While using && (logical AND), we must put the condition first whose probability of getting false is high so that compiler doesn’t need to check the second condition if the first condition is false.
C++14
#include <bits/stdc++.h>
using namespace std;
bool isOdd(int n) { return (n & 1); }
bool isPrime(int n)
{
if (n <= 1)
return false;
for (int i = 2; i <= sqrt(n); i++)
if ((n % i) == 0)
return false;
return true;
}
int main()
{
int cnt = 0, n = 10;
for (int i = 2; i <= n; i++) {
if (isOdd(i) && isPrime(i))
cnt++;
}
cnt = 0;
n = 10;
for (int i = 2; i <= n; i++) {
if (isPrime(i) && isOdd(i))
cnt++;
}
}
|
Java
public class Main {
public static boolean isOdd(int n) {
return n % 2 != 0;
}
public static boolean isPrime(int n) {
if (n <= 1) {
return false;
}
for (int i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
public static void main(String[] args) {
int cnt = 0;
int n = 10;
for (int i = 2; i < n; i++) {
if (isOdd(i) && isPrime(i)) {
cnt++;
}
}
System.out.println("Implementation 1: " + cnt);
cnt = 0;
n = 10;
for (int i = 2; i < n; i++) {
if (isPrime(i) && isOdd(i)) {
cnt++;
}
}
System.out.println("Implementation 2: " + cnt);
}
}
|
Python3
def isOdd(n):
pass
def isPrime(n):
pass
if __name__ == '__main__':
cnt = 0; n = 10
for i in range(2,n) :
if isOdd(i) and isPrime(i):
cnt+=1
cnt = 0
n = 10
for i in range(2,n):
if isPrime(i) and isOdd(i):
cnt+=1
|
C#
using System;
class MainClass {
public static bool isOdd(int n) {
return n % 2 != 0;
}
public static bool isPrime(int n) {
if (n <= 1) {
return false;
}
for (int i = 2; i <= Math.Sqrt(n); i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
public static void Main() {
int cnt = 0;
int n = 10;
for (int i = 2; i < n; i++) {
if (isOdd(i) && isPrime(i)) {
cnt++;
}
}
Console.WriteLine("Implementation 1: " + cnt);
cnt = 0;
n = 10;
for (int i = 2; i < n; i++) {
if (isPrime(i) && isOdd(i)) {
cnt++;
}
}
Console.WriteLine("Implementation 2: " + cnt);
}
}
|
Javascript
<script>
function isOdd(n) { return (n & 1); }
function isPrime(n)
{
if (n <= 1)
return false;
for (let i = 2; i <= Math.floor(Math.sqrt(n)); i++)
if ((n % i) == 0)
return false;
return true;
}
let cnt = 0, n = 10;
for (let i = 2; i <= n; i++) {
if (isOdd(i) && isPrime(i))
cnt++;
}
cnt = 0;
n = 10;
for (let i = 2; i <= n; i++) {
if (isPrime(i) && isOdd(i))
cnt++;
}
</script>
|
Consider the above implementation:
In implementation 1, we avoid checking even numbers whether they are prime or not as primality test requires more computation than checking a number for even/odd.
Probability of a number getting odd is more than of it being a prime that’s why we first check whether the number is odd before checking it for prime.
On the other hand in implementation 2, we are checking whether the number is prime or not before checking whether it is odd which makes unnecessary computation as all even numbers other than 2 are not prime but the implementation still checks them for prime.
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