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12 def primes(): 13 yield 2 14 yield 3 15 16 for i in itertools.count(start=5, step=2): 17 if is_prime(i): 18 yield i
746 def prime_number_factorisation(n): 747 if n < 2: 748 return [n] 749 i = 2 750 factors = [] 751 while i * i <= n: 752 if n % i: 753 i += 1 754 else: 755 n //= i 756 factors.append(i) 757 if n > 1: 758 factors.append(n) 759 return factors
50 def prime_factors(n): 51 """Lists prime factors of a given natural integer, from greatest to smallest 52 :param n: Natural integer 53 :rtype : list of all prime factors of the given natural n 54 """ 55 i = 2 56 while i <= sqrt(n): 57 if n % i == 0: 58 l = prime_factors(n/i) 59 l.append(i) 60 return l 61 i += 1 62 return [n] # n is prime
7 def isPrime(num): 8 # Returns True if num is a prime number, otherwise False. 9 10 # Note: Generally, isPrime() is slower than primeSieve(). 11 12 # all numbers less than 2 are not prime 13 if num < 2: 14 return False 15 16 # see if num is divisible by any number up to the square root of num 17 for i in range(2, int(math.sqrt(num)) + 1): 18 if num % i == 0: 19 return False 20 return True
16 def primeGen(n): 17 for i in xrange(2, n): 18 prime = True 19 if i % 2 == 0 and i != 2: 20 continue 21 sqrtp = int(i ** 1 / 2) 22 for j in xrange(2, sqrtp): 23 if j % 2 == 0: 24 continue 25 if i % j == 0: 26 prime = False 27 break 28 if prime: 29 yield i
20 def prime(num): 21 # num is actually a string because input() returns strings. We'll convert it to int 22 num = int(num) 23 24 if num < 0: 25 print("Negative integers can not be prime") 26 quit() 27 if num is 1: 28 print("1 is neither prime nor composite") 29 # See how I lazily terminated program otherwise it'd forward "None"(default behaviour of python when function 30 # returns nothing) rather than True or False. Which could mess up the program. 31 # If we hit this if statement above statement is printed then program exits. 32 quit() # Now you don't need to get sys.exit() to exit python has quit to handle the same thing 33 if num in [2, 3]: 34 # if given argument is 2 or 3, it is prime. We used list without defining a variable which is perfectly valid 35 return True 36 if num % 2 == 0: # excluding all even numbers except two. 37 return False 38 else: 39 # Here we are starting counter variable from 3 in range. Second argument excludes numbers above one third 40 # of the given argument. Third argument in range sets steps to take to 2. This makes loop to iterate odds 41 for x in range(3, int(num/3), 2): 42 # Checking if argument is divisible by counter. % is modulus operator which returns remainder of division 43 if num % x == 0: 44 return False 45 # It's okay to have more than one return statement when program hits return statement it exits the function. 46 return True
17 def isprime(no): 18 if no == 2: 19 return True 20 elif no % 2 == 0: 21 return False 22 sq = int(math.sqrt(no)) + 1 23 for i in range(3, sq, 2): 24 if no % i == 0: 25 return False 26 return True
23 def findprimes(start, end): 24 for i in range(start, end): 25 if i not in Checked: 26 Checked.append(i) 27 if is_prime(i): 28 Primes.append(i)
7 def prime(): 8 D = {9: 3, 25: 5} 9 yield 2 10 yield 3 11 yield 5 12 MASK = 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 13 MODULOS = frozenset((1, 7, 11, 13, 17, 19, 23, 29)) 14 15 for q in it.compress( 16 it.islice(it.count(7), 0, None, 2), 17 it.cycle(MASK)): 18 p = D.pop(q, None) 19 if p is None: 20 D[q * q] = q 21 yield q 22 else: 23 x = q + 2 * p 24 while x in D or (x % 30) not in MODULOS: 25 x += 2 * p 26 D[x] = p
1 def is_prime(n): 2 if n < 2: 3 return False 4 5 i = 2 6 7 while i * i <= n: 8 if n % i == 0: 9 return False 10 11 i += 1 12 13 return True