Why my function return an array with only one element?

Question

I am writing a function that will return an array with prime numbers.

The function should return an array with n elements. (n is a parameter) But it returns only one element. Why?

My codes:

function findPrimes(n)
{
  var arr = [];
  var currIndex = 0;
  var sqrtNum;
  var ceiledNum;
  var ceiledIndex = 0;
  var currCompose;
  var res;
  for (initNum = 2; arr.length < n; ++initNum)
  {
    sqrtNum = Math.sqrt(initNum);
    ceiledNum = Math.ceil(sqrtNum);
      for (currCompose = 2; currCompose <= ceiledNum; ++currCompose)
       {
         res = initNum % currCompose;
         if (res == 0 && initNum != currCompose)
         {
          break;
         }
         else if (res == 0 && initNum == currCompose)
         {
          arr[currIndex] = initNum;
          ++currIndex;
          break;
         }
         else if (res != 0 && initNum != currCompose)
         {
          continue;
         }
         else
         {
          console.log("Impossible result!");
         }
       }
  }
 return arr;
}

findPrimes(2); //return 2
findPrimes(10); //return 2 too

Jsbin

One gotcha with the JSBin tool you are using is that it shows [2] as 2, which I find a little bit confusing. Your above code is returning a list with 1 item in it, which is not the correct answer (see solutions below), but at least it is the right datatype. — Greg Prisament
– Greg Prisament, Commented Mar 16, 2015 at 8:27

Greg Prisament · Accepted Answer · 2015-03-16 08:14:26Z

You should not be comparing initNum to currCompose. Keep in mind that initNum is the number you are checking (say, 71), and currCompose will be at most ceil(sqrt(initNum)) (say 9), so the two will never be equal.

Also note that it is best to append to the list and verify that no divisors where found only after the inner loop has finished.

This modified version works.

function findPrimes(n)
{
    var arr = [];
    var currIndex = 0;
    var sqrtNum;
    var ceiledNum;
    var ceiledIndex = 0;
    var currCompose;
    var res;
    var initNum;



    for (initNum = 2; arr.length < n; ++initNum)
    {
        sqrtNum = Math.sqrt(initNum);
        ceiledNum = Math.ceil(sqrtNum);
        for (currCompose = 2; currCompose <= ceiledNum; ++currCompose)
        {
            res = initNum % currCompose;
            if (res == 0 && initNum != currCompose)
            {
                break;
            }
        }
        if (currCompose == ceiledNum+1)
        {
            arr[currIndex] = initNum;
            ++currIndex;
        }
    }
    return arr;
}

var primes = findPrimes(6);
document.write(primes);

AnilSh · Accepted Answer · 2015-03-16 07:53:24Z

1

correct Line 14 of your code as follows and it works like charm.
for (currCompose = 2; currCompose <= initNum; ++currCompose)

answered Mar 16, 2015 at 7:53

AnilSh

363 bronze badges

4 Comments

Greg Prisament Over a year ago

This solution does not make use of the optimization that you only need to check for divisors up to sqrt(initNum).

Anakin Over a year ago

@GregPrisament What do you mean??

Greg Prisament Over a year ago

The solution given by @AnilSh checks for divisors in the range 2...initNum. But an efficient implementation should only check for divisors in the range 2...sqrt(initNum). That is to say, you want to keep: for (currCompose = 2; currCompose <= ceiledNum; ++currCompose)

Anakin Over a year ago

Oh, I agreed that. Using initNum is too broad. Both your answer and @AnilSh answer work, but yours could speed up process.

Rex Adrivan · Accepted Answer · 2015-03-16 07:55:42Z

0

function FindPrime(numbers){
  if(numbers.constructor === Array){
    output = [];
    for (var i = numbers.length - 1; i >= 0; i--) {
      if(isPrime(numbers[i]) == true){
        output.push(numbers[i]);
      };
    };
    return output;
  }


}

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

numbers = [1,2,3,4,5];
test = FindPrime(numbers);
console.log('testing', test);

answered Mar 16, 2015 at 7:55

Rex Adrivan

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