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

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 &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

Python3

 # 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#

 // 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