Part of Advances in Neural Information Processing Systems 32 (NeurIPS 2019)

*Huaian Diao, Zhao Song, David Woodruff, Xin Yang*

In the total least squares problem, one is given an $m \times n$ matrix $A$, and an $m \times d$ matrix $B$, and one seeks to ``correct'' both $A$ and $B$, obtaining matrices $\hat{A}$ and $\hat{B}$, so that there exists an $X$ satisfying the equation $\hat{A}X = \hat{B}$. Typically the problem is overconstrained, meaning that $m \gg \max(n,d)$. The cost of the solution $\hat{A}, \hat{B}$ is given by $\|A-\hat{A}\|_F^2 + \|B - \hat{B}\|_F^2$. We give an algorithm for finding a solution $X$ to the linear system $\hat{A}X=\hat{B}$ for which the cost $\|A-\hat{A}\|_F^2 + \|B-\hat{B}\|_F^2$ is at most a multiplicative $(1+\epsilon)$ factor times the optimal cost, up to an additive error $\eta$ that may be an arbitrarily small function of $n$. Importantly, our running time is $\tilde{O}(\nnz(A) + \nnz(B)) + \poly(n/\epsilon) \cdot d$, where for a matrix $C$, $\nnz(C)$ denotes its number of non-zero entries. Importantly, our running time does not directly depend on the large parameter $m$. As total least squares regression is known to be solvable via low rank approximation, a natural approach is to invoke fast algorithms for approximate low rank approximation, obtaining matrices $\hat{A}$ and $\hat{B}$ from this low rank approximation, and then solving for $X$ so that $\hat{A}X = \hat{B}$. However, existing algorithms do not apply since in total least squares the rank of the low rank approximation needs to be $n$, and so the running time of known methods would be at least $mn^2$. In contrast, we are able to achieve a much faster running time for finding $X$ by never explicitly forming the equation $\hat{A} X = \hat{B}$, but instead solving for an $X$ which is a solution to an implicit such equation. Finally, we generalize our algorithm to the total least squares problem with regularization.

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