Nearly linear-time packing and covering LP solvers

Abstract

Packing and covering linear programs (PC-LP s) form an important class of linear programs (LPs) across computer science, operations research, and optimization. Luby and Nisan (in: STOC, ACM Press, New York, 1993) constructed an iterative algorithm for approximately solving PC-LP s in nearly linear time, where the time complexity scales nearly linearly in N, the number of nonzero entries of the matrix, and polynomially in $\varepsilon$, the (multiplicative) approximation error. Unfortunately, existing nearly linear-time algorithms (Plotkin et al. in Math Oper Res 20(2):257–301, 1995; Bartal et al., in: Proceedings 38th annual symposium on foundations of computer science, IEEE Computer Society, 1997; Young, in: 42nd annual IEEE symposium on foundations of computer science (FOCS’01), IEEE Computer Society, 2001; Koufogiannakis and Young in Algorithmica 70:494–506, 2013; Young in Nearly linear-time approximation schemes for mixed packing/covering and facility-location linear programs, 2014. arXiv:1407.3015; Allen-Zhu and Orecchia, in: SODA, 2015) for solving PC-LP s require time at least proportional to $\varepsilon ^{-2}$. In this paper, we break this longstanding barrier by designing a packing solver that runs in time $\widetilde{O}(N \varepsilon ^{-1})$ and covering LP solver that runs in time $\widetilde{O}(N \varepsilon ^{-1.5})$. Our packing solver can be extended to run in time $\widetilde{O}(N \varepsilon ^{-1})$ for a class of well-behaved covering programs. In a follow-up work, Wang et al. (in: ICALP, 2016) showed that all covering LPs can be converted into well-behaved ones by a reduction that blows up the problem size only logarithmically.

Appendix

1.1 Proof of Lemma 3.6

Lemma 3.6

We have $\mathsf {x}_k,\mathsf {y}_k,\mathsf {z}_k\in \varDelta _{\mathsf {box}}$ for all $k=0,1,\dots ,T$.

Proof

This is true at the beginning as $\mathsf {x}_0 = \mathsf {y}_0 = x^{\mathsf {start}}\in \varDelta _{\mathsf {box}}$ (see Fact 2.8) and $\mathsf {z}_0 = 0 \in \varDelta _{\mathsf {box}}$. In fact, it suffices for us to show that for every $k\ge 1$, $\mathsf {y}_k = \sum _{l=0}^k \gamma _k^l \mathsf {z}_l$ for some scalars $\gamma _k^l$ satisfying $\sum _l \gamma _k^l = 1$ and $\gamma _k^l \ge 0$ for each $l = 0,\dots ,k$. If this is true, we can prove the lemma by induction: at each iteration $k \ge 1$,

• 1. $\mathsf {x}_k = \tau \mathsf {z}_{k-1} + (1-\tau )\mathsf {y}_{k-1}$ must be in $\varDelta _{\mathsf {box}}$ because $\mathsf {y}_{k-1}$ and $\mathsf {z}_{k-1}$ are and $\tau \in [0,1]$,

• 2. $\mathsf {z}_k$ is in $\varDelta _{\mathsf {box}}$ by the definition that $\mathsf {z}_k = {\text {arg min}}_{z\in \varDelta _{\mathsf {box}}}\{\cdots \}$, and

• 3. $\mathsf {y}_k$ is also in $\varDelta _{\mathsf {box}}$ because $\mathsf {y}_k= \sum _{l=0}^k \gamma _k^l \mathsf {z}_l$ is a convex combination of the $\mathsf {z}_l$’s and $\varDelta _{\mathsf {box}}$ is convex.

For the rest of the proof, we show that $\mathsf {y}_k = \sum _{l=0}^k \gamma _k^l \mathsf {z}_l$ for every $k\ge 1$ with coefficients¹³

$$\begin{aligned} \gamma _{k}^l = \left\{ \begin{array}{ll} (1-\tau )\gamma _{k-1}^l, &{}\quad l = 0,\ldots ,k-2; \\ \big (\frac{1}{n \alpha _{k-1} L} - \frac{1}{n \alpha _k L}\big ) + \tau \big (1 - \frac{1}{n \alpha _{k-1} L}\big ), &{}\quad l=k-1; \\ \frac{1}{n\alpha _k L}, &{}\quad l=k. \end{array} \right. \end{aligned}$$

This is true at the base case $k=1$ because $\mathsf {y}_1 = \mathsf {x}_1 + \frac{1}{n\alpha _1 L}(\mathsf {z}_1 - \mathsf {z}_0) = \frac{1}{n\alpha _1 L} \mathsf {z}_1 + \big (1-\frac{1}{n\alpha _1 L}\big ) \mathsf {z}_0$. For the general $k\ge 2$, we have

$$\begin{aligned} \mathsf {y}_k&= \mathsf {x}_k + \frac{1}{n \alpha _k L}(\mathsf {z}_k - \mathsf {z}_{k-1}) \\&= \tau \mathsf {z}_{k-1} + (1-\tau ) \mathsf {y}_{k-1} + \frac{1}{n \alpha _k L}(\mathsf {z}_k - \mathsf {z}_{k-1}) \\&= \tau \mathsf {z}_{k-1} + (1-\tau ) \left( \sum _{l=0}^{k-2} \gamma _{k-1}^l \mathsf {z}_l + \frac{1}{n \alpha _{k-1} L} \mathsf {z}_{k-1}\right) + \frac{1}{n \alpha _k L}(\mathsf {z}_k - \mathsf {z}_{k-1}) \\&= \left( \sum _{l=0}^{k-2} (1-\tau ) \gamma _{k-1}^l \mathsf {z}_l \right) + \bigg (\left( \frac{1}{n \alpha _{k-1} L} - \frac{1}{n \alpha _k L}\right) + \tau \left( 1 - \frac{1}{n \alpha _{k-1} L}\right) \bigg ) \mathsf {z}_{k-1} \\&\quad + \frac{1}{n\alpha _k L} \mathsf {z}_k . \end{aligned}$$

Therefore, we obtain $\mathsf {y}_k = \sum _{l=0}^k \gamma _k^l \mathsf {z}_l$ as desired.

It is now easy to check that under our definition of $\alpha _k$ (which satisfies $\alpha _k\ge \alpha _{k-1}$ and $\alpha _k\ge \alpha _0 = \frac{1}{nL}$, we must have $\gamma _k^l \ge 0$ for all k and l. Also,

$$\begin{aligned} \sum _l \gamma _k^l&= \sum _{l=0}^{k-2} (1-\tau ) \gamma _{k-1}^l + \bigg (\left( \frac{1}{n \alpha _{k-1} L} - \frac{1}{n \alpha _k L}\right) + \tau \left( 1 - \frac{1}{n \alpha _{k-1} L}\right) \bigg ) + \frac{1}{n\alpha _k L} \\&= (1-\tau ) \left( 1-\frac{1}{n\alpha _{k-1}L}\right) + \bigg (\left( \frac{1}{n \alpha _{k-1} L} - \frac{1}{n \alpha _k L}\right) + \tau \left( 1 - \frac{1}{n \alpha _{k-1} L}\right) \bigg ) \\&\quad + \frac{1}{n\alpha _k L} = 1 . \end{aligned}$$

$\square$

1.2 Proof of Proposition 4.5

Proposition 4.5

• (a) $f_\mu (u^*) \le (1+\varepsilon )\mathsf {OPT}$ for $u^* {\mathop {=}\limits ^{\mathrm {\scriptscriptstyle def}}}(1+\varepsilon /2) x^*$.

• (b) $f_\mu (x) \ge (1-\varepsilon )\mathsf {OPT}$ for every $x \ge 0$.

• (c) For any $x \ge 0$ satisfying $f_\mu (x) \le 2\mathsf {OPT}$, we must have $Ax \ge (1-\varepsilon ){\mathbbm {1}}$.

• (d) If $x\ge 0$ satisfies $f_\mu (x) \le (1+\delta )\mathsf {OPT}$ for some $\delta \in [0,1]$, then $\frac{1}{1-\varepsilon } x$ is a $\frac{1+\delta }{1-\varepsilon }$-approximate solution to the covering LP.

Proof

• (a) We have ${\mathbbm {1}}^{T} u^* = (1+\varepsilon /2)\mathsf {OPT}$ by the definition of $\mathsf {OPT}$. Also, from the feasibility constraint $A x^* \ge {\mathbbm {1}}$ in the covering LP, we have $A u^* - {\mathbbm {1}}\ge \varepsilon /2 \cdot {\mathbbm {1}}$, and can compute $f_\mu (u^*)$ as follows:

  $$\begin{aligned} f_\mu (u^*) = \mu \sum _j e^{\frac{1}{\mu } (1 - (A u^*)_j)} + {\mathbbm {1}}^{T} u^*&\le \mu \sum _j e^{\frac{-\varepsilon /2}{\mu }} + (1+\varepsilon /2)\mathsf {OPT}\\&\le \frac{\mu m}{(nm)^2} + (1+\varepsilon /2)\mathsf {OPT}\le (1+\varepsilon )\mathsf {OPT}. \end{aligned}$$

• (b) Suppose towards contradiction that $f_\mu (x) < (1-\varepsilon )\mathsf {OPT}$. Since $f_\mu (x) <\mathsf {OPT}\le m$, we must have that for every $j\in [m]$, it satisfies that $e^{\frac{1}{\mu }(1-(Ax)_j)} \le f_\mu (x) / \mu \le m / \mu$. This further implies $(Ax)_j \ge 1-\varepsilon$ by the definition of $\mu$. In other words, $Ax \ge (1-\varepsilon ){\mathbbm {1}}$. By the definition of $\mathsf {OPT}$, we must then have ${\mathbbm {1}}^{T} x \ge (1-\varepsilon )\mathsf {OPT}$, finishing the proof that $f_\mu (x) \ge {\mathbbm {1}}^{T} x \ge (1-\varepsilon )\mathsf {OPT}$, giving a contradiction.

• (c) To show $Ax \ge (1-\varepsilon ){\mathbbm {1}}$, we can assume that $v = \max _j (1 - (Ax)_j) > \varepsilon$ because otherwise we are done. Under this definition, we have

  $$\begin{aligned} \textstyle f_\mu (x) \ge \mu e^{\frac{v}{\mu }} = \mu \big ((\frac{nm}{\varepsilon })^4\big )^{v/\varepsilon } \ge \frac{\varepsilon }{4\log (nm/\varepsilon )} (\frac{nm}{\varepsilon })^4 \gg 2\mathsf {OPT}, \end{aligned}$$

  contradicting to our assumption that $f_\mu (x) \le 2\mathsf {OPT}$. Therefore, we must have $v \le \varepsilon$, that is, $Ax \ge (1-\varepsilon ){\mathbbm {1}}$.

• (d) For any x satisfying $f_\mu (x) \le (1+\theta )\mathsf {OPT}\le 2 \mathsf {OPT}$, owing to Proposition 4.5c, we first have that x is approximately feasible, i.e., $Ax \ge (1-\varepsilon ){\mathbbm {1}}$. Next, because ${\mathbbm {1}}^{T} x \le f_\mu (x) \le (1+\theta )\mathsf {OPT}$, we know that x yields an objective ${\mathbbm {1}}^{T} x \le (1+\theta )\mathsf {OPT}$. Letting $x' = \frac{1}{1-\varepsilon } x$, we both have that $x'$ is feasible (i.e., $Ax' \ge {\mathbbm {1}}$), and $x'$ has an objective ${\mathbbm {1}}^{T} x'$ at most $\frac{1+\delta }{1-\varepsilon } \mathsf {OPT}$.

$\square$

1.3 Missing proofs for Sect. 5

In this section we prove Theorem 5.3. Because the proof structure is almost identical to that of Theorem 3.4, we spend most of the discussions only pointing out the difference rather than repeating the proofs. The following three lemmas are completely identical to the ones in the packing LP case, so we restate them below:

Lemma C.1

(cf. Lemma 3.3) Each iteration of $CovLPSolver^{\mathsf {wb}}$ can be implemented to run in expected O(N / n) time.

Lemma C.2

(cf. Lemma 3.6) We have $\mathsf {x}_k,\mathsf {y}_k,\mathsf {z}_k\in \varDelta _{\mathsf {box}}$ for all $k=0,1,\dots ,T$.

Lemma C.3

(cf. Lemma 3.7) For every $u\in \varDelta _{\mathsf {box}}$, it satisfies $\big \langle n\alpha _k \xi _{k}^{(i)}, \mathsf {z}_{k-1} - u \big \rangle \le n^2 \alpha _k^2 L \cdot \big \langle \xi _{k}^{(i)}, \mathsf {x}_{k} - \mathsf {y}_k^{(i)} \big \rangle + \frac{1}{2}\Vert \mathsf {z}_{k-1} - u\Vert _A^2 - \frac{1}{2}\Vert \mathsf {z}_k^{(i)} - u\Vert _A^2 .$

For the gradient descent guarantee of Sect. 3.3, one can first note that Lemma 2.7 remains true: this can be verified by replacing $\nabla _i f_\mu (x) + 1$ in its proof with $1 - \nabla _i f_\mu (x)$. For this reason, Lemma 3.9 (which is built on Lemma 2.7) also remains true. We state it below:

Lemma C.4

(cf. Lemma 3.9) We have $f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)}) \ge \frac{1}{2} \langle \nabla f_\mu (\mathsf {x}_k), \mathsf {x}_k - \mathsf {y}_k^{(i)}\rangle \ge 0$.

Putting all together Denote by $\eta _{k}^{(i)} \in \mathbb {R}_{\le 0}^n$ the vector that is only non-zero at coordinate i, and satisfies $\eta _{k,i}^{(i)} = \nabla _i f_\mu (\mathsf {x}_k) - \xi _{k,i}^{(i)} \in (-\infty , 0]$. In other words, the full gradient

$$\begin{aligned} \nabla f_\mu (\mathsf {x}_k) = \mathbf {E}_i[(0,\dots ,n \nabla _i f_\mu (\mathsf {x}_k),\dots ,0)] = \mathbf {E}_i[ n \eta _{k}^{(i)} + n \xi _k^{(i)} ] \end{aligned}$$

can be (in expectation) decomposed into the a large but non-positive component $\eta _k^{(i)} \in (-\infty , 0]^n$ and a small component $\xi _k^{(i)} \in [-1,1]^n$. Similar as Sect. 3.4, for any $u \in \varDelta _{\mathsf {box}}$, we can use a basic convexity argument and the mirror descent lemma to compute that

$$\begin{aligned}&\alpha _k (f_\mu (\mathsf {x}_k) - f_\mu (u)) \le \big \langle \alpha _k \nabla f_\mu (\mathsf {x}_k), \mathsf {x}_k-u \big \rangle \nonumber \\&\quad = \big \langle \alpha _k \nabla f_\mu (\mathsf {x}_k), \mathsf {x}_k - \mathsf {z}_{k-1} \big \rangle + \big \langle \alpha _k \nabla f_\mu (\mathsf {x}_k), \mathsf {z}_{k-1}-u \big \rangle \nonumber \\&\quad = \big \langle \alpha _k \nabla f_\mu (\mathsf {x}_k), \mathsf {x}_k - \mathsf {z}_{k-1} \big \rangle + \mathbf {E}_{i} \left[ \big \langle n \alpha _k \eta _{k}^{(i)}, \mathsf {z}_{k-1}-u \big \rangle + \big \langle n \alpha _k\xi _{k}^{(i)}, \mathsf {z}_{k-1}-u \big \rangle \right] \nonumber \\&\quad \overset{\textcircled {{\small 1}}}{=} \frac{(1-\tau ) \alpha _k}{\tau } \big \langle \nabla f_\mu (\mathsf {x}_k), \mathsf {y}_{k-1} - \mathsf {x}_k \big \rangle \nonumber \\&\qquad + \mathbf {E}_{i} \left[ \big \langle n \alpha _k \eta _{k}^{(i)}, \mathsf {z}_{k-1}-u \big \rangle + \big \langle n \alpha _k\xi _{k}^{(i)}, \mathsf {z}_{k-1}-u \big \rangle \right] \end{aligned}$$

(C.1)

$$\begin{aligned}&\quad \overset{\textcircled {{\small 2}}}{\le }\frac{(1-\tau )\alpha _k}{\tau } (f_\mu (\mathsf {y}_{k-1})-f_\mu (\mathsf {x}_k)) \nonumber \\&\qquad + \mathbf {E}_{i} \Big [\; \boxed {\big \langle n \alpha _k \eta _{k}^{(i)}, \mathsf {z}_{k-1}-u \big \rangle + n^2 \alpha _k^2 L \cdot \big \langle \xi _{k}^{(i)}, \mathsf {x}_{k} - \mathsf {y}_k^{(i)} \big \rangle } \nonumber \\&\qquad + \frac{1}{2}\Vert \mathsf {z}_{k-1} - u\Vert _A^2 - \frac{1}{2}\Vert \mathsf {z}_k^{(i)} - u\Vert _A^2 \Big ] \end{aligned}$$

(C.2)

Above, $\textcircled {{\small 1}}$is because $\mathsf {x}_{k} = \tau \mathsf {z}_{k-1} + (1-\tau ) \mathsf {y}_{k-1}$, which implies that $\tau (\mathsf {x}_k - \mathsf {z}_{k-1}) = (1-\tau ) (\mathsf {y}_{k-1} - \mathsf {x}_k)$. $\textcircled {{\small 2}}$uses convexity and Lemma C.3. We can establish the following lemma to upper bound the boxed term in (C.2). Its proof is in the same spirit to that of Lemma 3.10, and is the only place that we require all vectors to reside in $\varDelta _{\mathsf {box}}$.

Lemma C.5

(cf. Lemma 3.10) For every $u \in \varDelta _{\mathsf {box}}$,

$$\begin{aligned} \big \langle n \alpha _k \eta _{k}^{(i)}, \mathsf {z}_{k-1}-u \big \rangle + n^2 \alpha _k^2 L \cdot \big \langle \xi _{k}^{(i)}, \mathsf {x}_{k} - \mathsf {y}_k^{(i)} \big \rangle \le 21n \alpha _k L \cdot (f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)})). \end{aligned}$$

Proof of Lemma C.5

Now there are three possibilities:

• If $\eta _{k,i}^{(i)}=0$, then we must have $\xi _{k,i}^{(i)} = \nabla _i f_\mu (\mathsf {x}_k) \in [-1,1]$. Lemma C.4 implies

  $$\begin{aligned}&\big \langle n \alpha _k \eta _{k}^{(i)}, \mathsf {z}_{k-1}-u \big \rangle + n^2 \alpha _k^2 L \cdot \big \langle \xi _{k}^{(i)}, \mathsf {x}_{k} - \mathsf {y}_k^{(i)} \big \rangle \\&\quad = n^2 \alpha _k^2 L \cdot \big \langle \nabla f_\mu (\mathsf {x}_k), \mathsf {x}_{k} - \mathsf {y}_k^{(i)} \big \rangle \le 2 n^2 \alpha _k^2 L \cdot ( f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)}) ) \end{aligned}$$

• If $\eta _{k,i}^{(i)} < 0$ and $\mathsf {z}_{k,i}^{(i)} < \frac{10}{\Vert A_{:i}\Vert _{\infty }}$ (thus $\mathsf {z}_{k}^{(i)}$ is not on the boundary of $\varDelta _{\mathsf {box}}$), then we precisely have $\mathsf {z}_{k,i}^{(i)} = \mathsf {z}_{k-1,i} + \frac{n\alpha _k}{\Vert A_{:i}\Vert _{\infty }}$, and accordingly $\mathsf {y}_{k,i}^{(i)} = \mathsf {x}_{k,i} + \frac{1}{L \Vert A_{:i}\Vert _{\infty }} > \mathsf {x}_{k,i}$. In this case,

  $$\begin{aligned}&\big \langle n \alpha _k \eta _{k}^{(i)}, \mathsf {z}_{k-1}-u \big \rangle + n^2 \alpha _k^2 L \cdot \big \langle \xi _{k}^{(i)}, \mathsf {x}_{k} - \mathsf {y}_k^{(i)} \big \rangle \\&\quad \overset{\textcircled {{\small 1}}}{\le }n \alpha _k \cdot \nabla _i f_\mu (\mathsf {x}_k) \cdot \frac{-10}{\Vert A_{:i}\Vert _{\infty }} + n^2 \alpha _k^2 L \cdot \big \langle \xi _{k}^{(i)}, \mathsf {x}_{k} - \mathsf {y}_k^{(i)} \big \rangle \\&\quad \overset{\textcircled {{\small 2}}}{<} n \alpha _k \cdot \nabla _i f_\mu (\mathsf {x}_k) \cdot \frac{-10}{\Vert A_{:i}\Vert _{\infty }} + n^2 \alpha _k^2 L \cdot \big \langle \nabla f_\mu (\mathsf {x}_k), \mathsf {x}_{k} - \mathsf {y}_k^{(i)} \big \rangle \\&\quad \overset{\textcircled {{\small 3}}}{=} 10 n \alpha _k L \cdot \big \langle \nabla f_\mu (\mathsf {x}_k), \mathsf {x}_k - \mathsf {y}_k^{(i)} \big \rangle + n^2 \alpha _k^2 L \cdot \big \langle \nabla f_\mu (\mathsf {x}_k), \mathsf {x}_{k} - \mathsf {y}_k^{(i)} \big \rangle \\&\quad \overset{\textcircled {{\small 4}}}{\le }\big ( 20 n \alpha _k L + 2 n^2 \alpha _k^2 L \big ) \cdot (f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)})) . \end{aligned}$$

  Above, $\textcircled {{\small 1}}$follows from the fact that $\mathsf {z}_{k-1},u \in \varDelta _{\mathsf {box}}$ and therefore $z_{k-1,i}\ge 0$ and $u_i \le \frac{10}{\Vert A_{:i}\Vert _{\infty }}$ by the definition of $\varDelta _{\mathsf {box}}$, and $u\ge 0$; $\textcircled {{\small 2}}$follows from the fact that $\mathsf {x}_k$ and $\mathsf {y}_k^{(i)}$ are only different at coordinate i, and $\xi _{k,i}^{(i)}=-1 > \nabla _i f_\mu (\mathsf {x}_k)$ (since $\eta _{k,i}^{(i)}<0$); $\textcircled {{\small 3}}$follows from the fact that $\mathsf {y}_{k}^{(i)} = \mathsf {x}_{k} + \frac{\mathbf {e}_i}{L \Vert A_{:i}\Vert _{\infty }}$; and $\textcircled {{\small 4}}$uses Lemma C.4.

• If $\eta _{k,i}^{(i)} < 0$ and $\mathsf {z}_{k,i}^{(i)} = \frac{10}{\Vert A_{:i}\Vert _{\infty }}$, then we have

  $$\begin{aligned}&\big \langle n \alpha _k \eta _{k}^{(i)}, \mathsf {z}_{k-1}-u \big \rangle + n^2 \alpha _k^2 L \cdot \big \langle \xi _{k}^{(i)}, \mathsf {x}_{k} - \mathsf {y}_k^{(i)} \big \rangle \\&\quad \overset{\textcircled {{\small 1}}}{\le }\big \langle n \alpha _k \eta _{k}^{(i)}, \mathsf {z}_{k-1}- \mathsf {z}_k^{(i)} \big \rangle + n^2 \alpha _k^2 L \cdot \big \langle \nabla f_\mu (\mathsf {x}_k), \mathsf {x}_{k} - \mathsf {y}_k^{(i)} \big \rangle \\&\quad \overset{\textcircled {{\small 2}}}{\le }\big \langle n \alpha _k \nabla f_\mu (\mathsf {x}_k), \mathsf {z}_{k-1} - \mathsf {z}_k^{(i)} \big \rangle + n^2 \alpha _k^2 L \cdot \big \langle \nabla f_\mu (\mathsf {x}_k), \mathsf {x}_{k} - \mathsf {y}_k^{(i)} \big \rangle \\&\quad \overset{\textcircled {{\small 3}}}{=} n^2 \alpha _k^2 L \cdot \big \langle \nabla f_\mu (\mathsf {x}_k), \mathsf {x}_k - \mathsf {y}_k^{(i)} \big \rangle + n^2 \alpha _k^2 L \cdot \big \langle \nabla f_\mu (\mathsf {x}_k), \mathsf {x}_{k} - \mathsf {y}_k^{(i)} \big \rangle \\&\quad \overset{\textcircled {{\small 4}}}{\le }4 n^2 \alpha _k^2 L \cdot (f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)})) . \end{aligned}$$

  Above, $\textcircled {{\small 1}}$is because $u_i \le \frac{10}{\Vert A_{:i}\Vert _{\infty }} = \mathsf {z}_{k,i}^{(i)}$ and $\eta _{k,i}^{(i)} < 0$, together with $\nabla _i f_\mu (\mathsf {x}_k) < \xi _{k,i}^{(i)}$ and $\mathsf {x}_{k,i} \le \mathsf {y}_{k,i}^{(i)}$; $\textcircled {{\small 2}}$uses $\nabla _i f_\mu (\mathsf {x}_k) = \eta _{k,i}^{(i)} - 1 < \eta _{k,i}^{(i)}$ and $\mathsf {z}_{k,i}^{(i)} \ge \mathsf {z}_{k-1,i}$; $\textcircled {{\small 3}}$is from our choice of $\mathsf {y}_k$ which satisfies that $\mathsf {z}_{k-1} -\mathsf {z}_k^{(i)} = n \alpha _k L (\mathsf {x}_k - \mathsf {y}_k^{(i)})$; and $\textcircled {{\small 4}}$uses Lemma C.4.

Combining the three cases, and using the fact that $f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)}) \ge 0$, we conclude that

$$\begin{aligned}&\big \langle n \alpha _k \eta _{k}^{(i)}, \mathsf {z}_{k-1}-u \big \rangle + n^2 \alpha _k^2 L \cdot \big \langle \xi _{k}^{(i)}, \mathsf {x}_{k} - \mathsf {y}_k^{(i)} \big \rangle \\&\le (20n \alpha _k L + 4n^2 \alpha _k^2 L) \cdot (f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)})) \\&\le 21 n \alpha _k L \cdot (f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)})) . \end{aligned}$$

Above, the last inequality uses our choice of $\alpha _k$, which implies $n \alpha _k \le n \alpha _T = \frac{1}{\varepsilon L} \le \frac{1}{4}$.

Plugging Lemma C.5 back to (C.2), we have

$$\begin{aligned}&\alpha _k (f_\mu (\mathsf {x}_k) - f_\mu (u)) \le \big \langle \alpha _k \nabla f_\mu (\mathsf {x}_k), \mathsf {x}_k-u \big \rangle \nonumber \\&\quad \overset{\textcircled {{\small 1}}}{\le }\frac{(1-\tau )\alpha _k}{\tau } (f_\mu (\mathsf {y}_{k-1})-f_\mu (\mathsf {x}_k)) \nonumber \\&\qquad + \mathbf {E}_{i} \Big [ 21 n \alpha _k L \cdot (f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)})) + \frac{1}{2}\Vert \mathsf {z}_{k-1} - u\Vert _A^2 - \frac{1}{2}\Vert \mathsf {z}_k - u\Vert _A^2 \Big ] \nonumber \\&\quad \overset{\textcircled {{\small 2}}}{\le }\alpha _k f_\mu (\mathsf {x}_k) + \big (21n\alpha _k L - \alpha _k \big ) f_\mu (\mathsf {y}_{k-1}) \nonumber \\&\qquad + \mathbf {E}_{i} \Big [ -21 n \alpha _k L \cdot f_\mu (\mathsf {y}_k^{(i)}) + \frac{1}{2}\Vert \mathsf {z}_{k-1} - u\Vert _A^2 - \frac{1}{2}\Vert \mathsf {z}_k - u\Vert _A^2 \Big ] . \end{aligned}$$

(C.3)

Above, $\textcircled {{\small 1}}$uses Lemma C.5; and $\textcircled {{\small 2}}$is because we have chosen $\tau$ to satisfy $\frac{1}{\tau } = 21 n L$.

Next, recall that we have picked $\alpha _{k}$ so that $(21n L - 1) \alpha _k = 21n L \cdot \alpha _{k-1}$ in $CovLPSolver^{\mathsf {wb}}$. Telescoping (C.3) for $k=1,\dots ,T$ and choosing $u^*=(1+\varepsilon /2) x^*$, we have

$$\begin{aligned}&- \sum _{k=1}^T \alpha _k f_\mu (u^*) \le 21 f_\mu (\mathsf {y}_0) - 21 n\alpha _T L\\&\quad \cdot \mathbf {E}[f_\mu (\mathsf {y}_T)] + \Vert \mathsf {z}_0 - u^*\Vert _A^2 \le - 21 n\alpha _T L \cdot \mathbf {E}[f_\mu (\mathsf {y}_T)] + 75\mathsf {OPT}. \end{aligned}$$

Here, the second inequality is due to $f_\mu (\mathsf {y}_0) = f_\mu (x^{\mathsf {start}}) \le 3\mathsf {OPT}$ from Fact 5.2, and the fact that

$$\begin{aligned}&\Vert \mathsf {z}_0 - u^*\Vert _A^2 = \Vert u^*\Vert _A^2 = \sum _{i=1}^n (u^*_i)^2 \cdot \Vert A_{:i}\Vert _{\infty }\le (1+\varepsilon /2)^2\sum _{i=1}^n (x^*_i)^2 \cdot \Vert A_{:i}\Vert _{\infty }\\&\quad \le 10(1+\varepsilon /2)^2 \sum _{i=1}^n x^*_i < 12\mathsf {OPT}. \end{aligned}$$

Finally, using the fact that $\sum _{k=1}^T \alpha _k = \alpha _T \cdot \sum _{k=0}^{T-1} \big (1 - \frac{1}{21 nL}\big )^k = 21 n \alpha _T L \big (1-(1-\frac{1}{21 nL})^T\big )$, we rearrange and obtain that

$$\begin{aligned} \mathbf {E}[f_\mu (\mathsf {y}_T)] \le \frac{\sum _k \alpha _k}{21 n\alpha _T L} f_\mu (u^*) + \frac{75}{21 n\alpha _T L} \mathsf {OPT}=&\, \big (1-(1-\frac{1}{21 nL})^T\big ) f_\mu (u^*) \nonumber \\&+ \frac{75}{21 n\alpha _T L} \mathsf {OPT}. \end{aligned}$$

We choose $T = \lceil 21 n L \log (1/\varepsilon ) \rceil$ so that $\frac{1}{n\alpha _T L} = (1-\frac{1}{21 n L})^T \le \varepsilon$. Combining this with the fact that $f_\mu (u^*)\le (1+\varepsilon )\mathsf {OPT}$ (see Proposition 4.5a), we obtain

$$\begin{aligned} \mathbf {E}[f_\mu (\mathsf {y}_T)] \le (1+\varepsilon )\mathsf {OPT}+ 3.6\varepsilon \cdot \mathsf {OPT}< (1+4.6\varepsilon )\mathsf {OPT}. \end{aligned}$$

Therefore, we have finished proving Theorem 5.3. $\square$

1.4 Missing proofs for Sect. 6

Proposition 6.4

If $\mathsf {z}_{k-1}\in \varDelta _{\mathsf {simplex}}$ and $\mathsf {z}_{k-1}>0$, the minimizer $z = {\text {arg min}}_{z \in \varDelta _{\mathsf {simplex}}} \big \{ V_{\mathsf {z}_{k-1}}(z) + \langle \delta \mathbf {e}_i, z \rangle \big \}$ for any scalar $\delta \in \mathbb {R}$ and basis vector $\mathbf {e}_i$ can be computed as follows:

• 1. $z \leftarrow \mathsf {z}_{k-1}$.

• 2. $z_i \leftarrow z_i \cdot e^{-\delta }$.

• 3. If ${\mathbbm {1}}^{T} z > 2\mathsf {OPT}'$, $z \leftarrow \frac{2\mathsf {OPT}'}{{\mathbbm {1}}^{T} z} z$.

• 4. Return z.

Proof

Let us denote by z the returned value of the described procedure, and $g(u) {\mathop {=}\limits ^{\mathrm {\scriptscriptstyle def}}}V_{\mathsf {z}_{k-1}}(u) + \langle \delta \mathbf {e}_i, u \rangle$. Since $\varDelta _{\mathsf {simplex}}$ is a convex body and $g(\cdot )$ is convex, to show $z = {\text {arg min}}_{z \in \varDelta _{\mathsf {simplex}}} \{g(u)\}$, it suffices for us to prove that for every $u \in \varDelta _{\mathsf {simplex}}$, $\langle \nabla g(z), u-z \rangle \ge 0$. Since the gradient $\nabla g(z)$ can be written explicitly, this is equivalent to

$$\begin{aligned} \textstyle \delta (u_i - z_i) + \sum _{\ell =1}^n \log \frac{z_{\ell }}{\mathsf {z}_{k-1,\ell }} \cdot (u_\ell - z_\ell ) \ge 0 . \end{aligned}$$

If the re-scaling in step 3 is not executed, then we have $z_\ell = \mathsf {z}_{k-1,\ell }$ for every $\ell \ne i$, and $z_i = \mathsf {z}_{k-1,i} \cdot e^{-\delta }$; thus, the left-hand side is zero so the above inequality is true for every $u\in \varDelta _{\mathsf {simplex}}$.

Otherwise, we have ${\mathbbm {1}}^{T} z = 2\mathsf {OPT}'$ and there exists some constant $Z>1$ such that, $z_\ell = \mathsf {z}_{k-1,\ell } / Z$ for every $\ell \ne i$, and $z_i = \mathsf {z}_{k-1,i} \cdot e^{-\delta } / Z$. In such a case, the left-hand side equals to

$$\begin{aligned} \textstyle (u_i - z_i) \cdot (\delta - \delta ) + \sum _{\ell =1}^n -\log Z \cdot (u_\ell - z_\ell ) . \end{aligned}$$

It is clear at this moment that since $\log Z > 0$ and ${\mathbbm {1}}^{T} u \le 2\mathsf {OPT}' = {\mathbbm {1}}^{T} z$, the above quantity is always non-negative, finishing the proof. $\square$

Lemma 6.13

Denoting by $\gamma {\mathop {=}\limits ^{\mathrm {\scriptscriptstyle def}}}2\alpha _T n$, we have

$$\begin{aligned} \mathbf {E}_i \big [ \alpha _k \big \langle n \xi _k^{(i)}, \mathsf {z}_{k-1} - u^* \big \rangle \big ] \le V_{\mathsf {z}_{k-1}}\big (\frac{u^*}{1+\gamma }\big ) - \mathbf {E}_i \Big [ V_{\mathsf {z}^{(i)}_k}\big (\frac{u^*}{1+\gamma }\big ) \Big ] + 12 \mathsf {OPT}\cdot \gamma \alpha _k \beta . \end{aligned}$$

Proof

Define $w(x){\mathop {=}\limits ^{\mathrm {\scriptscriptstyle def}}}\sum _i x_i \log (x_i) - x_i$ and accordingly, $V_x(y) = w(y) - \langle \nabla w(x), y-x \rangle - w(x) = \sum _i y_i \log \frac{y_i}{x_i} + x_i - y_i$. We first compute using the classical analysis of mirror descent step as follows:

$$\begin{aligned}&\gamma \alpha _k \big \langle n \xi _k^{(i)}, \mathsf {z}_{k-1} \big \rangle + \alpha _k \big \langle n \xi _k^{(i)}, \mathsf {z}_{k-1} - u^* \big \rangle \nonumber \\&\quad = (1+\gamma )\alpha _k \Big \langle n \xi _k^{(i)}, \mathsf {z}^{(i)}_k - \frac{u^*}{1+\gamma } \Big \rangle + (1+\gamma )\alpha _k \big \langle n \xi _k^{(i)}, \mathsf {z}_{k-1} - \mathsf {z}^{(i)}_k \big \rangle \nonumber \\&\quad \overset{\textcircled {{\small 1}}}{\le } \Big \langle \nabla w(\mathsf {z}_{k-1}) - \nabla w(\mathsf {z}^{(i)}_k), \mathsf {z}^{(i)}_k - \frac{u^*}{1+\gamma } \Big \rangle + (1+\gamma )\alpha _k \big \langle n \xi _k^{(i)}, \mathsf {z}_{k-1} - \mathsf {z}^{(i)}_k \big \rangle \nonumber \\&\quad = \left( w\big (\frac{u^*}{1+\gamma }\big ) - w(\mathsf {z}_{k-1}) - \Big \langle \nabla w(\mathsf {z}_{k-1}), \frac{u^*}{1+\gamma } - \mathsf {z}_{k-1}\Big \rangle \right) \nonumber \\&\qquad - \left( w\big (\frac{u^*}{1+\gamma }\big ) - w(\mathsf {z}^{(i)}_k) - \Big \langle \nabla w(\mathsf {z}^{(i)}_k), \frac{u^*}{1+\gamma } - \mathsf {z}^{(i)}_k\Big \rangle \right) \nonumber \\&\qquad + \left( w(\mathsf {z}_{k-1}) - w(\mathsf {z}^{(i)}_k) - \big \langle \nabla w(\mathsf {z}_{k-1}), \mathsf {z}_{k-1} - \mathsf {z}^{(i)}_k\big \rangle \right) \nonumber \\&\qquad + (1+\gamma )\alpha _k \big \langle n \xi _k^{(i)}, \mathsf {z}_{k-1} - \mathsf {z}^{(i)}_k \big \rangle \nonumber \\&\quad = V_{\mathsf {z}_{k-1}}\big (\frac{u^*}{1+\gamma }\big ) - V_{\mathsf {z}^{(i)}_k}\big (\frac{u^*}{1+\gamma }\big ) + \boxed {(1+\gamma )\alpha _k\big \langle n \xi _k^{(i)}, \mathsf {z}_{k-1} - \mathsf {z}^{(i)}_k \big \rangle - V_{\mathsf {z}_{k-1}}(\mathsf {z}^{(i)}_k)} . \end{aligned}$$

(D.1)

Above, $\textcircled {{\small 1}}$is because $\mathsf {z}^{(i)}_k = {\text {arg min}}_{z \in \varDelta _{\mathsf {simplex}}}\big \{V_{\mathsf {z}_{k-1}}(z) + \langle (1+\gamma )\alpha _k n \xi _k^{(i)}, z \rangle \big \}$, which is equivalent to saying

$$\begin{aligned}&\forall u \in \varDelta _{\mathsf {simplex}},\quad \langle \nabla V_{\mathsf {z}_{k-1}}(\mathsf {z}^{(i)}_k) + (1+\gamma )\alpha _k n \xi _k^{(i)}, u - \mathsf {z}^{(i)}_k \rangle \ge 0 \\ \Longleftrightarrow \quad&\forall u \in \varDelta _{\mathsf {simplex}},\quad \langle \nabla w(\mathsf {z}^{(i)}_k) - \nabla w(\mathsf {z}_{k-1}) + (1+\gamma )\alpha _k n \xi _k^{(i)}, u - \mathsf {z}^{(i)}_k \rangle \ge 0 . \end{aligned}$$

In particular, we have ${\mathbbm {1}}^{T} \frac{u^*}{1+\gamma } = {\mathbbm {1}}^{T} \frac{(1+\varepsilon /2)x^*}{1+\gamma } < 2\mathsf {OPT}\le 2\mathsf {OPT}'$ and therefore $\frac{u^*}{1+\gamma } \in \varDelta _{\mathsf {simplex}}$. Substituting $u = \frac{u^*}{1+\gamma }$ into the above inequality we get $\textcircled {{\small 1}}$.

Next, we upper bound the term in the box:

$$\begin{aligned}&(1+\gamma )\alpha _k\langle n \xi _k^{(i)}, \mathsf {z}_{k-1} - \mathsf {z}^{(i)}_k \rangle - V_{\mathsf {z}_{k-1}}(\mathsf {z}^{(i)}_k) \nonumber \\&\quad \overset{\textcircled {{\small 1}}}{\le }(1+\gamma )\alpha _k n \xi _{k,i} \cdot (\mathsf {z}_{k-1,i} - \mathsf {z}^{(i)}_{k,i}) - \left( \mathsf {z}^{(i)}_{k,i} \log \frac{\mathsf {z}^{(i)}_{k,i}}{\mathsf {z}_{k-1,i}} + \mathsf {z}_{k-1,i} - \mathsf {z}^{(i)}_{k,i} \right) \nonumber \\&\quad \overset{\textcircled {{\small 2}}}{\le }(1+\gamma )\alpha _k n \xi _{k,i} \cdot (\mathsf {z}_{k-1,i} - \mathsf {z}^{(i)}_{k,i}) - \frac{|\mathsf {z}^{(i)}_{k,i} - \mathsf {z}_{k-1,i}|^2}{2\max \{\mathsf {z}^{(i)}_{k,i}, \mathsf {z}_{k-1,i}\}} \nonumber \\&\quad \overset{\textcircled {{\small 3}}}{\le }(1+\gamma )\alpha _k n \xi _{k,i} \cdot (\mathsf {z}_{k-1,i} - \mathsf {z}^{(i)}_{k,i}) - \frac{|\mathsf {z}^{(i)}_{k,i} - \mathsf {z}_{k-1,i}|^2}{4 \mathsf {z}_{k-1,i}} \nonumber \\&\quad \overset{\textcircled {{\small 4}}}{\le }(1+\gamma )^2\mathsf {z}_{k-1,i} \cdot (\alpha _k n \xi _{k,i})^2 \overset{\textcircled {{\small 5}}}{\le }2\mathsf {z}_{k-1,i} \cdot (\alpha _k n \xi _{k,i})^2 \overset{\textcircled {{\small 6}}}{\le }\mathsf {z}_{k-1,i} \cdot \gamma \alpha _k n |\xi _{k,i}| \nonumber \\&\quad \overset{\textcircled {{\small 7}}}{\le }\mathsf {z}_{k-1,i} \cdot \gamma \alpha _k n \xi _{k,i} + 2 \mathsf {z}_{k-1,i} \cdot \gamma \alpha _k n \beta = \gamma \alpha _k \langle n \xi _k^{(i)}, \mathsf {z}_{k-1} \rangle + 2 \mathsf {z}_{k-1,i} \cdot \gamma \alpha _k n \beta . \end{aligned}$$

(D.2)

Above, $\textcircled {{\small 1}}$uses the facts (i) $a \log \frac{a}{b} + b - a \ge 0$ for any $a,b>0$, (ii) $\mathsf {z}_{k-1,i}-\mathsf {z}_k^{(i)}$ and $\xi _{k,i}$ have the same sign, and (iii) $\xi _{k,i'}^{(i)}=0$ for every $i' \ne i$; $\textcircled {{\small 2}}$uses the inequality that for every $a,b>0$, we have $a \log \frac{a}{b} + b - a \ge \frac{(a-b)^2}{2\max \{a,b\}}$. $\textcircled {{\small 3}}$uses the fact that $\mathsf {z}^{(i)}_{k,i} \le 2\mathsf {z}_{k-1,i}$.¹⁴$\textcircled {{\small 4}}$uses Cauchy-Shwarz: $ab - b^2/4 \le a^2$. $\textcircled {{\small 5}}$uses $(1+\gamma )^2 < 2$. $\textcircled {{\small 6}}$uses $|\xi _{k,i}| \le 1$ and $\gamma = 2\alpha _T n \ge 2\alpha _k n$. $\textcircled {{\small 7}}$uses $\xi _{k,i} \ge -\beta$.

Next, we combine (D.1) and (D.2) to conclude that

$$\begin{aligned} \alpha _k \big \langle n \xi _k^{(i)}, \mathsf {z}_{k-1} - u^* \big \rangle \le V_{\mathsf {z}_{k-1}}\big (\frac{u^*}{1+\gamma }\big ) - V_{\mathsf {z}^{(i)}_k}\big (\frac{u^*}{1+\gamma }\big ) + 2 \mathsf {z}_{k-1,i} \cdot \gamma \alpha _k n \beta . \end{aligned}$$

Taking expectation on both sides with respect to i, and using the property that ${\mathbbm {1}}^{T} \mathsf {z}_{k-1} \le 3\mathsf {OPT}' \le 6\mathsf {OPT}$, we obtain that

$$\begin{aligned} \mathbf {E}_i \big [ \alpha _k \big \langle n \xi _k^{(i)}, \mathsf {z}_{k-1} - u^* \big \rangle \big ] \le V_{\mathsf {z}_{k-1}}\big (\frac{u^*}{1+\gamma }\big ) - \mathbf {E}_i \Big [ V_{\mathsf {z}^{(i)}_k}\big (\frac{u^*}{1+\gamma }\big ) \Big ] + 12 \mathsf {OPT}\cdot \gamma \alpha _k \beta . \end{aligned}$$

$\square$

Lemma 6.14

For every $i\in [n]$, we have

• (a) $f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)}) \ge 0$, and

• (b) $f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)}) \ge \frac{\mu \beta }{12} \cdot \langle - \widetilde{\eta }_{k}^{(i)}, u^* \rangle .$

Proof of Lemma 6.14 part (a)

Since if $i\not \in B_k$ is not a large index we have $\mathsf {y}_k^{(i)} = \mathsf {x}_k$ and the claim is trivial, we focus on $i\in B_k$ in the remaining proof. Recall that $\mathsf {y}_k^{(i)} = \mathsf {x}_k + \delta \mathbf {e}_i$ for some $\delta >0$ defined in Algorithm 3, so we have

$$\begin{aligned} f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)}) = \int _{\tau = 0}^{\delta } \langle - \nabla f_\mu (\mathsf {x}_k + \tau \mathbf {e}_i), \mathbf {e}_i \rangle d \tau = \int _{\tau = 0}^{\delta } \big ( \langle A_{:i}, p(\mathsf {x}_k + \tau \mathbf {e}_i) \rangle - 1 \big ) d \tau . \end{aligned}$$

It is clear that $\langle A_{:i}, p(\mathsf {x}_k + \tau \mathbf {e}_i) \rangle$ decreases as $\tau$ increases, and therefore it suffices to prove that $\langle A_{:i}, p(\mathsf {x}_k + \delta \mathbf {e}_i) \rangle \ge 1$.

Suppose that the rows of $A_{:i}$ are sorted (for the simplicity of notation) by the increasing order of $A_{j,i}$. Now, by the definition of the algorithm (recall (6.1)), there exists some $j^* \in [m]$ satisfying that

$$\begin{aligned} \sum _{j< j^*} A_{j,i} \cdot p_j(\mathsf {x}_k) < 1+\beta \quad \text {and}\quad \sum _{j \le j^*} A_{j,i} \cdot p_j(\mathsf {x}_k) \ge 1+\beta . \end{aligned}$$

Next, by our choice of $\delta$ which satisfies $\delta = \frac{\mu \beta }{2A_{j^*,i}} \le \frac{\mu \beta }{2A_{j,i}}$ for every $j \le j^*$, we have for every $j\le j^*$:

$$\begin{aligned} p_j(\mathsf {x}_k + \delta \mathbf {e}_i) = p_j(\mathsf {x}_k) \cdot e^{-\frac{A_{j,i} \delta }{\mu }} \ge p_j(\mathsf {x}_k) \cdot e^{-\beta /2} \ge p_j(\mathsf {x}_k) \cdot (1-\beta /2) , \end{aligned}$$

and as a result,

$$\begin{aligned}&\langle A_{:i}, p(\mathsf {x}_k + \delta \mathbf {e}_i) \ge \sum _{j\le j^*} A_{j,i} \cdot p_j (\mathsf {x}_k + \delta \mathbf {e}_i) \ge (1-\beta /2) \sum _{j\le j^*} A_{j,i} \cdot p_j (\mathsf {x}_k) \\&\quad \ge (1-\beta /2) (1+\beta ) \ge 1 . \end{aligned}$$

$\square$

Proof of Lemma 6.14 part (b)

Owing to part (a), for every coordinate i such that $\widetilde{\eta }_{k,i}\ge 0$, we automatically have $f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)}) \ge 0$ so the lemma is obvious. Therefore, let us focus only on coordinates i such that $\widetilde{\eta }_{k,i}<0$; these are necessarily large indices $i\in B$. Recall from Definition 6.11 that $\widetilde{\eta }_{k,i} = (1+\beta ) - (\widetilde{A}^T p(\mathsf {x}_k))_i$, so we have

$$\begin{aligned} \textstyle \sum _{j=1}^m \widetilde{A}_{j,i} \cdot p_j(\mathsf {x}_k) - (1+\beta ) > 0 .\end{aligned}$$

For the simplicity of description, suppose again that each i-th column is sorted in non-decreasing order, that is, $A_{1,i}\le \cdots \le A_{m,i}$. The definition of $j^*$ can be simplified as

$$\begin{aligned} \textstyle \sum _{j< j^*} A_{j,i} \cdot p_j(\mathsf {x}_k) < 1+\beta \quad \text {and}\quad \sum _{j \le j^*} A_{j,i} \cdot p_j(\mathsf {x}_k) \ge 1+\beta .\end{aligned}$$

Let $j^{\flat } \in [m]$ be the row such that

$$\begin{aligned} \textstyle \sum _{j< j^{\flat }} \widetilde{A}_{j,i} \cdot p_j(\mathsf {x}_k) < 1+\beta \quad \text {and}\quad \sum _{j \le j^{\flat }} \widetilde{A}_{j,i} \cdot p_j(\mathsf {x}_k) \ge 1+\beta .\end{aligned}$$

Note that such a $j^{\flat }$ must exist because $\sum _{j=1}^m \widetilde{A}_{j,i} \cdot p_j > 1+\beta$. It is clear that $j^{\flat } \ge j^*$, owing to the definition that $\widetilde{A}_{ji} \le A_{ji}$ for all $i\in [n], j\in [m]$. Defining $\delta ^{\flat } = \frac{\mu \beta }{2A_{j^{\flat },i}} \le \delta$, the objective decrease is lower bounded as

$$\begin{aligned} f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)})&= \int _{\tau = 0}^{\delta } \langle - \nabla f_\mu (\mathsf {x}_k + \tau \mathbf {e}_i), \mathbf {e}_i \rangle d \tau = \int _{\tau = 0}^{\delta } \big ( \langle A_{:i}, p(\mathsf {x}_k + \tau \mathbf {e}_i) \rangle - 1 \big ) d \tau \\&\ge \int _{\tau = 0}^{\delta ^{\flat }} \big ( \langle A_{:i}, p(\mathsf {x}_k + \tau \mathbf {e}_i) \rangle - 1 \big ) d \tau \\&= \underbrace{\int _{\tau =0}^{\delta ^{\flat }} \left( -1 + \sum _{j\le j^{\flat }} A_{j,i} \cdot p_j(\mathsf {x}_k + \tau \mathbf {e}_i) \right) d\tau }_{I}\\&\quad + \underbrace{\sum _{j>j^{\flat }} \int _{\tau =0}^{\delta ^{\flat }} A_{j,i} \cdot p_j(\mathsf {x}_k + \tau \mathbf {e}_i) d\tau }_{I'} \end{aligned}$$

where the inequality is because $\delta ^{\flat } \le \delta$ and $\langle A_{:i}, p(\mathsf {x}_k + \tau \mathbf {e}_i) \rangle \ge 1$ for all $\tau \le \delta$ (see the proof of part (a)).

PartI To lower bound I, we use the monotonicity of $p_j(\cdot )$ and obtain that

$$\begin{aligned} I =&\int _{\tau =0}^{\delta ^{\flat }} \left( -1 + \sum _{j\le j^{\flat }} A_{j,i} \cdot p_j(\mathsf {x}_k + \tau \mathbf {e}_i) \right) d\tau \ge \delta ^{\flat }\\&\cdot \left( -1 + \sum _{j\le j^{\flat }} A_{j,i} \cdot p_j(\mathsf {x}_k + \delta ^{\flat } \mathbf {e}_i) \right) . \end{aligned}$$

However, our choice of $\delta ^{\flat } = \frac{\mu \beta }{2A_{j^{\flat },i}} \le \frac{\mu \beta }{2A_{j,i}}$ for all $j\le j^{\flat }$ ensures that

$$\begin{aligned} \sum _{j\le j^{\flat }} A_{j,i} \cdot p_j(\mathsf {x}_k + \delta ^{\flat } \mathbf {e}_i)&\ge \sum _{j\le j^{\flat }} A_{j,i} \cdot p_j(\mathsf {x}_k) \cdot e^{\frac{-A_{j,i} \cdot \delta ^{\flat }}{\mu }} \\&\ge \sum _{j\le j^{\flat }} A_{j,i} \cdot p_j(\mathsf {x}_k) \cdot (1-\beta /2) . \end{aligned}$$

Therefore, we obtain that

$$\begin{aligned} I \ge \delta ^{\flat } \left( -1 + (1-\beta /2)\sum _{j\le j^{\flat }} A_{j,i} \cdot p_j(\mathsf {x}_k)\right) \ge \frac{\delta ^{\flat }}{3} \left( -1 + \sum _{j\le j^{\flat }} A_{j,i} \cdot p_j(\mathsf {x}_k) \right) , \end{aligned}$$

where the inequality is because $\big (\frac{2}{3}-\frac{\beta }{2} \big ) \sum _{j\le j^{\flat }} A_{j,i} \cdot p_j(\mathsf {x}_k) \ge \frac{4-3\beta }{6} \cdot (1+\beta ) \ge \frac{2}{3}$ whenever $\beta \le \frac{1}{3}$ (or equivalently, whenever $\varepsilon \le 1/9$).

Now, suppose that $\sum _{j \le j^{\flat }} \widetilde{A}_{j,i} \cdot p_j(\mathsf {x}_k) - (1+\beta ) = b \cdot \widetilde{A}_{j^{\flat },i} \cdot p_{j^{\flat }}(\mathsf {x}_k)$ for some $b \in [0,1]$. Note that we can do so by the very definition of $j^{\flat }$. Then, we must have

$$\begin{aligned} -1 + \sum _{j\le j^{\flat }} A_{j,i} \cdot p_j(\mathsf {x}_k)&\ge -1 + \sum _{j< j^{\flat }} \widetilde{A}_{j,i} \cdot p_j(\mathsf {x}_k) + A_{j^{\flat },i} \cdot p_{j^{\flat }}(\mathsf {x}_k) \\&= -1 + (1+\beta ) - (1-b) \widetilde{A}_{j^{\flat },i} \cdot p_{j^{\flat }}(\mathsf {x}_k) + A_{j^{\flat },i} \cdot p_{j^{\flat }}(\mathsf {x}_k) \\&\ge \beta + b \cdot A_{j^{\flat },i} \cdot p_{j^{\flat }}(\mathsf {x}_k) . \end{aligned}$$

Therefore, we conclude that

$$\begin{aligned} I&\ge \frac{\delta ^{\flat }}{3} \left( -1 + \sum _{j\le j^{\flat }} A_{j,i} \cdot p_j(\mathsf {x}_k) \right) > \frac{\delta ^{\flat }}{3} \cdot b \cdot A_{j^{\flat },i} \cdot p_{j^{\flat }}(\mathsf {x}_k)\\&= \frac{\mu \beta }{6\widetilde{A}_{j^{\flat },i}} \cdot b \cdot \widetilde{A}_{j^{\flat },i} \cdot p_{j^{\flat }}(\mathsf {x}_k) \\&= \frac{\mu \beta }{6\widetilde{A}_{j^{\flat },i}} \cdot \left( - (1+\beta ) + \sum _{j \le j^{\flat }} \widetilde{A}_{j,i} \cdot p_j(\mathsf {x}_k)\right) \\&\ge \frac{\mu \beta }{12} \cdot u^*_i \cdot \left( - (1+\beta ) + \sum _{j \le j^{\flat }} \widetilde{A}_{j,i} \cdot p_j(\mathsf {x}_k) \right) . \end{aligned}$$

Above, the last inequality is because $u^*_i \cdot \widetilde{A}_{j^{\flat },i} \le \langle \widetilde{A}_{j^{\flat } :}, u^* \rangle \le 2$ by our definition of $\widetilde{A}$.

Part$I'$ To lower bound $I'$, consider every $j> j^{\flat }$ and the integral

$$\begin{aligned} \int _{\tau =0}^{\delta ^{\flat }} A_{j,i} \cdot p_j(\mathsf {x}_k + \tau \mathbf {e}_i) d\tau . \end{aligned}$$

Note that whenever $\tau \le \frac{\mu \beta }{2A_{j,i}} \le \frac{\mu \beta }{2A_{j^{\flat },i}} = \delta ^{\flat }$, we have that $p_j(\mathsf {x}_k + \tau \mathbf {e}_i) \ge p_j(\mathsf {x}_k) \cdot e^{-\beta /2} \ge \frac{1}{2} p_j(\mathsf {x}_k)$. Therefore,

$$\begin{aligned} \int _{\tau =0}^{\delta ^{\flat }} A_{j,i} \cdot p_j(\mathsf {x}_k + \tau \mathbf {e}_i) d\tau \ge \int _{\tau =0}^{\frac{\mu \beta }{2A_{j,i}}} A_{j,i} \cdot p_j(\mathsf {x}_k + \tau \mathbf {e}_i) d\tau \ge \frac{\mu \beta }{2A_{j,i}} \cdot A_{j,i} \cdot \frac{1}{2}p_j (\mathsf {x}_k) . \end{aligned}$$

This implies a lower bound on $I'$:

$$\begin{aligned} I' \ge \sum _{j>j^{\flat }} \frac{\mu \beta }{4A_{j,i}} \cdot A_{j,i} \cdot p_j(\mathsf {x}_k) \ge \frac{\mu \beta }{8} \cdot \sum _{j>j^{\flat }} u^*_i \cdot \widetilde{A}_{j,i} \cdot p_j(\mathsf {x}_k) ,\end{aligned}$$

where again in the last inequality we have used $u^*_i \cdot \widetilde{A}_{j^{\flat },i} \le \langle \widetilde{A}_{j^{\flat } :}, u^* \rangle \le 2$ by our definition of $\widetilde{A}$.

Together Combining the lower bounds on I and $I'$, we obtain

$$\begin{aligned} f_\mu (\mathsf {x}_k) - f_\mu (\mathsf {y}_k^{(i)})\ge & {} I + I' \ge \frac{\mu \beta }{12} \cdot u^*_i \cdot \left( - (1+\beta ) + \sum _{j=1}^m \widetilde{A}_{j,i} \cdot p_j(\mathsf {x}_k)\right) \\= & {} \frac{\mu \beta }{12} \cdot \langle -\widetilde{\eta }_{k}^{(i)}, u^* \rangle . \end{aligned}$$

$\square$

1.5 Proof of Lemma 3.3: Efficient Implementation of $PacLPSolver$

In this section, we illustrate how to implement each iteration of $PacLPSolver$ to run in an expected O(N / n) time. We maintain the following quantities

$$\begin{aligned} \mathsf {z}_k \in \mathbb {R}_{\ge 0}^n, \quad \mathsf {az}_k \in \mathbb {R}_{\ge 0}^m, \quad \mathsf {y}'_k \in \mathbb {R}^n, \quad \mathsf {ay}'_k \in \mathbb {R}^m, \quad B_{k,1},B_{k,2} \in \mathbb {R}_+ \end{aligned}$$

throughout the algorithm, so as to ensure the following invariants are always satisfied

$$\begin{aligned}&A \mathsf {z}_k = \mathsf {az}_k , \end{aligned}$$

(E.1)

$$\begin{aligned}&\mathsf {y}_k = B_{k,1} \cdot \mathsf {z}_k + B_{k,2} \cdot \mathsf {y}'_k , \quad A \mathsf {y}_k' = \mathsf {ay}'_k . \end{aligned}$$

(E.2)

It is clear that when $k=0$, letting $\mathsf {az}_k = A \mathsf {z}_0$, $\mathsf {y}'_k = \mathsf {y}_0$, $\mathsf {ay}'_k = A\mathsf {y}_0$, $B_{k,1}=0$, and $B_{k,2}=1$, we can ensure that all the invariants are satisfied initially. We denote $\Vert A_{:i}\Vert _0$ the number of nonzeros elements in vector $A_{:i}$. In each iteration $k=1,2,\dots ,T$:

• The step $\mathsf {x}_k = \tau \mathsf {z}_{k-1} + (1-\tau ) \mathsf {y}_{k-1}$ does not need to be implemented.

• The value $\nabla _i f(\mathsf {x}_k)$ requires the knowledge of $p_j(\mathsf {x}_k) = e^{\frac{1}{\mu } ((A\mathsf {x}_k)_j - 1)}$ for each j such that $A_{ij}\ne 0$. Accordingly, for each j, we need to know the value

  $$\begin{aligned} (A \mathsf {x}_k)_j= & {} \tau (A \mathsf {z}_{k-1})_j + (1-\tau ) (A \mathsf {y}_{k-1})_j \\= & {} \big (\tau + (1-\tau ) B_{k-1,1} \big ) \mathsf {az}_{k-1,j} + (1-\tau ) B_{k-1,2} \mathsf {ay}'_{k-1,j} . \end{aligned}$$

  This can be computed in O(1) time for each j, and $O(\Vert A_{:i}\Vert _0)$ time in total.

• Recall the step $\mathsf {z}_k \leftarrow {\text {arg min}}_{z \in \varDelta _{\mathsf {box}}} \big \{\frac{1}{2}\Vert z - \mathsf {z}_{k-1}\Vert _A^2 + \langle n \alpha _k \xi _{k}^{(i)}, z \rangle \big \}$ can be written as $\mathsf {z}_{k} = \mathsf {z}_{k-1} + \delta \mathbf {e}_i$ for some $\delta \in \mathbb {R}$ that can be computed in O(1) time (see Proposition 3.2). Observe also $\mathsf {z}_k = \mathsf {z}_{k-1} + \delta \mathbf {e}_i$ yields $\mathsf {y}_k = \tau \mathsf {z}_{k-1} + (1-\tau ) \mathsf {y}_{k-1} + \frac{\delta \mathbf {e}_i}{n \alpha _k L}$ due to Line 6 and Line 10 of Algorithm 1. Therefore, we perform two explicit updates on $\mathsf {z}_k$ and $\mathsf {az}_k$ as

  $$\begin{aligned}\mathsf {z}_k \leftarrow \mathsf {z}_{k-1} + \delta \mathbf {e}_i , \quad \mathsf {az}_k \leftarrow \mathsf {az}_{k-1} + \delta A_{:i}\end{aligned}$$

  and two implicit updates on $\mathsf {y}_k$ as

  $$\begin{aligned} \begin{array}{lll} &{}&{}B_{k,1} = \tau + (1-\tau ) B_{k-1,1} , B_{k,2} = (1-\tau ) B_{k-1,2} , \\ &{}&{}\mathsf {y}'_k \leftarrow \mathsf {y}'_{k-1} + \delta \mathbf {e}_i \cdot \left( - \frac{B_{k,1}}{B_{k,2}} + \frac{1}{n\alpha _k L} \frac{1}{B_{k,2}}\right) , \mathsf {ay}'_k \leftarrow \mathsf {ay}'_{k-1} + \delta A_{:i}\cdot \left( - \frac{B_{k,1}}{B_{k,2}} + \frac{1}{n\alpha _k L} \frac{1}{B_{k,2}}\right) \end{array} \end{aligned}$$

  It is not hard to verify that after these updates, $A \mathsf {y}_k' = \mathsf {ay}'_k$ and we have

  $$\begin{aligned}&B_{k,1} \cdot \mathsf {z}_k + B_{k,2} \cdot \mathsf {y}'_k = B_{k,1} \cdot \big (\mathsf {z}_{k-1} + \delta \mathbf {e}_i\big ) \\&\quad + B_{k,2} \cdot \left( \mathsf {y}'_{k-1} + \delta \mathbf {e}_i \cdot \left( - \frac{B_{k,1}}{B_{k,2}} + \frac{1}{n\alpha _k L} \frac{1}{B_{k,2}}\right) \right) \\&\quad = B_{k,1} \cdot \mathsf {z}_{k-1} + B_{k,2} \cdot \left( \mathsf {y}'_{k-1} + \delta \mathbf {e}_i \cdot \left( \frac{1}{n\alpha _k L} \frac{1}{B_{k,2}}\right) \right) \\&\quad = B_{k,1} \cdot \mathsf {z}_{k-1} + B_{k,2} \cdot \mathsf {y}'_{k-1} + \frac{\delta \mathbf {e}_i}{n\alpha _k L} \\&\quad = \big (\tau + (1-\tau ) B_{k-1,1}\big ) \cdot \mathsf {z}_{k-1} + \big ((1-\tau ) B_{k-1,2}\big ) \cdot \mathsf {y}'_{k-1} + \frac{\delta \mathbf {e}_i}{n\alpha _k L} \\&\quad = \tau \mathsf {z}_{k-1} + (1-\tau ) \mathsf {y}_{k-1} + \frac{\delta \mathbf {e}_i}{n\alpha _k L} = \mathsf {y}_k, \end{aligned}$$

  so the invariant $\mathsf {y}_k = B_{k,1} \cdot \mathsf {z}_k + B_{k,2} \cdot \mathsf {y}'_k$ also holds. In sum, after performing updates on $A\mathsf {z}_k$ and $\mathsf {ay}'_k$ in time $O(\Vert A_{:i}\Vert _0)$, we can ensure that the invariants in (E.1) and (E.2) are satisfied at iteration k.

In sum, we only need $O(\Vert A_{:i}\Vert _0)$ time to perform the updates in $PacLPSolver$ for an iteration k if the coordinate i is selected. Therefore, each iteration of $PacLPSolver$ can be implemented to run in an expected $O(\mathbf {E}_i[\Vert A_{:i}\Vert _0]) = O(N/n)$ time.

1.6 Proof of Lemma 6.5: Efficient Implementation of $CovLPSolver$

In this section we illustrate how to implement each iteration of $CovLPSolver$ to run in an expected O(N / n) time. We maintain the following quantities

$$\begin{aligned}&\mathsf {z}_k' \in \mathbb {R}_+^n, \quad \mathsf {sz}_k \in \mathbb {R}_+, \quad \mathsf {sumz}_k \in \mathbb {R}_+, \quad \mathsf {az}'_k \in \mathbb {R}_{\ge 0}^m, \quad \mathsf {y}'_k \in \mathbb {R}^n, \\&\mathsf {ay}'_k \in \mathbb {R}^m, \quad B_{k,1},B_{k,2} \in \mathbb {R}_+ \end{aligned}$$

throughout the algorithm, so as to maintain the following invariants

$$\begin{aligned}&\mathsf {z}_k = \mathsf {z}_k' / \mathsf {sz}_k, \quad&\mathsf {sumz}_k = {\mathbbm {1}}^{T} \mathsf {z}'_k, \quad&A \mathsf {z}_k = \mathsf {az}'_k / \mathsf {sz}_k, \end{aligned}$$

(F.1)

$$\begin{aligned}&\mathsf {y}_k = B_{k,1} \cdot \mathsf {z}_k' + B_{k,2} \cdot \mathsf {y}'_k, \quad&A \mathsf {y}_k = \mathsf {ay}'_k . \end{aligned}$$

(F.2)

It is clear that when $k=0$, letting $\mathsf {z}_k' = \mathsf {z}_0$, $\mathsf {sz}_k=1$, $\mathsf {sumz}_k = {\mathbbm {1}}^{T} \mathsf {z}_0$, $\mathsf {az}'_k = A\mathsf {z}_0$, $\mathsf {y}'_k = \mathsf {y}_0$, $\mathsf {ay}'_k = A\mathsf {y}_0$, $B_{k,1}=0$, and $B_{k,2}=1$, we can ensure that all the invariants are satisfied initially.

We denote by $\Vert A_{:i}\Vert _0$ the number of nonzero elements in vector $A_{:i}$. In each iteration $k=1,2,\dots ,T$:

• The step $\mathsf {x}_k = \tau \mathsf {z}_{k-1} + (1-\tau ) \mathsf {y}_{k-1}$ does not need to be implemented.

• The value $p_j(\mathsf {x}_k) = e^{\frac{1}{\mu } (1 - (A\mathsf {x}_k)_j)}$ for each j only requires the knowledge of

  $$\begin{aligned} (A \mathsf {x}_k)_j= & {} \tau (A \mathsf {z}_{k-1})_j + (1-\tau ) (A \mathsf {y}_{k-1})_j \\= & {} \big (\tau + (1-\tau ) B_{k-1,1} \big ) \frac{\mathsf {az}'_{k-1,j}}{\mathsf {sz}_{k-1}} + (1-\tau ) B_{k-1,2} \mathsf {ay}'_{k-1,j} . \end{aligned}$$

  This can be computed in O(1) time.

• The value $\nabla _i f(\mathsf {x}_k)$ requires the knowledge of $p_j(\mathsf {x}_k)$ for each $j\in [m]$ such that $A_{ij}\ne 0$. Since we have $\Vert A_{:i}\Vert _0$ such j’s, we can compute $\nabla _i f(\mathsf {x}_k)$ in $O(\Vert A_{:i}\Vert _0)$ time.

• Letting $\delta = (1+\gamma ) n \alpha _k \xi _{k,i}^{(i)}$, recall that the mirror step $\mathsf {z}_k \leftarrow {\text {arg min}}_{z \in \varDelta _{\mathsf {simplex}}} \big \{ V_{\mathsf {z}_{k-1}}(z) + \langle \delta \mathbf {e}_i, z \rangle \big \}$ has a very simple form (see Proposition 6.4): first multiply the i-th coordinate of $\mathsf {z}_{k-1}$ by $e^{-\delta }$ and then, if the sum of all coordinates have exceeded $2\mathsf {OPT}'$, scale everything down so as to sum up to $2\mathsf {OPT}'$. This can be implemented as follows: setting $\delta _1 = \mathsf {z}'_{k-1,i} (e^{-\delta }-1)$,

  $$\begin{aligned} \begin{array}{l} \mathsf {z}'_{k} \leftarrow \mathsf {z}'_{k-1} + \delta _1 \mathbf {e}_i , \mathsf {az}'_k \leftarrow \mathsf {az}'_{k-1} + \delta _1 A_{:i}, \\ \mathsf {sumz}_{k} \leftarrow \mathsf {sumz}_{k-1} + \delta _1 , \mathsf {sz}_k \leftarrow \mathsf {sz}_k \cdot \max \Big \{1, \frac{\mathsf {sumz}_k}{\mathsf {sz}_{k-1}\cdot 2\mathsf {OPT}'} \Big \} . \end{array} \end{aligned}$$

  These updates can be implemented to run in $O(\Vert A_{:i}\Vert _0)$ time, and they together ensure that the invariants in (F.1) are satisfied at iteration k.

• Recall that the gradient step is of the form $\mathsf {y}_k \leftarrow \mathsf {x}_k + \delta _2 \cdot \mathbf {e}_i$ for some value $\delta _2 \ge 0$. This value $\delta _2$ can be computed in $O(\Vert A_{:i}\Vert _0)$ time, since each $p_j(\mathsf {x}_k)$ can be computed in O(1) time, and we can sort the rows of each column of A by preprocessing.

  Since $\mathsf {y}_k = \mathsf {x}_k + \delta _2 \cdot \mathbf {e}_i = \tau \mathsf {z}_{k-1} + (1-\tau ) \mathsf {y}_{k-1} + \delta _2 \mathbf {e}_i$, we can implement this update by letting

  $$\begin{aligned} \begin{array}{l} B_{k,1} = \frac{\tau }{\mathsf {sz}_{k-1}} + (1-\tau ) B_{k-1,1} , B_{k,2} = (1-\tau ) B_{k-1,2} \\ \mathsf {y}'_k \leftarrow \mathsf {y}'_{k-1} + \mathbf {e}_i \cdot \left( - \frac{B_{k,1} \delta _1}{B_{k,2}} + \frac{\delta _2}{B_{k,2}}\right) , \mathsf {ay}'_k \leftarrow \mathsf {ay}'_{k-1} + A_{:i}\cdot \left( - \frac{B_{k,1} \delta _1}{B_{k,2}} + \frac{\delta _2}{B_{k,2}}\right) \end{array}\end{aligned}$$

  It is not hard to verify that after these updates, $\mathsf {ay}'_k = A \mathsf {y}'_k$ and we have

  $$\begin{aligned}&B_{k,1} \cdot \mathsf {z}'_k + B_{k,2} \cdot \mathsf {y}'_k = B_{k,1} \cdot \big (\mathsf {z}'_{k-1} + \delta _1 \mathbf {e}_i\big ) \\&\qquad + B_{k,2} \cdot \left( \mathsf {y}'_{k-1} + \mathbf {e}_i \cdot \left( - \frac{B_{k,1}\delta _1}{B_{k,2}} + \frac{\delta _2}{B_{k,2}}\right) \right) \\&\quad = B_{k,1} \cdot \mathsf {z}'_{k-1} + B_{k,2} \cdot \big (\mathsf {y}'_{k-1} + \delta _2 \mathbf {e}_i / B_{k,2} \big ) \\&\quad = B_{k,1} \cdot \mathsf {z}'_{k-1} + B_{k,2} \cdot \mathsf {y}'_{k-1} + \delta _2 \mathbf {e}_i \\&\quad = \big (\frac{\tau }{\mathsf {sz}_{k-1}} + (1-\tau ) B_{k-1,1}\big ) \cdot \mathsf {z}'_{k-1} + \big ((1-\tau ) B_{k-1,2}\big ) \cdot \mathsf {y}'_{k-1} + \delta _2 \mathbf {e}_i \\&\quad = \tau \mathsf {z}_{k-1} + (1-\tau ) \mathsf {y}_{k-1} + \delta _2 \mathbf {e}_i = \mathsf {y}_k , \end{aligned}$$

  so that the invariant $\mathsf {y}_k = B_{k,1} \cdot \mathsf {z}_k' + B_{k,2} \cdot \mathsf {y}'_k$ is also satisfied. In sum, after running time $O(\Vert A_{:i}\Vert _0)$, we can ensure that the invariants in (F.2) are satisfied at iteration k.

In sum, we only need $O(\Vert A_{:i}\Vert _0)$ time to perform the updates in $CovLPSolver$ for an iteration k if the coordinate i is selected. Therefore, each iteration of $CovLPSolver$ can be implemented to run in an expected $O(\mathbf {E}_i[\Vert A_{:i}\Vert _0]) = O(N/n)$ time.