# Number prime test in JavaScript

I'm trying to complete the codewars challenge that asks you to check if a number is a prime number. For whatever reason, my solution doesn't seem to work for the square of odd prime numbers (e.g. 9 returns true instead of false).

``````function isPrime(num) {

if (num === 2) {
return true;
}
else if(num > 1){
for (var i = 2;  i < num; i++) {

if (num % i !== 0 ) {
return true;
}

else if (num === i * i) {
return false
}

else {
return false;
}
}
}
else {
return false;
}

}

console.log(isPrime(121));
``````

P.s. I included that second else/if statement because I was trying to solve the problem.

As simple as possible:

``````function isPrime(num) {
for(var i = 2; i < num; i++)
if(num % i === 0) return false;
return num > 1;
}
``````

With the ES6 syntax:

``````const isPrime = num => {
for(let i = 2; i < num; i++)
if(num % i === 0) return false;
return num > 1;
}
``````

You can also decrease the complexity of the algorithm from `O(n)` to `O(sqrt(n))` if you run the loop until square root of a number:

``````const isPrime = num => {
for(let i = 2, s = Math.sqrt(num); i <= s; i++)
if(num % i === 0) return false;
return num > 1;
}
``````

A small suggestion here, why do you want to run the loop for whole n numbers?

If a number is prime it will have 2 factors (1 and number itself). If it's not a prime they will have 1, number itself and more, you need not run the loop till the number, may be you can consider running it till the square root of the number.

You can either do it by euler's prime logic. Check following snippet:

``````function isPrime(num) {
var sqrtnum=Math.floor(Math.sqrt(num));
var prime = num != 1;
for(var i=2; i<sqrtnum+1; i++) { // sqrtnum+1
if(num % i == 0) {
prime = false;
break;
}
}
return prime;
}
``````

Now the complexity is O(sqrt(n))

Hope it helps

Cool version:

``````const isPrime = n => ![...Array(n).keys()].slice(2).map(i => !(n%i)).includes(true) && ![0,1].includes(n)
``````

Prime numbers are of the form 6f ± 1, excluding 2 and 3 where f is any integer

`````` function isPrime(number)
{
if (number <= 1)
return false;

// The check for the number 2 and 3
if (number <= 3)
return true;

if (number%2 == 0 || number%3 == 0)
return false;

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

return true;
}
``````

Time Complexity of the solution: O(sqrt(n))

I think this question is lacking a recursive solution:

``````// Preliminary screen to save our beloved CPUs from unneccessary labour

const isPrime = n => {
if (n === 2 || n === 3) return true;
if (n < 2 || n % 2 === 0) return false;

return isPrimeRecursive(n);
}

// The recursive function itself, tail-call optimized.
// Iterate only over odd divisors (there's no point to iterate over even ones).

const isPrimeRecursive = (n, i = 3, limit = Math.floor(Math.sqrt(n))) => {
if (n % i === 0) return false;
if (i >= limit) return true; // Heureka, we have a prime here!
return isPrimeRecursive(n, i += 2, limit);
}

// Usage example

for (i = 0; i <= 50; i++) {
console.log(`\${i} is \${isPrime(i) ? `a` : `not a` } prime`);
}``````

This approach have it's downside – since browser engines are (written 11/2018) still not TC optimized, you'd probably get a literal stack overflow error if testing primes in order of tens lower hundreds of millions or higher (may vary, depends on an actual browser and free memory).

you can use below code in javascript for checking number is prime or not. It will reduce no of iteration and get the result fast.

``````function testPrime(num) {
var isPrime = true;
if (num >= 2) {
if(num == 2 || num == 3){
isPrime = true;
}
else if (num % 2 == 0) {
isPrime = false;
}
else {
for (i = 3; i <= Math.floor(Math.sqrt(num)); i += 2) {
if (num % i == 0) {
isPrime = false;
break;
}
}
}
}
else {
isPrime = false;
}
return isPrime;
}
``````

//testPrime(21) false

I think a better way to find a prime number is with this logic:

``````var p=prompt("input numeric value","10"); // input your number
for(j=2;j<p;j++){
if(isPrimes(j)){
document.write(j+", "); // for output the value
} // end if
}// end for loop
function isPrimes(n) {
var primes = true;// let prime is true
for (i=2;i<n;i++) {
if(n%i==0) {
primes= false; // return prime is false
break; // break the loop
}// end if inner
}// end inner loop
return primes; // return the prime true or false
}// end the function``````

You can try this one

``````function isPrime(num){

// Less than or equal to 1 are not prime
if (num<=1) return false;

// 2 and 3 are prime, so no calculations
if (num==2 || num==3 ) return true;

// If mod with square root is zero then its not prime
if (num % Math.sqrt(num)==0 ) return false;

// Run loop till square root
for(let i = 2, sqrt = Math.sqrt(num); i <= sqrt; i++) {

// If mod is zero then its not prime
if(num % i === 0) return false;
}

// Otherwise the number is prime
return true;
}

for(let i=-2; i <= 35; i++) {
console.log(`\${i} is\${isPrime(i) ? '' : ' not'} prime`);
}``````

One of the shortest version

``````isPrime=(n)=>[...Array(n-2)].map((_,i)=>i+2).filter(i=>n%i==0).length==0
``````

You are trying to check too much conditions. just one loop is required to check for a prime no.

``````function isPrime(num){
if(num==2)
return true;
for(i=2;i<Math.sqrt(num);i++) // mathematical property-no number has both of its factors greater than the square root
{
if(num % i==0)
return false; // otherwise it's a prime no.
}
return true;
}
``````

You have to consider every no. a prime no. unless it is divisible by some no. less than or equal to the square root.

Your solution has got a return statement for every case,thus it stops execution before it should.It doesn't check any number more than once.It gives wrong answer for multiple cases-- 15,35.. in fact for every no. that is odd.

It looks like your first if statement within the first 'if' statement within the for loop. Since if num = 9 and i = 2, 9 % i !== 0 but 9 is not prime since on the next iteration where i = 3, 9 % i === 0.

Here would be my answer to that question.

``````var isPrime = function(n) {
if(typeof n !== 'number' || n <= 1 || n % 1 !== 0){
return false;
}
for(var i = 2; i <= Math.sqrt(n); i += 1){
if(n % i === 0){
return false;
}
}
return true;
};
``````

The first if statement catches the edge cases. The for loop then checks from 2 up to the square root of n because of the mathematical property where no number has both of its factors greater than the square root of that number.

Hope this helps!

This one is I think more efficient to check prime number :

``````function prime(num){
if(num == 1) return true;
var t = num / 2;
var k = 2;
while(k <= t) {
if(num % k == 0) {
return false
} else {
k++;
}
}
return true;
}
console.log(prime(37))
``````

Simple version:

``````function isPrime(num) {
if (num <= 1) {
return false;
} else {
for (var i = 2; i < num; i++) {
if (num % i === 0) {
return false;
}
}
return true;
}
}

console.log(isPrime(9));
``````

This is how I'd do it:

``````function isPrime(num) {
if(num < 2) return false;
if(num == 2) return true;
for(var i = 2; i < num; i++) {
if(num % i === 0) return false;
}
return true;
}
``````

very simple

``````const isPrime = num => {
for (var i = 2; i < num; i++) if (num % i == 0) return false;
return num >= 2;
}
``````
``````(function(value){
var primeArray = [];
for(var i = 2; i <= value; i++){
if((i === 2) || (i === 3) || (i === 5) || (i === 7)){
primeArray.push(i);
}
else if((i % 2 !== 0) && (i % 3 !== 0) && (i % 5 !== 0) && (i % 7 !== 0)){
primeArray.push(i);
}
}
console.log(primeArray);
}(100));
``````

Using Ticked solution Ihor Sakaylyuk

``````const isPrime = num => {
for(let i = 2, s = Math.sqrt(num); i <= s; i++)
if(num % i === 0) return false;
return num !== 1 && num !== 0;
}
``````

Gives in console

isPrime( -100 ) true

``````const isPrime = num => {
// if not is_number num return false

if (num < 2) return false

for(let i = 2, s = Math.sqrt(num); i <= s; i++) {
if(num % i === 0) return false
}

return true
}
``````

Gives in console

isPrime( 1 ) false

isPrime( 100 ) false

isPrime( -100 ) false

First 6 primes ? 2 3 5 7 11 13 ?

isPrime( 1 ) false

isPrime( 2 ) true // Prime 1

isPrime( 3 ) true // Prime 2

isPrime( 4 ) false

isPrime( 5 ) true // Prime 3

isPrime( 6 ) false

isPrime( 7 ) true // Prime 4

isPrime( 8 ) false

isPrime( 9 ) false

isPrime( 10 ) false

isPrime( 11 ) true // Prime 5

isPrime( 12 ) false

isPrime( 13 ) true // Prime 6

This answer is based on the answer by Ihor Sakaylyuk. But instead of checking for all numbers, I am checking only the odd numbers. Doing so I reduced the time complexity of the solution to O(sqrt(n)/2).

``````function isPrime(num) {
if (num > 2 && num % 2 === 0) return false;
for (var i = 3; i < Math.sqrt(num); i += 2) {
if (num % i === 0) return false;
}
return num > 1;
}``````

My Solution,

``````function isPrimeNumber(number){
if(number <= 1) return false;
if(number <= 3) return true;
for(let i = 2; i < 9; i++) {
if(number === i) continue;
if(number % i === 0 ) return false;
}
return true;
}

for(let i = 0; i <= 100; i++){
}``````

This will cover all the possibility of a prime number . (order of the last 3 if statements is important)

``````   function isPrime(num){
if (num==0 || num==1) return false;
if (num==2 || num==3 ) return true;
if (num % Math.sqrt(num)==0 ) return false;
for (let i=2;i< Math.floor(Math.sqrt(num));i++) if ( num % i==0 ) return false;
if ((num * num - 1) % 24 == 0) return true;
}
``````

Following code uses most efficient way of looping to check if a given number is prime.

``````function checkPrime(number){
if (number==2 || number==3) {
return true
}
if(number<2 ||number % 2===0){
return false
}
else{
for (let index = 3; index <= Math.sqrt(number); index=index+2) {
if (number%index===0) {
return false
}
}
}
return true
}
``````
1. First check rules out 2 and lesser numbers and all even numbers
2. Using Math.sqrt() to minimize the looping count
3. Incrementing loop counter by 2, skipping all even numbers, further reducing loop count by half

Why are you trying to use loops!? Iterating is such a waste of computing power here...

This can be done using math:

``````function isPrime(num) {
if ( num !=1 && num%3 != 0 && num%5 != 0 && num%7 != 0 && num%9 != 0 && num%11 != 0 && Math.sign(num) == 1 && Math.round(num) == num) {

if ( (num-1)%6 == 0 || (num+1)%6 == 0 ) {
return true;
}

} // no need for else statement since if true, then will do return
return num==11||x==9||num==5||num==3||num==2; // will return T/F;
}
``````

# Steps:

1. Check if the number is divisible by 5 (evenly)
2. Check if the number is positive (negative numbers are not prime)
3. Check if the number is a whole number (5.236 by rule not prime)
4. Check if the number ± 1 is divisible by 6 (evenly)
5. Check if the number is 2, 3, 9 or 11 (outliers with the rule)
6. Check if the number is 7 or 5 (due to attempt at false-positive reduction)

Always try to do the mathematical way, rather than iterating over loops. Math is almost always the most efficient way to do something. Yes, this equation might be confusing... However, we don't need much readability when it comes to checking if something is prime or not... It likely isn't going to need to be maintained by future code editors.

EDIT:
Optimized code version:

``````function isPrime(x=0) {
const m = Math;
return (x%3!=0&&x%5!=0&&x%7!=0&&x%9!=0&&x%11!=0&&x!=1&&m.sign(x)*m.round(x)==x&&(!!!((x-1)%6)||!!!((x+1)%6)))||x==11||x==9||x==7||x==5||x==3||x==2;
}
``````

EDIT:
As it turns out... there's more to do with false-positive detection, because they're randomly distributed (at least seemingly) so the +-1 %6 stuff really isn't going to work for everything... But I'm definitely on to something here...

The only even prime number is 2, so we should exclude even numbers from the loop.

``````function isPrime(a) {
if (a < 2) return false;
if (a > 2) {
if (a % 2 == 0) return false;
for (let x = 3; x <= Math.sqrt(a); x += 2) {
if (a % x == 0) return false;
}
}
return true;
}
``````
``````function isPrime(num) {
if (num <= 1) return false; // negatives
if (num % 2 == 0 && num > 2) return false; // even numbers
let s = Math.sqrt(num); // store the square to loop faster
for(let i = 3; i <= s; i++) { // start from 3, stop at the square, increment
if(num % i === 0) return false; // modulo shows a divisor was found
}
return true;
}
``````

``````// A list prime numbers

function* Prime(number) {
const infinit = !number && number !== 0;
const re = /^.?\$|^(..+?)\1+\$/;
let actual = 1;

while (infinit || number-- ) {
if(!re.test('1'.repeat(actual)) == true) yield actual;
actual++
};
};

let [...primers] = Prime(101); //Example
console.log(primers);``````

``````function isPrime(num) {
var prime = num != 1;
for(var i=2; i<num; i++) {
if(num % i == 0) {
prime = false;
break;
}
}
return prime;
}
``````

DEMO

``````function isPrimeNumber(n) {
for (var i = 2; i < n; i++) { // i will always be less than the parameter so the condition below will never allow parameter to be divisible by itself ex. (7 % 7 = 0) which would return true
if(n % i === 0) return false; // when parameter is divisible by i, it's not a prime number so return false
}
return n > 1; // otherwise it's a prime number so return true (it also must be greater than 1, reason for the n > 1 instead of true)
}

``````
``````function isAPrimeNumber(num){
var counter = 0;
//loop will go k equals to \$num
for (k = 1; k <= num; k++) {
//check if the num is divisible by itself and 1
// `%` modulus gives the reminder of the value, so if it gives the reminder `0` then it is divisible by the value
if (num % k == 0) {
//increment counter value 1
counter  = counter  + 1;
}
}
//if the value of the `counter is 2` then it is a `prime number`
//A prime number is exactly divisible by 2 times only (itself and 1)
if (counter == 2) {
return num + ' is a Prime Number';
}else{
return num + ' is nota Prime Number';
}
}
``````

Now call `isAPrimeNumber()` function by passing a value.

``````var resp = isAPrimeNumber(5);
console.log(resp);
``````

Output:

``````5 is a Prime Number
``````
``````function isPrime(num) {
if(num < 2) return false;
for (var i = 2; i <= num/2; i++) {
if(num%i==0)
return false;
}
return true;
}
``````

If we want the prime number between two number we have to add this code only

``````for(var i = 0; i < 100; i++){
if(isPrime(i))
console.log(i);
}
``````

``````function isPrime(n){
if (isNaN(n) || !isFinite(n) || n%1 || n<2) {
return false;
}

if (n%2==0){
return (n==2);
}

var sqrt = Math.sqrt(n);
for (var i = 3; i < sqrt; i+=2) {
if(n%i == 0){
return false;
}
}

return true;
}``````

@IhorSakaylyuk's answer can be improved further down to `O(sqrt(n/2))` by skipping all of the even numbers except 2:

``````const isPrime = num => {
if(num % 2 === 0 && number !== 2) return false
for(let i = 3, s = Math.sqrt(num); i <= s; i += 2)
if(num % i === 0) return false;
return num > 1;
}
``````

You can also use the npm package `pial`.

I would do it like this:

``` const isPrime = (num) => num < 10 ? [2, 3, 5, 7].includes(num) : ![2, 3, 5, 7].some(i => !(num % i)); ```

Might be useful for some people: An implementation of the Miller Rabin primality test. Works for all positive integers less than Number.MAX_SAFE_INTEGER.

Try on JSFiddle: https://jsfiddle.net/4rxhas2o/

``````
let unsafeToSquare = Math.floor(Math.sqrt(Number.MAX_SAFE_INTEGER))

// Returns (a + b) % m

let sum = a + b

let result = sum % m

if (sum < Number.MAX_SAFE_INTEGER)
return result

let signature = ((a % 8) + (b % 8)) % 8

let sumMod = sum % 8

for (let i = -2; i <= 2; ++i) {
if ((sumMod + i) % 8 === signature) {
let ret = result + i

if (ret > m)
ret = (result - m) + i // prevent overflow

return ret
}
}
}

function mulMod(a, b, m) {
if (m === 0)
return 0

let prod = a * b

if (prod < Number.MAX_SAFE_INTEGER)
return prod % m

let y = 0
let result = a

while (b > 1) {
if (b % 2 === 0) {

b /= 2
} else {

b = (b - 1) / 2
}
}

}

function squareMod(b, m) {
// Computes (b * b % m)

return mulMod(b, b, m)
}

function expModLargeB(b, exponent, m) {
let y = 1

while (exponent > 1) {
if (exponent % 2 === 0) {
b = squareMod(b, m)

exponent /= 2
} else {
y = mulMod(y, b, m)
b = squareMod(b, m)

exponent = (exponent - 1) / 2
}
}

return mulMod(b, y, m)
}

function expMod(b, exponent, m) {
if (exponent === 0)
return 1

if (b >= unsafeToSquare || m >= unsafeToSquare) {
return expModLargeB(b, exponent, m)
}

let y = 1

while (exponent > 1) {
if (exponent % 2 === 0) {
b *= b
b %= m

exponent /= 2
} else {
y *= b
b *= b

y %= m
b %= m

exponent = (exponent - 1) / 2
}
}

return (b * y) % m
}

function _isPrimeTrialDivision(p) {
let sqrtP = Math.ceil(Math.sqrt(p))

for (let i = 23; i <= sqrtP + 1; i += 2) {
if (p % i === 0)
return false
}

return true
}

function _isProbablePrimeMillerRabin(p, base=2) {
let pm1 = p - 1
let pm1div = pm1
let d, r = 0

while (true) {
if (pm1div % 2 === 0) {
pm1div /= 2

r++
} else {
d = pm1div
break
}
}

let x = expMod(base, d, p)

if (x === 1 || x === pm1)
return true

for (let i = 0; i < r - 1; ++i) {
x = squareMod(x, p)

if (x === pm1)
return true
}

return false
}

function _isPrimeLarge(p) {
let bases

if (p < 2047)
bases = [2]
else if (p < 1373653)
bases = [2, 3]
else if (p < 9080191)
bases = [31, 73]
else if (p < 25326001)
bases = [2, 3, 5]
else if (p < 3215031751)
bases = [2, 3, 5, 7]
else if (p < 4759123141)
bases = [2, 7, 61]
else if (p < 1122004669633)
bases = [2, 13, 23, 1662803]
else if (p < 2152302898747)
bases = [2, 3, 5, 7, 11]
else if (p < 3474749660383)
bases = [2, 3, 5, 7, 11, 13]
else if (p < 341550071728321)
bases = [2, 3, 5, 7, 11, 13, 17]
else
bases = [2, 3, 5, 7, 11, 13, 17, 19, 23]

return bases.every(base => _isProbablePrimeMillerRabin(p, base))
}

let smallPrimes = [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, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223]

function isPrime(p) {
if (!Number.isInteger(p) || p < 2)
return false

// Test for small primes
for (let i = 0; i < smallPrimes.length; ++i) {
let prime = smallPrimes[i]

if (p === prime)
return true
if (p % prime === 0)
return false
}

if (p < 150) {
return _isPrimeTrialDivision(p)
} else {
return _isPrimeLarge(p)
}
}

const tests = [1, 2, 3, 10, 100, 100019, 10000000019, 100000000003, 10000000000037]
let start = performance.now()

tests.forEach(test => {
console.log(`\${test} is \${ isPrime(test) ? "" : "not " }prime`)
})

let end = performance.now()
console.log("Tests completed in " + (end - start) + " ms.")
``````

This calculates square differently and skips even numbers.

``````const isPrime = (n) => {
if (n <= 1) return false;
if (n === 2) return true;
if (n % 2 === 0) return false;
//goto square root of number
for (let i = 3, s = n ** 0.5; i < s; i += 2) {
if (n % i == 0) return false;
}
return true;
};
``````