Add prime factors calculation.

Author: trekhleb

README.md

@@ -66,7 +66,7 @@ a set of rules that precisely define a sequence of operations.
   * `B` [Bit Manipulation](src/algorithms/math/bits) - set/get/update/clear bits, multiplication/division by two, make negative etc.
   * `B` [Factorial](src/algorithms/math/factorial) 
   * `B` [Fibonacci Number](src/algorithms/math/fibonacci) - classic and closed-form versions
-  * `B` [Prime Factors](src/algorithms/math/prime-factors) - finding distinct prime-factor count using both accurate & Hardy-Ramanujan's Algorithm   
+  * `B` [Prime Factors](src/algorithms/math/prime-factors) - finding prime factors and counting them using Hardy-Ramanujan's theorem  
   * `B` [Primality Test](src/algorithms/math/primality-test) (trial division method)
   * `B` [Euclidean Algorithm](src/algorithms/math/euclidean-algorithm) - calculate the Greatest Common Divisor (GCD)
   * `B` [Least Common Multiple](src/algorithms/math/least-common-multiple) (LCM)

src/algorithms/math/prime-factors/README.md

@@ -1,34 +1,34 @@
 # Prime Factors
 
-Prime factors are basically those prime numbers which multiply together to give the orignal number. For ex: 39 will have prime factors as 3 and 13 which are also prime numbers. Another example is 15 whose prime factors are 3 and 5.
+**Prime number** is a whole number greater than `1` that **cannot** be made by multiplying other whole numbers. The first few prime numbers are: `2`, `3`, `5`, `7`, `11`, `13`, `17`, `19` and so on.
 
-#### Method for finding the prime factors and their count accurately
+If we **can** make it by multiplying other whole numbers it is a **Composite Number**.
 
-The approach is to basically keep on dividing the  natural number 'n' by indexes from i = 2 to i = n by prime indexes only. This is ensured by an 'if' check. Then value of 'n' keeps on overriding by (n/i).
-The time complexity till now is O(n) in worst case since the loop run from index i = 2 to i = n even when no index 'i' is left to be divided by 'n' other than n itself. This time complexity can be reduced  to O(sqrt(n)) from O(n). This optimisation is acheivable when loop is ran from i = 2 to i = sqrt(n). Now, we go only till O(sqrt(n)) because when 'i' becomes greater than sqrt(n), we now have the confirmation there is no index 'i' left which can divide 'n' completely other than n itself.
+![Composite numbers](https://www.mathsisfun.com/numbers/images/prime-composite.svg)
 
-##### Optimised Time Complexity: O(sqrt(n))
+_Image source: [Math is Fun](https://www.mathsisfun.com/prime-factorization.html)_
 
+**Prime factors** are those [prime numbers](https://en.wikipedia.org/wiki/Prime_number) which multiply together to give the original number. For example `39` will have prime factors of `3` and `13` which are also prime numbers. Another example is `15` whose prime factors are `3` and `5`.
 
-#### Hardy-Ramanujan formula for approximate calculation of prime-factor count
+![Factors](https://www.mathsisfun.com/numbers/images/factor-2x3.svg)
 
-In 1917, a theorem was formulated by G.H Hardy and Srinivasa Ramanujan which approximately tells the total count of distinct prime factors of most 'n' natural numbers.
-The fomula is given by ln(ln(n)).
+_Image source: [Math is Fun](https://www.mathsisfun.com/prime-factorization.html)_
 
-#### Code Explaiation
+## Finding the prime factors and their count accurately
 
-There are on 4 functions used:
+The approach is to keep on dividing the natural number `n` by indexes from `i = 2` to `i = n` (by prime indexes only). The value of `n` is being overridden by `(n / i)` on each iteration.
 
-- getPrimeFactors : returns array containing all distinct prime factors for given input n.
+The time complexity till now is `O(n)` in the worst case scenario since the loop runs from index `i = 2` to `i = n`. This time complexity can be reduced from `O(n)` to `O(sqrt(n))`. The optimisation is achievable when loop runs from `i = 2` to `i = sqrt(n)`. Now, we go only till `O(sqrt(n))` because when `i` becomes greater than `sqrt(n)`, we have the confirmation that there is no index `i` left which can divide `n` completely other than `n` itself.
 
-- getPrimeFactorsCount: returns accurate total count of distinct prime factors of given input n.
+## Hardy-Ramanujan formula for approximate calculation of prime-factor count
 
-- hardyRamanujanApprox:  returns approximate total count of distinct prime factors of given input n using Hardy-Ramanujan formula.
+In 1917, a theorem was formulated by G.H Hardy and Srinivasa Ramanujan which states that the normal order of the number `ω(n)` of distinct prime factors of a number `n` is `log(log(n))`.
 
-- errorPercent : returns %age of error in approximation using formula to that of accurate result. The formula used is:   **[Modulus(accurate_val - approximate_val) / accurate_val ] * 100**. This shows deviation from accurate result.
- 
+Roughly speaking, this means that most numbers have about this number of distinct prime factors.
 
 ## References
 
-- [Youtube](https://www.youtube.com/watch?v=6PDtgHhpCHo)
-- [Wikipedia](https://en.wikipedia.org/wiki/Hardy%E2%80%93Ramanujan_theorem)
+- [Prime numbers on Math is Fun](https://www.mathsisfun.com/prime-factorization.html)
+- [Prime numbers on Wikipedia](https://en.wikipedia.org/wiki/Prime_number)
+- [Hardy–Ramanujan theorem on Wikipedia](https://en.wikipedia.org/wiki/Hardy%E2%80%93Ramanujan_theorem)
+- [Prime factorization of a number on Youtube](https://www.youtube.com/watch?v=6PDtgHhpCHo&list=PLLXdhg_r2hKA7DPDsunoDZ-Z769jWn4R8&index=82)

src/algorithms/math/prime-factors/__test__/primeFactors.test.js

@@ -0,0 +1,87 @@
+import {
+  primeFactors,
+  hardyRamanujan,
+} from '../primeFactors';
+
+/**
+ * Calculates the error between exact and approximate prime factor counts.
+ * @param {number} exactCount
+ * @param {number} approximateCount
+ * @returns {number} - approximation error (percentage).
+ */
+function approximationError(exactCount, approximateCount) {
+  return (Math.abs((exactCount - approximateCount) / exactCount) * 100);
+}
+
+describe('primeFactors', () => {
+  it('should find prime factors', () => {
+    expect(primeFactors(1)).toEqual([]);
+    expect(primeFactors(2)).toEqual([2]);
+    expect(primeFactors(3)).toEqual([3]);
+    expect(primeFactors(4)).toEqual([2, 2]);
+    expect(primeFactors(14)).toEqual([2, 7]);
+    expect(primeFactors(40)).toEqual([2, 2, 2, 5]);
+    expect(primeFactors(54)).toEqual([2, 3, 3, 3]);
+    expect(primeFactors(100)).toEqual([2, 2, 5, 5]);
+    expect(primeFactors(156)).toEqual([2, 2, 3, 13]);
+    expect(primeFactors(273)).toEqual([3, 7, 13]);
+    expect(primeFactors(300)).toEqual([2, 2, 3, 5, 5]);
+    expect(primeFactors(980)).toEqual([2, 2, 5, 7, 7]);
+    expect(primeFactors(1000)).toEqual([2, 2, 2, 5, 5, 5]);
+    expect(primeFactors(52734)).toEqual([2, 3, 11, 17, 47]);
+    expect(primeFactors(343434)).toEqual([2, 3, 7, 13, 17, 37]);
+    expect(primeFactors(456745)).toEqual([5, 167, 547]);
+    expect(primeFactors(510510)).toEqual([2, 3, 5, 7, 11, 13, 17]);
+    expect(primeFactors(8735463)).toEqual([3, 3, 11, 88237]);
+    expect(primeFactors(873452453)).toEqual([149, 1637, 3581]);
+  });
+
+  it('should give approximate prime factors count using Hardy-Ramanujan theorem', () => {
+    expect(hardyRamanujan(2)).toBeCloseTo(-0.366, 2);
+    expect(hardyRamanujan(4)).toBeCloseTo(0.326, 2);
+    expect(hardyRamanujan(40)).toBeCloseTo(1.305, 2);
+    expect(hardyRamanujan(156)).toBeCloseTo(1.6193, 2);
+    expect(hardyRamanujan(980)).toBeCloseTo(1.929, 2);
+    expect(hardyRamanujan(52734)).toBeCloseTo(2.386, 2);
+    expect(hardyRamanujan(343434)).toBeCloseTo(2.545, 2);
+    expect(hardyRamanujan(456745)).toBeCloseTo(2.567, 2);
+    expect(hardyRamanujan(510510)).toBeCloseTo(2.575, 2);
+    expect(hardyRamanujan(8735463)).toBeCloseTo(2.771, 2);
+    expect(hardyRamanujan(873452453)).toBeCloseTo(3.024, 2);
+  });
+
+  it('should give correct deviation between exact and approx counts', () => {
+    expect(approximationError(primeFactors(2).length, hardyRamanujan(2)))
+      .toBeCloseTo(136.651, 2);
+
+    expect(approximationError(primeFactors(4).length, hardyRamanujan(2)))
+      .toBeCloseTo(118.325, 2);
+
+    expect(approximationError(primeFactors(40).length, hardyRamanujan(2)))
+      .toBeCloseTo(109.162, 2);
+
+    expect(approximationError(primeFactors(156).length, hardyRamanujan(2)))
+      .toBeCloseTo(109.162, 2);
+
+    expect(approximationError(primeFactors(980).length, hardyRamanujan(2)))
+      .toBeCloseTo(107.330, 2);
+
+    expect(approximationError(primeFactors(52734).length, hardyRamanujan(52734)))
+      .toBeCloseTo(52.274, 2);
+
+    expect(approximationError(primeFactors(343434).length, hardyRamanujan(343434)))
+      .toBeCloseTo(57.578, 2);
+
+    expect(approximationError(primeFactors(456745).length, hardyRamanujan(456745)))
+      .toBeCloseTo(14.420, 2);
+
+    expect(approximationError(primeFactors(510510).length, hardyRamanujan(510510)))
+      .toBeCloseTo(63.201, 2);
+
+    expect(approximationError(primeFactors(8735463).length, hardyRamanujan(8735463)))
+      .toBeCloseTo(30.712, 2);
+
+    expect(approximationError(primeFactors(873452453).length, hardyRamanujan(873452453)))
+      .toBeCloseTo(0.823, 2);
+  });
+});

src/algorithms/math/prime-factors/__test__/primefactors.test.js

@@ -0,0 +1,42 @@
+/**
+ * Finds prime factors of a number.
+ *
+ * @param {number} n - the number that is going to be split into prime factors.
+ * @returns {number[]} - array of prime factors.
+ */
+export function primeFactors(n) {
+  // Clone n to avoid function arguments override.
+  let nn = n;
+
+  // Array that stores the all the prime factors.
+  const factors = [];
+
+  // Running the loop till sqrt(n) instead of n to optimise time complexity from O(n) to O(sqrt(n)).
+  for (let factor = 2; factor <= Math.sqrt(nn); factor += 1) {
+    // Check that factor divides n without a reminder.
+    while (nn % factor === 0) {
+      // Overriding the value of n.
+      nn /= factor;
+      // Saving the factor.
+      factors.push(factor);
+    }
+  }
+
+  // The ultimate reminder should be a last prime factor,
+  // unless it is not 1 (since 1 is not a prime number).
+  if (nn !== 1) {
+    factors.push(nn);
+  }
+
+  return factors;
+}
+
+/**
+ * Hardy-Ramanujan approximation of prime factors count.
+ *
+ * @param {number} n
+ * @returns {number} - approximate number of prime factors.
+ */
+export function hardyRamanujan(n) {
+  return Math.log(Math.log(n));
+}