Robust monotone submodular function maximization

Abstract

We consider a robust formulation, introduced by Krause et al. (J Artif Intell Res 42:427–486, 2011), of the classical cardinality constrained monotone submodular function maximization problem, and give the first constant factor approximation results. The robustness considered is w.r.t. adversarial removal of up to $\tau$ elements from the chosen set. For the fundamental case of $\tau =1$, we give a deterministic $(1-1/e)-1/\varTheta (m)$ approximation algorithm, where m is an input parameter and number of queries scale as $O(n^{m+1})$. In the process, we develop a deterministic $(1-1/e)-1/\varTheta (m)$ approximate greedy algorithm for bi-objective maximization of (two) monotone submodular functions. Generalizing the ideas and using a result from Chekuri et al. (in: FOCS 10, IEEE, pp 575–584, 2010), we show a randomized $(1-1/e)-\epsilon$ approximation for constant $\tau$ and $\epsilon \le \frac{1}{\tilde{\varOmega }(\tau )}$, making $O(n^{1/\epsilon ^3})$ queries. Further, for $\tau \ll \sqrt{k}$, we give a fast and practical 0.387 algorithm. Finally, we also give a black box result result for the much more general setting of robust maximization subject to an Independence System.

A Appendix

1.1 A.1

Lemma 17

(Nemhauser, Wolsey [25, 26]) For all $\alpha \ge 0$, greedy algorithm terminated after $\alpha k$ steps yields a set A with $f(A)\ge \beta (0,\alpha )f({ OPT}(k,N,0))$.

Proof

Let $A_i$ be the set at iteration i of the greedy algorithm. Then by monotonicity, we have,

$$\begin{aligned} f(A_i\cup { OPT}(k,N,0))\ge f({ OPT}(k,N,0)) \end{aligned}$$

and by submodularity,

$$\begin{aligned} \sum _{e\in { OPT}(k,N,0)-A_i} f(e|A_i)\ge f({ OPT}(k,N,0)|A_i)\ge f({ OPT}(N,k,0))-f(A_i) \end{aligned}$$

Hence, there exists an element e in ${ OPT}(k,N,0)-A_i$ such that,

$$\begin{aligned} f(e|A_i)\ge (f({ OPT}(k,N,0))-f(A_i))/k \end{aligned}$$

Hence, we get the recurring inequality,

$$\begin{aligned} f(A_{i+1})\ge f(A_{i}+e)\ge f(A_i)+(f({ OPT}(k,N,0))-f(A_i))/k \end{aligned}$$

The above implies that the difference between the value of the greedy set and optimal solution decreases by a factor of $(1-1/k)$ at each step, so after $\alpha k$ steps,

$$\begin{aligned} f(A_{\alpha k})&\ge (1-(1-1/k)^{\alpha k})f({ OPT}(k,N,0))\\ \implies f(A_{\alpha k})&\ge \beta (0,\alpha ) f({ OPT}(k,N,0)) \end{aligned}$$

$\square$

1.2 A.2

Lemma 18

There exists no polytime algorithm with approximation ratio greater than $(1-1/e)$ for P2 unless $P=NP$. For the value oracle model, we have the same threshold, but for algorithms that make only a polynomial number of queries.

Proof

We will give a strict reduction from the classical problem P1 (for which the above hardness result holds [12, 26]) to the robust problem P2. Consider an instance of P1, denoted by (k, N, 0). We intend to reduce this to an instance of P2 on an augmented ground set $N\cup X$ i.e. $(k+\tau ,N\cup X,\tau )$.

The set $X=\{x_1,\ldots ,x_\tau \}$ is such that $f(x_i)=(k+1) f(a_1)$ (and recall that $f(a_1)\ge f(a_i),\forall a_i\in N$) and $f(x_i|S)=f(x_i)$ for every i and $S\subset N\cup X$ not containing $x_i$. We will show that $g({ OPT}(k+\tau ,N\cup X,\tau ))=f({ OPT}(k,N,0))$.

First, note that for an arbitrary set $S=S_N \cup S_X$, such that $|S|=k+\tau$ and $S_X=S\cap X$, we have that every minimizer contains $S_X$. This follows by definition of X, since for any two subsets P, Q of S with $|P|=|Q|= k$ and P disjoint with X but $Q\cap X \ne \emptyset$, we have by monotonicity $f(Q)\ge f(x_i)= (k+1)f(a_1)> kf(a_1)$ and by submodularity $kf(a_1) \ge f(P)$. This implies that X is the minimizer of ${ OPT}(k,N,0)\cup X$ and hence $f({ OPT}(k,N,0))\le g({ OPT}(k+\tau ,N\cup X,\tau ))$

For the other direction, consider the set ${ OPT}(k+\tau ,N\cup X,\tau )$ and define,

$$\begin{aligned} M={ OPT}(k+\tau ,N\cup X,\tau ) \cap X \end{aligned}$$

Next, observe that carving out an arbitrary set B of size $\tau -|M|$ from ${ OPT}(k+\tau ,N\cup X,\tau )-M$ will give us the set

$$\begin{aligned} C={ OPT}(k+\tau ,N\cup X,\tau )-M-B \end{aligned}$$

of size $k+\tau -(|M| +\tau -|M|)=k$. Also note that by design, $C\subseteq N$ and hence $f(C)\le f({ OPT}(k,N,0))$, but by definition, we have that $g({ OPT}(k+\tau ,N\cup X,\tau ))\le f(C)$. This gives us the other direction and we have $g({ OPT}(k+\tau ,N\cup X,\tau ))=f({ OPT}(k,N,0))$.

To complete the reduction we need to show how to obtain an $\alpha$-approximate solution to (k, N, 0) given an $\alpha$-approximate solution to $(k+\tau ,N\cup X,\tau )$. Let $S=S_N \cup S_X$ be such a solution i.e. a set of size $k+\tau$ with $S_X=S\cap X$, such that $g(S)\ge \alpha g({ OPT}(k+\tau ,N\cup X,\tau ))$. Now consider an arbitrary subset $S'_N$ of $S_N$ of size $\tau -|S_X|$. Observe that $|S_N-S'_N|=|S|-|S_X|-(\tau -|S_X|)=k$ and further $f(S_N-S'_N)\ge g(S)\ge \alpha g({ OPT}(k+\tau ,N\cup X,\tau ))=\alpha f({ OPT}(k,N,0))$, by definition. Hence the set $S_N-S'_N\subseteq N$ is an $\alpha$-approximate solution to (k, N, 0) that, given S, can be obtained in polynomial time/queries. $\square$

1.3 A.3 Tight analysis of Algorithm 3

Theorem 19

The 0.387-algorithm is $\frac{1}{2}\beta (0.5,\frac{k-2}{k-1})(> 0.387$ asymptotically) approximate.

Proof

Let ${ OPT}=g({ OPT}(k,N,1))$, A be the output of the 0.387-algorithm and $a'_1$ be the first element added to A apart from $a_1$. The case $z=a_1$ is straightforward since $f(A-a_1)\ge \beta (0,1)f({ OPT}(k-1,N-a_1,0))\ge \beta (0,1){ OPT}$ where the last inequality follows from Lemma 2. So assume $z\ne a_1$. Further, let $f(z|A-a_1-z)=\eta f(A-a_1)$ which implies that $f(a'_1)\ge f(z)\ge f(z|A-a_1-z)= \eta f(A-a_1)$ and now from Lemma 5 with N replaced by $N-a_1$, A replaced by $A-a_1$ and thus k replaced by $k-1$, $S=a'_1$ with $s=1$ and $l=k-1-|S|=k-2$, we get,

$$\begin{aligned} f(A-a_1)\ge \beta \left( \eta ,\frac{k-2}{k-1}\right) f({ OPT}(k-1,N-a_1,0)) \end{aligned}$$

This together with Lemma 2 implies, $f(A-a_1)\ge \beta (\eta ,\frac{k-2}{k-1}){ OPT}$. Also, we have by definition,

$$\begin{aligned} f(A-a_1-z)=(1-\eta )f(A-a_1)\ge (1-\eta )\beta (\eta ,\frac{k-2}{k-1}){ OPT} \end{aligned}$$

Further, we have,

$$\begin{aligned} g(A)&\ge \max \{ f(a_1), f(A-a_1-z)\}\\&\ge \max \{ f(z|A-a_1-z), f(A-a_1-z)\}\\&\ge \max \left\{ \eta \beta \left( \eta ,\frac{k-2}{k-1}\right) , (1-\eta )\beta \left( \eta ,\frac{k-2}{k-1}\right) \right\} { OPT} \\&\ge 0.5\beta \left( 0.5,\frac{k-2}{k-1}\right) { OPT}\quad [\text {for } \eta =0.5]\\&\xrightarrow {k\rightarrow \infty } 0.387 { OPT} \end{aligned}$$

$\square$

We now give an instance where the above analysis is tight. Let the algorithm start with a maximum value element $a_1$, then pick $a_2$, and then add the set C, such that the output of the algorithm is $a_1\cup a_2 \cup C$, with C being a set of size $k-2$. Let $f(a_1)=1, f(a_2)=1, f(C)=1$ with $f(a_1+C)=1, f(a_1+a_2)=2, f(a_2+C)=2$ i.e. C copies $a_1$. Hence $f(a_1+a_2+C)=2$ and $g(a_1+a_2+C)=f(a_1+C)=1$.

Let ${ OPT}(k,N,1)$ include $a_2$, a copy $a'_2$ of $a_2$ (so $f(a'_2)=1, f(a_2+a'_2)=1$) and a set D of $k-2$ elements of value $\frac{1}{(k-2)\beta (0,1)}$ each, such that $f({ OPT}(k,N,1))=1+(k-2)\frac{1}{(k-2)\beta (0,1)}=1+\frac{e}{e-1}=\frac{2}{\beta (0.5,1)}$. Observe that the small value elements are all minimizers and $g({ OPT}(k,N,1))\approx \frac{2}{\beta (0.5,1)}$ as k becomes large. Note that $f(D)=\frac{f(C)}{\beta (0,1)}$ and we can have sets C and D as above based on the worst case example for the greedy algorithm given in [26].This proves that the inequality in Lemma 5 is tight.

1.4 A.4 Analysis of Algorithm 4

Theorem 20

Algorithm 4 is $0.5547-\varOmega (1/k)$ approximate.

Proof

Let A denote the output and $A_0\subset A$ denote $\{a_1,a_2\}$. Due to submodularity, there exists at most two distinct $x\in A$ with $f(x|A-x)>\frac{f(A)}{3}$. Additionally, for every $x\not \in A_0$, we have that $f(x|S)\le f(a_1)$ and $f(x|S)\le f(a_2|a_1)$ for arbitrary subset S of A containing $A_0$ and $x\not \in S$. This implies that that $2f(x|S)\le f(A_0)\le f(S)$, which gives us that $f(x|S)\le \frac{f(S+x)}{3}$.

Note that due to condition in Phase 1, the algorithm ignores $a_1$ even if it is not a minimizer, as long as its marginal is more than a third the value of the set at that iteration. At the end of Phase 1, if $a_2$ has marginal more than third of the set value, then it is ignored until its contribution/marginal decreases. Phase 3 adds greedily (without ignoring any element added). As argued above, no element other than $a_1,a_2$ can have marginal more than a third of the set value at any iteration, so during Phase 2 we have that $a_2$ is also a minimizer.

We will now proceed by splitting into cases. Denote $g({ OPT}(k,N,1))$ as ${ OPT}$ and recall from Lemma 2, ${ OPT}\le f({ OPT}(k-1,N-a_i,0))$ for $i\in \{1,2\}$. Also, let the set of elements added to $A_0$ during Phase 1 be $U=\{u_1,\ldots ,u_p\}$, similarly elements added during Phases 2 and 3 be $V=\{v_1,\ldots ,v_q\},W=\{w_1,\ldots ,w_r\}$ respectively, with indexing in order of addition to the set. Finally, let $\alpha _p=\frac{p-2}{k-1}\, ,\, \alpha _q=\frac{q-1}{k-1}\, ,\, \alpha _r=\frac{r}{k-1}\, ,\, \alpha =\alpha _p+\alpha _q+\alpha _r=\frac{k-5}{k-1}$, and assume $k\ge 8$.

Case 1 Phase 2,3 do not occur i.e. $p=k-2, q=r=0$.

Since we have,

$$\begin{aligned} f(A-a_1)&\overset{(a)}{\ge } f(a_2) + \beta \left( 0,\frac{k-2}{k-1}\right) (f({ OPT}(k-1,N-a_1,0))-f(a_2)) \nonumber \\&\ge \beta \left( 0,\frac{k-2}{k-1}\right) f({ OPT}(k-1,N-a_1,0)) \nonumber \\&\overset{(b)}{\ge } \beta \left( 0,\frac{k-2}{k-1}\right) { OPT} , \end{aligned}$$

(16)

where (a) follows from Lemma 1 and (b) from Lemma 2. This deals with the case $z=a_1$. If $z=u_p$, we have,

$$\begin{aligned} f(A-u_p)\ge f(A-u_p-a_1)&\overset{(c)}{\ge } \beta \left( 0,\frac{k-3}{k-1}\right) f({ OPT}(k-1,N-a_1,0))\\&\ge \beta \left( 0,\frac{k-3}{k-1}\right) { OPT}, \end{aligned}$$

where (c) is like (16) above but with $k-2$ replaced by $k-3$ when using Lemma 1. Finally, let $z\not \in \{a_1,u_p\}$, then due to the Phase 1 termination criteria, we have $f(a_1|A-a_1-u_p)\ge f(A-u_p)/3$, which implies that,

$$\begin{aligned} 2 f(a_1|A-a_1-u_p)&\ge f(A-a_1-u_p) \end{aligned}$$

(17)

Now letting $\eta =\frac{f(z|A-a_1-u_p-z)}{f(A-a_1-u_p)}$, we have by submodularity $f(z|A-z)\le \eta f(A-a_1-u_p)$ and by definition $f(A-a_1-u_p-z)=(1-\eta )f(A-a_1-u_p)$. Using the above we get,

$$\begin{aligned}&f(A-z)\overset{(d)}{\ge } \max \{f(a_1), f(A-u_p-z)\} \nonumber \\&\ge \max \{f(z|A-a_1-u_p-z), f(A-a_1-u_p-z)+ f(a_1|A-a_1-u_p-z)\} \nonumber \\&\ge \max \{\eta f(A-a_1-u_p), (1-\eta )f(A-a_1-u_p)+ f(a_1|A-a_1-u_p)\} \nonumber \\&\ge \max \left\{ \eta , (1-\eta )+ \frac{1}{2}\right\} f(A-a_1-u_p) \end{aligned}$$

(18)

where (d) follows from monotonicity and the fact that $a_1\in A-z$ and $A-u_p-z\subset A-z$. Now, from Lemma 5 with $S=\{a_2,u_1\}$, $l=p-2$, k replaced by $k-1$, N by $N-a_1$ and $s=1$, we have, $f(A-a_1-u_p)\ge \beta (\eta ,\alpha _p)f({ OPT}(k-1,N-a_1,0))\ge \beta (\eta ,\alpha _p){ OPT}$. Substituting this in (18) above we get,

$$\begin{aligned} f(A-z)&\max \left\{ \eta , \frac{3}{2}-\eta \right\} \beta (\eta ,\alpha _p) { OPT} \nonumber \\&\ge \frac{3}{4}\beta \left( \frac{3}{4},\alpha _p\right) { OPT}\quad [\eta =3/4] \nonumber \\&> \beta (0,\alpha _p) { OPT}=\beta \left( 0,\frac{k-4}{k-1}\right) { OPT} \end{aligned}$$

(19)

Case 2 Phase 2 occurs, 3 doesn’t i.e. $p+q=k-2$ and $q>0$.

As stated earlier, during Phase 2, $a_2$ is the minimizer of $A_0\cup U\cup (V-v_q)$. We have $g(A)\ge g(A-v_q)= f(A-v_q-a_2)=f(a_1+U) + f(V-v_q|a_1+U)$. Further, since the addition rule in Phase 2 ignores $a_2$, we have from Lemma 1, $f(V-v_q|a_1+U)\ge \beta (0,\alpha _q)({ OPT}-f(a_1+U))$, and $f(a_1+U)\ge \beta (0,\alpha _p){ OPT}$ follows from the previous case (to see this, suppose that the algorithm was terminated after Phase 1 and note that $z=a_2$ falls under the scenario $z\not \in \{a_1,u_p\}$). Using this,

$$\begin{aligned} g(A-v_q)&\ge f(a_1+U)+\beta (0,\alpha _q)({ OPT}-f(a_1+U)) \nonumber \\ \frac{f(A-z)}{{ OPT}}&\ge (1-\beta (0,\alpha _q))\beta (0,\alpha _p) +\beta (0,\alpha _q) \nonumber \\&= \beta (0,\alpha )= \beta \left( 0,\frac{k-5}{k-1}\right) \end{aligned}$$

(20)

Case 3 Phase 3 occurs i.e., $r>0$.

We consider two sub-cases, $z\in A-W$ and $z\in W$. Suppose $z\in A-W$. Due to $f(z|A-W-z)\le f(A-W)/3$ we have, $f(A-W)\le \frac{3}{2}f(A-W-z)$. Also, $f(W|A-W-z)\ge f(W|A-W)\ge \beta (0,\alpha _r)({ OPT}-f(A-W))$. Then using this along with the previous cases,

$$\begin{aligned} f(A-z)&= f(A-W-z)+ f(W|A-W-z)\\&\ge f(A-W-z)+ \beta (0,\alpha _r)({ OPT}-f(A-W))\\&\ge f(A-W-z)+ \beta (0,\alpha _r)\left( { OPT}-\frac{3}{2}f(A-W-z)\right) \\&\ge \left( 1-\frac{3}{2}\beta (0,\alpha _r)\right) f(A-W-z) + \beta (0,\alpha _r){ OPT}\\&\ge \left( 1-\frac{3}{2}\beta (0,\alpha _r)\right) \beta (0,\alpha _p+\alpha _q) { OPT} + \beta (0,\alpha _r){ OPT}\quad [\text {from } (19),(20)]\\ \frac{f(A-z)}{{ OPT}}&\ge 0.5-\frac{3}{2e^{\alpha }} + \frac{1}{2} (e^{-(\alpha _p+\alpha _q)} +e^{-\alpha _r})\\&\ge 0.5-\frac{3}{2e^\alpha }+e^{-\alpha /2}\quad [\text {for } \alpha _r=\alpha _p+\alpha _q=\alpha /2]\\&= 0.5- \frac{3}{2e^\frac{k-4}{k-1}}+e^{-\frac{k-4}{2(k-1)}}\xrightarrow {k\rightarrow \infty } 0.5547 \end{aligned}$$

Now, suppose $z\in W$, then note that for $p+q\ge 6$ we have either $p\ge 3$, and hence due to greedy additions $f(z|A-z)\le f(\{u_1,u_2,u_3\})/3\le f(A-W)/3$, or $q\ge 3$, and again due to greedy additions $f(z|A-z)\le f(a_1+U\cup \{v_1,v_2\})/3\le f(A-W)/3$.

If $q>0$, then note that $f(A-W)\ge f(A-W-v_q)\ge \frac{3}{2} f(A-W-v_q-a_2)$ due to the Phase 2 termination conditions. Now we reduce the analysis to look like the previous sub-case through the following,

$$\begin{aligned} f(A-z)&= f(A-W)+f(W|A-W)-f(z|A-z)\\&\ge f(A-W)+ \beta (0,\alpha _r) ({ OPT}-f(A-W)) -f(z|A-z)\\&\ge (1-\beta (0,\alpha _r))f(A-W) +\beta (0,\alpha _r){ OPT} -f(A-W)/3\\&\ge \left( 1-\frac{3}{2}\beta (0,\alpha _r)\right) \frac{2}{3}f(A-W) +\beta (0,\alpha _r){ OPT}\\&\ge \left( 1-\frac{3}{2}\beta (0,\alpha _r)\right) f(A-W-v_q-a_2) +\beta (0,\alpha _r){ OPT} \end{aligned}$$

Which since $f(A-W-v_q-a_2)\ge \beta \Big (0,\frac{p+q-3}{k-1}\Big ) { OPT}$ from (20), leads to the same ratio asymptotically as when $z\in A-W$. The case $q=0$ can be dealt with similarly by using $f(A-W)\ge f(A-W-u_p)\ge \frac{3}{2} f(A-W-u_p-a_1)$

If $p+q<6$, then let $f(z|A-z)=\eta f(A)$. Now we have, $f(A-W)\ge f(A_0)=f(a_1)+f(a_2|a_1)\ge 2f(z|A-z)= 2\eta f(A)$, which further implies that $f(A-W+w_1)\ge 3 \eta f(A)$ since $z\in W$. Then proceeding as in Lemma 5 with k replaced by $k-1$, $S=A-W+w_1$ and hence $s=3$ and finally $l=k-|S|=k-(p+q+2+1)\le k-8$ gives us,

$$\begin{aligned} f(A)\ge \beta (3\eta ,\frac{k-8}{k-1})f({ OPT}(k-1,N,0))\ge \beta (3\eta ,\frac{k-8}{k-1}) { OPT} \end{aligned}$$

Then using Lemma 6 we have, $f(A-z)\ge (1-\eta )\beta (3\eta ,\frac{k-8}{k-1}){ OPT} \ge \beta (0,\frac{k-8}{k-1}){ OPT}$. $\square$