I am having trouble with the prime number finder in my JavaScript program. It seems to not work for certain values. What could be

Question

I am having trouble with the prime number finder in my JavaScript program. It seems to not work for certain values. What could be

I am in the process of developing a code to identify prime numbers less than n. However, I encountered an issue where the code mistakenly flags 33 and 35 as prime numbers. I am puzzled by this unexpected outcome. Here is the code that I have been working on:

function primeFinder(n) {
  let prime = []
  let index = 0
  for (let i = 2; i <= n; i++) {
    let root = Math.floor(Math.sqrt(i))
    for (let j = 2; j <= root; j++) {
      if (i % j == 0) {
        i++
        break
      }
    }
    prime[index] = i
    index++
  }
  return (prime)
}

console.log(primeFinder(35))

Despite my efforts to accurately determine prime numbers, the code does not function as intended.

javascript

Answer 1

Answer №1

When your inner loop stops upon finding a divisor of i, it signifies that i is not prime. Afterwards, i gets incremented and added to the prime array. This process dictates that for every non-prime value, i+1 becomes the next prime number.

To determine if you traverse through the inner loop without spotting a divisor - indicating the number as prime - you must set a flag before initiating the loop, update it when encountering a divisor, and assess the flag post-loop.

function primeFinder(n) {
  let prime = [];
  for (let i = 2; i <= n; i++) {
    let root = Math.ceil(Math.sqrt(i))
    let primeFlag = true;
    for (let j = 2; j <= root; j++) {
      if (i % j == 0) {
        primeFlag = false;
        break;
      }
    }
    if (primeFlag) {
      prime.push(i)
    }
  }
  return (prime)
}

console.log(primeFinder(35))

Answer 2

When your inner loop stops upon finding a divisor of i, it signifies that i is not prime. Afterwards, i gets incremented and added to the prime array. This process dictates that for every non-prime value, i+1 becomes the next prime number.

To determine if you traverse through the inner loop without spotting a divisor - indicating the number as prime - you must set a flag before initiating the loop, update it when encountering a divisor, and assess the flag post-loop.

function primeFinder(n) {
  let prime = [];
  for (let i = 2; i <= n; i++) {
    let root = Math.ceil(Math.sqrt(i))
    let primeFlag = true;
    for (let j = 2; j <= root; j++) {
      if (i % j == 0) {
        primeFlag = false;
        break;
      }
    }
    if (primeFlag) {
      prime.push(i)
    }
  }
  return (prime)
}

console.log(primeFinder(35))

Answer 3

Answer №2

Here is a sample code for finding prime numbers:

function calculatePrimes(limit) {
  const primesList = [];

  for (let num = 2; num <= limit; num++) {
    let isPrime = true;
    const squareRoot = Math.floor(Math.sqrt(num));

    for (let divisor = 2; divisor <= squareRoot; divisor++) {
      if (num % divisor === 0) {
        isPrime = false;
        break;
      }
    }

    if (isPrime) {
      primesList.push(num);
    }
  }

  return primesList;
}

console.log(calculatePrimes(35));

Answer 4

Here is a sample code for finding prime numbers:

function calculatePrimes(limit) {
  const primesList = [];

  for (let num = 2; num <= limit; num++) {
    let isPrime = true;
    const squareRoot = Math.floor(Math.sqrt(num));

    for (let divisor = 2; divisor <= squareRoot; divisor++) {
      if (num % divisor === 0) {
        isPrime = false;
        break;
      }
    }

    if (isPrime) {
      primesList.push(num);
    }
  }

  return primesList;
}

console.log(calculatePrimes(35));

Answer 5

Answer №3

Breaking the for statement when i is divisible by j prevents adding i to the prime list.

    for (let j = 2; j <= root; j++) {
      if (i % j == 0) {
        i++
        break // The break statement doesn't affect pushing i into primes
      }
    }
    prime[index] = i // Regardless of divisibility, i is pushed to primes 
    index++

This results in all odd numbers being listed as prime.

To optimize performance and handle large n values, use the following code to find all primes under n:

function primeFinder(n) {
  const primes = [];
  const isPrime = [];
  for (let i = 0; i <= n; i += 1) {
    isPrime[i] = true;
  }
  const root = Math.floor(Math.sqrt(n));
  for (let i = 2; i <= root; i++) {
    if (isPrime[i] == false) continue;
    for (let j = i; j <= n; j += i) {
      isPrime[j] = false;
    }
  }
  for (let i = 2; i <= n; i += 1) {
    if (isPrime[i]) primes.push(i);
  }

  return primes;
}

Avoid using divisions whenever possible as they can significantly impact performance in various programming languages. The provided code eliminates the need for divisions.

Answer 6

Breaking the for statement when i is divisible by j prevents adding i to the prime list.

    for (let j = 2; j <= root; j++) {
      if (i % j == 0) {
        i++
        break // The break statement doesn't affect pushing i into primes
      }
    }
    prime[index] = i // Regardless of divisibility, i is pushed to primes 
    index++

This results in all odd numbers being listed as prime.

To optimize performance and handle large n values, use the following code to find all primes under n:

function primeFinder(n) {
  const primes = [];
  const isPrime = [];
  for (let i = 0; i <= n; i += 1) {
    isPrime[i] = true;
  }
  const root = Math.floor(Math.sqrt(n));
  for (let i = 2; i <= root; i++) {
    if (isPrime[i] == false) continue;
    for (let j = i; j <= n; j += i) {
      isPrime[j] = false;
    }
  }
  for (let i = 2; i <= n; i += 1) {
    if (isPrime[i]) primes.push(i);
  }

  return primes;
}

Avoid using divisions whenever possible as they can significantly impact performance in various programming languages. The provided code eliminates the need for divisions.

I am having trouble with the prime number finder in my JavaScript program. It seems to not work for certain values. What could be

Answer №1

Answer №2

Answer №3

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