self made pow() c++

Question

I was reading through How can I write a power function myself? and the answer given by dan04 caught my attention mainly because I am not sure about the answer given by fortran, but I took that and implemented this:

#include <iostream>
using namespace std;
float pow(float base, float ex){
    // power of 0
    if (ex == 0){
        return 1;
    // negative exponenet
    }else if( ex < 0){
        return 1 / pow(base, -ex);
    // even exponenet
    }else if ((int)ex % 2 == 0){
        float half_pow = pow(base, ex/2);
        return half_pow * half_pow;
    //integer exponenet
    }else{
        return base * pow(base, ex - 1);
    }
}
int main(){
    for (int ii = 0; ii< 10; ii++){\
        cout << "pow(" << ii << ".5) = " << pow(ii, .5) << endl;
        cout << "pow(" << ii << ",2) = " << pow(ii,  2) << endl;
        cout << "pow(" << ii << ",3) = " << pow(ii,  3) << endl;
    }
}

though I am not sure if I translated this right because all of the calls giving .5 as the exponent return 0. In the answer it states that it might need a log2(x) based on a^b = 2^(b * log2(a)), but I am unsure about putting that in as I am unsure where to put it, or if I am even thinking about this right.

NOTE: I know that this might be defined in a math library, but I don't need all the added expense of an entire math library for a few functions.

EDIT: does anyone know a floating-point implementation for fractional exponents? (I have seen a double implementation, but that was using a trick with registers, and I need floating-point, and adding a library just to do a trick I would be better off just including the math library)

Answer 1

I have looked at this paper here which describes how to approximate the exponential function for double precision. After a little research on Wikipedia about single precision floating point representation I have worked out the equivalent algorithms. They only implemented the exp function, so I found an inverse function for the log and then simply did

    POW(a, b) = EXP(LOG(a) * b).

compiling this gcc4.6.2 yields a pow function almost 4 times faster than the standard library's implementation (compiling with O2).

Note: the code for EXP is copied almost verbatim from the paper I read and the LOG function is copied from here.

Here is the relevant code:

    #define EXP_A 184
    #define EXP_C 16249 

    float EXP(float y)
    {
      union
      {
        float d;
        struct
        {
    #ifdef LITTLE_ENDIAN
          short j, i;
    #else
          short i, j;
    #endif
        } n;
      } eco;
      eco.n.i = EXP_A*(y) + (EXP_C);
      eco.n.j = 0;
      return eco.d;
    }

    float LOG(float y)
    {
      int * nTemp = (int*)&y;
      y = (*nTemp) >> 16;
      return (y - EXP_C) / EXP_A;
    }

    float POW(float b, float p)
    {
      return EXP(LOG(b) * p);
    }

There is still some optimization you can do here, or perhaps that is good enough. This is a rough approximation but if you would have been satisfied with the errors introduced using the double representation, I imagine this will be satisfactory.

Answer 2

I think the algorithm you're looking for could be 'nth root'. With an initial guess of 1 (for k == 0):

#include <iostream>
using namespace std;


float pow(float base, float ex);

float nth_root(float A, int n) {
    const int K = 6;
    float x[K] = {1};
    for (int k = 0; k < K - 1; k++)
        x[k + 1] = (1.0 / n) * ((n - 1) * x[k] + A / pow(x[k], n - 1));
    return x[K-1];
}

float pow(float base, float ex){
    if (base == 0)
        return 0;
    // power of 0
    if (ex == 0){
        return 1;
    // negative exponenet
    }else if( ex < 0){
        return 1 / pow(base, -ex);
    // fractional exponent
    }else if (ex > 0 && ex < 1){
        return nth_root(base, 1/ex);
    }else if ((int)ex % 2 == 0){
        float half_pow = pow(base, ex/2);
        return half_pow * half_pow;
    //integer exponenet
    }else{
        return base * pow(base, ex - 1);
    }
}
int main_pow(int, char **){
    for (int ii = 0; ii< 10; ii++){\
        cout << "pow(" << ii << ", .5) = " << pow(ii, .5) << endl;
        cout << "pow(" << ii << ",  2) = " << pow(ii,  2) << endl;
        cout << "pow(" << ii << ",  3) = " << pow(ii,  3) << endl;
    }
    return 0;
}

test:

pow(0, .5) = 0.03125
pow(0,  2) = 0
pow(0,  3) = 0
pow(1, .5) = 1
pow(1,  2) = 1
pow(1,  3) = 1
pow(2, .5) = 1.41421
pow(2,  2) = 4
pow(2,  3) = 8
pow(3, .5) = 1.73205
pow(3,  2) = 9
pow(3,  3) = 27
pow(4, .5) = 2
pow(4,  2) = 16
pow(4,  3) = 64
pow(5, .5) = 2.23607
pow(5,  2) = 25
pow(5,  3) = 125
pow(6, .5) = 2.44949
pow(6,  2) = 36
pow(6,  3) = 216
pow(7, .5) = 2.64575
pow(7,  2) = 49
pow(7,  3) = 343
pow(8, .5) = 2.82843
pow(8,  2) = 64
pow(8,  3) = 512
pow(9, .5) = 3
pow(9,  2) = 81
pow(9,  3) = 729

Answer 3

I think that you could try to solve it by using the Taylor's series, check this. http://en.wikipedia.org/wiki/Taylor_series

With the Taylor's series you can solve any difficult to solve calculation such as 3^3.8 by using the already known results such as 3^4. In this case you have 3^4 = 81 so

3^3.8 = 81 + 3.8*3( 3.8 - 4) +..+.. and so on depend on how big is your n you will get the closer solution of your problem.

Answer 4

I and my friend faced similar problem while we're on an OpenGL project and math.h didn't suffice in some cases. Our instructor also had the same problem and he told us to seperate power to integer and floating parts. For example, if you are to calculate x^11.5 you may calculate sqrt(x^115, 10) which may result more accurate result.

Answer 5

Reworked on @capellic answer, so that nth_root works with bigger values as well.

Without the limitation of an array that is allocated for no reason:

#include <iostream>

float pow(float base, float ex);

inline float fabs(float a) {
    return a > 0 ? a : -a;
}

float nth_root(float A, int n, unsigned max_iterations = 500, float epsilon = std::numeric_limits<float>::epsilon()) {
    if (n < 0)
        throw "Invalid value";

    if (n == 1 || A == 0)
        return A;

    float old_value = 1;
    float value;
    for (int k = 0; k < max_iterations; k++) {
        value = (1.0 / n) * ((n - 1) * old_value + A / pow(old_value, n - 1));

        if (fabs(old_value - value) < epsilon)
            return value;

        old_value = value;
    }

    return value;
}

float pow(float base, float ex) {
    if (base == 0)
        return 0;

    if (ex == 0){
        // power of 0
        return 1;
    } else if( ex < 0) {
        // negative exponent
        return 1 / pow(base, -ex);
    } else if (ex > 0 && ex < 1) {
        // fractional exponent
        return nth_root(base, 1/ex);
    } else if ((int)ex % 2 == 0) {
        // even exponent
        float half_pow = pow(base, ex/2);
        return half_pow * half_pow;
    } else {
        // integer exponent
        return base * pow(base, ex - 1);
    }
}

int main () {
    for (int i = 0; i <= 128; i++) {
        std::cout << "pow(" << i << ", .5) = " << pow(i, .5) << std::endl;
        std::cout << "pow(" << i << ", .3) = " << pow(i, .3) << std::endl;
        std::cout << "pow(" << i << ",  2) = " << pow(i,  2) << std::endl;
        std::cout << "pow(" << i << ",  3) = " << pow(i,  3) << std::endl;
    }

    std::cout << "pow(" << 74088 << ",  .3) = " << pow(74088,  .3) << std::endl;

    return 0;
}

Answer 6

This solution of MINE will be accepted upto O(n) time complexity utpo input less then 2^30 or 10^8 IT will not accept more then these inputs It WILL GIVE TIME LIMIT EXCEED warning but easy understandable solution

#include<bits/stdc++.h>
using namespace std;

double recursive(double x,int n)
{
    // static is important here
    // other wise it will store same  values while multiplying 
    
    double p = x;
    
    double ans;
   
    // as we multiple p it will multiply it with q which has the 
    //previous value of this ans latter we will update the q
    // so that q has fresh value for further test cases here 
    
  static double q=1; // important 
    
    if(n==0){ ans = q; q=1; return ans;}
   
    if(n>0)
    {
        p *= q;
        // stored value got multiply by p
        q=p;
        // and again updated to q 
        p=x;
        //to update the value to the same value of that number 
        
        
        // cout<<q<<" ";
        recursive(p,n-1);
        
    }

    return ans;
}

class Solution {
public:
    double myPow(double x, int n) {
        

        
        // double q=x;double N=n;
        // return pow(q,N);
        
        
        // when both sides are double this function works 
        
        
        if(n==0)return 1;
        
   
         x = recursive(x,abs(n));
        
 
        
        if(n<0) return double(1/x);
        
        // else 
        return x;
        
    }
};

For More help you may try LEETCODE QUESTION NUMBER 50

Answer 7

**NOW the Second most optimize code pow(x,n) ** logic is that we have to solve it in O(logN) so we devide the n by 2 when we have even power n=4 , 4/2 is 2 means we have to just square it (22)(22) but when we have odd value of power like n=5, 5/2 here we have square it to get also the the number itself to it like (22)(2*2)*2 to get 2^5 = 32 HOPE YOU UNDERSTAND FOR MORE YOU CAN VISIT POW(x,n) question on leetcode below the optimized code and above code was for O(n) only *

#include<bits/stdc++.h>
using namespace std;
double recursive(double x,int n)
{
   
 // recursive calls will return the whole value of the program at every calls  

    if(n==0){return 1;}
    
    // 1 is multiplied when the last value we get as we don't have to multiply further 
    
    double store;
    store = recursive(x,n/2);
    
    // call the function after the base condtion you have given to  it here
    
    if(n%2==0)return store*store;
    else
    {
        return store*store*x;
        // odd power we have the perfect square multiply the value;
    }

}

// main function or the function for indirect call to recursive function 
 double myPow(double x, int n) {

        if(n==0)return 1;
        
         x = recursive(x,abs(n));
     
        // for negatives powers
        
        if(n<0) return double(1/x);
        // else for positves 
        return x;    
    }