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 import math
def is_prime(num: int) -> bool: def is_prime(number: int) -> bool:
""" """Checks to see if a number is a prime in O(sqrt(n)).
Returns boolean representing primality of given number num. 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) >>> is_prime(2)
True True
@ -26,23 +28,21 @@ def is_prime(num: int) -> bool:
>>> is_prime(2999) >>> is_prime(2999)
True True
>>> is_prime(0) >>> is_prime(0)
Traceback (most recent call last): False
...
ValueError: Parameter num must be greater than or equal to two.
>>> is_prime(1) >>> is_prime(1)
Traceback (most recent call last): False
...
ValueError: Parameter num must be greater than or equal to two.
""" """
if num <= 1: if 1 < number < 4:
raise ValueError("Parameter num must be greater than or equal to two.") # 2 and 3 are primes
if num == 2:
return True 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 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 False
return True return True

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@ -15,29 +15,37 @@ References:
from math import sqrt from math import sqrt
def is_prime(num: int) -> bool: def is_prime(number: int) -> bool:
""" """Checks to see if a number is a prime in O(sqrt(n)).
Determines whether the given number is prime or not 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) >>> is_prime(2)
True True
>>> is_prime(15) >>> is_prime(3)
True
>>> is_prime(27)
False False
>>> is_prime(29) >>> is_prime(2999)
True True
>>> is_prime(0) >>> is_prime(0)
False False
>>> is_prime(1)
False
""" """
if num == 2: if 1 < number < 4:
# 2 and 3 are primes
return True 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 return False
else:
sq = int(sqrt(num)) + 1 # All primes number are in format of 6k +/- 1
for i in range(3, sq, 2): for i in range(5, int(sqrt(number) + 1), 6):
if num % i == 0: if number % i == 0 or number % (i + 2) == 0:
return False return False
return True return True

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@ -11,22 +11,39 @@ What is the 10001st prime number?
References: References:
- https://en.wikipedia.org/wiki/Prime_number - https://en.wikipedia.org/wiki/Prime_number
""" """
import math
def is_prime(number: int) -> bool: def is_prime(number: int) -> bool:
""" """Checks to see if a number is a prime in O(sqrt(n)).
Determines whether the given number is prime or not 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) >>> is_prime(2)
True True
>>> is_prime(15) >>> is_prime(3)
False
>>> is_prime(29)
True 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 1 < number < 4:
if number % i == 0: # 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 False
return True return True

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@ -16,20 +16,37 @@ import math
def is_prime(number: int) -> bool: def is_prime(number: int) -> bool:
""" """Checks to see if a number is a prime in O(sqrt(n)).
Determines whether a given number is prime or not 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) >>> is_prime(2)
True True
>>> is_prime(15) >>> is_prime(3)
False
>>> is_prime(29)
True 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 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(): def prime_generator():

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@ -11,12 +11,14 @@ References:
- https://en.wikipedia.org/wiki/Prime_number - https://en.wikipedia.org/wiki/Prime_number
""" """
from math import sqrt import math
def is_prime(n: int) -> bool: def is_prime(number: int) -> bool:
""" """Checks to see if a number is a prime in O(sqrt(n)).
Returns boolean representing primality of given number num. 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) >>> is_prime(2)
True True
@ -26,13 +28,24 @@ def is_prime(n: int) -> bool:
False False
>>> is_prime(2999) >>> is_prime(2999)
True True
>>> is_prime(0)
False
>>> is_prime(1)
False
""" """
if 1 < n < 4: if 1 < number < 4:
# 2 and 3 are primes
return True 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 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: def solution(n: int = 2000000) -> int:

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@ -16,8 +16,10 @@ from itertools import takewhile
def is_prime(number: int) -> bool: def is_prime(number: int) -> bool:
""" """Checks to see if a number is a prime in O(sqrt(n)).
Returns boolean representing primality of given number num. 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) >>> is_prime(2)
True True
@ -27,11 +29,24 @@ def is_prime(number: int) -> bool:
False False
>>> is_prime(2999) >>> is_prime(2999)
True 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 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]: def prime_generator() -> Iterator[int]:

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@ -23,22 +23,39 @@ n = 0.
import math import math
def is_prime(k: int) -> bool: def is_prime(number: int) -> bool:
""" """Checks to see if a number is a prime in O(sqrt(n)).
Determine if a number is prime A number is prime if it has exactly two factors: 1 and itself.
>>> is_prime(10) Returns boolean representing primality of given number num (i.e., if the
False result is true, then the number is indeed prime else it is not).
>>> is_prime(11)
>>> is_prime(2)
True 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 if 1 < number < 4:
elif k == 2: # 2 and 3 are primes
return True return True
else: elif number < 2 or number % 2 == 0 or number % 3 == 0:
for x in range(3, int(math.sqrt(k) + 1), 2): # Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
if k % x == 0: return False
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 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 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: 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. 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 from __future__ import annotations
seive = [True] * 1000001 import math
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
def is_prime(n: int) -> bool: def is_prime(number: int) -> bool:
""" """Checks to see if a number is a prime in O(sqrt(n)).
Returns True if n is prime,
False otherwise, for 1 <= n <= 1000000 A number is prime if it has exactly two factors: 1 and itself.
>>> is_prime(87)
>>> is_prime(0)
False False
>>> is_prime(1) >>> is_prime(1)
False 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 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]: def list_truncated_nums(n: int) -> list[int]:

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@ -12,25 +12,45 @@ pandigital prime.
""" """
from __future__ import annotations from __future__ import annotations
import math
from itertools import permutations from itertools import permutations
from math import sqrt
def is_prime(n: int) -> bool: def is_prime(number: int) -> bool:
""" """Checks to see if a number is a prime in O(sqrt(n)).
Returns True if n is prime,
False otherwise. A number is prime if it has exactly two factors: 1 and itself.
>>> is_prime(67483)
>>> is_prime(0)
False
>>> is_prime(1)
False
>>> is_prime(2)
True
>>> is_prime(3)
True
>>> is_prime(27)
False
>>> is_prime(87)
False False
>>> is_prime(563) >>> is_prime(563)
True True
>>> is_prime(87) >>> is_prime(2999)
True
>>> is_prime(67483)
False 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 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 False
return True return True

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@ -19,30 +19,49 @@ prime and twice a square?
from __future__ import annotations from __future__ import annotations
seive = [True] * 100001 import math
i = 2
while i * i <= 100000:
if seive[i]:
for j in range(i * i, 100001, i):
seive[j] = False
i += 1
def is_prime(n: int) -> bool: def is_prime(number: int) -> bool:
""" """Checks to see if a number is a prime in O(sqrt(n)).
Returns True if n is prime,
False otherwise, for 2 <= n <= 100000 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) >>> is_prime(87)
False False
>>> is_prime(23) >>> is_prime(563)
True True
>>> is_prime(25363) >>> is_prime(2999)
True
>>> is_prime(67483)
False 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]: 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. The bruteforce of this solution will be about 1 sec.
""" """
import math
from itertools import permutations from itertools import permutations
from math import floor, sqrt
def is_prime(number: int) -> bool: def is_prime(number: int) -> bool:
""" """Checks to see if a number is a prime in O(sqrt(n)).
function to check whether the number is prime or not.
>>> is_prime(2) A number is prime if it has exactly two factors: 1 and itself.
True
>>> is_prime(6) >>> is_prime(0)
False False
>>> is_prime(1) >>> is_prime(1)
False False
>>> is_prime(-800) >>> is_prime(2)
False
>>> is_prime(104729)
True 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 return False
for i in range(2, floor(sqrt(number)) + 1): # All primes number are in format of 6k +/- 1
if number % i == 0: for i in range(5, int(math.sqrt(number) + 1), 6):
if number % i == 0 or number % (i + 2) == 0:
return False return False
return True 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. count of current primes.
""" """
from math import isqrt import math
def is_prime(number: int) -> int: def is_prime(number: int) -> bool:
""" """Checks to see if a number is a prime in O(sqrt(n)).
Returns whether the given number is prime or not
A number is prime if it has exactly two factors: 1 and itself.
>>> is_prime(0)
False
>>> is_prime(1) >>> is_prime(1)
0 False
>>> is_prime(17) >>> is_prime(2)
1 True
>>> is_prime(10000) >>> is_prime(3)
0 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: if 1 < number < 4:
return 0 # 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): # All primes number are in format of 6k +/- 1
if number % i == 0: for i in range(5, int(math.sqrt(number) + 1), 6):
return 0 if number % i == 0 or number % (i + 2) == 0:
return 1 return False
return True
def solution(ratio: float = 0.1) -> int: def solution(ratio: float = 0.1) -> int: