Reputation: 1239
I am having some trouble translating a pseudocode for Horner's algorithm into a proper code in MatLab. I think my confusion stems from the fact that the code assumes that the first vector entry can be referred to by 0, whereas in MatLab, this has to be 1. I have tried to modify my code accordingly, but I don't get it to work properly. The pseducode is as follows:
input n, (a_i, : 0 ≤ i ≤ n), z_0
for k = 0 to n-1 do
for j = n-1 to k step -1 do
a_j = a_j + z_0*a_(j+1)
end do
end do
output (a_i: 0 ≤ i ≤ n)
Here is my attempt at writing this in MatLab, where a
is an input vector representing coefficients in a polynomial:
function x = horner(a,z_0)
n = length(a);
for k = 1:n-1
for j = n-1:-1:k
a(j) = a(j) + (z_0)*a(j+1);
end
end
x = a;
I tried this on the vector a = [1 -4 7 -5 -2]
which represents coefficients in a polynomial. I also set z_0 = 3
. According to my book, I should have received the output vecor a = [1 8 25 37 19]
, but my code gives the output vector a = [-245 -313 -146 -29 -2]
.
If anyone can help me clear up this code, I would be very grateful!
Upvotes: 1
Views: 25449
Reputation: 21351
Try this - here a
is the vector of polynomial coefficients listed with a(1) as the coefficient of the highest degree term in your polynomial. If your vector is the opposite way round, simply set
b = fliplr(a)
and call the function using vector b
. This function will evaluate the polynomial using Horners algorithm. Note that this assumes z_0
is the value that you want the polynomial evaluated at, hence a single value is returned (not a vector)
function x = horner(a,z_0)
n = length(a);
result = a(1);
for j = 2:n
result = result*z_0 + a(j);
end
x = result;
If you want to pass in a vector of values z
to evaluate so you can evaluate multiple points (the elements of z
) at the same time, you can pass them in via a vector:
function x = horner(a,z)
n = length(a);
m = length(z);
result = a(1)*ones(1,m);
for j = 2:n
result = result.*z + a(j);
end
x = result;
now the returned x
will be your vector of results
Upvotes: 5