Prime Numbers on kenji.blog

C Language Code to Generate Prime Numbers

Sat, 24 Aug 2024 09:38:10 +0900

Here is a simple C language code that generates prime numbers within a specified range. In this example, it enumerates prime numbers from 1 to n.

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#include <stdio.h>
#include <stdbool.h>

bool isPrime(int num) {
 if (num <= 1) return false;
 if (num <= 3) return true;
 if (num % 2 == 0 || num % 3 == 0) return false;

 for (int i = 5; i * i <= num; i += 6) {
 if (num % i == 0 || num % (i + 2) == 0) return false;
 }
 return true;
}

void printPrimes(int n) {
 printf("2 ");
 for (int i = 3; i <= n; i += 2) {
 if (isPrime(i)) {
 printf("%d ", i);
 }
 }
 printf("\n");
}

int main() {
 int n;
 printf("Please enter the maximum value of the range to generate prime numbers: ");
 scanf("%d", &n);
 printf("The prime numbers from 1 to %d are as follows:\n", n);
 printPrimes(n);
 return 0;
}

This code works as follows:

isPrime function: Determines whether a given number is a prime number. For efficiency, it first checks if it is divisible by 2 or 3, and then proceeds to check using steps of 6.
printPrimes function: Outputs the prime numbers within the specified range. It outputs 2 first, and then checks only odd numbers.
main function: Prompts the user to input the maximum value of the range and outputs the prime numbers within that range.

When you compile and run this code, it will display the prime numbers within the specified range.

How to Enumerate Prime Numbers Up to 1000 Using the Sieve of Eratosthenes

Sun, 09 Apr 2023 12:54:24 +0900

What is the Sieve of Eratosthenes?

The Sieve of Eratosthenes is an algorithm for finding all prime numbers up to any given limit. The algorithm is simple and can be implemented with the following steps:

Create an array of boolean values with N elements, and initialize all elements to true.
Set the 0th and 1st elements of the array to false (because 0 and 1 are not prime numbers).
If the 2nd element of the array is true, output 2 as a prime number.
Set all multiples of 2 from $2^2$ onwards in the array to false.*
If the 3rd element of the array is true, output 3 as a prime number.
Set all multiples of 3 from $3^2$ onwards in the array to false.
Repeat the same process for the 4th, 5th, …, Nth elements.

*The reason for targeting elements from the square of the number onwards to become false is because the numbers smaller than the square have already been processed (enumeration is complete).

Implementation in Rust

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fn main() {
 let n = 1000;
 let mut is_prime = vec![true; n+1];
 is_prime[0] = false;
 is_prime[1] = false;
 for i in 2..=n {
 if is_prime[i] {
 println!("{}", i);
 let mut j = i * i;
 while j <= n {
 is_prime[j] = false;
 j += i;
 }
 }
 }
}

Slightly Faster Version

Considering the following points, we implement a slightly faster version:

Instead of initializing the array with true, initialize it with false (this is faster).
Since multiples of 2 are not prime numbers, omit the process of setting the elements of multiples of 2 to false.
There is no need to loop up to n; by enumerating primes up to the square root of n, you can enumerate primes up to n.

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fn main() {
 let n = 1000;
 let mut is_prime = vec![false; n+1];
 is_prime[2] = true;
 for i in (3..=n).step_by(2) {
 is_prime[i] = true;
 }
 for i in 3..=((n as f64).sqrt() as usize) {
 if is_prime[i] {
 let mut j = i * i;
 while j <= n {
 is_prime[j] = false;
 j += i * 2;
 }
 }
 }
 for i in (2..=n).filter(|&x| is_prime[x]) {
 println!("{}", i);
 }
}

References

Sieve of Eratosthenes

Enumerate Prime Numbers in Rust

Fri, 09 Sep 2022 07:08:49 +0900

I wrote a program to enumerate prime numbers in Rust.

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fn main() {
 let max = 1000;
 let mut primes = vec![2];
 let mut n = 3;
 loop {
 let mut is_prime = true;
 for p in &primes {
 if n % p == 0 {
 is_prime = false;
 break;
 }
 }
 if is_prime {
 primes.push(n);
 }
 n += 2;
 if n > max {
 break;
 }
 }
 for p in &primes {
 println!("{}", p);
 }
}