Sublinear time algorithms for approximate semidefinite programming

Abstract

We consider semidefinite optimization in a saddle point formulation where the primal solution is in the spectrahedron and the dual solution is a distribution over affine functions. We present an approximation algorithm for this problem that runs in sublinear time in the size of the data. To the best of our knowledge, this is the first algorithm to achieve this. Our algorithm is also guaranteed to produce low-rank solutions. We further prove lower bounds on the running time of any algorithm for this problem, showing that certain terms in the running time of our algorithm cannot be further improved. Finally, we consider a non-affine version of the saddle point problem and give an algorithm that under certain assumptions runs in sublinear time.

Appendix: Auxiliary lemmas used in the proof of the main theorem

Most of the proofs given below are adopted from [8] to our needs and are brought here in full detail for completeness.

We begin by proving Lemma 5.

Proof

As a first step, note that for $x>C$ we have $x-\mathbb {E}[X] \ge C/2$, so that

$$\begin{aligned} C(x-C) \le 2(x-\mathbb {E}[X])(x-C) \le 2(x-\mathbb {E}[X])^2 . \end{aligned}$$

Hence, we obtain,

$$\begin{aligned} \mathbb {E}[X] - \mathbb {E}[\bar{X}]&= \int _{x<-C} (x+C) d\mu _X + \int _{x>C} (x-C) d\mu _X \\&\le \int _{x>C} (x-C) d\mu _X \\&\le \frac{2}{C} \int _{x>C} (x-\mathbb {E}[X])^2 d\mu _X \\&\le \frac{2}{C} \text{ Var }[X] . \end{aligned}$$

Similarly one can prove that $\mathbb {E}[X] - \mathbb {E}[\bar{X}] \ge -2\text{ Var }[X]/C$, and the result follows.

In the following lemmas we assume only that $v_t(i) = {{\mathrm{clip}}}(\tilde{v}_t(i),1/\eta )$ is the clipping of a random variable $\tilde{v}_t(i)$, the variance of $\tilde{v}_t(i)$ is at most one ($\text{ Var }[\tilde{v}_t(i)] \le 1$) and we use the notation $\mu _t(i) = \mathbb {E}[\tilde{v}_t(i)]$. We also assume that the expectations of $\tilde{v}_t(i)$ are bounded in absolute value by a constant, $|\mu _t(i)| \le C$, such that $1 \le 2C \le 1/\eta$.

The following lemma is a Bernstein-type inequality for martingales. For a proof see [8], Lemma B.3.

Lemma 22

Let $\{Z_t\}$ be a martingale difference sequence with respect to filtration $\{S_t\}$ (i.e., $\mathbb {E}[Z_t|S_1,\ldots ,S_t] = 0$). Assume the filtration $\{S_t\}$ is such that the values in $S_t$ are determined using only those in $S_{t-1}$, and not any previous history, and so the joint probability distribution satisfies:

$$\begin{aligned} {{\mathrm{Pr}}}\left( S_1=s_1, S_2=s_2, \ldots , S_T=s_t\right) = \prod _{t\in [T-1]} {{\mathrm{Pr}}}\left( S_{t+1}=s_{t+1}\mid S_t=s_t\right) . \end{aligned}$$

In addition, assume for all t, $\mathbb {E}[Z_t^2 | S_1,\ldots ,S_t]\le s$, and $|Z_t| \le V$. Then

$$\begin{aligned} \log {{\mathrm{Pr}}}\left( \sum \nolimits _{t\in [T]} Z_t \ge \alpha \right) \le -\frac{\alpha ^2/2}{Ts + \alpha V/3}. \end{aligned}$$

The following three lemmas prove Lemmas 12, 17.

Lemma 23

For $\sqrt{\frac{4\log {m}}{3T}} \le \eta \le 1/2C$, with probability at least $1-O(1/m)$ it holds that

$$\begin{aligned} \max _{i\in [m]}\sum _{t\in [T]}[v_t(i) - \mu _t(i)] \le 5 \eta T. \end{aligned}$$

Proof

Lemma 5 implies that $|\mathbb {E}[{v}_t(i)] - \mu _t(i)| \le 2\eta$, since $\text{ Var }[\tilde{v}_t(i)]\le 1$.

We show that for given $i\in [m]$, with probability $1 - O(1/m^{2})$, $\sum _{t\in [T]}[v_t(i) - \mathbb {E}[{v}_t(i)]] \le 3 \eta T$, and then apply the union bound over all $i\in [m]$. This together with the above bound on $|\mathbb {E}[{v}_t(i)] - \mu _t(i)|$ implies the lemma via the triangle inequality.

Fixing i, let $Z_t^i \equiv v_t(i) - \mathbb {E}[{v}_t(i)]$, and consider the filtration given by

$$\begin{aligned} S_t \equiv (x_t, p_t, w_t, v_{t-1}, i_{t-1}, j_{t-1}, v_{t-1} - \mathbb {E}[{v}_{t-1}]). \end{aligned}$$

Using the notation $\mathbb {E}_t[\cdot ] = \mathbb {E}[\cdot |S_t]$, observe that

• 1. $\forall t \ . \ \mathbb {E}_t[(Z_t^i) ^2 ] = \mathbb {E}_t[v_t(i)^2] - \mathbb {E}_t[v_t(i)]^2 = \text{ Var }(v_t(i)) \le 1$.

• 2. $|Z_t^i|\le 2/\eta$. This holds since by construction, $|v_t(i)|\le 1/\eta$, and hence

  $$\begin{aligned} |Z_t^i|&= | v_t(i) - \mathbb {E}[v_t(i)]| \le | v_t(i)| + |\mathbb {E}[v_t(i)]| \le \frac{2}{\eta }. \end{aligned}$$

Using these conditions, despite the fact that the $Z_t^i$ are not independent, we can use Lemma 22, and conclude that $Z\equiv \sum _{t\in T} Z_t^i$ satisfies the Bernstein-type inequality with $s=1$ and $V=2/\eta$, and so

$$\begin{aligned} -\log {{\mathrm{Pr}}}\left( Z \ge \alpha \right) \ge \frac{\alpha ^2/2}{Ts + \alpha V/3} \ge \frac{\alpha ^2/2}{T + 2\alpha /3\eta } . \end{aligned}$$

Hence, for $a \ge 3$ we have that

$$\begin{aligned} -\log {{\mathrm{Pr}}}\left( Z \ge a \eta T\right) \ge \frac{a^2/2}{1 + 2a/3} \eta ^2 T \ge \frac{a}{2} \eta ^2 T. \end{aligned}$$

For $\eta \ge \sqrt{4\log {m}/aT}$, the above probability is at most $e^{- 2 \log {m}} = 1/m^2$. Letting $a=3$ we obtain the statement of the lemma.

Lemma 24

For $\sqrt{\log {m}/T} \le \eta \le 1/2C$, with probability at least $1-O(1/m)$,

$$\begin{aligned} \Big | \sum _{t\in [T]}p_t ^{\top }v_t - \sum _{t\in [T]}{}p_t^{\top }\mu _t \Big | \le 4 \eta T . \end{aligned}$$

Proof

This Lemma is proven in essentially the same manner as Lemma 23, and proven below for completeness.

Lemma 5 implies that $|\mathbb {E}[{v}_t(i)] - \mu _t(i)| \le 2\eta ,$ using $\text{ Var }[\tilde{v}_t(i)]\le 1$. Since $p_t$ is a distribution, it follows that $|\mathbb {E}[p_t ^{\top }{v}_t] - p_t ^{\top }\mu _t| \le \eta .$ Let $Z_t \equiv p_t ^{\top }v_t - \mathbb {E}[p_t ^{\top }{v}_t] = \sum _i p_t(i) Z_t^i$, where $Z_t^i = v_t(i) - \mathbb {E}[v_t(i)]$. Consider the filtration given by

$$\begin{aligned} S_t \equiv (x_t, p_t, w_t, v_{t-1}, i_{t-1}, j_{t-1}, v_{t-1} - \mathbb {E}[{v}_{t-1}]) . \end{aligned}$$

Using the notation $\mathbb {E}_t[\cdot ] = \mathbb {E}[\cdot |S_t]$, the quantities $|Z_t|$ and $\mathbb {E}_t[Z_t^2]$ can be bounded as follows:

$$\begin{aligned} |Z_t|&= \!\left| \sum _{i\in [n]}p_t(i) Z_t^i\right| \le \sum _{i\in [n]}p_t(i) \left| Z_t^i\right| \!\le \! 2 \eta ^{-1},&\text{ using }\, \left| Z_t^i\right| \!\le \! 2\eta ^{-1}\, \text{ as } \text{ in } \text{ Lemma } \text{23 }. \end{aligned}$$

Also, using properties of variance, we have that

$$\begin{aligned} \mathbb {E}[Z_t^2] = {{\mathrm{Var}}}[p_t ^{\top }v_t] \le \max _i \text{ Var }[v_t(i)] \le 1. \end{aligned}$$

With these conditions, we can use Lemma 22 and conclude that $Z\equiv \sum _{t\in T} Z_t$ satisfies the Bernstein-type inequality with $s=1$ and $V=2/\eta$, and so

$$\begin{aligned} -\log {{\mathrm{Pr}}}\left( Z \ge \alpha \right) \ge \frac{\alpha ^2/2}{Ts + \alpha V/3} \ge \frac{\alpha ^2/2}{T + 2\alpha /3\eta } . \end{aligned}$$

Hence, for $a \ge 3$ we have that

$$\begin{aligned} -\log {{\mathrm{Pr}}}\left( Z \ge a \eta T\right) \ge \frac{a^2/2}{1 + 2a/3} \eta ^2 T \ge \frac{a}{2} \eta ^2 T. \end{aligned}$$

For $\eta \ge \sqrt{2\log {m}/aT}$, the above probability is at most $e^{-\log {m}} = 1/m$. Letting $a=2$ we obtain the statement of the lemma.

Lemma 25

For $\sqrt{\log {m}/T} \le \eta \le 1/4$, with probability at least $1-O(1/m)$,

$$\begin{aligned} \Big | \sum _{t\in [T]}\mu _t(i_t) - \sum _{t\in [T]}p_t ^{\top }\mu _t \Big | \le 4 C \eta T . \end{aligned}$$

Proof

Let $Z_t \equiv \mu _t(i_t) - p_t ^{\top }\mu _t$, where now $\mu _t$ is a constant vector and $i_t$ is the random variable, and consider the filtration given by

$$\begin{aligned} S_t \equiv (x_t, p_t, w_t, y_t, v_{t-1}, i_{t-1}, j_{t-1}, Z_{t-1}). \end{aligned}$$

The expectation of $\mu _t(i_t)$, conditioning on $S_t$ with respect to the random choice of $i_t$, is $p_t ^{\top }\mu _t$. Hence $\mathbb {E}_t[Z_t] = 0$, where $\mathbb {E}_t[\cdot ]$ denotes $\mathbb {E}[\cdot |S_t]$. The parameters $|Z_t|$ and $\mathbb {E}[Z_t^2]$ can be bounded as follows:

$$\begin{aligned} |Z_t|&\le |\mu _t(i)| + \left| p_t ^{\top }\mu _t\right| \le 2C,\\ \mathbb {E}\left[ Z_t^2\right]&= \mathbb {E}\left[ \left( \mu _t(i) - p_t ^{\top }\mu _t\right) ^2 \right] \le 2 \mathbb {E}\left[ \mu _t(i)^2\right] + 2 \left( p_t ^{\top }\mu _t\right) ^2 \le 4C^2 . \end{aligned}$$

Applying Lemma 22 to $Z\equiv \sum _{t\in T} Z_t$, with parameters $s = 4C^2,\ V = 2C$, we obtain

$$\begin{aligned} -\log {{\mathrm{Pr}}}\left( Z \ge \alpha \right) \ge \frac{\alpha ^2/2}{4C^2T + 2C\alpha /3}. \end{aligned}$$

Hence, for $\eta \le 1/a$ we have that

$$\begin{aligned} -\log {{\mathrm{Pr}}}\left( Z \ge a C \eta T\right) \ge \frac{a^2 \eta ^2 T/2}{4 + 2 a \eta /3} \ge \frac{a^2}{10} \eta ^2 T \end{aligned}$$

and if $\eta \ge \sqrt{10 \log {m}/a^2 T}$, the above probability is no more than 1 / m. Letting $a=4$ we obtain the lemma.

The following is a proof of Lemma 13.

Proof

For all $i\in [m]$ it holds that $\mathbb {E}[v_t(i)^2] \le \mathbb {E}[\tilde{v}_t(i)^2] \le 4$. Thus since $p_t$ is a distribution we have that $\mathbb {E}\left[ {\sum _{t\in {[T]}}p_t^{\top }v_t^2}\right] \le 4T$.

The result follows from applying Markov’s inequality to the random variable $\sum _{t\in {[T]}}p_t^{\top }v_t^2$.

The following is a proof of Lemma 11.

Proof

The proof relies on the analysis for the Lanczos method in [18], Theorem 4.2.

According to [18], given a positive semi-definite matrix M such that $\Vert {M}\Vert _2 \le \rho$ and parameter $\epsilon , \delta > 0$, the Lanczos method returns in time $O\left( {\frac{N}{\sqrt{\epsilon }}\log \frac{n}{\delta }}\right)$ and with probability at least $1-\delta$ a vector x such that

$$\begin{aligned} x^{\top }Mx \ge \lambda _{\max }(M)(1-\epsilon ) . \end{aligned}$$

In our case M need not be positive-semidefinite. Given M such that $\Vert {M}\Vert _2 \le \rho$, we define $M' = M + \rho {}\mathbf I$. Thus $M'$ is positive-semidefinite and it holds that $\Vert {M'}\Vert _2 \le 2\rho$. Thus if we apply the Lanczos procedure with error parameter $\epsilon ' = \epsilon /(2\rho )$ we get a unit vector x such that

$$\begin{aligned} x^{\top }M'x \ge \lambda _{\max }(M') - \frac{\epsilon }{2\rho }\lambda _{\max }(M') \ge \lambda _{\max }(M') - \epsilon . \end{aligned}$$

Now it holds that

$$\begin{aligned} x^{\top }M'x= & {} x^{\top }Mx + x^{\top }\rho {}\mathbf I x = x^{\top }Mx + \rho \\\ge & {} \lambda _{\max }(M' ) - \epsilon = \lambda _{\max }(M) + \rho - \epsilon . \end{aligned}$$

Thus,

$$\begin{aligned} x^{\top }Mx \ge \lambda _{\max }(M) - \epsilon . \end{aligned}$$