Project Euler & HackerRank Problem 3 Solution
Largest prime factor
by {BetaProjects} | MAY 17, 2009 | Project Euler & HackerRankProject Euler Problem 3 Statement
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143?
Solution
A reasonable way to solve this problem is to use trial division to factor an integer, n. In this instance, we create a set of possible integer factors, d, from the set {2} ∪ {3, 5, 7, 9, 11, …, √n} and try to divide n.
For all d that divide n, we remove d and all its factors. Once our set of factors are exhausted, the remainder becomes the largest prime factor.
Below is a walk–through of this algorithm to show the process of removing factors and leaving the remainder as the largest prime factor of n.
Finding the largest prime factor
n = 6435 d = 2 while (d*d <= n): if (n % d == 0): n //= d else: d += 2 if d>2 else 1 # after 2, consider only odd d
Example: n = 6435 Step d d2 n 1 2 and 4 ≤ 6435: 2 is not a factor of 6435, incrementd
to 3 2 3 and 9 ≤ 6435: 3 is a factor of 6435:n=n//3
= 2145 3 3 and 9 ≤ 2145: and 3 is a factor of 2145:n=n//3
= 715 4 3 and 9 ≤ 715: but, 3 is not a factor of 715, incrementd
to 5 5 5 and 25 ≤ 715: 5 is a factor of 715:n=n//5
= 143 6 5 and 25 ≤ 143: but, 5 is not a factor of 143, incrementd
to 7 7 7 and 49 ≤ 143: but, 7 is not a factor of 143, incrementd
to 9 8 9 and 81 ≤ 143: but, 9 is not a factor of 143, incrementd
to 111 9 11 and 121 ≤ 143: 11 is a factor of 143:n=n//11
= 13 10 11 and 121 > 13: thewhile
loop terminates and 13 is the largest prime factor2 1Needless overhead to check 9 as all multiples of 3 have already been removed, but such is the cost for checking all odd numbers after 2 instead of building a list of prime numbers. 2Thewhile
’s conditional, d2 ≤ n, is checked after executing the loop’s last statement.
If the given number, n, is already prime then n is returned as the largest prime factor, which is nice, because it is, in fact, itself the largest prime factor.
HackerRank version
HackerRank’s Project Euler Problem 3 runs up to 10 test cases with 10 ≤ N ≤ 1012
Python Source Code
Last Word
Trial division, used here, works well for smaller numbers less than 1021, for larger numbers there are several efficient algorithms available.
A more favorable method for factoring numbers under 100 digits long is the Quadratic sieve. It is a much simpler algorithm to implement compared to others and lends itself to parallelization. Because it is a sieve, large amounts of memory may be required.
Another is Pollard’s Rho algorithm with improvements by Brent. It’s easy to program and fast for numbers with small factors.
For a practical application of this algorithm see the JavaScript prime factor calculator. It finds all the prime factors of a number.