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>

project_euler/problem_003/sol1.py

@@ -13,9 +13,11 @@ References:
 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)
     True
@@ -26,23 +28,21 @@ def is_prime(num: int) -> bool:
     >>> is_prime(2999)
     True
     >>> is_prime(0)
-    Traceback (most recent call last):
+    False
-        ...
-    ValueError: Parameter num must be greater than or equal to two.
     >>> 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
-    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
 

project_euler/problem_007/sol1.py

@@ -15,28 +15,36 @@ References:
 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)
     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
+    # 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 True
 

project_euler/problem_007/sol2.py

@@ -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:
-    """
+    """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)
     True
-    >>> is_prime(15)
+    >>> is_prime(3)
-    False
-    >>> is_prime(29)
     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 True
 

project_euler/problem_007/sol3.py

@@ -16,20 +16,37 @@ import math
 
 
 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)
     True
-    >>> is_prime(15)
+    >>> is_prime(3)
-    False
-    >>> is_prime(29)
     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():

project_euler/problem_010/sol1.py

@@ -11,12 +11,14 @@ References:
     - 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)
     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:

project_euler/problem_010/sol2.py

@@ -16,8 +16,10 @@ from itertools import takewhile
 
 
 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)
     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]:

project_euler/problem_027/sol1.py

@@ -23,21 +23,38 @@ n = 0.
 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
+    >>> 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
-    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
+
+    # 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
 

project_euler/problem_037/sol1.py

@@ -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
+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
     >>> 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]:

project_euler/problem_041/sol1.py

@@ -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:
+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
     >>> 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
 

project_euler/problem_046/sol1.py

@@ -19,30 +19,49 @@ prime and twice a square?
 
 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)
     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]:

project_euler/problem_049/sol1.py

@@ -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:
-    """
+    """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
     >>> is_prime(1)
     False
-    >>> is_prime(-800)
+    >>> is_prime(2)
-    False
-    >>> is_prime(104729)
     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):
+    # 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 True
 
 

project_euler/problem_058/sol1.py

@@ -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:
+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)
-    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: