Gradient Descent implementation in octave
I've actually been struggling against this for like 2 months now. What is it that makes these different?
hypotheses= X * theta
temp=(hypotheses-y)'
temp=X(:,1) * temp
temp=temp * (1 / m)
temp=temp * alpha
theta(1)=theta(1)-temp
hypotheses= X * theta
temp=(hypotheses-y)'
temp=temp * (1 / m)
temp=temp * alpha
theta(2)=theta(2)-temp
theta(1) = theta(1) - alpha * (1/m) * ((X * theta) - y)' * X(:, 1);
theta(2) = theta(2) - alpha * (1/m) * ((X * theta) - y)' * X(:, 2);
The latter works. I'm just not sure why..I struggle to understand the need for the matrix inverse .
Answer
What you're doing in the first example in the second block you've missed out a step haven't you? I am assuming you concatenated X with a vector of ones.
temp=X(:,2) * temp
The last example will work but can be vectorized even more to be more simple and efficient.
I've assumed you only have 1 feature. it will work the same with multiple features since all that happens is you add an extra column to your X matrix for each feature. Basically you add a vector of ones to x to vectorize the intercept.
You can update a 2x1 matrix of thetas in one line of code. With x concatenate a vector of ones making it a nx2 matrix then you can calculate h(x) by multiplying by the theta vector (2x1), this is (X * theta) bit.
The second part of the vectorization is to transpose (X * theta) - y) which gives you a 1*n matrix which when multiplied by X (an n*2 matrix) will basically aggregate both (h(x)-y)x0 and (h(x)-y)x1. By definition both thetas are done at the same time. This results in a 1*2 matrix of my new theta's which I just transpose again to flip around the vector to be the same dimensions as the theta vector. I can then do a simple scalar multiplication by alpha and vector subtraction with theta.
X = data(:, 1); y = data(:, 2);
m = length(y);
X = [ones(m, 1), data(:,1)];
theta = zeros(2, 1);
iterations = 2000;
alpha = 0.001;
for iter = 1:iterations
theta = theta -((1/m) * ((X * theta) - y)' * X)' * alpha;
end