Code styling fixes for Sieve of Eratosthenes.

Author: trekhleb

README.md

@@ -49,6 +49,7 @@ a set of rules that precisely define a sequence of operations.
   * [Euclidean Algorithm](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/math/euclidean-algorithm) - calculate the Greatest Common Divisor (GCD)
   * [Least Common Multiple](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/math/least-common-multiple) (LCM)
   * [Integer Partition](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/math/integer-partition)
+  * [Sieve of Eratosthenes](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/math/sieve-of-eratosthenes) - finding all prime numbers up to any given limit
 * **Sets**
   * [Cartesian Product](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/cartesian-product) - product of multiple sets
   * [Power Set](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/power-set) - all subsets of a set

src/algorithms/math/sieve-of-eratosthenes/README.md

@@ -6,15 +6,18 @@ It is attributed to Eratosthenes of Cyrene, an ancient Greek mathematician.
 
 ## How it works
 
-1. Create a boolean array of `n+1` positions (to represent the numbers `0` through `n`)
+1. Create a boolean array of `n + 1` positions (to represent the numbers `0` through `n`)
 2. Set positions `0` and `1` to `false`, and the rest to `true`
 3. Start at position `p = 2` (the first prime number)
-4. Mark as `false` all the multiples of `p` (that is, positions `2*p`, `3*p`, `4*p`... until you reach the end of the array)
+4. Mark as `false` all the multiples of `p` (that is, positions `2 * p`, `3 * p`, `4 * p`... until you reach the end of the array)
 5. Find the first position greater than `p` that is `true` in the array. If there is no such position, stop. Otherwise, let `p` equal this new number (which is the next prime), and repeat from step 4
 
-When the algorithm terminates, the numbers remaining `true` in the array are all the primes below `n`.
+When the algorithm terminates, the numbers remaining `true` in the array are all 
+the primes below `n`.
 
-An improvement of this algorithm is, in step 4, start marking multiples of `p` from `p*p`, and not from `2*p`. The reason why this works is because, at that point, smaller multiples of `p` will have already been marked `false`.
+An improvement of this algorithm is, in step 4, start marking multiples 
+of `p` from `p * p`, and not from `2 * p`. The reason why this works is because, 
+at that point, smaller multiples of `p` will have already been marked `false`.
 
 ## Example
 
@@ -26,4 +29,4 @@ The algorithm has a complexity of `O(n log(log n))`.
 
 ## References
 
-[Wikipedia](https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes)
+- [Wikipedia](https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes)

src/algorithms/math/sieve-of-eratosthenes/__test__/sieveOfEratosthenes.test.js

@@ -1,6 +1,6 @@
 import sieveOfEratosthenes from '../sieveOfEratosthenes';
 
-describe('factorial', () => {
+describe('sieveOfEratosthenes', () => {
   it('should find all primes less than or equal to n', () => {
     expect(sieveOfEratosthenes(5)).toEqual([2, 3, 5]);
     expect(sieveOfEratosthenes(10)).toEqual([2, 3, 5, 7]);

src/algorithms/math/sieve-of-eratosthenes/sieveOfEratosthenes.js

@@ -1,25 +1,33 @@
 /**
- * @param {number} n
+ * @param {number} maxNumber
  * @return {number[]}
  */
-export default function sieveOfEratosthenes(n) {
-  const isPrime = new Array(n + 1).fill(true);
+export default function sieveOfEratosthenes(maxNumber) {
+  const isPrime = new Array(maxNumber + 1).fill(true);
   isPrime[0] = false;
   isPrime[1] = false;
+
   const primes = [];
 
-  for (let i = 2; i <= n; i += 1) {
-    if (isPrime[i] === true) {
-      primes.push(i);
+  for (let number = 2; number <= maxNumber; number += 1) {
+    if (isPrime[number] === true) {
+      primes.push(number);
 
-      // Warning: When working with really big numbers, the following line may cause overflow
-      // In that case, it can be changed to:
-      // let j = 2 * i;
-      let j = i * i;
+      /*
+       * Optimisation.
+       * Start marking multiples of `p` from `p * p`, and not from `2 * p`.
+       * The reason why this works is because, at that point, smaller multiples
+       * of `p` will have already been marked `false`.
+       *
+       * Warning: When working with really big numbers, the following line may cause overflow
+       * In that case, it can be changed to:
+       * let nextNumber = 2 * number;
+       */
+      let nextNumber = number * number;
 
-      while (j <= n) {
-        isPrime[j] = false;
-        j += i;
+      while (nextNumber <= maxNumber) {
+        isPrime[nextNumber] = false;
+        nextNumber += number;
       }
     }
   }