Hello,
I am having a lot of trouble trying to understand how the matrix A, in the solutions to parts a) and b), is determined ...
In a) it seems we have a 1 in aij if user i has rated item j but for part b) I am confused.
To write the minimization problem in the standard form, we need to collect all of the parameters over which we optimize into the vector \(b\), so in (b), \(b\) is the concatenation of all the user biases \(b_u\) followed by the item biases \(b_i\). Then each row of \(A\) captures one term \(r_{ui}-\bar{r}- b_u - b_i \), so has two ones, one at position \(u\), and one at position \(n+i\).
HW 4 - ex2
Hello,
I am having a lot of trouble trying to understand how the matrix A, in the solutions to parts a) and b), is determined ...
In a) it seems we have a 1 in aij if user i has rated item j but for part b) I am confused.
Thank you
To write the minimization problem in the standard form, we need to collect all of the parameters over which we optimize into the vector \(b\), so in (b), \(b\) is the concatenation of all the user biases \(b_u\) followed by the item biases \(b_i\). Then each row of \(A\) captures one term \(r_{ui}-\bar{r}- b_u - b_i \), so has two ones, one at position \(u\), and one at position \(n+i\).
I hope this clarifies it?
I think the problem lies in the fact that the solution places the second one at position u + i and not n + i. Is this a typo ?
Yes indeed, it's yet another typo - sorry, I didn't even catch it after writing the explanation above.
1
Add comment