Given a number n, write an efficient function to print all prime factors of n. For example, if the input number is 12, then output should be “2 2 3”. And if the input number is 315, then output should be “3 3 5 7”.

Following are the steps to find all prime factors.**1)** While n is divisible by 2, print 2 and divide n by 2.**2)** After step 1, n must be odd. Now start a loop from i = 3 to square root of n. While i divides n, print i and divide n by i, increment i by 2 and continue.**3)** If n is a prime number and is greater than 2, then n will not become 1 by above two steps. So print n if it is greater than 2.

`# Python program to print prime factors` ` ` `import` `math` ` ` `# A function to print all prime factors of ` `# a given number n` `def` `primeFactors(n):` ` ` ` ` `# Print the number of two\'s that divide n` ` ` `while` `n ` `%` `2` `=` `=` `0` `:` ` ` `print` `2` `,` ` ` `n ` `=` `n ` `/` `2` ` ` ` ` `# n must be odd at this point` ` ` `# so a skip of 2 ( i = i + 2) can be used` ` ` `for` `i ` `in` `range` `(` `3` `,` `int` `(math.sqrt(n))` `+` `1` `,` `2` `):` ` ` ` ` `# while i divides n , print i ad divide n` ` ` `while` `n ` `%` `i` `=` `=` `0` `:` ` ` `print` `i,` ` ` `n ` `=` `n ` `/` `i` ` ` ` ` `# Condition if n is a prime` ` ` `# number greater than 2` ` ` `if` `n > ` `2` `:` ` ` `print` `n` ` ` `# Driver Program to test above function` ` ` `n ` `=` `315` `primeFactors(n)` ` ` `# This code is contributed by Harshit Agrawal` |

**Output:**

3 3 5 7

**How does this work?**

The steps 1 and 2 take care of composite numbers and step 3 takes care of prime numbers. To prove that the complete algorithm works, we need to prove that steps 1 and 2 actually take care of composite numbers. This is clear that step 1 takes care of even numbers. And after step 1, all remaining prime factor must be odd (difference of two prime factors must be at least 2), this explains why i is incremented by 2.

Now the main part is, the loop runs till square root of n not till. To prove that this optimization works, let us consider the following property of composite numbers.*Every composite number has at least one prime factor less than or equal to square root of itself.*

This property can be proved using counter statement. Let a and b be two factors of n such that a*b = n. If both are greater than √n, then a.b > √n, * √n, which contradicts the expression “a * b = n”.

In step 2 of the above algorithm, we run a loop and do following in loop

a) Find the least prime factor i (must be less than √n,)

b) Remove all occurrences i from n by repeatedly dividing n by i.

c) Repeat steps a and b for divided n and i = i + 2. The steps a and b are repeated till n becomes either 1 or a prime number.

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