Hello Everyone,
Given a range [low, high], print all primes in this range? For example, if the given range is [10, 20], then output is 11, 13, 17, 19.
A Naive approach is to run a loop from low to high and check each number for primeness.
A Better Approach is to precalculate primes up to the maximum limit using Sieve of Eratosthenes, then print all prime numbers in range.
The above approach looks good, but consider the input range [50000, 55000]. the above Sieve approach would precalculate primes from 2 to 50100. This causes a waste of memory as well as time. Below is the Segmented Sieve based approach.
Segmented Sieve (Background)
Below are basic steps to get an idea of how Segmented Sieve works
- Use Simple Sieve to find all primes up to a predefined limit (square root of ‘high’ is used in below code) and store these primes in an array “prime[]”. Basically we call Simple Sieve for a limit and we not only find prime numbers, but also puts them in a separate array prime[].
- Create an array mark[high-low+1]. Here we need only O(n) space where n is number of elements in given range.
- Iterate through all primes found in step 1. For every prime, mark its multiples in given range [low…high].
So unlike simple sieve, we don’t check for all multiples of every number smaller than square root of high, we only check for multiples of primes found in step 1. And we don’t need O(high) space, we need O(sqrt(high) + n) space.
Below is the implementation of above idea.
#include <bits/stdc++.h>
using namespace std;
// fillPrime function fills primes from 2 to sqrt of high in chprime(vector) array
void fillPrimes(vector< int >& prime, int high)
{
bool ck[high + 1];
memset (ck, true , sizeof (ck));
ck[1] = false ;
ck[0] = false ;
for ( int i = 2; (i * i) <= high; i++) {
if (ck[i] == true ) {
for ( int j = i * i; j <= high; j = j + i) {
ck[j] = false ;
}
}
}
for ( int i = 2; i * i <= high; i++) {
if (ck[i] == true ) {
prime.push_back(i);
}
}
}
// in segmented sieve we check for prime from range [low, high]
void segmentedSieve( int low, int high)
{
bool prime[high - low + 1];
//here prime[0] indicates whether low is prime or not similarly prime[1] indicates whether low+1 is prime or not
memset (prime, true , sizeof (prime));
vector< int > chprime;
fillPrimes(chprime, high);
//chprimes has primes in range [2,sqrt(n)]
// we take primes from 2 to sqrt[n] because the multiples of all primes below high are marked by these
// primes in range 2 to sqrt[n] for eg: for number 7 its multiples i.e 14 is marked by 2, 21 is marked by 3,
// 28 is marked by 4, 35 is marked by 5, 42 is marked 6, so 49 will be first marked by 7 so all number before 49
// are marked by primes in range [2,sqrt(49)]
for ( int i : chprime) {
int lower = (low / i);
//here lower means the first multiple of prime which is in range [low,high]
//for eg: 3's first multiple in range [28,40] is 30
if (lower <= 1) {
lower = i + i;
}
else if (low % i) {
lower = (lower * i) + i;
}
else {
lower = (lower * i);
}
for ( int j = lower; j <= high; j = j + i) {
prime[j - low] = false ;
}
}
for ( int i = low; i <= high; i++) {
if (prime[i - low] == true ) {
cout << (i) << " " ;
}
}
}
int main()
{
// low should be greater than or equal to 2
int low = 2;
// low here is the lower limit
int high = 100;
// high here is the upper limit
// in segmented sieve we calculate primes in range [low,high]
// here we initially we find primes in range [2,sqrt(high)] to mark all their multiples as not prime
//then we mark all their(primes) multiples in range [low,high] as false
// this is a modified sieve of eratosthenes , in standard sieve of eratosthenes we find prime from 1 to n(given number)
// in segmented sieve we only find primes in a given interval
cout<< "Primes in range " <<low<< " to " << high<< " are\n" ;
segmentedSieve(low, high);
}
Output
11 13 15 17 19
Segmented Sieve (What if ‘high’ value of range is too high and range is also big)
Consider a situation where given high value is so high that neither sqrt(high) nor O(high-low+1) can fit in memory. How to find prims in range. For this situation, we run step 1 (Simple Sieve) only for a limit that can fit in memory. Then we divide given range in different segments. For every segment, we run step 2 and 3 considering low and high as end points of current segment. We add primes of current segment to prime[] before running the next segment.