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;
}
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 << " ";
}
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)
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
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#
// 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)
OutputPrimes 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.
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Sieve of Eratosthenes in 0(n) time complexity
The classical Sieve of Eratosthenes algorithm takes O(N log (log N)) time to find all prime numbers less than N. In this article, a modified Sieve is discussed that works in O(N) time.Example : Given a number N, print all prime numbers smaller than N Input : int N = 15 Output : 2 3 5 7 11 13 Input :
12 min read
Programs and Problems based on Sieve of Eratosthenes
C++ Program for Sieve of Eratosthenes
Given a number n, print all primes smaller than or equal to n. It is also given that n is a small number. For example, if n is 10, the output should be "2, 3, 5, 7". If n is 20, the output should be "2, 3, 5, 7, 11, 13, 17, 19".CPP// C++ program to print all primes smaller than or equal to // n usin
2 min read
Java Program for Sieve of Eratosthenes
Given a number n, print all primes smaller than or equal to n. It is also given that n is a small number. For example, if n is 10, the output should be "2, 3, 5, 7". If n is 20, the output should be "2, 3, 5, 7, 11, 13, 17, 19". Java // Java program to print all primes smaller than or equal to // n
2 min read
Scala | Sieve of Eratosthenes
Eratosthenes of Cyrene was a Greek mathematician, who discovered an amazing algorithm to find prime numbers. This article performs this algorithm in Scala. Step 1 : Creating an Int Stream Scala 1== def numberStream(n: Int): Stream[Int] = Stream.from(n) println(numberStream(10)) Output of above step
4 min read
Check if a number is Primorial Prime or not
Given a positive number N, the task is to check if N is a primorial prime number or not. Print 'YES' if N is a primorial prime number otherwise print 'NO.Primorial Prime: In Mathematics, A Primorial prime is a prime number of the form pn# + 1 or pn# - 1 , where pn# is the primorial of pn i.e the pro
10 min read
Sum of all Primes in a given range using Sieve of Eratosthenes
Given a range [l, r], the task is to find the sum of all the prime numbers in the given range from l to r both inclusive.Examples: Input : l = 10, r = 20Output : 60Explanation: Prime numbers between [10, 20] are: 11, 13, 17, 19Therefore, sum = 11 + 13 + 17 + 19 = 60Input : l = 15, r = 25Output : 59E
1 min read
Prime Factorization using Sieve O(log n) for multiple queries
We can calculate the prime factorization of a number "n" in O(sqrt(n)) as discussed here. But O(sqrt n) method times out when we need to answer multiple queries regarding prime factorization.In this article, we study an efficient method to calculate the prime factorization using O(n) space and O(log
11 min read
Java Program to Implement Sieve of Eratosthenes to Generate Prime Numbers Between Given Range
A number which is divisible by 1 and itself or a number which has factors as 1 and the number itself is called a prime number. The sieve of Eratosthenes is one of the most efficient ways to find all primes smaller than n when n is smaller than 10 million or so. Example: Input : from = 1, to = 20 Out
3 min read
Segmented Sieve
Given a number n, print all primes smaller than n. Input: N = 10Output: 2, 3, 5, 7Explanation : The output â2, 3, 5, 7â for input N = 10 represents the list of the prime numbers less than or equal to 10. Input: N = 5Output: 2, 3, 5 Explanation : The output â2, 3, 5â for input N = 5 represents the li
15+ min read
Segmented Sieve (Print Primes in a Range)
Given a range [low, high], print all primes in this range? For example, if the given range is [10, 20], then output is 11, 13, 17, 19. A Naive approach is to run a loop from low to high and check each number for primeness. A Better Approach is to precalculate primes up to the maximum limit using Sie
15 min read
Longest sub-array of Prime Numbers using Segmented Sieve
Given an array arr[] of N integers, the task is to find the longest subarray where all numbers in that subarray are prime. Examples: Input: arr[] = {3, 5, 2, 66, 7, 11, 8} Output: 3 Explanation: Maximum contiguous prime number sequence is {2, 3, 5} Input: arr[] = {1, 2, 11, 32, 8, 9} Output: 2 Expla
13 min read
Sieve of Sundaram to print all primes smaller than n
Given a number n, print all primes smaller than or equal to n.Examples: Input: n = 10Output: 2, 3, 5, 7Input: n = 20Output: 2, 3, 5, 7, 11, 13, 17, 19We have discussed Sieve of Eratosthenes algorithm for the above task. Below is Sieve of Sundaram algorithm.printPrimes(n)[Prints all prime numbers sma
10 min read