# 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.

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

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 ` `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 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) ` `    ``{ ` `        ``if` `(high >= n)  ` `           ``high = 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, ` `            ``// we start with 33. ` `            ``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

## 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 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 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) ` `        ``{ ` `            ``if` `(high >= n)  ` `                ``high = 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, ` `                ``// we start with 33. ` `                ``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

## C#

 `// C# program to print print ` `// all primes smaller than ` `// n using segmented sieve ` `using` `System; ` `using` `System.Collections; ` ` `  `class` `GFG ` `{ ` `    ``// 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, ` `                            ``ArrayList 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 = ``new` `bool``[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``) ` `            ``{ ` `                ``prime.Add(p); ` `                ``Console.Write(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``) (Math.Floor(Math.Sqrt(n)) + 1); ` `        ``ArrayList prime = ``new` `ArrayList();  ` `        ``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) ` `        ``{ ` `            ``if` `(high >= n)  ` `                ``high = 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 = ``new` `bool``[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.Count; 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, we start with 33. ` `                ``int` `loLim = ((``int``)Math.Floor((``double``)(low /  ` `                            ``(``int``)prime[i])) * (``int``)prime[i]); ` `                ``if` `(loLim < low) ` `                    ``loLim += (``int``)prime[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 += (``int``)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``) ` `                    ``Console.Write(i + ``" "``); ` `     `  `            ``// Update low and high for next segment ` `            ``low = low + limit; ` `            ``high = high + limit; ` `        ``} ` `    ``} ` `     `  `    ``// Driver code ` `    ``static` `void` `Main()  ` `    ``{ ` `        ``int` `n = 100; ` `        ``Console.WriteLine(``"Primes smaller than "` `+ n + ``":"``); ` `        ``segmentedSieve(n); ` `    ``} ` `} ` ` `  `// This code is contributed by mits `

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