Output of loop is 2x1 vector

I am trying to perform a manual computation of gradient descent, but having trouble with the for loop in storing the output as a vector (2x1 in this case).

With my inputs:

m=20x2 vector

f=20x1 vector

I have set up this for loop:

length= 10000; %k steps
c = 0.000001; %constant
a0 = [1;1] %first guess

a(1) = a0 -c * M' * (M* a0 -f)  %first computation

for k=2:length
    a(k) = a(k-1) - c * M' * (M * a(k-1) - f);
end

Unfortunately, I'm seeing `Unable to perform assignment because the left and right sides have a different number

of elements.`

I've verified that the output of the single-case computation a = a0 - c * M' * (M* a0 -f) is [a1;a2]. This is correct.

My question is how one stores an 'a1, a2' vector as the loop continuously iterates and updates the 'a1, a2' ?

Answer 1

[Jon]

If you want to store the results of each loop in a 2 by numSteps matrix you can do this

numSteps= 10000; %k steps
c = 0.000001; %constant
a0 = [1;1] %first guess
% preallocate array to hold results
a = zeros(2,numSteps)
a(:,1) = a0 -c * M' * (M* a0 -f)  %first computation
for k=2:length
    a(:,k) = a(k-1) - c * M' * (M * a(k-1) - f);
end

Note, I changed your variable name from length to numSteps, length is a Matlab command that gives the maximum dimension of a matrix. Using it as a variable name will work, the variable assignment supersedes the MATLAB function, but is confusing if you think length is a function

[mgc872]

Thanks for taking a shot, Jon!

I've implemented your code and removed 'length' given the possible confusion.

numSteps= 10000; %k steps
c = 0.000001; %constant
a0 = [1;1] %first guess
a = zeros(2,numSteps); % preallocate array to hold results

a(:,1) = a0 -c * M' * (M* a0 -f)  %first computation

for k=2:numSteps
    a(:,k) = a(k-1) - c * M' * (M * a(k-1) - f);
end

WIth this code, though, we still get an assignment error: Unable to perform assignment because the size of the left side is 2-by-1 and the size of the right side is 2-by-2.

Ultimately, we are solving this system for a* which will result in [a1, a2] (from Ma=b) given the size of the inputs M (20x2) and b(20x2). Seems the loop still does not want to update a1, a2.

[Jon]

oops, need to pull vectors out of a matrix on right hand side too does this help?

a(:,k) = a(:,k-1) - c * M' * (M * a(:,k-1) - f);

[Jon]

If this doesn't work, could you please provide a little more of your example so that I can actually run it (I think I need something that defines M and -f to run it)

Answer 2

[Solomon]

no

[Walter Roberson]

I am having trouble figuring out how "no" is an Answer to the problem the user posed? It certainly is possible to store output vectors, so I am having trouble figuring out what the "no" is in response to?