Sequence independent lifting for a set of submodular maximization problems

Abstract

We study the polyhedral structure of a mixed 0-1 set arising from the submodular maximization problem, given by $P = \{(w,x)\in {\mathbb {R}}\times \{0,1\}^n: w\le f(x), x\in {\mathcal {X}}\}$, where submodular function f(x) is represented by a concave function composed with an affine function, and ${\mathcal {X}}$ is the feasible region of binary variables x. For ${\mathcal {X}}= \{0,1\}^n$, two families of facet-defining inequalities are proposed for the convex hull of P through restriction and lifting using submodular inequalities. When ${\mathcal {X}}$ involves multiple disjoint cardinality constraints, we propose a new class of facet-defining inequalities for the convex hull of P through multidimensional sequence independent lifting. The derived polyhedral results not only strengthen and generalize some existing developments in the literature, but are also linked to the classical results for the mixed 0-1 knapsack and single-node flow sets. Our computational study on a set of randomly generated submodular maximization instances demonstrates the superiority of the proposed facet-defining inequalities within a branch-and-cut scheme.

Appendices

Appendix

A Proof of Theorem 6

Let us first consider the function $\gamma _T(z)$ (defined in equation (22)). As shown in Example 3, $\gamma _T(z)$ is not subadditive on $z\in {\mathbb {R}}_+$ in general for $T \ne \emptyset$. However, we can still find the subadditive structure inside $\gamma _T(z)$ in the following discussion. Consider a new function ${\tilde{\gamma }}_T(z)$ as follows:

$$\begin{aligned}{\tilde{\gamma }}_T(z) = \gamma _T(z + a(T)) - \sum _{j\in T} \rho _j(S\setminus j)\quad \forall z \ge 0.\end{aligned}$$

Recall that we assume $S = \{1, \ldots , s\}$ such that $a_1\ge \cdots \ge a_s$. Let ${\tilde{A}}_k = \sum _{j\in [k]\setminus T} a_j$ for $k\in S$ and ${\tilde{A}}_0 = 0$. Note that if $k \in T$, then ${\tilde{A}}_k = {\tilde{A}}_{k-1}$. Then following Lemma 8, we have ${\tilde{\gamma }}_T({\tilde{A}}_k) = \sum _{j\in [k]\setminus T} \rho _j(S\setminus j)$ and

$$\begin{aligned} {\tilde{\gamma }}_T(z) = {\left\{ \begin{array}{ll} g(a(S) - {\tilde{A}}_{k+1} + z) + {\tilde{\gamma }}_T({\tilde{A}}_{k+1}) - f(S) \quad &{} \text {if } {\tilde{A}}_{k} \le z \le {\tilde{A}}_{k+1}, \\ &{} \;\;k= 0, \ldots , s-2, \\ g(z + a(T)) + {\tilde{\gamma }}_T({\tilde{A}}_{s}) - f(S) \quad &{} \text {if } z \ge {\tilde{A}}_{s-1}. \end{array}\right. } \end{aligned}$$

(29)

Consequently, based on Proposition 4, we can verify that function ${\tilde{\gamma }}_T(z)$ is subadditive on $z \in {\mathbb {R}}_+$ for a given $T\subseteq S$.

To establish the proof of Theorem 6, our basic idea is to exploit inequalities (20) in Lemma 7. To show (20), we first establish that inequalities similar in spirit to (20) hold for $\gamma _T(z)$ in Lemmata 13 and 14, which, in turn, rely on the technical results in Lemmata 11 and 12. Then we use the fact that $\gamma {\left( {\begin{array}{c}z\\ {\mathbf {u}}\end{array}}\right) }$ can be computed through equation (23) and $\gamma _T(z)$ to complete the proof of Theorem 6.

Lemma 11

Function ${\tilde{\gamma }}_T(z)$ is nonincreasing on T, that is ${\tilde{\gamma }}_T(z) \le {\tilde{\gamma }}_{T'}(z) \le {\gamma _0(z)}$ for any $T'\subseteq T \subseteq S.$

Proof

Note that if $T=\emptyset$, then ${\tilde{\gamma }}_\emptyset (z) = \gamma _\emptyset (z) = \gamma _0(z)$. Thus, it suffices to show that ${\tilde{\gamma }}_T(z) \le {\tilde{\gamma }}_{T\setminus j_0}(z)$ for any $j_0\in T$. Let $T' = T\setminus j_0$ and ${\tilde{A}}'_k = \sum _{j\in [k]\setminus T'} a_j$ for $k\in S$.

If $z \le {\tilde{A}}_{j_0-1}$, then ${\tilde{A}}'_k = {\tilde{A}}_k$ for $k=0, \ldots , j_0-1$. By (29), it is easy to verify that ${\tilde{\gamma }}_T(z) = {\tilde{\gamma }}_{T'}(z)$.

If $z > {\tilde{A}}_{j_0-1}$, then ${\tilde{A}}'_k = {\tilde{A}}_k + a_{j_0}$ for $k=j_0, \ldots , s$. By (29), it is easy to verify that ${\tilde{\gamma }}_{T'}(z+a_{j_0}) - {\tilde{\gamma }}_{T'}({\tilde{A}}'_{j_0}) = {\tilde{\gamma }}_T(z) - {\tilde{\gamma }}_T({\tilde{A}}_{j_0-1}).$ Meantime, we can verify that ${\tilde{\gamma }}_{T'}(z)$ has the form of (12) in Sect. 2.3. Then by Lemma 2 we have

$$\begin{aligned} {\tilde{\gamma }}_{T'}(z) - {\tilde{\gamma }}_{T'}({\tilde{A}}'_{j_0-1})&\ge {\tilde{\gamma }}_{T'}(z+a_{j_0}) - {\tilde{\gamma }}_{T'}({\tilde{A}}'_{j_0}) \\&= {\tilde{\gamma }}_T(z) - {\tilde{\gamma }}_T({\tilde{A}}_{j_0-1}), \end{aligned}$$

which implies that ${\tilde{\gamma }}_{T'}(z) \ge {\tilde{\gamma }}_T(z)$ as ${\tilde{\gamma }}_{T'}({\tilde{A}}'_{j_0-1}) = {\tilde{\gamma }}_T({\tilde{A}}_{j_0-1})$. $\square$

Lemma 12

Let $\varDelta \in [0, a_j]$ for some $j\in T$. If $0\le z \le {\tilde{A}}_{j-1}$, then

$$\begin{aligned} \gamma _T(a(T)) - \gamma _T(a(T)-\varDelta ) \le \gamma _T(z+a(T)) - \gamma _T(z+a(T)-\varDelta ). \end{aligned}$$

Proof

Firstly, by Lemma 8, we have

$$\begin{aligned} \gamma _T(a(T)) - \gamma _T(a(T)-\varDelta ) = g(a(S)) - g(a(S)-\varDelta ).\end{aligned}$$

We assume that $z \in [{\tilde{A}}_{k-1}, {\tilde{A}}_{k}]$ for some $k\in S\setminus T$, then $k \le j$ and $a_k \ge a_j$. If $z - \varDelta \le 0$, then $k=1$. Let $\varOmega = a(S) + z \ge a(S)$. By Lemma 8, we have

$$\begin{aligned}&[\gamma _T(z+a(T)) - \gamma _T(z+a(T)-\varDelta )] - [\gamma _T(a(T)) - \gamma _T(a(T)-\varDelta )] \\&\qquad = [g(\varOmega - a_1) - g(\varOmega -\varDelta )] - [g(a(S) - a_1) - g(a(S)-\varDelta )] \\&\qquad \ge 0, \end{aligned}$$

where the inequality follows from $a_1 \ge a_j\ge \varDelta$ and the concavity of g.

If $z - \varDelta \ge 0$, then $\gamma _T(z+a(T)) - \gamma _T(z+a(T)-\varDelta ) ={\tilde{\gamma }}_T(z) - {\tilde{\gamma }}_T(z-\varDelta )$ based on the definition of ${\tilde{\gamma }}_T(z)$. Assume $z - \varDelta \in [{\tilde{A}}_{\ell -1}, {\tilde{A}}_{\ell }]$ for some $\ell \le k$. Since $\varDelta \le a_j \le a_{k}$, then either $\ell = k$ or $\ell =k-1$. If $\ell = k$, then

$$\begin{aligned} {\tilde{\gamma }}_T(z) - {\tilde{\gamma }}_T(z-\varDelta )&= g(a(S) - {\tilde{A}}_{k} + z) - g(a(S) - {\tilde{A}}_{k} + z - \varDelta ) \\&\ge g(a(S)) - g(a(S)-\varDelta ), \end{aligned}$$

where the inequality follows from $z \le {\tilde{A}}_{k}$ and the concavity of g. If $\ell =k-1$, let $\varOmega = a(S) - {\tilde{A}}_{k-1} + z \ge a(S)$. By equation (29), we have

$$\begin{aligned}&[{\tilde{\gamma }}_T(z) - {\tilde{\gamma }}_T(z-\varDelta )] - [\gamma _T(a(T)) - \gamma _T(a(T)-\varDelta )] \\&\quad = [g(\varOmega -a_k) - g(\varOmega - \varDelta )] - [g(a(S) - a_k) - g(a(S) - \varDelta )] \\&\quad \ge 0, \end{aligned}$$

where the inequality follows from $\varDelta \le a_k$ and the concavity of g. $\square$

Lemma 13

For any $T\subseteq S$, we have

$$\begin{aligned} \sum _{j\in T} \gamma _{\{j\}}(z_j) \ge \gamma _T(z(T)) \quad \forall z_j \ge 0, \end{aligned}$$

where $z(T) = \sum _{j\in T} z_j.$

Proof

We prove the result by induction. If $|T| = 1$, then the statement is trivial.

If the statement holds for $|T| = k-1$, then we establish that the statement still holds for $|T| = k$. Observe that it is sufficient to show that there exists some $\ell \in T$ such that

$$\begin{aligned} \gamma _{\{\ell \}}(z_\ell ) + \gamma _{T\setminus \ell }(z(T')) \ge \gamma _T(z(T)), \end{aligned}$$

where $T' = T\setminus \ell$. There are four possible cases to consider:

Case 1 $\exists \; \ell \in T$ such that $z_\ell \ge a_\ell , z(T') \ge a(T')$:

Based on the assumption, we have $z(T) \ge a(T)$, it follows that

$$\begin{aligned} \gamma _{\{\ell \}}(z_\ell ) + \gamma _{T'}(z(T'))&= {\tilde{\gamma }}_{\{\ell \}}(z_\ell - a_\ell ) + {\tilde{\gamma }}_{T'}(z(T') - a(T')) + \sum _{j\in T} \rho _j(S\setminus j) \\&\ge {\tilde{\gamma }}_T(z_\ell - a_\ell ) + {\tilde{\gamma }}_T(z(T') - a(T')) + \sum _{j\in T} \rho _j(S\setminus j) \\&\ge {\tilde{\gamma }}_T(z_\ell +z(T') - a(T)) + \sum _{j\in T} \rho _j(S\setminus j) \\&= \gamma _T(z(T)), \end{aligned}$$

where the first inequality is based on Lemma 11, and the second inequality follows from the fact that ${\tilde{\gamma }}_T(z)$ is subadditive on $z\in {\mathbb {R}}_+$ for a given T.

Case 2 $\exists \; \ell \in T$ such that $z_\ell \le a_\ell , z(T') \le a(T')$:

Based on the assumption, we have $z(T) \le a(T)$. Let $\varOmega = a(S) + z(T') - a(T')$, then $\varOmega \le a(S)$. By Lemma 8, it follows that

$$\begin{aligned}&\gamma _{\{\ell \}}(z_\ell ) + \gamma _{T'}(z(T')) - \gamma _T(z(T))\\&\quad = g(a(S) - a_\ell + z_\ell ) + g(\varOmega ) - g(a(S)) - g(\varOmega - a_\ell + z_\ell ) \\&\quad = [g(a(S) - a_\ell + z_\ell ) - g(a(S))] - [g(\varOmega - a_\ell + z_\ell ) - g(\varOmega )] \\&\quad \ge 0, \end{aligned}$$

where the inequality follows from the concavity of g and $z_\ell - a_\ell \le 0.$

Case 3 $\exists \; \ell \in T$ such that $z_\ell \le a_\ell , z(T') \ge a(T')$:

Based on Lemma 8, we have $\gamma _{\{\ell \}}(z_\ell ) = g(a(S) - a_\ell + z_\ell ) - g(a(S) - a_\ell )$. Without loss of generality, we assume $\ell$ is the smallest index in T. Then there have two cases to consider:

• if $z(T') \ge a(T') + {\tilde{A}}_{\ell -1}$, we have

  $$\begin{aligned}&\gamma _{T'}(z_\ell + z(T')) - \gamma _{T'}(z(T')) \\&\quad = {\tilde{\gamma }}_{T'}(z(T')-a(T')+z_\ell ) - {\tilde{\gamma }}_{T'}(z(T')-a(T')) \\&\quad \le {\tilde{\gamma }}_{T'}(A_{\ell -1} + z_\ell ) - {\tilde{\gamma }}_{T'}( A_{\ell -1}) = \gamma _{\{\ell \}}(z_\ell ), \end{aligned}$$

  where the inequality follows from Lemma 2 that ${\tilde{\gamma }}$ has the form of (11). By Lemma 11, we have that

  $$\begin{aligned} \gamma _{\{\ell \}}(z_\ell ) + \gamma _{T'}(z(T')) \ge \gamma _{T'}(z_\ell +z(T')) \ge \gamma _T(z(T)). \end{aligned}$$

• if $z(T') \le a(T') + {\tilde{A}}_{\ell -1}$, then by Lemma 8 it can be verified that

  $$\begin{aligned} \gamma _T(z(T') + a_\ell ) = \gamma _{T'}(z(T')) + \rho _\ell (S\setminus \ell ). \end{aligned}$$

  Let $\varDelta = a_\ell - z_\ell \ge 0$. Note that $\varDelta \le a_\ell$ and $z(T')-a(T')\le {\tilde{A}}_{\ell -1}$, it follows from Lemma 12 that

  $$\begin{aligned} \gamma _T(z(T') + a_\ell ) - \gamma _T(z(T')+z_\ell )&\ge \gamma _T(a(T)) - \gamma _T(a(T)-\varDelta ) \\&= g(a(S)) - g(a(S)-\varDelta ) \\&= \rho _\ell (S\setminus \ell ) - \gamma _{\{\ell \}}(z_\ell ). \end{aligned}$$

  Replacing $\gamma _T(z(T') + a_\ell )$ with $\gamma _{T'}(z(T')) + \rho _\ell (S\setminus \ell )$ in the above inequality, it yields the desired results.

Case 4 $\exists \; \ell \in T$ such that $z_\ell \ge a_\ell , z(T') \le a(T')$:

Since $z(T') \le a(T')$, then there exists $\ell ' \in T$ such that $z_{\ell '} \le a_{\ell '}$. We can reduce this case to either Case 2 or Case 3.

In summary, the claim holds for $|T| = k$ and we complete the proof. $\square$

Lemma 14

For any $j\in S$ and $z_0, z_1 \ge 0$, we have

$$\begin{aligned}\gamma _0(z_0) + \gamma _{\{j\}}(z_1) \ge \gamma _{\{j\}}(z_0+z_1).\end{aligned}$$

Proof

If $z_1 \ge A_{j-1}$, then $\gamma _{\{j\}}(z) = \gamma _0(z)$ and $\gamma _{\{j\}}(z_0+z_1) = \gamma _0(z_0+z_1)$ based on Lemma 8. It yields that the claim holds due to the subadditivity of $\gamma _0$ in this case.

If $a_{j} \le z_1 \le A_{j-1}$, then $\gamma _{\{j\}}(z_1) = {\tilde{\gamma }}(z_1 - a_{j}) + \rho _{j}(S\setminus j)$. Note that $\gamma _0(z) = {\tilde{\gamma }}_{\{j\}}(z)$ when $z \le A_{j-1}$ through (29), thus

$$\begin{aligned} \gamma _0(z_0) + \gamma _{\{j\}}(z_1)&= \gamma _0(z_0) + \gamma _0(z_1-a_{j}) + \rho _{j}(S\setminus j)\\&\ge \gamma _0(z_0 + z_1-a_{j}) + \rho _{j}(S\setminus j) \\&\ge {\tilde{\gamma }}_{\{j\}}(z_0+z_1-a_{j}) + \rho _{j}(S\setminus j) = \gamma _{\{j\}}(z_0+z_1), \end{aligned}$$

where the second inequality follows from Lemma 11 that ${\tilde{\gamma }}_{\{j\}}(z) \le \gamma _0(z)$.

If $z_1 \le a_{j}$ and $z_0+z_1 \le a_{j}$, then by Lemma 8, we have

$$\begin{aligned} \gamma _{\{j\}}(z_0+z_1) - \gamma _{\{j\}}(z_1)&= g(a(S)-a_{j}+z_0+z_1) - g(a(S)-a_{j}+z_1) \\&\le g(a(S)-a_1 + z_0+z_1) - g(a(S)-a_1+z_1) \\&= \gamma _0(z_0+z_1) - \gamma _0(z_1) \le \gamma _0(z_0), \end{aligned}$$

where the first inequality follows from $a_{j} \le a_1$, and the second inequality follows from the subadditivity of $\gamma _0$.

If $z_1 \le a_{j}$, $z_0+z_1 \ge a_j$ and $z_0 \le A_{j}$, then let $\varDelta = a_{j} - z_1 \le z_0$. By Lemmata 1 and 8, we have that

$$\begin{aligned} \gamma _0(z_0) - \gamma _0(z_0-\varDelta ) \ge \gamma _0(A_j) - \gamma _0(A_j - \varDelta ) = \gamma _{\{j\}}(a_{j}) - \gamma _{\{j\}}(z_1). \end{aligned}$$

Let $z_1'=a_{j}$ and $z_0' = z_0-\varDelta$, then $z_0'+z_1' = z_0 + z_1$. Since $a_{j}\le z_1' \le A_{j}$, based on the aforementioned case, we have

$$\begin{aligned} \gamma _0(z_0-\varDelta ) + \gamma _{\{j\}}(z_1') \ge \gamma _{\{j\}}(z_0+z_1). \end{aligned}$$

Summing the above two inequalities, we get the desired result.

If $z_1 \le a_{j}$ and $z_0\ge A_{j-1}$, then by Lemma 8, we have $\gamma _{\{j\}}(z_1) = \gamma _0(A_{j-1} + z_1) - \gamma _0(A_{j-1})$ and $\gamma _{\{j\}}(z_0+z_1) = \gamma _0(z_0+z_1)$. Then the result directly follows from Lemma 2. $\square$

We now come to prove Theorem 6.

Proof

(Theorem 6) It suffices to show that the lifting function $\gamma {\left( {\begin{array}{c}z\\ {\mathbf {u}}\end{array}}\right) }$ satisfies the SI condition (20). Let $\varGamma \subseteq {\bar{S}}$ and $z_j \ge 0$ for $j\in \varGamma$. Then for any $T\subseteq S$ such that $|T_i| = \max \{0, |S_i|+ |\varGamma _i| -d_i\}$ for all $i\in [r]$, we construct $\{T^j\}_{j\in \varGamma }$ as follows: for each $i\in [r]$, suppose $\varGamma _i = \{1, \ldots , |\varGamma _i|\}$ and $T_i = \{i_1, \ldots , i_{|T_i|}\}$,

• if $|S_i| < d_i$, then $|T_i| < |\varGamma _i|$. Let $T^j = \{i_j\}$ for $j=1,\ldots , |T_i|$, and $T^j = \emptyset$ for $j=|T_i|+1, \ldots , |\varGamma _i|$.

• if $|S_i| = d_i$, then $|T_i| = |\varGamma _i|$. Let $T^j = \{i_j\}$ for $j\in \varGamma _i$.

Therefore, by Lemmata 13 and 14, we have

$$\begin{aligned} \sum _{j\in \varGamma } \gamma _{T^j}(z_j) \ge \gamma _T(z(\varGamma )). \end{aligned}$$

Recall equation (23), we have

$$\begin{aligned}&\gamma {z(\varGamma ) \atopwithdelims ()\sum _{j\in \varGamma } {\mathbf {e}}_{\sigma (j)} } = \max _{T\subseteq S} \{ \gamma _T(z(\varGamma )): |T_i| = \max \{0, |S_i|+ |\varGamma _i| -d_i\}\; \forall i\in [r]\} \\&\quad \le \max _{\{T^j\}_{j\in \varGamma }} \left\{ \sum _{j\in \varGamma } \gamma _{T^j}(z_j): |T^j| = \max \{0, |S_{i}|+1-d_{i}\}, T^j \subseteq S_{i}, i=\sigma (j) \right\} \\&\quad = \sum _{j\in \varGamma } \max _{T^j} \{ \gamma _{T^j}(z_j): |T^j| = \max \{0, |S_{i}|+1-d_{i}\}, T^j \subseteq S_{i}, i=\sigma (j) \}\\&\quad = \sum _{j\in \varGamma }\gamma {\left( {\begin{array}{c}z_j\\ {\mathbf {e}}_{\sigma (j)}\end{array}}\right) }. \square \end{aligned}$$