Segmented Sieve

Last Updated : 19 Nov, 2024

Given a number n, print all primes smaller than n.

Input: N = 10
Output: 2, 3, 5, 7
Explanation : The output “2, 3, 5, 7” for input N = 10 represents the list of the prime numbers less than or equal to 10.

Input: N = 5
Output: 2, 3, 5
Explanation : The output “2, 3, 5” for input N = 5 represents the list of the prime numbers less than or equal to 5.

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

C++

#include <iostream> #include <vector>  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.     std::vector<bool> mark(limit, true);      // 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 * p; 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)             std::cout << p << " "; }  int main() {     int limit = 100;      simpleSieve(limit);     return 0; }

// 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];     for(int i = 0; i<limit; i++) {       mark[i] = true;     }       // 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*p; 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 << "  "; }

Java

import java.util.Arrays;  public class Main {     public static void main(String[] args) {         int limit = 100;           simpleSieve(limit);     }      // This function finds all primes smaller than 'limit'     // using simple sieve of eratosthenes.     static 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.         boolean []mark = new boolean[limit];         Arrays.fill(mark, true);          // 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 * p; 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 + " ");     } }

Python

def simple_sieve(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.     mark = [True for _ in range(limit)]      # One by one traverse all numbers so that their     # multiples can be marked as composite.     for p in range(2, int(limit**0.5) + 1):         # If p is not changed, then it is a prime         if mark[p] == True:             # Update all multiples of p             for i in range(p * p, limit, p):                 mark[i] = False      # Print all prime numbers and store them in prime     for p in range(2, limit):         if mark[p] == True:             print(p, end=" ")  limit = 100  simple_sieve(limit)

// This functions finds all primes smaller than 'limit' // using simple sieve of eratosthenes. static 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 = new bool[limit];     Array.Fill(mark, true);       // 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 * p; 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)             Console.Write(p + " ");  }  // This code is contributed by pratham76.

JavaScript

function simpleSieve(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.     let mark = new Array(limit).fill(true);      // One by one traverse all numbers so that their     // multiples can be marked as composite.     for (let 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 (let i = p * p; i < limit; i += p)                 mark[i] = false;         }     }      // Print all prime numbers and store them in prime     for (let p = 2; p < limit; p++)         if (mark[p] === true)             console.log(p + " "); }  let limit = 100;   simpleSieve(limit);

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

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

Segmented Sieve
The idea of a 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.

Use Simple Sieve to find all primes up to the square root of ‘n’ and store these primes in an array “prime[]”. Store the found primes in an array ‘prime[]’.
We need all primes in the range [0..n-1]. We divide this range into different segments such that the size of every segment is at-most ?n
Do following for every segment [low..high]
- Create an array mark[high-low+1]. Here we need only O(x) space where x is a number of elements in a given range.
- Iterate through all primes found in step 1. For every prime, mark its multiples in the 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 the implementation of the above idea.

C++

// C++ program to 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.     vector<bool> mark(limit + 1, 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*p; 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;      prime.reserve(limit);     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<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;     } }  // 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 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 method 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*p; 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);                 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)         {             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<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;         }     }          // Driver method     public static void main(String args[])      {         int n = 100;         System.out.println("Primes smaller than " + n + ":");         segmentedSieve(n);     } }

Python

# Python3 program to print all primes  # smaller than n, using segmented sieve  import math prime = []  # This method finds all primes  # smaller than 'limit' using  # simple sieve of eratosthenes.  # It also stores found primes in list prime def simpleSieve(limit):          # Create a boolean list "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.      mark = [True for i in range(limit + 1)]     p = 2     while (p * p <= limit):                  # If p is not changed, then it is a prime          if (mark[p] == True):                           # Update all multiples of p              for i in range(p * p, limit + 1, p):                  mark[i] = False           p += 1              # Print all prime numbers      # and store them in prime      for p in range(2, limit):          if mark[p]:             prime.append(p)             print(p,end = " ")              # Prints all prime numbers smaller than 'n'  def segmentedSieve(n):          # Compute all primes smaller than or equal      # to square root of n using simple sieve     limit = int(math.floor(math.sqrt(n)) + 1)     simpleSieve(limit)          # Divide the range [0..n-1] in different segments      # We have chosen segment size as sqrt(n).      low = limit     high = limit * 2          # 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.          mark = [True for i in range(limit + 1)]                  # Use the found primes by simpleSieve()          # to find primes in current range          for i in range(len(prime)):                          # 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.              loLim = int(math.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 j in range(loLim, high, prime[i]):                 mark[j - low] = False                          # Numbers which are not marked as False are prime          for i in range(low, high):             if mark[i - low]:                 print(i, end = " ")                          # Update low and high for next segment          low = low + limit         high = high + limit  # Driver Code n = 100 print("Primes smaller than", n, ":") segmentedSieve(100)  # This code is contributed by bhavyadeep

// C# program to print  // all primes smaller than // n using segmented sieve using System; using System.Collections;  class GFG {     // This method 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 * p; 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

JavaScript

// JavaSCript program to print all primes smaller than // n using segmented sieve  // This functions finds all primes smaller than 'limit' // using simple sieve of eratosthenes. It also stores // found primes in vector prime[]  let res = "";  function simpleSieve(limit, 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.     let mark = new Array(limit+1).fill(true);          for (let 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 (let i=p*p; i<limit; i+=p){                 mark[i] = false;             }         }     }      // Print all prime numbers and store them in prime     for (let p=2; p<limit; p++)     {         if (mark[p] === true)         {             prime.push(p);             res = res + p + " ";         }     } }  // Prints all prime numbers smaller than 'n' function segmentedSieve(n) {     // Compute all primes smaller than or equal     // to square root of n using simple sieve     let limit = Math.floor(Math.sqrt(n))+1;     let prime = new Array(limit);          simpleSieve(limit, prime);      // Divide the range [0..n-1] in different segments     // We have chosen segment size as sqrt(n).     let low = limit;     let 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.         let mark = new Array(limit+1).fill(true);          // Use the found primes by simpleSieve() to find         // primes in current range         for (let i = 0; i < prime.length; 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.             let loLim = Math.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 (let j=loLim; j<high; j+=prime[i]){                 mark[j-low] = false;             }                     }          // Numbers which are not marked as false are prime         for (let i = low; i<high; i++){               if (mark[i - low] == true){                 res = res + i + " ";             }           }         // Update low and high for next segment         low = low + limit;         high = high + limit;     }     console.log(res); }  // Driver program to test above function let n = 100; console.log("Primes smaller than", n); segmentedSieve(n);  // The code is contributed by Gautam goel (gautamgoel962)

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

Time Complexity : O(n * ln(sqrt(n)))

Auxiliary Space: O(sqrt(n))

Note that time complexity (or a number of operations) by Segmented Sieve is the same as Simple Sieve. It has advantages for large 'n' as it has better locality of reference thus allowing better caching by the CPU and also requires less memory space.

Segmented Sieve (Print Primes in a Range)

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