Sum-of-squares: proofs, beliefs, and algorithms — Boaz Barak and David Steurer

Tensor decomposition via sum of squares

One of the most basic tools in data analysis is least squares regression. The basic setting is where we are given points $$v_1,\ldots,v_m \in \R^2$$ and we wish to find the line $$L$$ that minimizes the sum of square distances of the $$v_i$$’s to $$L$$. More generally, we can think of case that $$v_1,\ldots,v_m \in \R^n$$ and we wish to find a $$k$$ dimensional subspace $$L$$ that minimizes the sum of squares of distances to $$L$$. Another way to think about this is that we are given the matrix $$M= \sum_{i=1}^m v_iv_i^\top$$ and we want to find a rank $$r$$ matrix $$L = \sum_{i=1}^r w_iw_i^\top$$ that minimizes $$\norm{M-L}_F^2$$ (where $$\norm{A}_F=\Tr(AA^\top)^{1/2}$$ denotes the Frobenius norm of a the matrix $$A$$).

This problem can be efficiently solved by doing a singular value decomposition (SVD) of the p.s.d. matrix $$M$$ as $$M=\sum_{i=1}^n \lambda_i w_iw_i^\top$$ and taking the vectors corresponding to the largest $$k$$ eigenvalues. However, SVD is not always sufficient to recover information we are interested in it. For example, if the $$v_i$$’s come from two separate “clusters” centered at $$w_1$$ and $$w_2$$ respectively, then SVD can recover the subspace $$Span\{w_1,w_2\}$$ but not the individual centers.

The underlying difficulty comes from the fact that very different configurations of vectors $$v_1,\ldots,v_m$$ can yield the same second moment matrix $$\sum v_iv_i^\top$$. It turns out that in many cases one can break this symmetry by looking at the third (or higher) moment matrix $$\sum_i v_i^{\otimes 3}$$. This suggest the following task of tensor decomposition:

Let $$T\in\R^{n^d}$$ be a tensor. A rank $$r$$ decomposition of $$T$$ is a set of vectors $$\{ a_{i,j} \}_{i\in[r],j\in[d]}$$ such that $T = \sum_{i=1}^r a_{i,1}\otimes \cdots \otimes a_{i,d}$ where for $$v \in R^n$$ and $$w\in \R^m$$, $$v\otimes w$$ is the $$mn$$ dimensional vector where $$(v\otimes w)_{i,j}=v_iw_j$$.

The rank of a tensor $$T$$ is the minimum $$r$$ such that $$T$$ has a rank $$r$$ decomposition. Let $$\norm{\cdot}$$ be some norm on $$\R^{n^d}$$. An $$\epsilon$$-approximate rank $$r$$ decomposition of $$T$$ w.r.t. the norm $$\norm{\cdot}$$ is a set of vectors as above such that $\norm{T -\sum_{i=1}^r a_{i,1}\otimes \cdots \otimes a_{i,d}} \leq \epsilon\norm{T}$

The tensor decomposition problem is to recover, given a rank $$r$$ tensor $$T$$, a rank $$r$$ decomposition of $$T$$. Noisy tensor decomposition is the task of recovering an approximate decomposition of $$T$$ when such a decomposition exists. Tensor decomposition is an extremely useful primitive Kolda and Bader (2009) but unfortunately it is NP-hard in general (Hillar and Lim 2013). However, there are certain conditions under which a decomposition can be recovered, and the sos algorithm turns out to be extremely useful in this setting.

Classical tensor decomposition algorithms

While tensor decomposition is a hard problem in general, there are algorithm that succeed for it in certain regimes. The philosophy behind them is that a la Halevy, Norvig, and Pereira (2009), a sufficient amount of data can compensate for computational difficulty. Specifically in the context of tensor decomposition the data is the tensor $$T\in \R^{n^d}$$ and so we have a “blessing of dimensionality” where the problem actually becomes easier as $$d$$ grows for a fixed rank $$r$$.

Let’s restrict attention to the symmetric noiseless case, where the tensor $$T$$ has a decomposition of the form $$T = \sum_{i=1}^r a_i^{\otimes d}$$. Note that $$a_1,\ldots,a_r$$ can be described using $$r n$$ numbers, and so one might hope to be able to recover these from the $$n^d$$ dimensional tensor as long as $$r \ll n^{d-1}$$. Indeed, under certain natural conditions this would be possible information theoretically. However, known efficient algorithms require (in addition to other conditions) the rank $$r$$ to be much smaller.
A classical algorithm, attributed to Jennrich (Harshman 1970,Leurgans, Ross, and Abel (1993)) can recover the decomposition in the case that $$d=3$$ and the vectors are $$a_1,\ldots,a_r$$ are linearly independent (and so in particular $$r<n$$). It works as follows:

The input is a 3-tensor $$T\in \R^{n^3}$$ such that $$T= \sum_{i=1}^r a_i^{\otimes 3}$$ for linearly independent $$a_1,\ldots,a_r$$. For simplicity assume that the $$a_i$$’s are orthogonal to one another. (We will see how to relax this assumption later.) Compute a decomposition of $$T$$ as follows:

1. Pick $$v\in\R^n$$ at random, and compute $$M = Tv$$, where we think of $$T$$ here as an $$n^2\times n$$ matrix.
2. Do a singular value decomposition of $$M = \sum_{i=1}^{r'} \lambda_i w_iw_i^\top$$.
3. Solve a least squares problem to find the $$\alpha_1,\ldots,\alpha_r$$ that minimize $$\norm{T-\sum_{i=1}^r \alpha_i w_i^{\otimes 3}}_2^2$$. (Note that the only unknown here are the variables $$\alpha_1,\ldots,\alpha_r$$.)

To see that this algorithm works, note that $M = \sum_{i=1}^r \iprod{a_i,v}a_ia_i^\top\,.$ and so since the $$a_i$$’s are orthogonal this is going to be a singular value decomposition. Moreover, since $$v$$ as random the values $$\iprod{a_i,v}$$ are going to be distinct with probability one, and hence this singular value decomposition is unique. Thus if we run SVD we will get vectors $$w_1,\ldots,w_r$$ that are (up to permuting the indices) equal up to scale to $$a_1,\ldots,a_r$$ and hence there would be a solution to the $$\alpha_1,\ldots,\alpha_r$$ that makes $$T = \sum_{i=1}^r \alpha_i w_i^{\otimes 3}$$. (Note that in practice we need to worry about highly non-trivial issues of noise and numerical stability, but we ignore these at the moment for the purposes of the current discussion.)

Whitening transformation: Suppose that the $$a_i$$’s are linearly independent but not orthogonal to one another. If we are given the second moment matrix $$M_2 = \sum_{i=1}^r a_ia_i^\top$$ then by applying the transformation $$M_2^{-1/2}$$ (restricted to the image space of $$M_2$$ if $$r<n$$) we can reduce to the orthogonal case.

Recovering higher rank tensors: One limitation of Jennrich’s algorithm is that it only works in the case that the rank $$r$$ of the tensor is at most $$n$$, while often we are interested in the so called “overcomplete” case when $$r\gg n$$. Indeed, as mentioned above, one could hope to recover a rank $$r$$ decomposition of a $$d$$-dimensional tensor as long as $$r \ll n^{d-1}$$. Jennrich’s algorithm can scale to higher rank, at the expense of larger dimension. For example, if $$r$$ is some constant times $$n^2$$ and the vectors are “generic” then we would expect the vectors $$a_1^{\otimes 2},\ldots,a_r^{\otimes 2}$$ to be linearly independent, which means we can run Jennrich’s algorithm on the tensor $$\sum a_i^{\otimes 6}$$, thinking of it as a $$3$$-tensor over $$n^2$$. A closer examination shows that a dimension $$5$$ tensor suffices (can you see why?). More generally, we can get an algorithm for a rank $$\Omega(n^{\ceil{d/2}+1})$$ decomposition of generic $$d$$-tensors using Jennrich’s algorithm. The best sos based algorithm achieve rank $$\Omega(n^{d/2})$$ which (depending on whether $$d$$ is odd or even) saves a multiplicative factor of about $$n$$ or $$n^2$$ in the number of observations. This multiplicative factor is non-trivial, but it is not the only advantage of these sos-based algorithms. They also enjoy better robustness to noise, which can be even more crucial than saving data for certain applications.

The “brute data algorithm”

We saw that if we have enough data (i.e., about $$r^3$$ observations instead of the minimum of $$O(rn)$$), we can use Jennrich’s algorithm to recover the decomposition. It turns out that if we have even more data, we can use a simpler algorithm:This simpler algorithm is not more efficient, but it has a conceptually simpler and more robust analysis, which will be useful for us later on.

Suppose that $$d> O(1/\epsilon^2)\log r$$ and $$T = \sum_{i=1}^r a_i^{\otimes d}$$ and that $$\norm{a_i}=1$$ for all $$i$$. Then for every $$i$$, with probability at least $$1/n^{O(1/\epsilon^2)}$$ over a Gaussian $$v\in\R^n$$, $a_i^{\otimes 2} T v^{\otimes d-2} \geq (1-\epsilon) \norm{a_i^{\otimes 2}}\norm{Tv^{\otimes d-2}} \;,$ treating $$T$$ as an $$n^2\times n^{d-2}$$ matrix.

Reference:lem-brute-data yields an algorithm that can iteratively learn the $$a_i$$’s vectors one by one, since if we choose a random $$v$$ then with high probability the matrix $$Tv^{\otimes d-2}$$ will be close to a rank one matrix of the form $$aa^\top$$ where $$a$$ is (up to scaling) equal to one of the $$a_i$$’s.

For any $$i_0$$, with probability at least $$n^{-O(1/\epsilon^2)}$$, we will get that $$(*)$$ $$\iprod{a_{i_0},v}=\alpha$$ for $$\alpha\geq (100/\epsilon)\sqrt{\log n}$$ standard deviations. If we condition on this event, then the distribution of $$\iprod{u,v}$$ for $$u\perp a_{i_0}$$ is a standard Gaussian and so by the union bound with high probability in this conditional space, $$(**)$$ for every unit vector $$u$$ orthogonal to $$a_{i_0}$$, $$\iprod{a_{i_0},v}$$ is at most $$2\sqrt{\log n} = (\epsilon/50)\alpha$$ standard deviations.

Suppose that that both $$(*)$$ and $$(**)$$ happen. Then, for every vector $$a = \sqrt{1-\epsilon^2}a_i + \epsilon u$$, $\iprod{a,v} \leq \alpha\sqrt{1-\epsilon^2} + \epsilon (\epsilon/ 50)\alpha \leq (1-\epsilon^2/3)\alpha$ And hence in particular if $$\iprod{a_i,a} \leq 1-\epsilon^2$$ then when this event occurs then $$\iprod{a_i,v}^{d-2} \ll \iprod{a_i,v}/r$$. Thus, in the tensor $$Tv^{\otimes d-2} = \sum_{i=1}^r \iprod{a_i,v}^{d-2}a_i^{2}$$ the contributions of $$a_i$$’s where $$\iprod{a_i,a_{i_0}} \leq 1-\epsilon^2$$ are negligible compared to the contributions of the $$a_i$$’s (of which there is at least one, namely $$i_0$$) where $$\iprod{a_i,a_{i_0}} \geq 1-\epsilon^2$$. This completes the proof.

We have seen that in the context of tensor decomposition, we can trade time for data. While we don’t know of subexponential algorithms that can recover a decomposition from the information theoretically minimal number of moments (e.g., rank $$\Omega(n^{d-1})$$ for $$d$$-tensors), we can get faster algorithms if we have access to more observations. Somewhat surpisingly, the sos algorithm allows us to make the reverse tradeoff. That is, it allows us to use computation time to compensate for the lack of observations. The idea is that we use our pseudo-distributions to produce “fake moments” of higher degree than the ones we’ve truly observed. We then run a simple algorithm such as Jennrich’s or the “brute data algorithm” on the fake moments. As long as the analysis of these simple algorithms can be embedded in the low degree sos proof system then we can show that recovery will still succeed even if they are given fake moments. One can think of this approach as “dreaming” of moments that we don’t have, and then running the algorithm on those “illusionary” moments and bringing back the solution from the dream.

Applying this methodology to the “brute data algorithm”, we obtain the following tensor decomposition algorithm (Barak, Kelner, and Steurer 2015).The algorithm makes sense for $$3$$-tensors, but the analysis is somewhat simpler for $$4$$ tensors, so we describe it in this setting here.

Given input $$T\in \R^{n^4} = \sum_{i=1}^r a_i^{\otimes 4}$$, operate as follows:

1. Use the sum-of-squares algorithm to compute a degree $$d=O(\log n / \epsilon^2)$$ pseudo-distribution $$\mu$$ such that $$\pE_{\mu(a)} a^{\otimes 3} = \tfrac{1}{r}T$$.
2. Compute $$\hat{T} = \pE_{\mu(a)} a^{\otimes d}$$ and apply the “brute data algorithm” on it:
• Pick $$v$$ to be a random Gaussian.
• Pick $$w$$ using the quadratic sampling lemma from a distribution matching $$\pE_{\mu(a)} \iprod{v,a}^{d-2}a^{\otimes 2}$$.Note that if $$M = \pE_{\mu(a)} \iprod{v,a^{d-2}}a^{\otimes 2}$$ is close to a rank one matrix $$uu^\top$$ then the vector $$w$$ we output will be highly correlated with $$u$$.
• Output $$w$$

This algorithm outputs a single vector which we can repeat to get a tensor decomposition. Thus showing that this algorithm succeeds in quasipolynomial time boils down to the following theorem:

Suppose that $$r= n^{2-\epsilon}$$, then there is $$d=O(\log n/\epsilon^2)$$ such that if $$a_i$$’s are random Gaussian vectors then with probability $$1/qpoly(n)$$, the vector $$w$$ that is output satisfies $$\iprod{w,a_1}^2 \geq 0.99\norm{w}\norm{a_1}$$.

The main property we will use about random Gaussian vectors $$a_1,\ldots,a_r$$, for $$r\ll n^2$$, is that with high probability…

Making it more efficient: This algorithm runs in quasipolynomial time, but Ma, Shi, and Steurer (2016) have shown an analog sos algorithm that “dreams” of using Jennrich’s algorithm instead of the brute data one, and can run in polynomial time. Also, Hopkins et al. (2016) used the analysis of sos-based tensor decomposition to extract a direct tensor decomposition algorithm that does not go through solving general semidefinite programs, and hence is much more efficient.

References

Barak, Boaz, Jonathan A. Kelner, and David Steurer. 2015. “Dictionary Learning and Tensor Decomposition via the Sum-of-Squares Method.” In STOC, 143–51. ACM.

Halevy, Alon, Peter Norvig, and Fernando Pereira. 2009. “The Unreasonable Effectiveness of Data.” IEEE Intelligent Systems 24 (2). IEEE: 8–12.

Harshman, Richard A. 1970. “Foundations of the Parafac Procedure: Models and Conditions for an‘ Explanatory’ Multi-Modal Factor Analysis.” University of California at Los Angeles Los Angeles.

Hillar, Christopher J., and Lek-Heng Lim. 2013. “Most Tensor Problems Are NP-Hard.” J. ACM 60 (6): Art. 45, 39. doi:10.1145/2512329.

Hopkins, Samuel B., Tselil Schramm, Jonathan Shi, and David Steurer. 2016. “Fast Spectral Algorithms from Sum-of-Squares Proofs: Tensor Decomposition and Planted Sparse Vectors.” In STOC, 178–91. ACM.

Kolda, Tamara G, and Brett W Bader. 2009. “Tensor Decompositions and Applications.” SIAM Review 51 (3). SIAM: 455–500.

Leurgans, S. E., R. T. Ross, and R. B. Abel. 1993. “A Decomposition for Three-Way Arrays.” SIAM J. Matrix Anal. Appl. 14 (4): 1064–83. doi:10.1137/0614071.

Ma, Tengyu, Jonathan Shi, and David Steurer. 2016. “Polynomial-Time Tensor Decompositions with Sum-of-Squares.” ArXiv Preprint ArXiv:1610.01980.