Sieve of Eratosthenes, What is it?

Image source : famousmathematicians.net

What is it?

Sieve of Eratosthenes is an algorithm devised by the Eratosthenes of Cyrene. It does the job of finding all the prime numbers within a given upper limit. This ancient algorithm is efficient and smart till the upper limit is a few billions. So we'll discuss the process and JavaScript code for the same, below.

How it works?

The algorithm starts with generating a list of all numbers starting from 2 to n (where n is the upper limit), with the assumption of all the numbers in the list are prime. It starts from 2 and removes all the multiples of 2 in the list by traversing the list in the interval of 2.

So, now we consider n as 10

let sample_array = [2, 3, 4, 5, 6, 7, 8, 9, 10];

Starting from 2, it removes the multiples of 2 by traversing the above list in a step count of 2.

Note: '*' below means removed from list.

let sample_array = [2, 3, 4*, 5, 6*, 7, 8*, 9, 10*];

After removing all the multiples of 2, we move to the next non-removed number (that is 3), now from 3, we traverse the list with the step count of 3 and remove its multiples.

let sample_array = [2, 3, 4*, 5, 6*, 7, 8*, 9*, 10*];

We then proceed to the next non-removed number, which is 5. But here's the thing, the multiples of 5 are already removed from the list. We just make sure when to end this cycle of traversal and removal by calculating the square of 5, that is 5*5 = 25, which is obviously greater than n that is 10. So we stop the process and get the remaining elements, which are prime.

Here's the final list we get,

let sample_array = [2, 3, 5, 7];

Hurray!, we've done with the theory part, let's get our hands dirty with some JS to actually code it.

Execution in JS 💻

Let's start by creating an empty array called Boolarray, why naming 'Bool', because we are going for a Boolean array. We also initialize the value of n as 20.

let Boolarray = [];
let n = 20;

Remember, we first made an assumption that all the numbers in the list (here array) are prime. So we use true for is prime and false for not a prime, with this in mind we first fill the empty array with boolean values of all True (based on our assumption). We use a for loop with iterator i to iterate from 1 to n and fill the array with True.

let Boolarray = [];
let n = 20;
for (var i = 0; i < n; i++) {
   Boolarray.push(true);
 }

Now, we have an array of length 20 with true on all indexes. We now follow the procedure of Sieve of Eratosthenes by starting the for with iterator j from 2 to j*j<=n (j*j<=n checks when to end the looping). If the current element in the array is true, we then loop over its multiples with iterator kand a step count, (according to the current element) and mark them false.

let Boolarray = [];
 let n = 20;
 for (var i = 0; i < n; i++) {
   Boolarray.push(true);
 }
 for (let j = 2; j * j <= n; j++) {
   if (Boolarray[j] == true) {
     for (let k = 2 * j; k <= n; k += j) {
       Boolarray[k] = false;
     }
   }
 }

After this execution, we are left with a Boolean array, which contains true in places of prime (remember true → is prime) and false in places of non-prime numbers in the array.

Now it's all logging it onto the console 🎉

We use another for loop to iterate on Boolarray with iterator num, starting from 2 to num<=n. We console log only the num's which contains true in the Boolarray.

for (let num = 2; num <= n; num++) {
   if (Boolarray[num] == true) {
     console.log(num);
   }
 }

So, we end with this final code,

	let Boolarray = [];
	let n = 100;
	for (var i = 0; i < n; i++) {
	Boolarray.push(true);
	}
	for (let j = 2; j * j <= n; j++) {
	if (Boolarray[j] == true) {
	for (let k = 2 * j; k <= n; k += j) {
	Boolarray[k] = false;
	}
	}
	}

	console.log(Boolarray)
	for (let num = 2; num <= n; num++) {
	if (Boolarray[num] == true) {
	console.log(num);
	}
	}