# Computational complexity of least square regression operation

In a least square regression algorithm, I have to do the following operations to compute regression coefficients:

• Matrix multiplication, complexity: $O(C^2N)$
• Matrix inversion, complexity: $O(C^3)$
• Matrix multiplication, complexity: $O(C^2N)$
• Matrix multiplication, complexity: $O(CN)$

How can I determine the overall computational complexity of this algorithm?

EDIT: I studied least square regression from the book Introduction to Data Mining by Pang Ning Tan. The explanation about linear least square regression is available in the appendix, where a solution by the use of normal equation is provided (something of the form $a=(X^TX)^{-1}X^Ty)$, which involves 3 matrix multiplications and 1 matrix inversion).

My goal is to determine the overall computational complexity of the algorithm. Above, I have listed the 4 operations needed to compute the regression coefficients with their own complexity. Based on this information, can we determine the overall complexity of the algorithm?

Thanks!

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Could you expand further? You give three different measures of effort for matrix multiplication, and I'm not sure which is right. Also, there are at least three methods I know of for doing linear least squares (and a bit more for nonlinear least squares). What are you trying to do? Where did you get the algorithm you currently have? –  Ｊ. Ｍ. Nov 22 '11 at 7:46
@J.M. I have provided more elaboration in my question. Please let me know if you need more information. –  Andree Nov 22 '11 at 8:21
Okay... forming $\mathbf M=\mathbf X^\top\mathbf X$ is a matrix multiplication. Forming $\mathbf b=\mathbf X^\top \mathbf y$ is a matrix-vector multiplication. But, please, please, please, do not use inversion+multiplication to compute $\mathbf c=\mathbf M^{-1}\mathbf b$! It is better computational practice to form the Cholesky decomposition of $\mathbf M$ and use that in the computation of $\mathbf c$. –  Ｊ. Ｍ. Nov 22 '11 at 8:28
Sure, thanks for the advice! However, I still need to know about how to determine the computational complexity of my current implementation. Can it be inferred from the information I provided above? –  Andree Nov 22 '11 at 8:35
Actually, all I want to know is this: From the 4 matrix operations I listed above (with their own complexity), which one has the highest degree of complexity? 3 of them have the same degree of complexity, so I'm not sure which one that I can assign as the algorithm's overall complexity. –  Andree Nov 22 '11 at 8:41

For a least squares regression with $N$ training examples and $C$ features, it takes:

• $O(C^2N)$ to multiply $X^T$ by $X$
• $O(CN)$ to multiply $X^T$ by $Y$
• $O(C^3)$ to compute the LU (or Cholesky) factorization of $X^TX$ and use that to compute the product $(X^TX)^{-1} (X^TY)$

Asymptotically, $O(C^2N)$ dominates $O(CN)$ so we can forget the $O(CN)$ part.

Since you're using the normal equation I will assume that $N>C$ - otherwise the matrix $X^T X$ would be singular (and hence non-invertible), which means that $O(C^2N)$ asymptotically dominates $O(C^3)$.

Therefore the total time complexity is $O(C^2N)$. You should note that this is only the asymptotic complexity - in particular, for $C$, $N$ smallish you may find that computing the LU or Cholesky decomposition of $X^T X$ takes significantly longer than multiplying $X^T$ by $X$.

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Thanks for the answer Chris! I'm really impressed about how you notice that N > C, thus O(C^2N) dominates O(C^3). –  Andree Nov 22 '11 at 8:50