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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
99 lines
2.6 KiB
Python
99 lines
2.6 KiB
Python
"""
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The number 3797 has an interesting property. Being prime itself, it is possible
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to continuously remove digits from left to right, and remain prime at each stage:
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3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.
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Find the sum of the only eleven primes that are both truncatable from left to right
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and right to left.
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NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
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"""
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from __future__ import annotations
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seive = [True] * 1000001
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seive[1] = False
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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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Returns True if n is prime,
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False otherwise, for 1 <= n <= 1000000
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>>> is_prime(87)
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False
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>>> is_prime(1)
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False
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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 list_truncated_nums(n: int) -> list[int]:
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"""
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Returns a list of all left and right truncated numbers of n
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>>> list_truncated_nums(927628)
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[927628, 27628, 92762, 7628, 9276, 628, 927, 28, 92, 8, 9]
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>>> list_truncated_nums(467)
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[467, 67, 46, 7, 4]
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>>> list_truncated_nums(58)
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[58, 8, 5]
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"""
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str_num = str(n)
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list_nums = [n]
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for i in range(1, len(str_num)):
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list_nums.append(int(str_num[i:]))
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list_nums.append(int(str_num[:-i]))
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return list_nums
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def validate(n: int) -> bool:
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"""
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To optimize the approach, we will rule out the numbers above 1000,
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whose first or last three digits are not prime
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>>> validate(74679)
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False
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>>> validate(235693)
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False
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>>> validate(3797)
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True
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"""
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if len(str(n)) > 3:
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if not is_prime(int(str(n)[-3:])) or not is_prime(int(str(n)[:3])):
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return False
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return True
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def compute_truncated_primes(count: int = 11) -> list[int]:
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"""
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Returns the list of truncated primes
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>>> compute_truncated_primes(11)
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[23, 37, 53, 73, 313, 317, 373, 797, 3137, 3797, 739397]
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"""
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list_truncated_primes: list[int] = []
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num = 13
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while len(list_truncated_primes) != count:
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if validate(num):
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list_nums = list_truncated_nums(num)
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if all(is_prime(i) for i in list_nums):
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list_truncated_primes.append(num)
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num += 2
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return list_truncated_primes
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def solution() -> int:
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"""
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Returns the sum of truncated primes
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"""
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return sum(compute_truncated_primes(11))
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if __name__ == "__main__":
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print(f"{sum(compute_truncated_primes(11)) = }")
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