machine learning - gradient descent seems to fail

Question

I implemented a gradient descent algorithm to minimize a cost function in order to gain a hypothesis for determining whether an image has a good quality. I did that in Octave. The idea is somehow based on the algorithm from the machine learning class by Andrew Ng

Therefore I have 880 values "y" that contains values from 0.5 to ~12. And I have 880 values from 50 to 300 in "X" that should predict the image's quality.

Sadly the algorithm seems to fail, after some iterations the value for theta is so small, that theta0 and theta1 become "NaN". And my linear regression curve has strange values...

here is the code for the gradient descent algorithm: (theta = zeros(2, 1);, alpha= 0.01, iterations=1500)

function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)

m = length(y); % number of training examples
J_history = zeros(num_iters, 1);

for iter = 1:num_iters


    tmp_j1=0;
for i=1:m, 
    tmp_j1 = tmp_j1+ ((theta (1,1) + theta (2,1)*X(i,2)) - y(i));
end

    tmp_j2=0;
for i=1:m, 
    tmp_j2 = tmp_j2+ (((theta (1,1) + theta (2,1)*X(i,2)) - y(i)) *X(i,2)); 
end

    tmp1= theta(1,1) - (alpha *  ((1/m) * tmp_j1))  
    tmp2= theta(2,1) - (alpha *  ((1/m) * tmp_j2))  

    theta(1,1)=tmp1
    theta(2,1)=tmp2

    % ============================================================

    % Save the cost J in every iteration    
    J_history(iter) = computeCost(X, y, theta);
end
end

And here is the computation for the costfunction:

function J = computeCost(X, y, theta)   %

m = length(y); % number of training examples
J = 0;
tmp=0;
for i=1:m, 
    tmp = tmp+ (theta (1,1) + theta (2,1)*X(i,2) - y(i))^2; %differenzberechnung
end
J= (1/(2*m)) * tmp
end

Answer 1

If you are wondering how the seemingly complex looking for loop can be vectorized and cramped into a single one line expression, then please read on. The vectorized form is:

theta = theta - (alpha/m) * (X' * (X * theta - y))

Given below is a detailed explanation for how we arrive at this vectorized expression using gradient descent algorithm:

This is the gradient descent algorithm to fine tune the value of θ:

Assume that the following values of X, y and θ are given:

• m = number of training examples
• n = number of features + 1

Here

• m = 5 (training examples)
• n = 4 (features+1)
• X = m x n matrix
• y = m x 1 vector matrix
• θ = n x 1 vector matrix
• xⁱ is the i^th training example
• x_j is the j^th feature in a given training example

Further,

• h(x) = ([X] * [θ]) (m x 1 matrix of predicted values for our training set)
• h(x)-y = ([X] * [θ] - [y]) (m x 1 matrix of Errors in our predictions)

whole objective of machine learning is to minimize Errors in predictions. Based on the above corollary, our Errors matrix is m x 1 vector matrix as follows:

To calculate new value of θ_j, we have to get a summation of all errors (m rows) multiplied by j^th feature value of the training set X. That is, take all the values in E, individually multiply them with j^th feature of the corresponding training example, and add them all together. This will help us in getting the new (and hopefully better) value of θ_j. Repeat this process for all j or the number of features. In matrix form, this can be written as:

This can be simplified as:

• [E]' x [X] will give us a row vector matrix, since E' is 1 x m matrix and X is m x n matrix. But we are interested in getting a column matrix, hence we transpose the resultant matrix.

More succinctly, it can be written as:

Since (A * B)' = (B' * A'), and A'' = A, we can also write the above as

This is the original expression we started out with:

theta = theta - (alpha/m) * (X' * (X * theta - y))