Unify O(sqrt(N)) is_prime functions under project_euler (#6258)

* 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>
This commit is contained in:
Nikos Giachoudis 2022-09-14 11:40:04 +03:00 committed by GitHub
parent 81e30fd33c
commit 2104fa7aeb
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12 changed files with 313 additions and 135 deletions

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@ -13,9 +13,11 @@ References:
import math
def is_prime(num: int) -> bool:
"""
Returns boolean representing primality of given number num.
def is_prime(number: int) -> bool:
"""Checks to see if a number is a prime in O(sqrt(n)).
A number is prime if it has exactly two factors: 1 and itself.
Returns boolean representing primality of given number (i.e., if the
result is true, then the number is indeed prime else it is not).
>>> is_prime(2)
True
@ -26,23 +28,21 @@ def is_prime(num: int) -> bool:
>>> is_prime(2999)
True
>>> is_prime(0)
Traceback (most recent call last):
...
ValueError: Parameter num must be greater than or equal to two.
False
>>> is_prime(1)
Traceback (most recent call last):
...
ValueError: Parameter num must be greater than or equal to two.
False
"""
if num <= 1:
raise ValueError("Parameter num must be greater than or equal to two.")
if num == 2:
if 1 < number < 4:
# 2 and 3 are primes
return True
elif num % 2 == 0:
elif number < 2 or number % 2 == 0 or number % 3 == 0:
# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
return False
for i in range(3, int(math.sqrt(num)) + 1, 2):
if num % i == 0:
# All primes number are in format of 6k +/- 1
for i in range(5, int(math.sqrt(number) + 1), 6):
if number % i == 0 or number % (i + 2) == 0:
return False
return True

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@ -15,29 +15,37 @@ References:
from math import sqrt
def is_prime(num: int) -> bool:
"""
Determines whether the given number is prime or not
def is_prime(number: int) -> bool:
"""Checks to see if a number is a prime in O(sqrt(n)).
A number is prime if it has exactly two factors: 1 and itself.
Returns boolean representing primality of given number (i.e., if the
result is true, then the number is indeed prime else it is not).
>>> is_prime(2)
True
>>> is_prime(15)
>>> is_prime(3)
True
>>> is_prime(27)
False
>>> is_prime(29)
>>> is_prime(2999)
True
>>> is_prime(0)
False
>>> is_prime(1)
False
"""
if num == 2:
if 1 < number < 4:
# 2 and 3 are primes
return True
elif num % 2 == 0:
elif number < 2 or number % 2 == 0 or number % 3 == 0:
# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
return False
else:
sq = int(sqrt(num)) + 1
for i in range(3, sq, 2):
if num % i == 0:
return False
# All primes number are in format of 6k +/- 1
for i in range(5, int(sqrt(number) + 1), 6):
if number % i == 0 or number % (i + 2) == 0:
return False
return True

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@ -11,22 +11,39 @@ What is the 10001st prime number?
References:
- https://en.wikipedia.org/wiki/Prime_number
"""
import math
def is_prime(number: int) -> bool:
"""
Determines whether the given number is prime or not
"""Checks to see if a number is a prime in O(sqrt(n)).
A number is prime if it has exactly two factors: 1 and itself.
Returns boolean representing primality of given number (i.e., if the
result is true, then the number is indeed prime else it is not).
>>> is_prime(2)
True
>>> is_prime(15)
False
>>> is_prime(29)
>>> is_prime(3)
True
>>> is_prime(27)
False
>>> is_prime(2999)
True
>>> is_prime(0)
False
>>> is_prime(1)
False
"""
for i in range(2, int(number**0.5) + 1):
if number % i == 0:
if 1 < number < 4:
# 2 and 3 are primes
return True
elif number < 2 or number % 2 == 0 or number % 3 == 0:
# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
return False
# All primes number are in format of 6k +/- 1
for i in range(5, int(math.sqrt(number) + 1), 6):
if number % i == 0 or number % (i + 2) == 0:
return False
return True

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@ -16,20 +16,37 @@ import math
def is_prime(number: int) -> bool:
"""
Determines whether a given number is prime or not
"""Checks to see if a number is a prime in O(sqrt(n)).
A number is prime if it has exactly two factors: 1 and itself.
Returns boolean representing primality of given number (i.e., if the
result is true, then the number is indeed prime else it is not).
>>> is_prime(2)
True
>>> is_prime(15)
False
>>> is_prime(29)
>>> is_prime(3)
True
>>> is_prime(27)
False
>>> is_prime(2999)
True
>>> is_prime(0)
False
>>> is_prime(1)
False
"""
if number % 2 == 0 and number > 2:
if 1 < number < 4:
# 2 and 3 are primes
return True
elif number < 2 or number % 2 == 0 or number % 3 == 0:
# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
return False
return all(number % i for i in range(3, int(math.sqrt(number)) + 1, 2))
# All primes number are in format of 6k +/- 1
for i in range(5, int(math.sqrt(number) + 1), 6):
if number % i == 0 or number % (i + 2) == 0:
return False
return True
def prime_generator():

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@ -11,12 +11,14 @@ References:
- https://en.wikipedia.org/wiki/Prime_number
"""
from math import sqrt
import math
def is_prime(n: int) -> bool:
"""
Returns boolean representing primality of given number num.
def is_prime(number: int) -> bool:
"""Checks to see if a number is a prime in O(sqrt(n)).
A number is prime if it has exactly two factors: 1 and itself.
Returns boolean representing primality of given number num (i.e., if the
result is true, then the number is indeed prime else it is not).
>>> is_prime(2)
True
@ -26,13 +28,24 @@ def is_prime(n: int) -> bool:
False
>>> is_prime(2999)
True
>>> is_prime(0)
False
>>> is_prime(1)
False
"""
if 1 < n < 4:
if 1 < number < 4:
# 2 and 3 are primes
return True
elif n < 2 or not n % 2:
elif number < 2 or number % 2 == 0 or number % 3 == 0:
# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
return False
return not any(not n % i for i in range(3, int(sqrt(n) + 1), 2))
# All primes number are in format of 6k +/- 1
for i in range(5, int(math.sqrt(number) + 1), 6):
if number % i == 0 or number % (i + 2) == 0:
return False
return True
def solution(n: int = 2000000) -> int:

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@ -16,8 +16,10 @@ from itertools import takewhile
def is_prime(number: int) -> bool:
"""
Returns boolean representing primality of given number num.
"""Checks to see if a number is a prime in O(sqrt(n)).
A number is prime if it has exactly two factors: 1 and itself.
Returns boolean representing primality of given number num (i.e., if the
result is true, then the number is indeed prime else it is not).
>>> is_prime(2)
True
@ -27,11 +29,24 @@ def is_prime(number: int) -> bool:
False
>>> is_prime(2999)
True
>>> is_prime(0)
False
>>> is_prime(1)
False
"""
if number % 2 == 0 and number > 2:
if 1 < number < 4:
# 2 and 3 are primes
return True
elif number < 2 or number % 2 == 0 or number % 3 == 0:
# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
return False
return all(number % i for i in range(3, int(math.sqrt(number)) + 1, 2))
# All primes number are in format of 6k +/- 1
for i in range(5, int(math.sqrt(number) + 1), 6):
if number % i == 0 or number % (i + 2) == 0:
return False
return True
def prime_generator() -> Iterator[int]:

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@ -23,22 +23,39 @@ n = 0.
import math
def is_prime(k: int) -> bool:
"""
Determine if a number is prime
>>> is_prime(10)
False
>>> is_prime(11)
def is_prime(number: int) -> bool:
"""Checks to see if a number is a prime in O(sqrt(n)).
A number is prime if it has exactly two factors: 1 and itself.
Returns boolean representing primality of given number num (i.e., if the
result is true, then the number is indeed prime else it is not).
>>> is_prime(2)
True
>>> is_prime(3)
True
>>> is_prime(27)
False
>>> is_prime(2999)
True
>>> is_prime(0)
False
>>> is_prime(1)
False
>>> is_prime(-10)
False
"""
if k < 2 or k % 2 == 0:
return False
elif k == 2:
if 1 < number < 4:
# 2 and 3 are primes
return True
else:
for x in range(3, int(math.sqrt(k) + 1), 2):
if k % x == 0:
return False
elif number < 2 or number % 2 == 0 or number % 3 == 0:
# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
return False
# All primes number are in format of 6k +/- 1
for i in range(5, int(math.sqrt(number) + 1), 6):
if number % i == 0 or number % (i + 2) == 0:
return False
return True

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@ -1,4 +1,7 @@
"""
Truncatable primes
Problem 37: https://projecteuler.net/problem=37
The number 3797 has an interesting property. Being prime itself, it is possible
to continuously remove digits from left to right, and remain prime at each stage:
3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.
@ -11,28 +14,46 @@ NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
from __future__ import annotations
seive = [True] * 1000001
seive[1] = False
i = 2
while i * i <= 1000000:
if seive[i]:
for j in range(i * i, 1000001, i):
seive[j] = False
i += 1
import math
def is_prime(n: int) -> bool:
"""
Returns True if n is prime,
False otherwise, for 1 <= n <= 1000000
>>> is_prime(87)
def is_prime(number: int) -> bool:
"""Checks to see if a number is a prime in O(sqrt(n)).
A number is prime if it has exactly two factors: 1 and itself.
>>> is_prime(0)
False
>>> is_prime(1)
False
>>> is_prime(25363)
>>> is_prime(2)
True
>>> is_prime(3)
True
>>> is_prime(27)
False
>>> is_prime(87)
False
>>> is_prime(563)
True
>>> is_prime(2999)
True
>>> is_prime(67483)
False
"""
return seive[n]
if 1 < number < 4:
# 2 and 3 are primes
return True
elif number < 2 or number % 2 == 0 or number % 3 == 0:
# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
return False
# All primes number are in format of 6k +/- 1
for i in range(5, int(math.sqrt(number) + 1), 6):
if number % i == 0 or number % (i + 2) == 0:
return False
return True
def list_truncated_nums(n: int) -> list[int]:

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@ -12,25 +12,45 @@ pandigital prime.
"""
from __future__ import annotations
import math
from itertools import permutations
from math import sqrt
def is_prime(n: int) -> bool:
"""
Returns True if n is prime,
False otherwise.
>>> is_prime(67483)
def is_prime(number: int) -> bool:
"""Checks to see if a number is a prime in O(sqrt(n)).
A number is prime if it has exactly two factors: 1 and itself.
>>> is_prime(0)
False
>>> is_prime(1)
False
>>> is_prime(2)
True
>>> is_prime(3)
True
>>> is_prime(27)
False
>>> is_prime(87)
False
>>> is_prime(563)
True
>>> is_prime(87)
>>> is_prime(2999)
True
>>> is_prime(67483)
False
"""
if n % 2 == 0:
if 1 < number < 4:
# 2 and 3 are primes
return True
elif number < 2 or number % 2 == 0 or number % 3 == 0:
# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
return False
for i in range(3, int(sqrt(n) + 1), 2):
if n % i == 0:
# All primes number are in format of 6k +/- 1
for i in range(5, int(math.sqrt(number) + 1), 6):
if number % i == 0 or number % (i + 2) == 0:
return False
return True

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@ -19,30 +19,49 @@ prime and twice a square?
from __future__ import annotations
seive = [True] * 100001
i = 2
while i * i <= 100000:
if seive[i]:
for j in range(i * i, 100001, i):
seive[j] = False
i += 1
import math
def is_prime(n: int) -> bool:
"""
Returns True if n is prime,
False otherwise, for 2 <= n <= 100000
def is_prime(number: int) -> bool:
"""Checks to see if a number is a prime in O(sqrt(n)).
A number is prime if it has exactly two factors: 1 and itself.
>>> is_prime(0)
False
>>> is_prime(1)
False
>>> is_prime(2)
True
>>> is_prime(3)
True
>>> is_prime(27)
False
>>> is_prime(87)
False
>>> is_prime(23)
>>> is_prime(563)
True
>>> is_prime(25363)
>>> is_prime(2999)
True
>>> is_prime(67483)
False
"""
return seive[n]
if 1 < number < 4:
# 2 and 3 are primes
return True
elif number < 2 or number % 2 == 0 or number % 3 == 0:
# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
return False
# All primes number are in format of 6k +/- 1
for i in range(5, int(math.sqrt(number) + 1), 6):
if number % i == 0 or number % (i + 2) == 0:
return False
return True
odd_composites = [num for num in range(3, len(seive), 2) if not is_prime(num)]
odd_composites = [num for num in range(3, 100001, 2) if not is_prime(num)]
def compute_nums(n: int) -> list[int]:

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@ -25,32 +25,46 @@ After that, bruteforce all passed candidates sequences using
The bruteforce of this solution will be about 1 sec.
"""
import math
from itertools import permutations
from math import floor, sqrt
def is_prime(number: int) -> bool:
"""
function to check whether the number is prime or not.
>>> is_prime(2)
True
>>> is_prime(6)
"""Checks to see if a number is a prime in O(sqrt(n)).
A number is prime if it has exactly two factors: 1 and itself.
>>> is_prime(0)
False
>>> is_prime(1)
False
>>> is_prime(-800)
False
>>> is_prime(104729)
>>> is_prime(2)
True
>>> is_prime(3)
True
>>> is_prime(27)
False
>>> is_prime(87)
False
>>> is_prime(563)
True
>>> is_prime(2999)
True
>>> is_prime(67483)
False
"""
if number < 2:
if 1 < number < 4:
# 2 and 3 are primes
return True
elif number < 2 or number % 2 == 0 or number % 3 == 0:
# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
return False
for i in range(2, floor(sqrt(number)) + 1):
if number % i == 0:
# All primes number are in format of 6k +/- 1
for i in range(5, int(math.sqrt(number) + 1), 6):
if number % i == 0 or number % (i + 2) == 0:
return False
return True

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@ -33,29 +33,46 @@ So we check individually each one of these before incrementing our
count of current primes.
"""
from math import isqrt
import math
def is_prime(number: int) -> int:
"""
Returns whether the given number is prime or not
def is_prime(number: int) -> bool:
"""Checks to see if a number is a prime in O(sqrt(n)).
A number is prime if it has exactly two factors: 1 and itself.
>>> is_prime(0)
False
>>> is_prime(1)
0
>>> is_prime(17)
1
>>> is_prime(10000)
0
False
>>> is_prime(2)
True
>>> is_prime(3)
True
>>> is_prime(27)
False
>>> is_prime(87)
False
>>> is_prime(563)
True
>>> is_prime(2999)
True
>>> is_prime(67483)
False
"""
if number == 1:
return 0
if number % 2 == 0 and number > 2:
return 0
if 1 < number < 4:
# 2 and 3 are primes
return True
elif number < 2 or number % 2 == 0 or number % 3 == 0:
# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
return False
for i in range(3, isqrt(number) + 1, 2):
if number % i == 0:
return 0
return 1
# All primes number are in format of 6k +/- 1
for i in range(5, int(math.sqrt(number) + 1), 6):
if number % i == 0 or number % (i + 2) == 0:
return False
return True
def solution(ratio: float = 0.1) -> int: