# Horner's Algorithm Implementation using MATLAB

Hello Friends,

While Gearing up for my exam scheduled tommorow, I just came across Horner's Rule or the so called Horner's Algorithm for evaluating polynomials.

In this post, I want to share with you a MATLAB program which I wrote for implementing Horner's Algorithm.

You can read more about Horner's Rule here and for a concise explanation, you can also read this. Horner's Algorithm is basically an algorithm which reduces the computation time for evaluating polynomials of higher degree.

Basically, if you consider a polynomial of degree

The short and simple MATLAB program is given below:

While Gearing up for my exam scheduled tommorow, I just came across Horner's Rule or the so called Horner's Algorithm for evaluating polynomials.

In this post, I want to share with you a MATLAB program which I wrote for implementing Horner's Algorithm.

You can read more about Horner's Rule here and for a concise explanation, you can also read this. Horner's Algorithm is basically an algorithm which reduces the computation time for evaluating polynomials of higher degree.

Basically, if you consider a polynomial of degree

**n**, then it would take nearly**n**additions and**(((n^2)+n)/2)**multiplications, when solved using conventional old school methods. But using Horner's Algorithm, we can achieve it using**n**additions and**n**multiplications, which would greatly reduce computation time. These kinds of algorithms can be best utilized in Electronics in VLSI Architecture related research work.The short and simple MATLAB program is given below:

```
%Horner's Algorithm Implementation in MATLAB
%Programmed By: Amit Biswal
%URL: www.123mylist.com
clc
clear all
fprintf('\n------SOLVING POLYNOMIAL USING HORNERS RULE-------');
n=input('\nEnter the Degree of Polynomial');
fprintf('\nEnter the coefficients from Lowest degree to Highest');
for i = n:-1:0
fprintf('\na%d = ',i);
temp=input('');
a(i+1)=temp;
end
x=input('\nEnter the value of x');
p(n+1)=a(n+1);
for i = n:-1:1
p(i)=(p(i+1)*x) + a(i);
end
fprintf('\n------ANSWER-------');
fprintf('\nf(x) at x=%0.2f is %0.2f',x,p(1));
```

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