A maximum hypergraph 3-cut problem with limited unbalance: approximation and analysis

Abstract

We consider the max hypergraph 3-cut problem with limited unbalance (MH3C-LU). The objective is to divide the vertex set of an edge-weighted hypergraph $H=(V,E,w)$ into three disjoint subsets $V_{1}$, $V_{2}$, and $V_{3}$ such that the sum of edge weights cross different parts is maximized subject to $||V_{i}|-|V_{l}||\le B$ ($\forall i\ne l\in \{1,2,3\}$) for a given parameter B. This problem is NP-hard because it includes some well-known problems like the max 3-section problem and the max 3-cut problem as special cases. We formulate the MH3C-LU as a ternary quadratic program and present a randomized approximation algorithm based on the complex semidefinite programming relaxation technique.

Appendix

In this appendix, we present a detailed proof of Lemma 5. As defined in Sect. 3.2, let $\lambda =\sum \limits _{i<k\in S_j}\mathrm{Re}(Y_{ik})$ and $N_{|S_j|}=\frac{|S_j|(|S_j|-1)}{2}$. Since

$$\begin{aligned} \sum _{i,k\in S_j}<y_i,y_k> \ge {\left\{ \begin{array}{ll} 0,&{}\hbox { if}\ |S_j|\equiv 0\;(\mod 3);\\ 1,&{}\text {otherwise}; \end{array}\right. } \end{aligned}$$

thus $\frac{-|S_j|}{2}\le \lambda \le N_{|S_j|}$.

Based on the range of $\lambda$, we can construct a new form for z in (4):

Lemma 9

$$\begin{aligned} z= {\left\{ \begin{array}{ll} 1,&{}\hbox { iff}\ \lambda \in \left[ \frac{-|S|}{2},\frac{(|S_j|-3)(|S_j|-1)}{2}\right] ;\\ \frac{2(N_{|S_j|}-\lambda )}{3(|S_j|-1)},&{}\text {iff } \lambda \in \left[ \frac{(|S_j|-3)(|S_j|-1)}{2},N_{|S_j|}\right) . \end{array}\right. } \end{aligned}$$

Proof

Consider the third constraint of (4):

$$\begin{aligned} z=\min \left\{ 1,\frac{1}{|S_j|-1}\sum _{i<k\in S_j}\frac{2}{3}(1-\mathrm{Re}({Y}_{ik}))\right\} >0. \end{aligned}$$

If $-|S_j|/2\le \lambda \le (|S_j|-3)(|S_j|-1)/2$, then $[1/(|S_j|-1)]\sum _{i<k}2(1-\mathrm{Re}(Y_{ik}))/3\ge 1$, implying that $z=1$. Otherwise, $[1/(|S_j|-1)]\sum _{i<k}2(1-\mathrm{Re}(Y_{ik}))/3\le 1$, implying that $z=2(N_{|S_j|}-\lambda )/3(|S_j|-1)$. $\square$

Proof of Lemma 5

Let $(Y^{*},x^{*},z^{*})$ be an optimal solution of (4). If $Y^{*}_{ik}<\beta$ for some $i<k\in S_j$, then according to the first constraint of (4), we have that:

$$\begin{aligned} \alpha _{|S_j|}\ge x^{*}\ge \frac{2(1-f(\theta \beta ))}{3}. \end{aligned}$$

Otherwise $Y^{*}_{ik}\ge \beta$, $\forall i<k,$ implying that $Y^{*}$ is feasible to (5) for some $\lambda (Y^{*})\in [-|S|/2,N_{|S_j|}]$. From the second constraint of (4), we have that:

$$\begin{aligned} \alpha _{|S_j|}&\ge x^{*}\\&\ge \frac{1}{\tau _{|S_j|}}\frac{2}{3z^{*}}\sum _{i<k\in S_j}\left( 1-f\left( \mathrm{Re}\left( \theta Y^{*}_{ik}\right) \right) \right) \\&\ge \frac{1}{\tau _{|S_j|}}\frac{2}{3z^{*}}\left( N_{|S_j|}-\lambda \left( Y^{*}\right) \right) \\&\ge \frac{1}{\tau _{|S_j|}}\frac{2}{3z^{*}}q\left( \lambda \left( Y^{*}\right) ,\theta ,\beta \right) \\&=\gamma _{\theta ,\beta }\left( \lambda \left( Y^{*}\right) \right) \\&\ge \gamma ^{*}_{\theta ,\beta }. \end{aligned}$$

$\square$

If we can choose a suitable $\beta \in [-0.5,1)$ for a give $\theta \in (0,1]$ such that $2(1-f(\theta \beta ))/3$ and $\gamma ^{*}_{\theta ,\beta }$ are numerically close, then we will obtain a good lower bound for $\alpha _{|S|}$. Note that $2(1-f(\theta \beta ))/3$ is easy to compute for any given $\theta$ and $\beta$. Thus our main attention is to eliminate the dependence of $\gamma ^{*}_{\theta ,\beta }$ on Y.

By observing the structure of $\gamma ^{*}_{\theta ,\beta }$, it can be derived that the key is to construct the analytic form of $q(\lambda ,\theta ,\beta )$ defined in (5). In the following, we will consider the structural feature of (5).

Lemma 10

There exists a minimal solution $Y^{*}$ of (5) such that every variable $Y^{*}_{ik}$ takes one of the three values, $\{\beta , \nu ,1\}$.

Proof

For every optimal solution $Y^{*}$ of (5), each variable $Y^{*}_{ik}$ may take the value $\beta$, 1, or some value in $(\beta ,1)$. Note that if $Y^{*}_{ik}$ takes value in $(\beta ,1)$, it must satisfy the KKT condition, which means there is a Lagrange multiplier $\mu$ for the first constraint of (5) such that every interior $Y^{*}_{ik}$ satisfies:

$$\begin{aligned} -\frac{\theta }{f^{\prime }(\mathrm{Re}(\theta Y^{*}_{ik}))}=\mu . \end{aligned}$$

Since $f^{\prime }$ is a monotone increasing function, thus the above equation has a unique solution. $\square$

Define a mixed integer and continuous function:

$$\begin{aligned} q_{\theta ,\beta }(\lambda ,\bar{N},\underline{N})=N_{|S_j|}-\bar{N}f(\theta )-\underline{N}f(\theta \beta )-(N_{|S_j|}-\bar{N}-\underline{N})f\left( \theta \frac{\lambda -\bar{N}-\underline{N}\beta }{N_{|S_j|}-\bar{N}-\underline{N}}\right) \end{aligned}$$

Lemma 11

For any $\lambda \in [-|S|/2,N_{|S_j|})$, there exist two non-negative integers $\bar{N}(\lambda )$ and $\underline{N}(\lambda )$, such that

• (i) $\bar{N}(\lambda )+\underline{N}(\lambda )\le N_{|S_j|}-1$;

• (ii) $N_{|S_j|}\beta +\bar{N}(\lambda )(1-\beta )\le \lambda \le N_{|S_j|}-\underline{N}(\lambda )(1-\beta )$;

• (iii) The minimum value of (5) $q(\lambda ,\beta ,\theta )=q_{\theta ,\beta }(\lambda ,\bar{N}(\lambda ),\underline{N}(\lambda )).$

Proof

Suppose that in the optimal solution of program (5), the numbers of variables $\mathrm{Re}(Y_{ik}^{*})$ with value 1 and $\beta$ are denoted by $\bar{N}^{\prime }$ and $\underline{N}^{\prime }$, respectively. It can be derived that $\bar{N}^{\prime }$ and $\underline{N}^{\prime }$ are both nonnegative and $\bar{N}^{\prime }+\underline{N}^{\prime }\le N_{|S_j|}$.

If $\bar{N}^{\prime }+\underline{N}^{\prime }=N_{|S_j|}$, then $\lambda <N_{|S_j|}$ implies that $\bar{N}^{\prime }<N_{|S_j|}$ and $\underline{N}^{\prime }\ge 1$. Let

$$\begin{aligned} \bar{N}(\lambda )=\bar{N}^{\prime }\ge 0,~\underline{N}(\lambda )=\underline{N}^{\prime }-1 \end{aligned}$$

such that $\bar{N}(\lambda )+\underline{N}(\lambda )\le N_{|S_j|}-1$. Then $N_{|S_j|}-\bar{N}(\lambda )-\underline{N}(\lambda )=1$ holds. In addition, it can be obtained from the linear constraint of the program (5) that

$$\begin{aligned} \frac{\lambda -\bar{N}(\lambda )-\underline{N}(\lambda )\beta }{N_{|S_j|}-\bar{N}(\lambda )-\underline{N}(\lambda )}=\beta . \end{aligned}$$

Thus, the optimum value of program (5) is

$$\begin{aligned} q(\lambda ,\theta ,\beta )=N_{|S_j|}-\bar{N}^{\prime }f(\theta )-\underline{N}^{\prime }f(\theta \beta )=q_{\theta ,\beta }(\lambda ,\bar{N}(\lambda ),\underline{N}(\lambda )). \end{aligned}$$

If $\bar{N}^{\prime }+\underline{N}^{\prime }<N_{|S_j|}$, then according to Lemma 10, all the variables $\mathrm{Re}(Y_{ik}^{*})$ equal to

$$\begin{aligned} \nu =\frac{\lambda -\bar{N}^{\prime }-\underline{N}^{\prime }\beta }{N_{|S_j|}-\bar{N}^{\prime }-\underline{N}^{\prime }}\in (\beta ,1). \end{aligned}$$

Now, let

$$\begin{aligned} \bar{N}(\lambda )=\bar{N}^{\prime }\ge 0,~\underline{N}(\lambda )=\underline{N}^{\prime }\ge 0 \end{aligned}$$

such that $\bar{N}(\lambda )+\underline{N}(\lambda )\le N_{|S_j|}-1$. Thus the optimal value function of program (5) an also be expressed as

$$\begin{aligned} q(\lambda ,\theta ,\beta )=q_{\theta ,\beta }(\lambda ,\bar{N}(\lambda ),\underline{N}(\lambda )). \end{aligned}$$

(i) and (iii) have been proved. In order to prove (ii), we notice that it happens in both cases

$$\begin{aligned} \beta \le \frac{\lambda -\bar{N}(\lambda )-\underline{N}(\lambda )\beta }{N_{|S_j|}-\bar{N}(\lambda )-\underline{N}(\lambda )}\le 1. \end{aligned}$$

Thus $N_{|S_j|}\beta +\bar{N}(\lambda )(1-\beta )\le \lambda \le N_{|S_j|}-\underline{N}(\lambda )(1-\beta )$. $\square$

Let

$$\begin{aligned} \gamma _{\theta ,\beta }(\lambda ,\bar{N},\underline{N})= {\left\{ \begin{array}{ll} \frac{2}{3\tau _{|S_j|}}q_{\theta ,\beta }(\lambda ,\bar{N},\underline{N}),&{}\hbox { if}\ \lambda \in \left[ \frac{-|S_j|}{2},\frac{(|S_j|-3)(|S_j|-1)}{2}\right] ;\\ \frac{|S_j|-1}{\tau _{|S_j|(N_{S_j}-\lambda )}}q_{\theta ,\beta }(\lambda ,\bar{N},\underline{N}),&{}\text {if } \lambda \in \left[ \frac{(|S_j|-3)(|S_j|-1)}{2},N_{|S_j|}\right) . \end{array}\right. } \end{aligned}$$

For a given pair of nonnegative integers $\bar{N}$ and $\underline{N}$, define a one-dimensional minimization problem

$$\begin{aligned} \gamma _{\theta ,\beta }(\bar{N},\underline{N})=\min \limits _{\lambda \in [-|S_j|/2,N_{|S_j|})\cap [N_{|S_j|}\beta +\bar{N}(\lambda )(1-\beta ), N_{|S_j|}-\underline{N}(\lambda )(1-\beta )]}\gamma _{\theta ,\beta }(\lambda ,\bar{N},\underline{N}). \end{aligned}$$

(25)

Corollary 3

There is a pair of nonnegative integers $\bar{N}^{*}$ and $\underline{N}^{*}$, such that $\bar{N}^{*}+\underline{N}^{*}\le N_{|S_j|}-1$ and $\gamma ^{*}_{\theta ,\beta }\ge \gamma _{\theta ,\beta }(\bar{N}^{*},\underline{N}^{*})$.

Proof

Let $\lambda ^{*}$ denote the optimal solution of (6), i.e. $\gamma ^{*}_{\theta ,\beta }=\gamma _{\theta ,\beta }(\lambda ^{*})$. According to Lemma 11, there exists a pair of nonnegative integers $\bar{N}^{*}:=\bar{N}(\lambda ^{*})$ and $\underline{N}^{*}:=\underline{N}(\lambda ^{*})$, such that:

$$\begin{aligned} q(\lambda ^{*},\theta ,\beta )=q_{\theta ,\beta }(\lambda ^{*},\bar{N}^{*},\underline{N}^{*}). \end{aligned}$$

By the definition of $\gamma _{\theta ,\beta }(\lambda ,\bar{N},\underline{N})$ and $\gamma _{\theta ,\beta }(\bar{N},\underline{N})$, we get $\gamma ^{*}_{\theta ,\beta }\ge \gamma _{\theta ,\beta }(\bar{N}^{*},\underline{N}^{*})$. $\square$

When we do not know $\bar{N}^{*}$ and $\underline{N}^{*}$, we can enumerate all possible pairs of nonnegative integers ($\bar{N}^{*}$, $\underline{N}^{*}$) with $\bar{N}^{*}+\underline{N}^{*}\le N_{|S_j|}-1$. Without loss of generality, suppose that $\bar{N}^{*}$ and $\underline{N}^{*}$ are known. Then the minimization problem (6) can be reduced to an easier problem (25). For each pair ($\bar{N}^{*}$, $\underline{N}^{*}$), we solve (25) to obtain $\gamma _{\theta ,\beta }(\bar{N}^{*},\underline{N}^{*})$. Then we select $\gamma _{\theta ,\beta }(\bar{N}^{*},\underline{N}^{*})$ among all candidate pairs as

$$\begin{aligned} \gamma ^{*}_{\theta ,\beta }=\min \limits _{(\bar{N}^{*}, \underline{N}^{*})|\bar{N}^{*}\ge 0,\underline{N}^{*}\ge 0,\bar{N}^{*}+\underline{N}^{*}\le N_{|S_j|}-1}\gamma _{\theta ,\beta }(\bar{N}^{*},\underline{N}^{*}). \end{aligned}$$

For any fixed pair ($\bar{N}^{*}$, $\underline{N}^{*}$), we can solve (25) straightforwardly. If $\lambda ^{*}\in [-|S_j|/2,(|S_j|-3)(|S_j|-1)/2]$, then

$$\begin{aligned} \gamma _{\theta ,\beta }(\lambda ,\bar{N}^{*},\underline{N}^{*})=\frac{2}{3\tau _{|S_j|}}q_{\theta ,\beta }(\lambda ,\bar{N}^{*},\underline{N}^{*}). \end{aligned}$$

Note that $q_{\theta ,\beta }(\lambda ,\bar{N}^{*},\underline{N}^{*})$ is decreasing in $\lambda$, implying the same for $\gamma _{\theta ,\beta }(\lambda ,\bar{N}^{*},\underline{N}^{*})$. Thus $\gamma _{\theta ,\beta }(\lambda ,\bar{N}^{*},\underline{N}^{*})$ obtains its minimum value at the upper limit of $\lambda ^{*}$, i.e.,

$$\begin{aligned} \lambda ^{*}=\min \left\{ \frac{(|S_j|-3)(|S_j|-1)}{2},N_{|S_j|}-\underline{N}^{*}(1-\beta )\right\} . \end{aligned}$$

If $\lambda ^{*}\in [(|S_j|-3)(|S_j|-1)/2,N_{|S_j|})$, we have that:

$$\begin{aligned}&\gamma _{\theta ,\beta }(\lambda ,\bar{N}^{*},\underline{N}^{*})=\frac{|S_j|-1}{\tau _{|S_j|}(N_{S_j}-\lambda )}q_{\theta ,\beta }(\lambda ,\bar{N}^{*},\underline{N}^{*})\\&\quad =\frac{|S_j|-1}{\tau _{|S_j|}}\frac{N_{|S_j|}-\bar{N}^{*}f(\theta )-\underline{N}^{*}f(\theta \beta )-(N_{|S_j|}-\bar{N}^{*}-\underline{N}^{*})f\left( \theta \frac{\lambda -\bar{N}^{*}-\underline{N}^{*}\beta }{N_{|S_j|}-\bar{N}^{*}-\underline{N}^{*}}\right) }{N_{|S_j|}-1}. \end{aligned}$$

Let

$$\begin{aligned} x=\frac{\lambda -\bar{N}^{*}-\underline{N}^{*}\beta }{N_{|S_j|}-\bar{N}^{*}-\underline{N}^{*}}\in [\beta ,1], \end{aligned}$$

then

$$\begin{aligned} \gamma _{\theta ,\beta }(\lambda ,\bar{N}^{*},\underline{N}^{*})=h(x):=\frac{|S_j|-1}{\tau _{|S_j|}}\frac{A-f(\theta x)}{B-x}, \end{aligned}$$

where

$$\begin{aligned} A=\frac{N_{|S_j|}-\bar{N}^{*}f(\theta )-\underline{N}^{*}f(\theta \beta )}{N_{|S_j|}-\bar{N}^{*}-\underline{N}^{*}} \end{aligned}$$

and

$$\begin{aligned} B=\frac{N_{|S_j|}-\bar{N}^{*}-\underline{N}^{*}\beta }{N_{|S_j|}-\bar{N}^{*}-\underline{N}^{*}}. \end{aligned}$$

Note that $A,B\ge 1$. Therefore, minimizing $\gamma _{\theta ,\beta }(\lambda )$ over $\lambda \in [N_{|S_j|}\beta +\bar{N}(\lambda )(1-\beta ), N_{|S_j|}-\underline{N}(\lambda )(1-\beta )]\cap [(|S_j|-3)(|S_j|-1)/2,N_{|S_j|})$ is equivalent to minimizing h(x) over

$$\begin{aligned} x\in [\beta ,1]\cap \left[ \frac{(|S_j|-3)(|S_j|-1)/2-\bar{N}^{*}-\underline{N}^{*}\beta }{N_{|S_j|}-\bar{N}^{*}-\underline{N}^{*}},\frac{N_{|S_j|}-\bar{N}^{*}-\underline{N}^{*}\beta }{N_{|S_j|}-\bar{N}^{*}-\underline{N}^{*}}\right) . \end{aligned}$$

Thus h(x) can be calculated numerically.