Min–Max Partitioning of Hypergraphs and Symmetric Submodular Functions

Abstract

We consider the complexity of minmax partitioning of graphs, hypergraphs and (symmetric) submodular functions. Our main result is an algorithm for the problem of partitioning the ground set of a given symmetric submodular function $f:2^V \rightarrow {\mathbb {R}}$ into k non-empty parts $V_1,V_2,\ldots ,V_k$ to minimize $\max _{i=1}^k f(V_i)$. Our algorithm runs in $n^{O(k^2)}T$ time, where $n=|V|$ and T is the time to evaluate f on a given set; hence, this yields a polynomial time algorithm for any fixed k in the evaluation oracle model. As an immediate corollary, for any fixed k, there is a polynomial-time algorithm for the problem of partitioning a given hypergraph $H=(V,E)$ into k non-empty parts to minimize the maximum capacity of the parts. The complexity of this problem, termed Minmax-Hypergraph- $k$ -Part, was raised by Lawler in 1973 (Networks 3:275–285, 1973). In contrast to our positive result, the reduction in Chekuri and Li (Theory Comput 16(14):1–8, 2020) implies that when k is part of the input, Minmax-Hypergraph- $k$ -Part is hard to approximate to within an almost polynomial factor under the Exponential Time Hypothesis (ETH).

Appendix A: Proof of Theorem 4.1

Our proof strategy for Theorem 4.1 is similar to the proof of Theorem 1.3: The first step shows a containment lemma similar to Lemma 2.1, but for the minsum objective. The second step will use the uncrossing result from Theorem 2.1. We begin with the containment lemma for the minsum objective.

Lemma A.1

Let $f:2^V\rightarrow \mathbb {R}$ be a symmetric submodular function, $k\ge 2$ be an integer, $(V_1, \ldots , V_k)$ be a $V_1$-maximal minsum k-partition with respect to f, and $S\subseteq V_1$, $T\subseteq \overline{V_1}$ such that $T\cap V_j\ne \emptyset$ for every $j\in \{ 2, \ldots , k \}$. Suppose $(U, \overline{U})$ is a minimum (S, T)-terminal cut. Then, $U\subseteq V_1$.

Proof

The proof strategy is identical to the proof of the containment lemma for the minmax objective (i.e., Lemma 2.1). For the sake of contradiction, suppose $U{\setminus } V_1\ne \emptyset$. Consider $W_1:=V_1\cup U$ and $W_j:=V_j-U$ for every $j\in \{ 2,\ldots , k \}$ (see Fig. 1).

Since $W_1\supseteq S\ne \emptyset$ and $W_j\supseteq T\cap V_j\ne \emptyset$ for all $j\in \{ 2,\ldots , k \}$, we have that $(W_1, \ldots , W_k)$ is a k-partition. We note that Claim 2.1 is still applicable for the minsum setting as well. As a consequence of this claim, the sum-objective value of the k-partition $(W_1, \ldots , W_k)$ is at most that of $(V_1, \ldots , V_k)$. Hence, $(W_1, \ldots , W_k)$ is a minsum k-partition. Moreover, $W_1$ is a strict superset of $V_1$ as $U\setminus V_1\ne \emptyset$ and hence, $(W_1, \ldots , W_k)$ contradicts $V_1$-maximality of the minsum k-partition $(V_1, \ldots , V_k)$. $\square$

We are now ready to prove Theorem 4.1.

Theorem 4.1

Let $f:2^{V}\rightarrow \mathbb {R}$ be a symmetric submodular function and let $k\ge 2$ be an integer. Let $(V_1,V_2,\ldots ,V_k)$ be a $V_1$-maximal minsum k-partition such that $V_1$ is the cheapest part. Then, for every subset $T\subseteq \overline{V_1}$ such that $T\cap V_j\ne \emptyset$ for every $j\in \{ 2, \ldots , k \}$, there exists a subset $S\subseteq V_1$ of size at most $k-1$ such that $(V_1, \overline{V_1})$ is the source maximal minimum (S, T)-terminal cut.

Proof

The proof is almost identical to the proof of Theorem 1.3 except for the last but one sentence.

Suppose $|V_1|\le k-1$. Then, we consider $S=V_1$. We have that $|S|\le k-1$ and moreover, using Lemma A.1, we have that $(V_1, \overline{V_1})$ is the source maximal minimum (S, T)-terminal cut for every $T\subseteq \overline{V_1}$ such that $T\cap V_j\ne \emptyset$ for every $j\in \{ 2,\ldots , k \}$, thus proving the theorem. We consider the case of $|V_1|\ge k$ in the rest of the proof.

For the sake of contradiction, suppose that the theorem is false for some subset $T\subseteq \overline{V_1}$ such that $T\cap V_j\ne \emptyset$ for all $j\in \{ 2,\ldots , k \}$. Our proof strategy is to obtain a cheaper k-partition than $(V_1, \ldots , V_k)$, thereby contradicting the optimality of $(V_1, \ldots , V_k)$. Let $OPT_k$ denote the sum-objective value of $(V_1, \ldots , V_k)$. For a subset $X\subseteq V_1$, let $(V_{X}, \overline{V_X})$ be the source maximal minimum (X, T)-terminal cut. By Lemma A.1, we have that $V_X\subseteq V_1$ for all $X\subseteq V_1$.

Among all possible subsets of $V_1$ of size $k-1$, pick a subset S such that $f(V_{S})$ is maximum. Then, by Lemma A.1 and assumption, we have that $V_S\subsetneq V_1$. By source maximality of the minimum (S, T)-terminal cut $(V_S, \overline{V_S})$, we have that $f(V_S)<f(V_1)$. Let $u_1,\ldots , u_{k-1}$ be the vertices in S. Since $V_{S}\subsetneq V_1$, there exists a vertex $u_{k}\in V_1{\setminus } V_{S}$. Let $C:=\{ u_1,\ldots , u_{k} \}=S\cup \{ u_{k} \}$. For $i\in [k]$, let $(B_i, \overline{B_i})$ be the source maximal minimum $(C-\{ u_i \},T)$-terminal cut. We note that $(B_{k}, \overline{B_{k}})=(V_{S}, \overline{V_{S}})$ and the size of $C-\{ u_i \}$ is $k-1$ for every $i\in [k]$. By Lemma 2.1 and assumption, we have that $B_i\subsetneq V_1$ for every $i\in [k]$. Hence, we have

$$\begin{aligned} f(B_i)&\le f(V_{S}) <f(V_1) \text { and } B_i\subsetneq V_1 \text { for every } i\in [k]. \end{aligned}$$

(2)

The next claim will set us up to apply Theorem 2.1.

Claim A.1

For every $i\in [k]$, we have that $u_i\in \overline{B_i}$.

Proof

The claim holds for $i=k$ by choice of $u_{k}$. For the sake of contradiction, suppose $u_i\in B_i$ for some $i\in [k-1]$. Then, the 2-partition $(V_{S}\cap B_i,\overline{V_{S}\cap B_i})$ is a (S, T)-terminal cut while $(V_S, \overline{V_S})$ is a minimum (S, T)-terminal cut and hence

$$\begin{aligned} f(V_{S}\cap B_i)\ge f(V_{S}). \end{aligned}$$

We also have that

$$\begin{aligned} f(V_{S}\cup B_i)\ge f(V_{S}) \end{aligned}$$

since $(V_{S}\cup B_i,\overline{V_S\cup B_i})$ is a (S, T)-terminal cut while $(V_S, \overline{V_S})$ is a minimum (S, T)-terminal cut. Thus,

$$\begin{aligned} 2f(V_S)&\ge f(V_S) + f(B_i) \quad \quad \quad \quad \quad \quad \text { (By choice of { S})}\\&\ge f(V_S \cup B_i) + f(V_S \cap B_i) \quad \quad \text {(By submodularity)}\\&\ge 2f(V_S). \quad \quad \quad \quad \quad \quad \quad \quad \quad \text {(By the inequalities above)} \end{aligned}$$

Therefore, all inequalities above should be equations and hence, $f(V_S\cup B_i)=f(V_S)$. Consequently, the 2-partition $(V_S\cup B_i, \overline{V_S\cup B_i})$ is a minimum (S, T)-terminal cut. However, this contradicts source maximality of the minimum (S, T)-terminal cut $(V_S, \overline{V_S})$ since $u_{k} \in B_i$ and $u_{k} \not \in V_S$. $\square$

We note that for every $i\in [k]$, the 2-partition $(B_i, \overline{B_i})$ is a minimum $(C-\{ u_i \}, \overline{V_1})$-terminal cut since $\overline{V_1}\subseteq \overline{B_i}$.

We will now apply Theorem 2.1. We consider $U:=V_1$ and $C=\{ u_1,\ldots , u_{k} \}\subseteq U$. Let $(\overline{A_i}, A_i):=(B_i, \overline{B_i})$ for every $i\in [k]$. The 2-partition $(\overline{A_i}, A_i)$ is a minimum $(C\setminus \{ u_i \}, \overline{U})$-terminal cut for every $i\in [k]$. By Claim A.1, we have that $u_i\in A_i$ for every $i\in [k]$. Since $(B_j,\overline{B_j})$ is a $(C-\{ u_j \},T)$-terminal cut, we have that $u_i\not \in \overline{B_j}$ for every distinct $i, j\in [k]$. Thus, $u_i\in A_i{\setminus } (\cup _{j\in [k]{\setminus } \{ i \}}A_j)$ for every $i\in [k]$. We may reindex the elements in C so that $f(A_1)\le f(A_2)\le \ldots \le f(A_k)$. Therefore, the sets U, C, and the 2-partitions $(\overline{A_i}, A_i)$ for $i\in [k]$ satisfy the conditions of Theorem 2.1. By Theorem 2.1 and statement (2), we obtain a k-partition $(P_1, \ldots , P_k)$ of V such that

$$\begin{aligned} \sum _{i=1}^k f(P_i)&\le k\max \{ f(A_i):i\in [k] \} = kf(V_S) <kf(V_1) \le \sum _{i=1}^k f(V_i) = OPT_k. \end{aligned}$$

Thus, we have obtained a k-partition whose sum-objective value is strictly smaller than $OPT_k$, a contradiction. $\square$