Published at Jul 13 2018
·
1 comment

Instructions

Test suite

Solution

Use the Sieve of Eratosthenes to find all the primes from 2 up to a given number.

The Sieve of Eratosthenes is a simple, ancient algorithm for finding all prime numbers up to any given limit. It does so by iteratively marking as composite (i.e. not prime) the multiples of each prime, starting with the multiples of 2.

Create your range, starting at two and continuing up to and including the given limit. (i.e. [2, limit])

The algorithm consists of repeating the following over and over:

- take the next available unmarked number in your list (it is prime)
- mark all the multiples of that number (they are not prime)

Repeat until you have processed each number in your range.

When the algorithm terminates, all the numbers in the list that have not been marked are prime.

The wikipedia article has a useful graphic that explains the algorithm: https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

Notice that this is a very specific algorithm, and the tests don't check that you've implemented the algorithm, only that you've come up with the correct list of primes.

Go through the setup instructions for JavaScript to install the necessary dependencies:

http://exercism.io/languages/javascript/installation

The provided test suite uses Jasmine. You can install it by opening a terminal window and running the following command:

```
npm install -g jasmine
```

Run the test suite from the exercise directory with:

```
jasmine sieve.spec.js
```

In many test suites all but the first test have been marked "pending".
Once you get a test passing, activate the next one by changing `xit`

to `it`

.

Sieve of Eratosthenes at Wikipedia http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

```
var Sieve = require('./sieve');
describe('Sieve', function () {
it('finds primes up to 10', function () {
expect(new Sieve(10).primes).toEqual([2, 3, 5, 7]);
});
xit('finds primes up to 13, and considers the limit passed in', function () {
expect(new Sieve(13).primes).toEqual([2, 3, 5, 7, 11, 13]);
});
xit('finds primes up to 1000', function () {
expect(new Sieve(1000).primes).toEqual([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, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997]);
});
});
```

```
function Sieve(max) {
var sieve=[];
for (var x=2; x<=Math.sqrt(max); ++x) {
if (sieve[x]) continue;
for (var y=x+x; y<=max; y+=x) {
sieve[y]=1;
}
}
this.primes = [];
for (var x=2; x<=max; ++x) {
if (!sieve[x]) this.primes.push(x) }
}
module.exports=Sieve;
```

A huge amount can be learned from reading other people’s code. This is why we wanted to give exercism users the option of making their solutions public.

Here are some questions to help you reflect on this solution and learn the most from it.

- What compromises have been made?
- Are there new concepts here that you could read more about to improve your understanding?

Level up your programming skills with 3,373 exercises across 50 languages, and insightful discussion with our volunteer team of welcoming mentors.
Exercism is
**100% free forever**.

## Community comments

It would be nice to use a filter, but I don't an easy way to pass back the index.