finding the nth prime javascript

Question

the functions below are supposed to spit out the nth prime number. However, it keeps on spitting out 3. Can somebody please help? Cheers, Anthony

function Prime(num) {
output = true  
    for (i=2 ; i<num ; i++) {
        if (num%i === 0)  {
           output = false ; break
        }
    }
return output
}

function PrimeMover(num) {
var count = 0
    for (i=2 ; i<10000 ; i++)  {
        if (Prime(i) === true) {
            count = count + 1 
        }
        if (count === num) {
            return i
            break
        } 
    }
}

You need to use local variables. Otherwise, all of your loops will overwrite each other! Take a look at my prime number programming interface here: http://www.myersdaily.org/joseph/javascript/primes-10k.2.html in which you can do what you want with code like this `var t = [], i=2, n = 300; while (t.length < n) { prime(i) && (t[t.length] = i); i++; } t[t.length-1]; // our list` then just click "Execute" (outputs 1987) — Joseph Myers, Mar 17 '15 at 04:54
PS typing 10000 into the above code, the 10000th prime is 104729. — Joseph Myers, Mar 17 '15 at 04:59
Related questions: [Writing first 100 prime numbers to a file using node.js](http://stackoverflow.com/questions/17382910/writing-first-100-prime-numbers-to-a-file-using-node-js) and [How to find prime numbers between 0 - 100?](http://stackoverflow.com/questions/11966520/how-to-find-prime-numbers-between-0-100) — jfriend00, Mar 17 '15 at 05:22

score 2 · Answer 1 · answered Mar 17 '15 at 05:25

2

You have created loop counter i in global scope.so both PrimeMover and Prime mutates same global i.In every iteration ,PrimeMover assigns i=2.After that Prime assigns i=2.your i variable's value will be changed between 2 and 3.use local loop counter variable var i=0;

function Prime(num) {
output = true  
for (var i=2 ; i<num ; i++) { //var i=2
    if (num%i === 0)  {
       output = false ; break
    }
}
return output
}

function PrimeMover(num) {
var count = 0
for (var i=2 ; i<10000 ; i++)  { //var i=2
    if (Prime(i) === true) {
        count = count + 1 
    }
    if (count === num) {
        return i
        break
    } 
}
}

answered Mar 17 '15 at 05:25

balajisoundar

581
2
11

@ Joseph and Balaji,Understood. Thank you so much. – Anthony Mar 17 '15 at 06:48
Instead of using an `output` variable, can you not simply return `false` in the loop and return `true` at the end? I think that would be easier. – Ethan McTague Oct 23 '15 at 12:48
Yes , that would be much more readable . But i was trying to explain the bug in the code.Thanks for your comment.:) – balajisoundar Oct 26 '15 at 05:41
Also, it is enough to look for dividers from 2 to square root of num. And in your main fn, if i is greather then or equal to 3, you can increase i again (skipp the even numbers)... – ppseprus Jan 05 '16 at 16:40

Won · Answer 2 · 2020-02-24T17:37:01.277

2

For minimal code lovers,

function nthprime(n)
{
  var prime=[], i=1
  while (i++ && prime.length<n) prime.reduce((a,c)=>(i%c)*a,2) && prime.push(i)
  return prime.length?prime.pop():-1
}
[-1,0,1,2,3,5,10,100].forEach(n=>console.log(`nthprime(${n})=${nthprime(n)}`))

edited Feb 24 '20 at 17:37

answered Jul 12 '19 at 18:04

Won

1,795
2
12
22

3

Nice answer, except that mathematicians exclude 1 from the list, and consider 2 as the smallest prime number (you have `nthprime(1)==1`). – Purplejacket Feb 16 '20 at 00:47

Ashish Yadav · Answer 3 · 2018-07-13T07:49:28.520

1

function main(inp) {
  var count = 0;
  for (var i = 2; i <= 100000; i++) {
   if (isPrime(i)) count = count + 1;
   if (count == inp) return i;
 }
}
function isPrime(i) {
 for (var j = 2; j < i; j++) {
  //instead of `j < i` it can be reduced using other conditions 
  if (i % j == 0) {
   return false
  }
 }
 return true
}
main(5) // any number

edited Jul 13 '18 at 07:49

answered May 18 '18 at 07:52

Ashish Yadav

3,039
1
16
23

score 0 · Answer 4 · answered Jan 05 '16 at 16:42

This might be a bit more optimal

function nthPrime(n) {
    var P = 0;

    function isPrime(x) {
        var isPrime= true;

        for (var d = 2; d <= Math.sqrt(x); d++) {
            if((x/d) % 1 == 0) {
                isPrime = false;
                break;
            }
        }

        return isPrime;
    }

    for (var i = 1; 0 < n; i++) {

        if(isPrime(i)) {
            P = i; n--;
        }

        // we can skip the even numbers
        if(3 <= i){
            i++;
        }

    }

    return P;
}

Sameer Safi · Answer 5 · 2021-06-06T13:47:36.880

0

Try this

var pos=10001;
console.log(primeNumforPos(pos));

function primeNumforPos(pos){
  var num=2,curPos=0;
  while(curPos<=pos){
    if(isPrime(num)){
      curPos++;
    }
    if(curPos==pos){
      return num;
    }else{
      num++;
    }
  }
}

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

edited Jun 06 '21 at 13:47

answered Jun 06 '21 at 13:35

Sameer Safi

1
2

Welcome to Stack Overflow! Code-only answers are not particularly helpful. Please add some descriptions of how this code solves the problem. – Sven Eberth Jun 06 '21 at 16:10

New contributor · Answer 6 · 2022-09-26T07:05:26.730

So, I decided to optimise the hell out of the code (cuz why not). It is almost 6 times as fast as that of ppseprus (297ms vs 1773ms in nth_prime(100000)).

let primes = [2, 3]; 

const nth_prime = (n) => {
    if (n <= primes.length) return primes[n - 1]; // handle values which have been cached

    let i = 1; 
    
    while (1){
        const a = 6 * i - 1, b = 6 * i + 1, a_limit = Math.sqrt(a), b_limit = Math.sqrt(b); // the 6n - 1 and 6n + 1 rule for primes
        let a_prime = true, b_prime = true; 

        i++; 
        
        // prime check

        for (const prime of primes){
            if (prime > a_limit) break; 
            if (a % prime == 0){
                a_prime = false; 
                break; 
            }
        }
        
        if (a_prime){
            if (primes.length + 1 == n) return a; 
            primes.push(a); // cache
        }
        
        for (const prime of primes){
            if (prime > b_limit) break; 
            if (b % prime == 0){
                b_prime = false; 
                break; 
            }
        }
        
        if (b_prime){
            if (primes.length + 1 == n) return b; 
            primes.push(b); // cache
        }
    }
}

score -1 · Answer 7 · answered Mar 19 '21 at 17:40

const findPrime = num => {
let i, primes = [2, 3], n = 5
const isPrime = n => {
    let i = 1, p = primes[i],
        limit = Math.ceil(Math.sqrt(n))
    while (p <= limit) {
        if (n % p === 0) {
            return false
        }
        i += 1
        p = primes[i]
    }
    return true
}
for (i = 2; i <= num; i += 1) {
    while (!isPrime(n)) {
        n += 2
    }
    primes.push(n)
    n += 2
};
return primes[num - 1]

} console.time('Time')

let x = findPrime(9999)

console.timeEnd('Time')

console.log(x)

finding the nth prime javascript

7 Answers7