Segmented Sieve (Print Primes in a Range)

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 upto 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 Segmeented Sieve based approach.

Segmented Sieve (Background)
Below are basic steps to get idea how Segmented Sieve works

  1. Use Simple Sieve to find all primes upto 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[].
  2. Create an array mark[high-low+1]. Here we need only O(n) space where n is number of elements in given range.
  3. 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 C++ implementation of above idea.





// C++ program to print print all primes in a range
// using concept of Segmented Sieve
#include <bits/stdc++.h>
using namespace std;
// This functions finds all primes smaller than limit
// using simple sieve of eratosthenes.  It stores found
// primes in vector prime[]
void simpleSieve(int limit, vector<int>& prime)
    bool mark[limit + 1];
    memset(mark, false, sizeof(mark));
    for (int i = 2; i <= limit; ++i) {
        if (mark[i] == false) {
            // If not marked yet, then its a prime
            for (int j = i; j <= limit; j += i)
                mark[j] = true;
// Finds all prime numbers in given range using
// segmented sieve
void primesInRange(int low, int high)
    // Comput all primes smaller or equal to
    // square root of high using simple sieve
    int limit = floor(sqrt(high)) + 1;
    vector<int> prime;
    simpleSieve(limit, prime);
    // Count of elements in given range
    int n = high - low + 1;
    // Declaring boolean only for [low, high]
    bool mark[n + 1];
    memset(mark, false, sizeof(mark));
    // Use the found primes by simpleSieve() to find
    // primes in given range
    for (int i = 0; i < prime.size(); i++) {
        // Find the minimum number in [low..high] that is
        // a multiple of prime[i] (divisible by prime[i])
        int loLim = floor(low / prime[i]) * prime[i];
        if (loLim < low)
            loLim += prime[i];
            loLim += prime[i];
        /*  Mark multiples of prime[i] in [low..high]:
            We are marking j - low for j, i.e. each number
            in range [low, high] is mapped to [0, high - low]
            so if range is [50, 100]  marking 50 corresponds
            to marking 0, marking 51 corresponds to 1 and
            so on. In this way we need to allocate space only
            for range  */
        for (int j = loLim; j <= high; j += prime[i])
            mark[j - low] = true;
    // Numbers which are not marked in range, are prime
    for (int i = low; i <= high; i++)
        if (!mark[i - low])
            cout << i << "  ";
// Driver program to test above function
int main()
    int low = 10, high = 100;
    primesInRange(low, high);
    return 0;



11  13  17  19  23  29  31  37  41  43  47  
53  59  61  67  71  73  79  83  89  97

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.

This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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Improved By : RajatSinghal

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