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 to use Simple Sieve of Eratosthenes.
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;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 << " ";}// This code is contributed by sanjoy_62. |
Java
// 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. 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 + " "); }// This code is contributed by rutvik_56. |
Python3
# This functions finds all primes smaller than 'limit'# using simple sieve of eratosthenes.def 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. mark = [True for i in range(limit)] # One by one traverse all numbers so that their # multiples can be marked as composite. for p in range(p * p, limit - 1, 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 - 1, p): mark[i] = False # Print all prime numbers and store them in prime for p in range(2, limit - 1, 1): if (mark[p] == True): print(p, end=" ")# This code is contributed by Dharanendra L V. |
C#
// 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
<script>// This functions finds all primes smaller than 'limit'// using simple sieve of eratosthenes.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. var mark = Array(limit).fill(true); // One by one traverse all numbers so that their // multiples can be marked as composite. for (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 (i = p * p; i < limit; i += p) mark[i] = false; } } // Print all prime numbers and store them in prime for (p = 2; p < limit; p++) if (mark[p] == true) document.write(p + " "); }// This code is contributed by todaysgaurav </script> |
C
// 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 << " ";} |
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 Θ(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 functionint 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 sieveimport 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); }} |
Python3
# Python3 program to print all primes # smaller than n, using segmented sieve import mathprime = []# This method finds all primes # smaller than 'limit' using # simple sieve of eratosthenes. # It also stores found primes in list primedef 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 Coden = 100print("Primes smaller than", n, ":")segmentedSieve(100)# This code is contributed by bhavyadeep |
C#
// C# program to print // all primes smaller than// n using segmented sieveusing 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 functionlet 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.
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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