Geometric Stabbing via Threshold Rounding and Factor Revealing LPs

Abstract

Kovaleva and Spieksma (SIAM J Discrete Math 20(3):48–768, 2006) considered the problem of stabbing a given set of horizontal line segments with the smallest number of horizontal and vertical lines. The standard LP relaxation for this problem is easily shown to have an integrality gap of at most 2 by treating the horizontal and vertical lines separately. However, Kovaleva and Spieksma observed that threshold rounding can be used to obtain an integrality gap of $e/(e-1) \approx 1.58$ which is also shown to be tight. This is one of the rare known examples where the obvious upper bound of 2 on the integrality gap of the standard LP relaxation can be improved. Our goal in this paper is to extend their proof to two other problems where the goal is to stab a set $\mathcal {R}$ of objects with horizontal and vertical lines: in the first problem $\mathcal {R}$ is a set of horizontal and vertical line segments, and in the second problem $\mathcal {R}$ is a set of unit sized squares. The proof of Kovaleva and Spieksma essentially shows the existence of an appropriate threshold which yields the improved approximation factor. We begin by showing that a random threshold picked from an appropriate distribution works. This reduces the problem to finding an appropriate distribution for a desired approximation ratio. In the first problem, we show that the required distribution can be found by solving a linear program. In the second problem, while it seems harder to find the optimal distribution, we show that using the uniform distribution an improved approximation factor can still be obtained by solving a number of linear programs.

Appendices

Appendix A: Analytical Upper Bound for Segment Stabbing

In this section we give an analytical proof that for $\textsc {SegStab}$ threshold rounding yields a solution which is at most 1.935 times the objective value of an optimal solution to the corresponding LP. As we did in Sect. 2.2, we define two functions h and $\rho$. However we choose these functions to be different from the what we chose there. These functions were also derived using the kind of experiments from which derived the functions in Sect. 2.2.

Let us recall our algorithm. We first pick a $\tau$ from a distribution defined in the interval $[\alpha , \beta ]$ where $\alpha = 0.25$ and $\beta = 0.45$ as follows. The density at any $\tau \in [\alpha , \beta ]$ is $\rho (\tau ) = \frac{2(\tau -\alpha )}{(\beta -\alpha )^2}$. We then set $\tau ' = h(\tau ):= 1 - (1-\beta )^2/(1-\tau )$. We note that for any $\tau \in [\alpha , \beta ]$, $\tau ' \ge \tau$ and as $\tau$ increases from $\alpha$ to $\beta$, $\tau '$ decreases from $\gamma := h(\alpha ) = 0.59\overline{6}$ to $\beta$. Finally, we either set ($\tau _1 = \tau$ and $\tau _2 = \tau '$) or ($\tau _1 = \tau '$ and $\tau _2 = \tau$) with equal probability.

The expected contribution of any line $\ell$ with LP value z (i.e., either $x^*_\ell = z$ or $y^*_\ell = z$ depending on whether $\ell$ is horizontal or vertical) is at most $z\cdot \textsc {k}(z)$ where $\textsc {k}(z):= \int _{\alpha }^{\beta } \psi (z, \tau ) \rho (\tau )\, \mathrm{{d}}\tau$ and $\psi (z, \tau ):= g(z, \tau , h(\tau ))$. Our goal now is to show that $\textsc {k}(z)$ is less that 1.935 for all $z \in [0,1]$.

Lemma 5.1

For any $z \in [0,1]$, $\textsc {k}(z) < 1.935$.

Proof

We do a case analysis based on the value of z. The cases below cover all possible values of z in [0, 1] but overlap on the boundaries. For any $\tau \in [\alpha , \beta ]$, we use $\tau '$ as a shorthand for $h(\tau )$.

• Case 1. $z \in [0,\alpha ]$. In this case for any choice of $\tau$, $z \le \tau , \tau '$ and therefore $\psi (z,t) = \frac{1}{2}\left( \frac{1}{1-\tau } + \frac{1}{1-\tau '}\right)$. Thus

  $$\begin{aligned} \textsc {k}(z){} & {} = \int _\alpha ^\beta \psi (z,\tau )\,\rho (\tau )\,\mathrm{{d}}\tau = \int _{\alpha }^\beta \left( \frac{1}{1-\tau } + \frac{1-\tau }{(1-\beta )^2} \right) \cdot \frac{\tau -\alpha }{(\beta - \alpha )^2}\,\mathrm{{d}}\tau \\{} & {} \quad = \int _{\alpha }^\beta \left( A_1\tau ^2 + B_1\tau + C_1 + \frac{D_1}{1-\tau } \right) \mathrm{{d}}\tau , \end{aligned}$$

  where $A_1 = -\frac{1}{(1-\beta )^2(\beta -\alpha )^2}, B_1 = \frac{1+\alpha }{(1-\beta )^2\cdot (\beta -\alpha )^2}, C_1 = -\frac{1}{(\beta -\alpha )^2} - \frac{\alpha }{(1-\beta )^2(\beta -\alpha )^2}$ and $D_1 = \frac{1-\alpha }{(\beta -\alpha )^2}$. Thus,

  $$\begin{aligned} \overline{\psi }(z) = \left[ A_1\frac{\tau ^3}{3} + B_1\frac{\tau ^2}{2} + C_1\tau - D_1 \ln (1-\tau ) \right] _\alpha ^\beta \approx 1.835. \end{aligned}$$

• Case 2. $z \in [\alpha , \beta ]$. In this case,

  $$\begin{aligned} \textsc {k}(z)&= \int _\alpha ^\beta \psi (z,\tau )\,\rho (\tau )\,\mathrm{{d}}\tau = \frac{1}{2}\int _\alpha ^z \left( \frac{1}{z} + \frac{1}{1-\tau }\right) \rho (\tau )\,\mathrm{{d}}\tau \ \\&\quad +\ \frac{1}{2} \int _z^\beta \left( \frac{1}{1-\tau } + \frac{1-\tau }{(1-\beta )^2} \right) \rho (\tau )\,\mathrm{{d}}\tau \end{aligned}$$

  The derivative of the above expression with respect to z (using Leibniz integral rule) is

  $$\begin{aligned}{} & {} \frac{1}{2}\left( \frac{1}{z} + \frac{1}{1-z}\right) \rho (z) -\frac{1}{2}\int _\alpha ^z\frac{\rho (\tau )}{z^2}\,\mathrm{{d}}\tau - \frac{1}{2}\left( \frac{1}{1-z} + \frac{1-z}{(1-\beta )^2} \right) \rho (z) \\{} & {} \quad = \frac{1}{2}\left( \frac{1}{z} - \frac{1-z}{(1-\beta )^2}-\frac{z-\alpha }{2z^2} \right) \rho (z). \end{aligned}$$

  Note that $f(z):=\frac{1}{z} - \frac{1-z}{(1-\beta )^2}-\frac{z-\alpha }{2z^2}$ has derivative $f'(z)= \frac{1}{(1-\beta )^2}-\frac{1}{2z^2}-\frac{\alpha }{z^3}$, which is non-decreasing in z, implying that $f'(z)\le f'(\beta )=\frac{1}{(1-\beta )^2}-\frac{2\alpha +\beta }{2\beta ^3}\approx -1.907$, for $z\in [\alpha ,\beta ]$. Hence, f(z) is decreasing with $f(\alpha )=\frac{1}{\alpha }-\frac{1-\alpha }{(1-\beta )^2}\approx 1.521$ and $f(\beta )=\frac{\alpha +\beta }{2\beta ^2}-\frac{1}{1-\beta }\approx -0.09$, which in turn implies that $\textsc {k}(z)$ has a unique maximum in $[\alpha ,\beta ]$ at the point $z_0\in [\alpha ,\beta ]$ satisfying $f(z_0)=0$. It can be be verified that $z_0\approx 0.414$. Thus, $\textsc {k}(z) \le \textsc {k}(z_0)$ for $z\in [\alpha ,\beta ]$. Now,

  $$\begin{aligned} \textsc {k}(z_0)&= \frac{1}{2}\int _\alpha ^{z_0} \left( \frac{1}{z_0} + \frac{1}{1-\tau }\right) \rho (\tau )\,\mathrm{{d}}\tau +\ \frac{1}{2} \int _{z_0}^\beta \left( \frac{1}{1-\tau } + \frac{1-\tau }{(1-\beta )^2} \right) \rho (\tau )\,\mathrm{{d}}\tau \\&= \int _\alpha ^{z_0} \left( \frac{1}{z_0} + \frac{1}{1-\tau }\right) \cdot \frac{\tau -\alpha }{(\beta -\alpha )^2}\,\mathrm{{d}}\tau \\&\quad +\ \int _{z_0}^\beta \left( \frac{1}{1-\tau } + \frac{1-\tau }{(1-\beta )^2} \right) \cdot \frac{\tau -\alpha }{(\beta - \alpha )^2}\,\mathrm{{d}}\tau \\&=\int _\alpha ^{z_0} \left( A_2\tau + B_2 + \frac{C_2}{1-\tau }\right) \mathrm{{d}}\tau \\&\quad +\ \int _{z_0}^\beta \left( A_1\tau ^2 + B_1\tau + C_1 + \frac{D_1}{1-\tau } \right) \mathrm{{d}}\tau , \end{aligned}$$

  where $A_2 =\frac{1}{z_0\cdot (\beta -\alpha )^2}, B_2 = - \frac{\alpha }{z_0 \cdot (\beta -\alpha )^2} - \frac{1}{(\beta -\alpha )^2}$, $C_2 =\frac{1-\alpha }{(\beta -\alpha )^2}$, and $A_1$, $B_1$, $C_1$, $D_1$ are as above. Thus

  $$\begin{aligned} \textsc {k}(z_0){} & {} = \left[ A_2\frac{\tau ^2}{2} + B_2\tau - C_2\ln (1-\tau )\right] _\alpha ^{z_0}+\left[ A_1\frac{\tau ^3}{3} + B_1\frac{\tau ^2}{2} \right. \\{} & {} \quad \left. + C_1\tau - D_1 \ln (1-\tau ) \right] _{z_0}^\beta \approx 1.9347. \end{aligned}$$

  This implies that $\textsc {k}(z) < 1.935$ for all $z \in [\alpha , \beta ]$.

• Case 3. $z \in [\beta , \gamma ]$. In this case,

  $$\begin{aligned} \textsc {k}(z)= & {} \int _\alpha ^\beta \psi (z,\tau )\,\rho (\tau )\,\mathrm{{d}}\tau \\= & {} \frac{1}{2}\int _\alpha ^{h(z)} \left( \frac{1}{z} + \frac{1}{1-\tau }\right) \rho (\tau )\,\mathrm{{d}}\tau \ +\ \frac{1}{2}\int _{h(z)}^\beta \frac{2}{z}\, \rho (\tau )\,\mathrm{{d}}\tau . \end{aligned}$$

  Let $w = h(z) \in [\alpha , \beta ]$. In terms of w, the above expression is

  $$\begin{aligned} \int _\alpha ^{w} \left( \frac{1}{h(w)} + \frac{1}{1-\tau }\right) \rho (\tau )\,\mathrm{{d}}\tau \ +\ \int _{w}^\beta \frac{2}{h(w)} \rho (\tau )\,\mathrm{{d}}\tau \end{aligned}$$

  The derivative of the above expression with respect to w is

  $$\begin{aligned}&\left( \frac{1}{h(w)} + \frac{1}{1-w}\right) \rho (w)-\frac{h'(w)}{h^2(w)}\int _\alpha ^w\rho (t)\, \mathrm{{d}}\tau \\ \nonumber&\quad -\frac{2}{h(w)}\rho (w)-\frac{2h'(w)}{h^2(w)}\int _w^\beta \rho (t)\, \mathrm{{d}}\tau \end{aligned}$$

  (44)
  $$\begin{aligned}&=\left[ \frac{1}{1-w} - \frac{1}{h(w)}+\frac{h'(w)}{2h^2(w)}(w-\alpha )\right] \rho (w)-\frac{2h'(w)}{h^2(w)}, \end{aligned}$$

  (45)

  where $h'$ denotes the derivative of h. It can be verified that the right-hand side of (44)–(45), call it $\sigma (w)$ is decreasing in $w\in [\alpha ,\beta ]$, and hence has value at least $\sigma (\beta )\approx 0.898$. This implies that the integral is maximized when $w = \beta$, i.e., when $z = \beta$. Similar to the previous case, we can estimate the integral by

  $$\begin{aligned} \textsc {k}(\beta )=\left[ A_3\frac{\tau ^2}{2} + B_3\tau - C_3\ln (1-\tau )\right] _\alpha ^{\beta } \approx 1.927, \end{aligned}$$

  where $A_3 =\frac{1}{\beta \cdot (\beta -\alpha )^2}, B_3 = - \frac{\alpha }{\beta } - \frac{1}{(\beta -\alpha )^2}$ and $C_3 =\frac{1-\alpha }{(\beta -\alpha )^2}$. This implies that $\psi (z) < 1.927$ for all $z \in [\beta , \gamma ]$.

• Case 4. $z \in [\gamma , 1]$. In this case,

  $$\begin{aligned} \textsc {k}(z) = \int _\alpha ^\beta \psi (z,\tau )\,\rho (\tau )\,\mathrm{{d}}\tau = \int _{\alpha }^{\beta } \frac{1}{z} \cdot \rho (\tau ) \,\mathrm{{d}}\tau \end{aligned}$$

  which is clearly maximized when $z = \gamma$ and $\textsc {k}(\gamma ) = 1/\gamma \approx 1.68$. Thus, $\textsc {k}(z) \le 1.7$ for all $z \in [\gamma ,1]$.

The lemma follows. $\square$

Appendix B: Limitation of Our Analysis for $\textsc {SegStab}$

In this section, we argue that our analysis in Sect. 2 cannot be improved significantly. We do this by showing that there exist $x^*$ and $y^*$ s.t. no matter what $\tau _1$ and $\tau _2$ are chosen, the size of the solution output by the algorithm is at least $1.89\,(\sum _{i}x^*_i + \sum _j y^*_j )$. Our choice of $x^*$ and $y^*$ will in fact identical, i.e., they will have the same set of values. We will thus only state what the values in $x^*$ are.

The values in $x^*$ are in (0, 0.5) and their distribution is chosen according to a density function f defined below (see Fig. 8). In other words, the number of values in $x^*$ that lie between a and $a+\delta$ for any $a \in (0,0.5-\delta )$ is proportional to $f(a) \cdot \delta$ for some sufficiently small $\delta$. The function f is defined in [0, 1] as follows:

$$\begin{aligned} f(x) =\left\{ \begin{array}{ll} 0.5 &{}\text { if } x \in (0, 0.2),\\ 18.75x - 3.25 &{}\text { if } x \in [0.2, 0.4),\\ 4.25 &{}\text { if } x \in [0.4, 0.5),\\ 0 &{}\text { otherwise }. \end{array} \right. \end{aligned}$$

Fig. 8

The density function f used in the tightness example

More specifically, $x^*$ consists of $\lfloor N\cdot f(\frac{i}{N}) \rfloor$ numbers uniformly distributed in the interval $(\frac{i}{N}, \frac{i+1}{N})$ for each $i \in \{0, \dots , N-1\}$, where N is a sufficiently large integer. For any choice of $\tau _1, \tau _2 \in (0,1)$, the weight of the solution returned by the algorithm is

$$\begin{aligned} W&= |\{i: x^*_i> \tau _1\}| + |\{j: x^*_j > \tau _2\}| + \frac{1}{1-\tau _1} \sum _{j:x^*_j \le \tau _2} x^*_j + \frac{1}{1-\tau _2}\sum _{i:x^*_i \le \tau _1} x^*_i\\&= N^2 \cdot \left[ \int _{\tau _1}^1 f(x)\,\mathrm{{d}}x \ +\ \int _{\tau _2}^1 f(t)\,\mathrm{{d}}t + \frac{1}{1-\tau _1} \int _0^{\tau _2} tf(t)\,\mathrm{{d}}t + \frac{1}{1-\tau _2}\int _0^{\tau _1} tf(t)\,\mathrm{{d}}t \right] \\&\quad \pm O(N). \end{aligned}$$

Recall that $y^*$ is the same as $x^*$. Since the “LP-solution" $(x^*,y^*)$ has value $V = 2\sum _{i} x^*_i = 2N^2 \int _{0}^1 t f(t)\,\mathrm{{d}}t \ \pm O(N)$, the ratio W/V can be made arbitrarily close to

$$\begin{aligned} \gamma (\tau _1, \tau _2){} & {} = \frac{1}{2\int _0^1 tf(t)\,\mathrm{{d}}t} \cdot \left[ \int _{\tau _1}^1 f(t)\,\mathrm{{d}}t \ +\ \int _{\tau _2}^1 f(t)\,\mathrm{{d}}t + \frac{1}{1-\tau _1} \int _0^{\tau _2} tf(t)\,\mathrm{{d}}x \right. \\{} & {} \quad \left. + \frac{1}{1-\tau _2}\int _0^{\tau _1} tf(t)\,\mathrm{{d}}t \right] \end{aligned}$$

by choosing a sufficiently large N. It can be shown that $\gamma (\tau _1, \tau _2) \ge 1.89$ for any $\tau _1, \tau _2 \in [0,1]$. We skip the technical proof since it does not yield much insight and since this can be easily checked numerically.

Remark 5.2

What we have shown above is that our method of analysis cannot yield a bound better than 1.89. Note however that this is not necessarily a lower bound on the integrality gap of the LP-relaxation or even on the approximation factor of our algorithm since the $x^*$ we used above need not correspond to the solution of the LP-relaxation for a valid instance of the problem. The currently best lower bound remains the one of $e/(e-1)$ shown in [25].

Appendix C: Proof of Lemma 3.9

Proof

We proceed as in lemma 3.4, we set $k=6$ and consider the dual LP (35)–(38). Again, it is enough to show that $\gamma ^*\le \frac{8}{5}$; we do this by constructing a feasible dual solution with $\gamma =\frac{8}{5}$. Note that, for any given value of $\beta$ satisfying (37) and (38), setting $\gamma :=\max _{i\in V}\sum _{I\in {{\mathcal {I}}}_\ell :~i\in I}\beta _I$ gives a feasible dual solution.

Let ${{\mathcal {I}}}_1:=\{I_1\}$, ${{\mathcal {I}}}_2=\{I_2,I_3\}$, ${{\mathcal {I}}}_3=\{I_4,I_5,I_6\}$, ${{\mathcal {I}}}_4=\{I_7,I_8,I_9,I_{10}\}$, and ${{\mathcal {I}}}_5:=\{I_{11},I_{12},I_{13},I_{14},I_{15}\}$, where intervals in a given ${{\mathcal {I}}}_\ell$ are ordered by their left endpoints. For simplicity, we sometimes write $\beta _j:=\beta _{I_j}$. We consider a number of cases:

• Case 1. There exists $I\in {{\mathcal {I}}}_2$ such that $I\cap I_1=\emptyset$. W.l.o.g., $I=I_2$. As $|{{\mathcal {I}}}_5|=5$, there also exists $I\in {{\mathcal {I}}}_5$ such that $I\cap I_1=\emptyset$ and $I\cap I_2=\emptyset$. In this case, we assign $\beta _1=\beta _2=\beta _I:=1$, $\beta _3:=0$, $\beta _4=\beta _5=\beta _6:=\frac{1}{3}$, and $\beta _7=\beta _8=\beta _9=\beta _{10}:=\frac{1}{4}$, and $\beta _{I'}=0$ for $I'\in {{\mathcal {I}}}_5{\setminus }\{I\}$. Then $\gamma \le 1+\frac{1}{3}+\frac{1}{4}=\frac{19}{12}<\frac{8}{5}$.

Thus we may assume in the following cases that $I_1\cap I_2\ne \emptyset$ and $I_1\cap I_3\ne \emptyset$. Furthermore, there exists $I\in {{\mathcal {I}}}_3$ such that $I\cap I_1=\emptyset$. W.l.o.g., $I=I_6$ (either $I=I_4$ or $I=I_6$; if $I=I_4$ we get a symmetric case).

In the following, we fix $\beta _1:=1$, $\beta _2=\beta _3:=\frac{1}{2}$, $\beta _4=\beta _5:=\frac{1}{10}$, and $\beta _6:=\frac{4}{5}$. We will argue that it is always possible to complete this assignment of the dual variables such that (37) holds and $\gamma \le \frac{8}{5}$. In fact, we will argue this can be done without assigning positive values to more than two variables at level 4 (i.e., corresponding to intervals in ${{\mathcal {I}}}_4$) and no more than three variables at level 5. We begin by assigning $\beta _I:=0$ for all $I\in {{\mathcal {I}}}_4\cup {{\mathcal {I}}}_5$ such that $I\cap I_1\ne \emptyset$. Note that at most two variables at level 4 and at most two variables at level 5 get assigned 0 this way (Fig. 13).

Fig. 9

Partial assignment before considering Case 2 in Claim 3.4. Numbers in red indicate the current partial assignment on the (variables corresponding to the) intervals. Numbers on the top indicate the (current) conservative depth $\bar{d}(i)$; for instance, if any of the assigned intervals, say $I_5$, is moved left or right, the current depth at any point remains within the conservative depth

Fig. 10

Assignment for Case 2.1 in Claim 3.4. Numbers in red indicate the current partial assignment on the (variables corresponding to the) intervals. Numbers on the top indicate the (current) conservative depth ${\bar{d}}(i)$. Numbers in blue indicate the completed assignment

Fig. 11

Assignment for Case 2.2 in Claim 3.4

Fig. 12

Assignment for Case 3.1 in Claim 3.4

Fig. 13

Assignment for Case 3.2 in Claim 3.4

For a given point $i\in V$, we denote by

$$\begin{aligned} d(i):=\sum _{I~:~\beta _I\text { has been assigned },~i\in I}\beta _I \end{aligned}$$

the current depth of i. For an interval I, define further $d(I):=\max _{i\in I}d(i)$, and call it the current depth of I. We also define the conservative depth of $i\in V$, denoted by ${\bar{d}}(i)$, as an upper bound on the depth of i, implied by the current assignment (which takes into consideration the different possible locations of the intervals at levels 1, 2 and 3).

• Case 2. $I_3\cap I_6\ne \emptyset$. The current configuration and partial assignment imply that there exist at most one unassigned interval $I\in {{\mathcal {I}}}_4$ and one unassigned interval $J\in {{\mathcal {I}}}_5$ such that $I\cap I_3\cap I_6\ne \emptyset$, and $J\cap I_3\cap I_6\ne \emptyset$. (If there are two intervals, say in ${{\mathcal {I}}}_4$, that overlap with $I_3\cap I_6$, then, by the assumptions that $I_1\cap I_3\ne \emptyset$ and $I_3\cap I_6\ne \emptyset$, the left-most one of these two intervals would overlap with $I_1$ and hence its corresponding dual variable would have already been set to 0.) Observe that, up to this point, we have at least two unassigned intervals $I,I'\in {{\mathcal {I}}}_4$, and at least three unassigned intervals $J,J',J''\in {{\mathcal {I}}}_5$. Among the intervals $I,I'$ (resp., $J,J',J"$), if there is one overlapping with $I_3\cap I_6$, we assume it is I (resp., J). It follows also from our current assumptions that $I_3$ does not overlap with any of the intervals $I'$, $J'$ and $J''$, and that either one of the latter two intervals, say $J'$, does not overlap with any of $I_1$, $I_3$ and $I_6$. Figure 9 shows the current configuration and partial assignment. We consider two further cases:

  • Case 2.1. $I\cap I_6\ne \emptyset$ (Fig. 10). Then, it is easy to verify that $d(I)\le 1.3$, $d(I')\le 0.8$, $d(J)\le 1.3$, $d(J')\le 0.7$ and $d(J'')\le 1.5$. In this case, setting $\beta _I:=0.3$, $\beta _{I'}:=0.7$, $\beta _{J}:=0$, $\beta _{J'}:=0.9$, and $\beta _{J''}:=0.1$, would give $\gamma \le 1.6$.

  • Case 2.2. $I\cap I_6=\emptyset$ (Fig. 11). Then, it is easy to verify that $d(I)\le 0.6$, $d(I')\le 0.8$, $d(J)\le 1.3$, $d(J')\le 0.6$ and $d(J'')\le 0.8$. In this case, setting $\beta _I:=0.5$, $\beta _{I'}:=0.5$, $\beta _{J}:=0.3$, $\beta _{J'}:=0.4$, and $\beta _{J''}:=0.3$, would give $\gamma \le 1.6$.

• Case 3. $I_3\cap I_6=\emptyset$. Up to this point, we have at least two unassigned intervals $I,I'\in {{\mathcal {I}}}_4$, and at least three unassigned intervals $J,J',J''\in {{\mathcal {I}}}_5$. Among the latter three intervals, at least two, say $J',J''$ overlap with neither $I_1$ nor $I_3.$ We consider two cases:

  • Case 3.1. Either $J'$ or $J''$ does not overlap with $I_6$. W.l.o.g., $J'\cap I_6=\emptyset$ (Fig. 12). Then, it is easy to verify that $d(I)\le 0.8$, $d(I')\le 0.8$, $d(J)\le 0.8$, $d(J')\le 0.8$ and $d(J'')\le 0.6$, and thus, setting $\beta _I=\beta _{I'}:=0.5$, $\beta _{J}=0.2$, $\beta _{J'}:=0.5$, and $\beta _{J''}:=0.3$, would give $\gamma \le 1.6$.

  • Case 3.2. Both $J'$ and $J''$ overlap with $I_6$. Then J does not overlap with $I_6$. Then, it is easy to verify that $d(I)\le 0.8$, $d(I')\le 0.8$, $d(J)\le 0.6$, $d(J')\le 0.8$ and $d(J'')\le 0.8$, and thus, setting $\beta _I=0.4$, $\beta _{I'}:=0.6$, $\beta _{J}=0.6$, $\beta _{J'}=\beta _{J''}:=0.2$, would give $\gamma \le 1.6$. $\square$