added prime-factors algo in src/algorithms/math (#532)

README.md

@@ -66,6 +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` [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

@@ -0,0 +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.
+
+#### Method for finding the prime factors and their count accurately
+
+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.
+
+##### Optimised Time Complexity: O(sqrt(n))
+
+
+#### Hardy-Ramanujan formula for approximate calculation of prime-factor count
+
+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)).
+
+#### Code Explaiation
+
+There are on 4 functions used:
+
+- getPrimeFactors : returns array containing all distinct prime factors for given input n.
+
+- getPrimeFactorsCount: returns accurate total count of distinct prime factors of given input n.
+
+- hardyRamanujanApprox:  returns approximate total count of distinct prime factors of given input n using Hardy-Ramanujan formula.
+
+- 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.
+ 
+
+## References
+
+- [Youtube](https://www.youtube.com/watch?v=6PDtgHhpCHo)
+- [Wikipedia](https://en.wikipedia.org/wiki/Hardy%E2%80%93Ramanujan_theorem)

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

@@ -0,0 +1,40 @@
+import primefactors from '../primefactors';
+
+describe('prime-factors', () => {
+      it('should give prime factors', () => {
+        expect(primefactors.getPrimeFactors(510510)).toEqual([2, 3, 5, 7, 11, 13, 17]);
+        expect(primefactors.getPrimeFactors(343434)).toEqual([2, 3, 7, 13, 17, 37]);
+        expect(primefactors.getPrimeFactors(456745)).toEqual([5, 167, 547]);
+        expect(primefactors.getPrimeFactors(8735463)).toEqual([3, 11, 88237]);
+        expect(primefactors.getPrimeFactors(873452453)).toEqual([149, 1637, 3581]);
+        expect(primefactors.getPrimeFactors(52734)).toEqual([2, 3, 11, 17, 47]);
+    });
+
+    it('should give prime factors count accurately', () => {
+        expect(primefactors.getPrimeFactorsCount(primefactors.getPrimeFactors(510510))).toEqual(7);
+        expect(primefactors.getPrimeFactorsCount(primefactors.getPrimeFactors(343434))).toEqual(6);
+        expect(primefactors.getPrimeFactorsCount(primefactors.getPrimeFactors(456745))).toEqual(3);
+        expect(primefactors.getPrimeFactorsCount(primefactors.getPrimeFactors(8735463))).toEqual(3);
+        expect(primefactors.getPrimeFactorsCount(primefactors.getPrimeFactors(873452453))).toEqual(3);
+        expect(primefactors.getPrimeFactorsCount(primefactors.getPrimeFactors(52734))).toEqual(5);
+    });
+
+    it('should give prime factors count approximately using Hardy-Ramanujan-Approx', () => {
+        expect(primefactors.hardyRamanujanApprox(510510)).toBeCloseTo(2.5759018900,5);
+        expect(primefactors.hardyRamanujanApprox(343434)).toBeCloseTo(2.54527635538,5);
+        expect(primefactors.hardyRamanujanApprox(456745)).toBeCloseTo(2.5673987036,5);
+        expect(primefactors.hardyRamanujanApprox(8735463)).toBeCloseTo(2.771519494900,5);
+        expect(primefactors.hardyRamanujanApprox(873452453)).toBeCloseTo(3.0247066455016,5);
+        expect(primefactors.hardyRamanujanApprox(52734)).toBeCloseTo(2.386284094835,5);
+    });
+
+    it('should give error percentage of deviation of Hardy-Ramanujan-Approx prime-factors count from accurate prime-factors count', () => {
+        expect(primefactors.errorPercent(primefactors.getPrimeFactorsCount(primefactors.getPrimeFactors(510510)),primefactors.hardyRamanujanApprox(510510))).toBeCloseTo(63.20140157059997,5);
+        expect(primefactors.errorPercent(primefactors.getPrimeFactorsCount(primefactors.getPrimeFactors(343434)),primefactors.hardyRamanujanApprox(343434))).toBeCloseTo(57.5787274,5);
+        expect(primefactors.errorPercent(primefactors.getPrimeFactorsCount(primefactors.getPrimeFactors(456745)),primefactors.hardyRamanujanApprox(456745))).toBeCloseTo(14.420043212851,5);
+        expect(primefactors.errorPercent(primefactors.getPrimeFactorsCount(primefactors.getPrimeFactors(8735463)),primefactors.hardyRamanujanApprox(8735463))).toBeCloseTo(7.61601683663378,5);
+        expect(primefactors.errorPercent(primefactors.getPrimeFactorsCount(primefactors.getPrimeFactors(873452453)),primefactors.hardyRamanujanApprox(873452453))).toBeCloseTo(0.8235548500,5);
+        expect(primefactors.errorPercent(primefactors.getPrimeFactorsCount(primefactors.getPrimeFactors(52734)),primefactors.hardyRamanujanApprox(52734))).toBeCloseTo(52.27431810328,5);
+    });
+});
+

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

@@ -0,0 +1,44 @@
+export default {
+
+  getPrimeFactors : (n) => {
+    let factorsArray = []; // an array where all the prime factors will be stored
+
+    //over here optimisation is made by running loop till sqrt(n) instead of n
+    for (let i = 2 ; i <= Math.sqrt(n); i++){
+      if(n % i === 0){  // if check to ensure i completely divides n
+        let count = 0;  // This count keeps track of number of times i divides n
+        while(n % i === 0){
+          n = n/i;       // override the value of n
+          count++;        // count value updated
+        }
+        factorsArray.push(i); // array gets populated
+      }
+    }
+    if(n !== 1){ // finally we cannot push 1 to array since it cannot be a prime-factor
+        factorsArray.push(n);
+    }
+  
+    return factorsArray;
+  },
+
+  //returns accurate prime-factors count
+  getPrimeFactorsCount : (factorsArray) => {
+    return factorsArray.length;
+  },
+  
+  
+  //returns Hardy-Ramanujan Approximation of prime-factors count 
+  hardyRamanujanApprox : (n) => {
+    return Math.log(Math.log(n));
+  },
+  
+  //returns %age of error in approximation using formula to that of accurate result. 
+  errorPercent : (exactFactorCount,approximateFactorCount) => {
+  let diff = exactFactorCount-approximateFactorCount > 0 ? exactFactorCount-approximateFactorCount: -(exactFactorCount-approximateFactorCount);
+  return (diff/exactFactorCount * 100);
+  }
+  
+
+}
+
+