# Segmented Sieve

Given a number n, print all primes smaller than n. For example, if the given number is 10, output 2, 3, 5, 7.

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

A Naive approach is to run a loop from 0 to n-1 and check each number for primeness. A Better Approach is use Simple Sieve of Eratosthenes.

```// This functions finds all primes smaller than 'limit'
// using simple sieve of eratosthenes.
void simpleSieve(int limit)
{
// Create a boolean array "mark[0..limit-1]" and
// initialize all entries of it as true. A value
// in mark[p] will finally be false if 'p' is Not
// a prime, else true.
bool mark[limit];
memset(mark, true, sizeof(mark));

// One by one traverse all numbers so that their
// multiples can be marked as composite.
for (int p=2; p*p<limit; p++)
{
// If p is not changed, then it is a prime
if (mark[p] == true)
{
// Update all multiples of p
for (int i=p*2; i<limit; i+=p)
mark[i] = false;
}
}

// Print all prime numbers and store them in prime
for (int p=2; p<limit; p++)
if (mark[p] == true)
cout << p << "  ";
}

```

Problems with Simple Sieve:
The Sieve of Eratosthenes looks good, but consider the situation when n is large, the Simple Sieve faces following issues.

• An array of size Θ(n) may not fit in memory
• The simple Sieve is not cache friendly even for slightly bigger n. The algorithm traverses the array without locality of reference

Segmented Sieve
The idea of segmented sieve is to divide the range [0..n-1] in different segments and compute primes in all segments one by one. This algorithm first uses Simple Sieve to find primes smaller than or equal to √(n). Below are steps used in Segmented Sieve.

1. Use Simple Sieve to find all primes upto square root of ‘n’ and store these primes in an array “prime[]”. Store the found primes in an array ‘prime[]’.
2. We need all primes in range [0..n-1]. We divide this range in different segments such that size of every segment is at-most √n
3. Do following for every segment [low..high]
• Create an array mark[high-low+1]. Here we need only O(x) space where x 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].

In Simple Sieve, we needed O(n) space which may not be feasible for large n. Here we need O(√n) space and we process smaller ranges at a time (locality of reference)

Below is implementation of above idea.

## C++

```// C++ program to print print all primes smaller than
// n using segmented sieve
#include <bits/stdc++.h>
using namespace std;

// This functions finds all primes smaller than 'limit'
// using simple sieve of eratosthenes. It also stores
// found primes in vector prime[]
void simpleSieve(int limit, vector<int> &prime)
{
// Create a boolean array "mark[0..n-1]" and initialize
// all entries of it as true. A value in mark[p] will
// finally be false if 'p' is Not a prime, else true.
bool mark[limit+1];
memset(mark, true, sizeof(mark));

for (int p=2; p*p<limit; p++)
{
// If p is not changed, then it is a prime
if (mark[p] == true)
{
// Update all multiples of p
for (int i=p*2; i<limit; i+=p)
mark[i] = false;
}
}

// Print all prime numbers and store them in prime
for (int p=2; p<limit; p++)
{
if (mark[p] == true)
{
prime.push_back(p);
cout << p << "  ";
}
}
}

// Prints all prime numbers smaller than 'n'
void segmentedSieve(int n)
{
// Compute all primes smaller than or equal
// to square root of n using simple sieve
int limit = floor(sqrt(n))+1;
vector<int> prime;
simpleSieve(limit, prime);

// Divide the range [0..n-1] in different segments
// We have chosen segment size as sqrt(n).
int low  = limit;
int high = 2*limit;

// While all segments of range [0..n-1] are not processed,
// process one segment at a time
while (low < n)
{
// To mark primes in current range. A value in mark[i]
// will finally be false if 'i-low' is Not a prime,
// else true.
bool mark[limit+1];
memset(mark, true, sizeof(mark));

// Use the found primes by simpleSieve() to find
// primes in current 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])
// For example, if low is 31 and prime[i] is 3,
int loLim = floor(low/prime[i]) * prime[i];
if (loLim < low)
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] = false;
}

// Numbers which are not marked as false are prime
for (int i = low; i<high; i++)
if (mark[i - low] == true)
cout << i << "  ";

// Update low and high for next segment
low  = low + limit;
high = high + limit;
if (high >= n) high = n;
}
}

// Driver program to test above function
int main()
{
int n = 100;
cout << "Primes smaller than " << n << ":n";
segmentedSieve(n);
return 0;
}
```

## Java

```// Java program to print print all primes smaller than
// n using segmented sieve

import java.util.Vector;
import static java.lang.Math.sqrt;
import static java.lang.Math.floor;

class Test
{
// This methid finds all primes smaller than 'limit'
// using simple sieve of eratosthenes. It also stores
// found primes in vector prime[]
static void simpleSieve(int limit, Vector<Integer> prime)
{
// Create a boolean array "mark[0..n-1]" and initialize
// all entries of it as true. A value in mark[p] will
// finally be false if 'p' is Not a prime, else true.
boolean mark[] = new boolean[limit+1];

for (int i = 0; i < mark.length; i++)
mark[i] = true;

for (int p=2; p*p<limit; p++)
{
// If p is not changed, then it is a prime
if (mark[p] == true)
{
// Update all multiples of p
for (int i=p*2; i<limit; i+=p)
mark[i] = false;
}
}

// Print all prime numbers and store them in prime
for (int p=2; p<limit; p++)
{
if (mark[p] == true)
{
System.out.print(p + "  ");
}
}
}

// Prints all prime numbers smaller than 'n'
static void segmentedSieve(int n)
{
// Compute all primes smaller than or equal
// to square root of n using simple sieve
int limit = (int) (floor(sqrt(n))+1);
Vector<Integer> prime = new Vector<>();
simpleSieve(limit, prime);

// Divide the range [0..n-1] in different segments
// We have chosen segment size as sqrt(n).
int low  = limit;
int high = 2*limit;

// While all segments of range [0..n-1] are not processed,
// process one segment at a time
while (low < n)
{
// To mark primes in current range. A value in mark[i]
// will finally be false if 'i-low' is Not a prime,
// else true.
boolean mark[] = new boolean[limit+1];

for (int i = 0; i < mark.length; i++)
mark[i] = true;

// Use the found primes by simpleSieve() to find
// primes in current range
for (int i = 0; i < prime.size(); i++)
{
// Find the minimum number in [low..high] that is
// a multiple of prime.get(i) (divisible by prime.get(i))
// For example, if low is 31 and prime.get(i) is 3,
int loLim = (int) (floor(low/prime.get(i)) * prime.get(i));
if (loLim < low)
loLim += prime.get(i);

/*  Mark multiples of prime.get(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.get(i))
mark[j-low] = false;
}

// Numbers which are not marked as false are prime
for (int i = low; i<high; i++)
if (mark[i - low] == true)
System.out.print(i + "  ");

// Update low and high for next segment
low  = low + limit;
high = high + limit;
if (high >= n) high = n;
}
}

// Driver method
public static void main(String args[])
{
int n = 100;
System.out.println("Primes smaller than " + n + ":");
segmentedSieve(n);
}
}
```

Output:
```Primes smaller than 100:
2 3 5 7 11 13 17 19 23 29 31 37 41
43 47 53 59 61 67 71 73 79 83 89 97  ```

Note that time complexity (or number of operations) by Segmented Sieve is same as Simple Sieve. It has advantages for large ‘n’ as it has better locality of reference and requires

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