A spectral partitioning algorithm for maximum directed cut problem

Abstract

We investigate the maximum directed cut (MaxDC) problem by designing a spectral partitioning algorithm. Given a directed graph with nonnegative arc weights, we wish to obtain a bipartition of the vertices such that the total weight of the directed cut arcs is maximized. Relaxing the MaxDC problem as a quadratic program allows us to explore combinatorial properties of the optimal solution, leading to a 0.272-approximation algorithm via the technique of spectral partitioning rounding.

Appendix

We give the proof of Lemma 2. The probability of a cut arc is given by

$$\begin{aligned} \begin{aligned} C(i,j)=&\mathbb {P}\left\{ \left( \begin{array}{c} y_{i} \\ y_{j} \\ \end{array} \right) \in \left\{ \left( \begin{array}{c} \mathrm {sgn}(x_{0}) \\ -\mathrm {sgn}(x_{0}) \\ \end{array} \right) ,\left( \begin{array}{c} -\mathrm {sgn}(x_{0}) \\ \mathrm {sgn}(x_{0}) \\ \end{array} \right) \right\} \right\} \\ =&\mathbb {P}\left\{ x_{0}x_{i}>\sqrt{t},x_{0}x_{j}<-\sqrt{t}~\mathrm {or}~ x_{0}x_{i}<-\sqrt{t},x_{0}x_{j}>\sqrt{t}\right\} \\ =&\mathbb {P}\left\{ \sqrt{t}<x_{0}x_{i},\sqrt{t}<-x_{0}x_{j}~\mathrm {or}~ \sqrt{t}<-x_{0}x_{i},\sqrt{t}<x_{0}x_{j}\right\} \\ =&\mathbb {P}\left\{ \sqrt{t}<\min \{x_{0}x_{i},-x_{0}x_{j}\}~\mathrm {or}~ \sqrt{t}<\min \{-x_{0}x_{i},x_{0}x_{j}\}\right\} .\\ \end{aligned} \end{aligned}$$

(23)

Similarly, the probability of a crossing arc can be obtained by

$$\begin{aligned} \begin{aligned} X(i,j)=&\mathbb {P}\left\{ \left( \begin{array}{c} y_{i} \\ y_{j} \\ \end{array} \right) \in \left\{ \left( \begin{array}{c} \mathrm {sgn}(x_{0}) \\ 0 \\ \end{array} \right) ,\left( \begin{array}{c} 0 \\ -\mathrm {sgn}(x_{0}) \\ \end{array} \right) ,\left( \begin{array}{c} -\mathrm {sgn}(x_{0}) \\ 0 \\ \end{array} \right) ,\left( \begin{array}{c} 0 \\ \mathrm {sgn}(x_{0}) \\ \end{array} \right) \right\} \right\} \\ =&\mathbb {P}\Big \{x_{0}x_{i}>\sqrt{t},|x_{0}x_{j}|\le \sqrt{t}~\mathrm {or}~ |x_{0}x_{i}|\le \sqrt{t},x_{0}x_{j}<-\sqrt{t}\\&~\mathrm {or}~x_{0}x_{i}<-\sqrt{t},|x_{0}x_{j}|\le \sqrt{t}~\mathrm {or}~ |x_{0}x_{i}|\le \sqrt{t},x_{0}x_{j}>\sqrt{t}\Big \}\\ =&\mathbb {P}\Big \{|x_{0}x_{j}|\le \sqrt{t}<x_{0}x_{i}~\mathrm {or}~|x_{0} x_{i}|\le \sqrt{t}<-x_{0}x_{j}\\&~\mathrm {or}~|x_{0}x_{j}|\le \sqrt{t}<-x_{0}x_{i}~\mathrm {or}~|x_{0}x_{i} |\le \sqrt{t}<x_{0}x_{j}\Big \}. \end{aligned} \end{aligned}$$

(24)

In the following, we prove equation (10). Note that the vector x is derived by Algorithm 1. We consider three cases.

• Case 1. If $x_{i}$ and $x_{j}\, (i,j\in \{1,2,\ldots ,n\})$ are both equal to zero, (10) holds obviously.

• Case 2. If exactly one of $x_{i}$ and $x_{j}$ is zero, then for symmetry, assume that $x_{i}$ is zero.

  • Case 2.1. $x_{j}\ne 0$ and $x_jx_0\ge 0$. We obtain $C(i,j)=0$ and $X(i,j)=x_{0}^{2}x_{j}^2$. Then, we have

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)&\ge \beta (1-\beta ) x_{0}^{2}x_{j}^2\\&=\beta (1-\beta ) x_{0}^{2}\left( x_{j}^2-2x_{0}x_{j}\right) +2\beta (1-\beta )x^3_{0}x_{j}\\&\ge \beta (1-\beta ) x_{0}^{2}\left( x_{j}^2-2x_{0}x_{j}\right) -2\beta (1-\beta )x^2_{0}x^2_{j}.\\ \end{aligned} \end{aligned}$$

  • Case 2.2. $x_{j}\ne 0$ and $x_jx_0\le 0$. We easily obtain $C(i,j)=0$ and $X(i,j)=x_{0}^2x_{j}^2$. Since $-x_{0}x_{j}\le x_{j}^{2}$, we get

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)&\ge \beta (1-\beta )x_{0}^{2}x_{j}^2\\&=\beta (1-\beta )x_{0}^{2}(x_{j}^{2}-2x_{0}^{2}x_{j})+2\beta (1-\beta )x_{0}^{3}x_{j}\\&\ge \beta (1-\beta )x_{0}^{2}(x_{j}^{2}-2x_{0}^{2}x_{j})-2\beta (1-\beta )x_{0}^{2}x_{j}^{2}. \end{aligned} \end{aligned}$$

  Equation (10) holds in these cases.

• Case 3. $x_{i}x_{j}\ne 0$. According to the signs and the sizes of $x_{i}$ and $x_{j}$, we consider eight cases in the following.

  • Case 3.1. $x_{0}x_{i}> 0$, $x_{0}x_{j}> 0$, and $|x_{0}x_{j}| \le |x_{0}x_{i}|$ ($x_{0}x_{j}\le x_{0}x_{i}$). Then we have

    $$\begin{aligned} \begin{aligned} C(i,j)=&0,\\ X(i,j)=&\mathbb {P}\left\{ |x_{0}x_{j}|\le \sqrt{t}<x_{0}x_{i}\right\} =x_{0}^2x_{i}^2-x_{0}^2x_{j}^2, \end{aligned} \end{aligned}$$

    and

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)=\,&\beta \left( x_{0}^2x_{i}^2-x_{0}^2x_{j}^2\right) \\ =\,&\beta \left( x_{0}x_{i}+x_{0}x_{j}\right) \left( x_{0}x_{i}-x_{0}x_{j}\right) \\ \ge \,&\beta (1-\beta )\left( x_{0}x_{i}-x_{0}x_{j}\right) ^{2}\\ =\,&\beta (1-\beta )\left( x_{0}^2x_{i}^2+x_{0}^2x_{j}^2-2x_{0}^2x_{i}x_{j}\right) \\ =\,&\beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2+2x_{0}x_{i}-2x_{0}x_{j} -2x_{i}x_{j}\right) \\ \,&-\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) . \end{aligned} \end{aligned}$$

    Since $x_{0}^2\le x_{0}x_{i}+x_{0}x_{j}$, we get

    $$\begin{aligned} \begin{aligned} 2x_{0}^2\left( x_{0}x_{i}-x_{0}x_{j}\right) \le \,&2\left( x_{0}x_{i} +x_{0}x_{j}\right) \left( x_{0}x_{i}-x_{0}x_{j}\right) \\ =\,&2\left( x_{0}^2x_{i}^2-x_{0}^2x_{j}^2\right) =2X(i,j). \end{aligned} \end{aligned}$$

    Therefore,

    $$\begin{aligned} \begin{aligned} -\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \ge&-\beta (1-\beta )2X(i,j)\\ =&-\beta (1-\beta )2X(i,j)-\beta (1-\beta )4C(i,j). \end{aligned} \end{aligned}$$

    Finally,

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)&\ge \beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2+2x_{0}x_{i}-2x_{0}x_{j}-2x_{i}x_{j}\right) \\&\quad -\beta (1-\beta )\left[ 2X(i,j)+4C(i,j)\right] . \end{aligned} \end{aligned}$$

  • Case 3.2. $x_{i}$, $x_{j}$ and $x_{0}$ all have the same sign, and $|x_{0}x_{i}|\le |x_{0}x_{j}|$ ($x_{0}x_{i}\le x_{0}x_{j}$). Then, we have

    $$\begin{aligned} \begin{aligned} C(i,j)&=0,\\ X(i,j)&=\mathbb {P}\left\{ |x_{0}x_{i}|\le \sqrt{t}<x_{0}x_{j}\right\} =x_{0}^2x_{j}^2-x_{0}^2x_{i}^2, \end{aligned} \end{aligned}$$

    and

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)=\,&\beta \left( x_{0}^2x_{j}^2-x_{0}^2x_{i}^2\right) \\ =\,&\beta \left( x_{0}x_{j}+x_{0}x_{i}\right) \left( x_{0}x_{j}-x_{0}x_{i}\right) \\ \ge \,&\beta (1-\beta )\left( x_{0}x_{j}-x_{0}x_{i}\right) ^{2}\\ =\,&\beta (1-\beta )\left( x_{0}^2x_{i}^2+x_{0}^2x_{j}^2-2x_{0}^2x_{i}x_{j}\right) \\ =\,&\beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2+2x_{0}x_{i}-2x_{0}x_{j} -2x_{i}x_{j}\right) \\ \,&-\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) . \end{aligned} \end{aligned}$$

    Since $x_{0}x_{i}-x_{0}x_{j}\le 0$, we have $-\beta (1-\beta )x_{0}^2 \left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \ge 0$. Moreover, $-\left( x_{0}x_{j}+x_{0}x_{i}\right) \le 0$. Therefore

    $$\begin{aligned} \begin{aligned}&-\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \ge 0\\&\quad \ge -\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \left[ -\left( x_{0} x_{i}+x_{0}x_{j}\right) \right] \\&\quad \ge -\beta (1-\beta )2\left( x_{0}^2x_{j}^2-x_{0}^2x_{i}^2\right) \\&\quad =-\beta (1-\beta )2X(i,j)\\&\quad =-\beta (1-\beta )\left[ 2X(i,j)+4C(i,j)\right] . \end{aligned} \end{aligned}$$

    Finally,

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)\ge&\,\beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2+2x_{0}x_{i}-2x_{0}x_{j}-2x_{i}x_{j}\right) \\&-\beta (1-\beta )\left[ 2X(i,j)+4C(i,j)\right] . \end{aligned} \end{aligned}$$

  • Case 3.3 $x_{0}x_{i}< 0$, $x_{0}x_{j}< 0$ and $|x_{0}x_{j}|\le |x_{0}x_{i}|$ ($x_{0}x_{i}\le x_{0}x_{j}$). Then, we have

    $$\begin{aligned} \begin{aligned} C(i,j)&= 0,\\ X(i,j)&= \mathbb {P}\left\{ |x_{0}x_{j}|\le \sqrt{t}<-x_{0}x_{i}\right\} =x_{0}^2x_{i}^2-x_{0}^2x_{j}^2, \end{aligned} \end{aligned}$$

    and

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)&= \beta \left( x_{0}^2x_{i}^2-x_{0}^2x_{j}^2\right) \\&\ge \beta (1-\beta )\left| x_{0}x_{i}+x_{0}x_{j}\right| \left| x_{0}x_{i} -x_{0}x_{j}\right| \\&\ge \beta (1-\beta )\left( x_{0}x_{i}-x_{0}x_{j}\right) ^{2}\\&= \beta (1-\beta )\left( x_{0}^2x_{i}^2+x_{0}^2x_{j}^2-2x_{0}^2x_{i}x_{j}\right) \\&= \beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2+2x_{0}x_{i}-2x_{0}x_{j} -2x_{i}x_{j}\right) \\&\quad -\,\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) . \end{aligned} \end{aligned}$$

    Because of $\left| x_{0}x_{i}-x_{0}x_{j}\right| \le \left| x_{0}x_{i} +x_{0}x_{j}\right|$, the third inequality follows from the second one. Since $\left( x_{0}x_{i}-x_{0}x_{j}\right) \le 0$, we have

    $$\begin{aligned} -\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \ge 0. \end{aligned}$$

    Moreover $\left( x_{0}x_{i}+x_{0}x_{j}\right) \le 0$, we have

    $$\begin{aligned} \begin{aligned}&-\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \ge 0\\&\quad \ge -\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \left( x_{0} x_{i}+x_{0}x_{j}\right) \\&\quad \ge -\beta (1-\beta )2\left( x_{0}^2x_{i}^2-x_{0}^2x_{j}^2\right) \\&\quad =-\beta (1-\beta )\left[ 2X(i,j)+4C(i,j)\right] . \end{aligned} \end{aligned}$$

    Therefore,

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)&\ge \beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2 +2x_{0}x_{i}-2x_{0}x_{j}-2x_{i}x_{j}\right) \\&\quad -\beta (1-\beta )\left[ 2X(i,j)+4C(i,j)\right] . \end{aligned} \end{aligned}$$

  • Case 3.4. $x_{0}x_{i}<0$, $x_{0}x_{j}<0$, and $|x_{0}x_{i}|\le |x_{0}x_{j}|$ ($x_{0}x_{j}\le x_{0}x_{i}$). Then, we have

    $$\begin{aligned} \begin{aligned} C(i,j)&=0,\\ X(i,j)&=\mathbb {P}\left\{ |x_{0}x_{i}|\le \sqrt{t}<-x_{0}x_{j}\right\} =x_{0}^2x_{j}^2-x_{0}^2x_{i}^2, \end{aligned} \end{aligned}$$

    and

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)&=\beta \left( x_{0}^2x_{j}^2-x_{0}^2x_{i}^2\right) \\&\ge \beta (1-\beta )\left| x_{0}x_{i}+x_{0}x_{j}\right| \left| x_{0}x_{i} -x_{0}x_{j}\right| \\&\ge \beta (1-\beta )\left( x_{0}x_{j}-x_{0}x_{i}\right) ^{2}\\&=\beta (1-\beta )\left( x_{0}^2x_{i}^2+x_{0}^2x_{j}^2-2x_{0}^2x_{i}x_{j}\right) \\&=\beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2+2x_{0}x_{i}-2x_{0}x_{j} -2x_{i}x_{j}\right) \\&\quad -\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) . \end{aligned} \end{aligned}$$

    Because of $\left| x_{0}x_{i}-x_{0}x_{j}\right| \le \left| x_{0}x_{i} +x_{0}x_{j}\right|$, the third inequality follows from the second one. Meanwhile, from $x_{0}^2\le -\left( x_{0}x_{i}+x_{0}x_{j}\right)$ and

    $$\begin{aligned} \begin{aligned} x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right)&\le -\left( x_{0}x_{i} +x_{0}x_{j}\right) \left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \\&=2\left( x_{0}^2x_{j}^2-x_{0}^2x_{i}^2\right) \\&=2X(i,j), \end{aligned} \end{aligned}$$

    we have

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)&\ge \beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2+2x_{0}x_{i}-2x_{0}x_{j}-2x_{i}x_{j}\right) \\&\quad -\beta (1-\beta )\left[ 2X(i,j)+4C(i,j)\right] . \end{aligned} \end{aligned}$$

  • Case 3.5. $x_{0}x_{i}> 0$, $x_{0}x_{j}<0$, and $|x_{0}x_{j}|\le |x_{0}x_{i}|$ ($-x_{0}x_{j}\le x_{0}x_{i}$). Then, we have

    $$\begin{aligned} \begin{aligned} C(i,j)=&x_{0}^2x_{j}^2,\\ X(i,j)=&\mathbb {P}\left\{ |x_{0}x_{j}|\le \sqrt{t}<x_{0}x_{i}\right\} =x_{0}^2x_{i}^2-x_{0}^2x_{j}^2. \end{aligned} \end{aligned}$$

    The inequality (Beckenbach and Bellman 1961)

    $$\begin{aligned} (1-\beta )a^2+\beta b^2\ge \beta (1-\beta )(a+b)^{2} \end{aligned}$$

    holds for $a,b\ge 0$, and $0\le \beta \le 1$. Since $x_{0}x_{i}>0$, $-x_{0}x_{j}>0$, we obtain that

    $$\begin{aligned} C(i,j)+\beta X(i,j)&=x_{0}^2x_{j}^2+\beta \left( x_{0}^2x_{i}^2-x_{0}^2x_{j}^2\right) \\&=\beta x_{0}^2x_{i}^2+(1-\beta )x_{0}^2x_{j}^2\\&\ge \beta (1-\beta )\left( x_{0}x_{i}-x_{0}x_{j}\right) ^{2}\\&=\beta (1-\beta )\left( x_{0}^2x_{i}^2+x_{0}^2x_{j}^2-2x_{0}^2x_{i}x_{j}\right) \\&=\beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2+2x_{0}x_{i}-2x_{0}x_{j}-2x_{i}x_{j}\right) \\&\quad -\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) . \end{aligned}$$

    Since $x_{0}x_{i}>0$ and $x_{0}x_{j}<0$, we have $x_{0}x_{i}-x_{0}x_{j}>0$, $x_{0}x_{i}+x_{0}x_{j}>0$, and

    $$\begin{aligned} \begin{aligned} x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right)&\le -x_{0}x_{j}\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \\&=2\left( x_{0}^2x_{j}^2+x_{0}x_{i}(-x_{0}x_{j})\right) \\&\le 2\left( x_{0}^2x_{j}^2+x_{0}^2x_{i}^2)\right) \\&=2\left( x_{0}^2x_{j}^2+x_{0}^2x_{i}^2-x_{0}^2x_{j}^2+x_{0}^2x_{j}^2)\right) \\&=2X(i,j)+4C(i,j). \end{aligned} \end{aligned}$$

    Therefore,

    $$\begin{aligned} \begin{aligned} -\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \ge -\beta (1-\beta )\left[ 2X(i,j)+4C(i,j)\right] . \end{aligned} \end{aligned}$$

    Finally,

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)&\ge \beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2+2x_{0}x_{i}-2x_{0}x_{j}-2x_{i} x_{j}\right) \\&\quad -\beta (1-\beta )\left[ 2X(i,j)+4C(i,j)\right] . \end{aligned} \end{aligned}$$

  • Case 3.6. $x_{0}x_{i}> 0$, $x_{0}x_{j}< 0$, and $|x_{0}x_{i}|\le |x_{0}x_{j}|$ ($x_{0}x_{i}\le -x_{0}x_{j}$). Then, we have

    $$\begin{aligned} \begin{aligned} C(i,j)=&x_{0}^2x_{i}^2,\\ X(i,j)=&\mathbb {P}\left\{ |x_{0}x_{i}|\le \sqrt{t}<-x_{0}x_{j}\right\} =x_{0}^2x_{j}^2-x_{0}^2x_{i}^2, \end{aligned} \end{aligned}$$

    and

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)&= x_{0}^2x_{i}^2+\beta \left( x_{0}^2x_{j}^2-x_{0}^2x_{i}^2\right) \\&=\beta x_{0}^2x_{j}^2+(1-\beta )x_{0}^2x_{i}^2\\&\ge \beta (1-\beta )\left( x_{0}x_{j}-x_{0}x_{i}\right) ^{2}\\&=\beta (1-\beta )\left( x_{0}^2x_{i}^2+x_{0}^2x_{j}^2-2x_{0}^2x_{i}x_{j}\right) \\&=\beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2+2x_{0}x_{i}-2x_{0}x_{j} -2x_{i}x_{j}\right) \\&\quad -\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) . \end{aligned} \end{aligned}$$

    It is easy to get $x_{0}x_{i}-x_{0}x_{j}>0$, $x_{0}x_{i}+x_{0}x_{j}>0$, and

    $$\begin{aligned} \begin{aligned} x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right)&\le x_{0}x_{i}\left( 2x_{0} x_{i}-2x_{0}x_{j}\right) \\&= 2\left( x_{0}^2x_{i}^2+x_{0}x_{i}(-x_{0}x_{j})\right) \\&\le 2\left( x_{0}^2x_{i}^2+x_{0}^2x_{j}^2\right) \\&= 2\left( x_{0}^2x_{i}^2+x_{0}^2x_{j}^2-x_{0}^2x_{i}^2+x_{0}^2x_{i}^2\right) \\&= 2X(i,j)+4C(i,j). \end{aligned} \end{aligned}$$

    Therefore,

    $$\begin{aligned} \begin{aligned} -\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \ge -\beta (1-\beta )\left[ 2X(i,j)+4C(i,j)\right] . \end{aligned} \end{aligned}$$

    Finally,

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)&\ge \beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2 +2x_{0}x_{i}-2x_{0}x_{j}-2x_{i}x_{j}\right) \\&\quad -\beta (1-\beta )\left[ 2X(i,j)+4C(i,j)\right] . \end{aligned} \end{aligned}$$

  • Case 3.7. $x_{0}x_{i}< 0$, $x_{0}x_{j}> 0$, and $|x_{0}x_{j}|\le |x_{0}x_{i}|$ ($x_{0}x_{j}\le -x_{0}x_{i}$). Then, we have

    $$\begin{aligned} \begin{aligned} C(i,j)&=x_{0}^2x_{j}^2,\\ X(i,j)&=\mathbb {P}\left\{ |x_{0}x_{j}|\le \sqrt{t}<-x_{0}x_{i}\right\} =x_{0}^2x_{i}^2-x_{0}^2x_{j}^2, \end{aligned} \end{aligned}$$

    and

    $$\begin{aligned} C(i,j)+\beta X(i,j)&=x_{0}^2x_{j}^2+\beta \left( x_{0}^2x_{i}^2 -x_{0}^2x_{j}^2\right) \\&=\beta x_{0}^2x_{i}^2+(1-\beta )x_{0}^2x_{j}^2\\&\ge \beta (1-\beta )\left( x_{0}x_{i}-x_{0}x_{j}\right) ^{2}\\&=\beta (1-\beta )\left( x_{0}^2x_{i}^2+x_{0}^2x_{j}^2-2x_{0}^2x_{i}x_{j}\right) \\&=\beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2+2x_{0}x_{i}-2x_{0} x_{j}-2x_{i}x_{j}\right) \\&\quad -\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) . \end{aligned}$$

    Since $x_{0}x_{i}+x_{0}x_{j}\le 0$, we have

    $$\begin{aligned}&-\beta (1-\beta ) x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \ge 0\\&\quad \ge -\beta (1-\beta ) x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \left( x_{0} x_{i}+x_{0}x_{j}\right) \\&\quad \ge -\beta (1-\beta )2\left( x_{0}^2x_{i}^2-x_{0}^2x_{j}^2\right) \\&\quad =-\beta (1-\beta )2X(i,j)\\&\quad \ge -\beta (1-\beta )\left[ 2X(i,j)+4C(i,j)\right] . \end{aligned}$$

    Therefore,

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)&\ge \beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2+2 x_{0}x_{i}-2x_{0}x_{j}-2x_{i}x_{j}\right) \\&\quad -\beta (1-\beta )\left[ 2X(i,j)+4C(i,j)\right] . \end{aligned} \end{aligned}$$

  • Case 3.8. $x_{i}x_{0}< 0$, $x_{j}x_{0}> 0$, and $|x_{0}x_{i}| \le |x_{0}x_{j}|$ ($-x_{0}x_{i}\le x_{0}x_{j}$). Then, we have

    $$\begin{aligned} \begin{aligned} C(i,j)&=x_{0}^2x_{i}^2,\\ X(i,j)&=\mathbb {P}\left\{ |x_{0}x_{i}|\le \sqrt{t}<-x_{0}x_{j}\right\} =x_{0}^2x_{j}^2-x_{0}^2x_{i}^2, \end{aligned} \end{aligned}$$

    and

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)&=x_{0}^2x_{i}^2+\beta \left( x_{0}^2x_{j}^2-x_{0}^2 x_{i}^2\right) \\&=\beta x_{0}^2x_{j}^2+(1-\beta )x_{0}^2x_{i}^2\\&\ge \beta (1-\beta )\left( x_{0}x_{j}-x_{0}x_{i}\right) ^{2}\\&=\beta (1-\beta )\left( x_{0}^2x_{i}^2+x_{0}^2x_{j}^2-2x_{0}^2x_{i}x_{j}\right) \\&=\beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2+2x_{0}x_{i}-2x_{0}x_{j}-2 x_{i}x_{j}\right) \\&\quad -\beta (1-\beta )x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) . \end{aligned} \end{aligned}$$

    Since $x_{0}x_{i}+x_{0}x_{j}\ge 0$ and $-\left( x_{0}x_{i}+x_{0}x_{j}\right) \le 0$, we have

    $$\begin{aligned}&-\beta (1-\beta ) x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \ge 0\\&\quad \ge -\beta (1-\beta ) x_{0}^2\left( 2x_{0}x_{i}-2x_{0}x_{j}\right) \left[ -\left( x_{0} x_{i}+x_{0}x_{j}\right) \right] \\&\quad \ge -\beta (1-\beta )2\left( x_{0}^2x_{j}^2-x_{0}^2x_{i}^2\right) \\&\quad =-\beta (1-\beta )2X(i,j)\\&\quad \ge -\beta (1-\beta )\left[ 2X(i,j)+4C(i,j)\right] . \end{aligned}$$

    Therefore,

    $$\begin{aligned} \begin{aligned} C(i,j)+\beta X(i,j)\ge \,&\beta (1-\beta )x_{0}^2\left( x_{i}^2+x_{j}^2+2x_{0} x_{i}-2x_{0}x_{j}-2x_{i}x_{j}\right) \\&-\,\beta (1-\beta )\left[ 2X(i,j)+4C(i,j)\right] . \end{aligned} \end{aligned}$$