Prime Factorization using Sieve O(log n) for multiple queries

Software Development
#1

Hello Everyone,

We can calculate the prime factorization of a number “n” in O(sqrt(n)) as discussed. But O(sqrt n) method times out when we need to answer multiple queries regarding prime factorization.
In this article we study an efficient method to calculate the prime factorization using O(n) space and O(log n) time complexity with per-computation allowed.

Key Concept: Our idea is to store the Smallest Prime Factor(SPF) for every number. Then to calculate the prime factorization of the given number by dividing the given number recursively with its smallest prime factor till it becomes 1.

To calculate to smallest prime factor for every number we will use the sieve of eratosthenes. In original Sieve, every time we mark a number as not-prime, we store the corresponding smallest prime factor for that number. Now, after we are done with precalculating the smallest prime factor for every number we will divide our number n (whose prime factorziation is to be calculated) by its corresponding smallest prime factor till n becomes 1.

Pseudo Code for prime factorization assuming SPFs are computed : PrimeFactors[] // To store result i = 0 // Index in PrimeFactors while n != 1 : // SPF : smallest prime factor PrimeFactors[i] = SPF[n] i++ n = n / SPF[n]

The implementation for the above method is given below :

// C++ program to find prime factorization of a

// number n in O(Log n) time with precomputation

// allowed.

#include "bits/stdc++.h"

using namespace std;

#define MAXN 100001

// stores smallest prime factor for every number

int spf[MAXN];

// Calculating SPF (Smallest Prime Factor) for every

// number till MAXN.

// Time Complexity : O(nloglogn)

void sieve()

{

spf[1] = 1;

for ( int i=2; i<MAXN; i++)

// marking smallest prime factor for every

// number to be itself.

spf[i] = i;

// separately marking spf for every even

// number as 2

for ( int i=4; i<MAXN; i+=2)

spf[i] = 2;

for ( int i=3; i*i<MAXN; i++)

{

// checking if i is prime

if (spf[i] == i)

{

// marking SPF for all numbers divisible by i

for ( int j=i*i; j<MAXN; j+=i)

// marking spf[j] if it is not

// previously marked

if (spf[j]==j)

spf[j] = i;

}

}

}

// A O(log n) function returning primefactorization

// by dividing by smallest prime factor at every step

vector< int > getFactorization( int x)

{

vector< int > ret;

while (x != 1)

{

ret.push_back(spf[x]);

x = x / spf[x];

}

return ret;

}

// driver program for above function

int main( int argc, char const *argv[])

{

// precalculating Smallest Prime Factor

sieve();

int x = 12246;

cout << "prime factorization for " << x << " : " ;

// calling getFactorization function

vector < int > p = getFactorization(x);

for ( int i=0; i<p.size(); i++)

cout << p[i] << " " ;

cout << endl;

return 0;

}

Output:

prime factorization for 12246 : 2 3 13 157

Note : The above code works well for n upto the order of 10^7. Beyond this we will face memory issues.

Time Complexity: The precomputation for smallest prime factor is done in O(n log log n) using sieve. Where as in the calculation step we are dividing the number every time by the smallest prime number till it becomes 1. So, let’s consider a worst case in which every time the SPF is 2 . Therefore will have log n division steps. Hence, We can say that our Time Complexity will be O(log n) in worst case.