
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if 1 < number < 4:
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# 2 and 3 are primes
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return True
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elif number < 2 or number % 2 == 0 or number % 3 == 0:
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# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
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return False
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# All primes number are in format of 6k +/- 1
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for i in range(5, int(math.sqrt(number) + 1), 6):
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if number % i == 0 or number % (i + 2) == 0:
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return False
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return True
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def search(target: int, prime_list: list) -> bool:
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"""
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function to search a number in a list using Binary Search.
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>>> search(3, [1, 2, 3])
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True
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>>> search(4, [1, 2, 3])
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False
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>>> search(101, list(range(-100, 100)))
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False
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"""
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left, right = 0, len(prime_list) - 1
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while left <= right:
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middle = (left + right) // 2
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if prime_list[middle] == target:
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return True
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elif prime_list[middle] < target:
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left = middle + 1
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else:
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right = middle - 1
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return False
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def solution():
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"""
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Return the solution of the problem.
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>>> solution()
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296962999629
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"""
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prime_list = [n for n in range(1001, 10000, 2) if is_prime(n)]
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candidates = []
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for number in prime_list:
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tmp_numbers = []
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for prime_member in permutations(list(str(number))):
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prime = int("".join(prime_member))
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if prime % 2 == 0:
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continue
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if search(prime, prime_list):
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tmp_numbers.append(prime)
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tmp_numbers.sort()
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if len(tmp_numbers) >= 3:
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candidates.append(tmp_numbers)
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passed = []
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for candidate in candidates:
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length = len(candidate)
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found = False
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for i in range(length):
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for j in range(i + 1, length):
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for k in range(j + 1, length):
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if (
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abs(candidate[i] - candidate[j])
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== abs(candidate[j] - candidate[k])
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and len({candidate[i], candidate[j], candidate[k]}) == 3
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):
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passed.append(
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sorted([candidate[i], candidate[j], candidate[k]])
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)
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found = True
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if found:
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break
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if found:
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break
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if found:
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break
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answer = set()
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for seq in passed:
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answer.add("".join([str(i) for i in seq]))
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return max(int(x) for x in answer)
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if __name__ == "__main__":
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print(solution())
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