Skip to main content

Binomial coefficient using C++

 We are aware of the Binomial coefficient, which is a mathematical term used to represent the number of ways to select a certain number of items/objects from a larger set, without regard to their order. It's denoted by a symbol C(n, k)(n choose k or n over k).

It gets computed by following formula: C(n, k) = n! / (k! * (n - k)!) 

The C++ implementation using the above formula is: 

int factorial(int n) {
    if (n <= 1)
        return 1;
    else
        return n * factorial(n - 1);
}

int binomialCoefficient(int n, int r)
{
    if (r > n)
        return 0;

    int numerator = factorial(n);
    int denominator = factorial(r) * factorial(n - r);
    int result = numerator / denominator;

    return result;
}
int main()
{
    std::cout << "Value of C(" << 5 << ", " << 2 << ") is "
         << binomialCoefficient(5, 2) << "\n";
}


So, the number of ways to choose two items/objects from a set of 5 objects is equal to 

            5!/2! * 3! = 5*4*3*2*1 / (2*1)(3*2*1) = 5 * 2 = 10


When dealing with combinations, the order doesn't matter, however, each item or object can be selected only once. In case we need to select items or objects multiple times then we need to use a multiset coefficient.

Anyway, this text is only on the Binomial coefficient. The very standard implementation of finding a Binomial coefficient is recursion using the following formula:

            C(n, k) = C(n-1, k-1) + C(n-1, k)

            C(n, 0) = C(n, n) = 1 // This is base case

The standard implementation of C++ code is like below:

#include <iostream>
#include <vector>

int binomCoefficient(int n, int k)
{
    if(k > n) return 0;
    if((k == 0) || (k == n)) return 1;

    return binomCoefficient(n - 1, k - 1) + binomCoefficient(n - 1, k);
}

int main()
{
    std::cout << "Value of C(" << 5 << ", " << 2 << ") is "
         << binomCoefficient(5, 2) << "\n";
}

So, the function int binomCoefficient(int n, int k) is getting called recursively and
we can see there are many overlapping subproblems that actually got solved but getting
calculated again over recursive calls.

The Time & space complexity is O(n*max(k,n-k)).

Can we improve this not by calling the same subproblem? And the answer is certainly we do. We can create a 2D array like any other DP problem and store already computed subproblems' results into it in a bottom-up manner. Let's see the code:

#include <iostream>
#include <vector>

int binomCoefficientOverlapping(int n, int k)
{
    std::vector<std::vector<int>> C(n + 1, std::vector<int>(k + 1));
   
    for(int i = 0; i <= n; ++i)
    {
        for(int j = 0; j <= std::min(i, k); ++j)
        {
            if(j == 0 or j == i) // This is for base cases
                C[i][j] = 1;
            else
                // Computing from already computed subproblems
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
    }

    return C[n][k];
}

int main()
{
    std::cout << "Value of C(" << 5 << ", " << 2 << ") is "
         << binomCoefficientOverlapping(5, 2) << "\n";
   
}

The Time & space complexity is O(n*k). This can be further optimized for space. However, I am not going in that direction, rather was curious to know if is there any standard C++ function that can calculate Binomial Coefficient for me. Then I found a function in the standard template library and it's named std::beta (source)[C++17].
Binomial coefficients can be expressed in terms of the beta function:

C(n, k) = 1/((n+1) * std::beta(n-k+1,k+1));

Now, in C++ with the help of the std::beta function calculating the
binomial coefficient became just a line of code. Let's have a look at the code:

#include <iostream>
#include <cmath>

double binom(int n, int k) {
    return 1/((n+1)*std::beta(n-k+1,k+1));
}

int main()
{
std::cout << "Value of C(" << 5 << ", " << 2 << ") is "
<< binom(5, 2);
}

The beta function implementation depends on the underlying gamma functions implementation.
The beta function implementation is like the below:

beta(x, y) = Γ(x) * Γ(y) / Γ(x + y), Γ(z)represents the gamma function.

Reference: Beta_Function
The Demo.
Labels:

Comments

Post a Comment

Popular posts from this blog

XOR (Exclusive OR) for branchless coding

The following example shows the array reversing using the  XOR operator . No need to take any additional variable to reverse the array.   int main(int argc, _TCHAR* argv[]) { char str[] = "I AM STUDENT"; int length = strlen(str); for(int i = 0; i < ((length/2)); i++) { str[i] ^= str[length - (1+i)]; str[length - (1+i)] ^= str[i]; str[i] ^= str[length - (1+i)]; } cout << str << endl; return 0; } The above example is one of the uses of XOR but XOR comes in handy when we can do branchless coding  methods like butterfly switch etc. Sometimes this is very effective in speeding up the execution.  Let's see one of the uses of XOR in branchless coding. I am taking a simple example of Y = | X |.  Yes, I am generating abs of a supplied number. So, my function signature/definition in C++ looks like below: int absoluteBranch( int x) {     if (x < 0 ) {         return ...
1 comment
Read more

Power of Two

  I n this post will be discussing how to calculate if a number is a power of two or not. As an example, 8 is a power of two but the number 10 is not. There are many ways we can solve this. First , we will take an approach which is simple and iterative. In this case, we will calculate the power of two one by one and check with the supplied number. The below code illustrates it. bool isPowerofTwo(unsigned num) { auto y = 1; while (0 != y) { if (num == y) return true; if (num < y) return false; y <<= 1; } return false; } Second , assuming, the number is a 32-bit number, this is also an iterative solution. In this scenario, iterating all bits and counting the set bits. Any number which is a power of 2 will have only one bit set and the rest will be zeros. As an example, 8 in binary representation is 1000. Using this observation, we can implement an iterative solution. bool isPowerofTwo(unsigned num) { auto one_count = 0; for (auto index = 0; index < ...
Post a Comment
Read more

Reversing char array without splitting the array to tokens

 I was reading about strdup, a C++ function and suddenly an idea came to my mind if this can be leveraged to aid in reversing a character array without splitting the array into words and reconstructing it again by placing spaces and removing trailing spaces. Again, I wanted an array to be passed as a function argument and an array size to be passed implicitly with the array to the function. Assumed, a well-formed char array has been passed into the function. No malformed array checking is done inside the function. So, the function signature and definition are like below: Below is the call from the client code to reverse the array without splitting tokens and reconstructing it. Finally, copy the reversed array to the destination.  For GNU C++, we should use strdup instead _strdup . On run, we get the following output: Demo code
Post a Comment
Read more