On the effect of the deployment setting on broadcasting in Euclidean radio networks

Abstract

The paper studies broadcasting in radio networks whose stations are represented by points in the Euclidean plane (each station knows its own coordinates). In any given time step, a station can either receive or transmit. A message transmitted from station v is delivered to every station u at distance at most 1 from v, but u successfully hears the message if and only if v is the only station at distance at most 1 from u that transmitted in this time step. A designated source station has a message that should be disseminated throughout the network. All stations other than the source are initially idle and wake up upon the first time they hear the source message. It is shown in [17] that the time complexity of deterministic broadcasting algorithms depends on two parameters of the network, namely, its diameter (in hops) D and a lower bound d on the Euclidean distance between any two stations. The inverse of d is called the granularity of the network, denoted by g. Specifically, the authors of [17] present a deterministic broadcasting algorithm that works in time O (Dg) and prove that every broadcasting algorithm requires $\varOmega \left( D \sqrt{g} \right)$ time. In this paper, we distinguish between the arbitrary deployment setting, originally studied in [17], in which stations can be placed everywhere in the plane, and the new grid deployment setting, in which stations are only allowed to be placed on a d-spaced grid. Does the latter (more restricted) setting provide any speedup in broadcasting time complexity? Although the O (Dg) broadcasting algorithm of [17] works under the (original) arbitrary deployment setting, it turns out that the $\varOmega \left( D \sqrt{g} \right)$ lower bound remains valid under the grid deployment setting. Still, the above question is left unanswered. The current paper answers this question affirmatively by presenting a provable separation between the two deployment settings. We establish a tight lower bound on the time complexity of deterministic broadcasting algorithms under the arbitrary deployment setting proving that broadcasting cannot be completed in less than $\varOmega (D g)$ time. For the grid deployment setting, we develop a deterministic broadcasting algorithm that runs in time $O \left( D g^{5 / 6} \log g \right)$, thus breaking the linear dependency on g.

Appendix 1: Proving Lemma 3

Our goal in this section is to prove Lemma 3 that we restate here for clarity of the exposition. Recall that $\bar{y} = 1/\sqrt{2}$ and fix two reals $0< \rho ,\tau < \bar{y}$, where $\rho \le \tau /4$. Let $r_1$, $r_2$, and $r_3$ be three points in $[0, \bar{y}]^2$ such that $x (r_1) + \tau \le x (r_2) \le x (r_3) - \tau$. It is clear that $\Vert {r_i, r_j} \Vert \le 1$ for every $i, j \in \{ 1, 2 ,3 \}$. Let $F _i$ (respectively, $\tilde{ F }_i$) denote the disk of radius 1 (resp., of radius $1 + \rho$) centered at $r_i$ and let $\mathcal{F}_i$ (resp., $\tilde{\mathcal{F}}_i$) be the intersection of $F _i$ (resp., $\tilde{ F }_i$) with the halfplane $y > \bar{y}$ for every $i = 1, 2, 3$. Define $V = \mathcal{F}_1 \cup \mathcal{F}_2 \cup \mathcal{F}_3$. For ease of reference, the reader may use Table 4 which summarizes the notations used throughout this proof.

Lemma 4

Suppose that $\tilde{\mathcal{F}}_2 - V \ne \emptyset$ and $(\tilde{\mathcal{F}}_1 \cap \tilde{\mathcal{F}}_3) - V \ne \emptyset$. Then there exist some reals $x_{\ell }, x_{r}$ such that (1) $x_{\ell } \le x_r \le x_{\ell } + 40 \pi \rho / \tau$; (2) $x \left( s\right) \in [x_\ell ,x_r]$ for all points $s \in \tilde{\mathcal{F}}_2 - V$; and (3) $x_{\ell } = x_{\ell }(r_1, r_3)$ and $x_{r} = x_{r}(r_1, r_3)$, i.e., $x_{\ell }$ and $x_{r}$ are determined independently of $r_2$.

Table 4 Notations for Appendix 1

Proof overview The proof of this lemma is organized as follows. First, in “Disk notation and definitions” section, we present some basic geometric definitions and notations. Then, in “Two disk toolbox” section, we develop a “two disk toolbox”, i.e., some technical geometric properties of two disks in the Euclidean plane. “Weaker version of the Lemma” section is dedicated to proving a “weaker” version of the lemma in which $x_\ell$ and $x_r$—denoted in that section by $\hat{x}_\ell$ and $\hat{x}_r$—depend also on $r_{2}$, rather than being fully determined by $r_{1}$ and $r_{3}$ as promised by Lemma 3. For these two reals, we show that

• (1) $\hat{x}_{\ell } \le \hat{x}_r \le \hat{x}_{\ell } + 16 \pi \rho / \tau$; and

• (2) $x(s) \in [\hat{x}_{\ell }, \hat{x}_{r}]$ for every $s \in \tilde{\mathcal{F}}_2 - V$.

Finally, in “Full version of the lemma” section, we prove the “full” version of the lemma. To do so we determine the real $x_{mid} = x_{mid}(r_{1}, r_{3})$ independently of $r_2$ and show that the distance of $x_{mid}$ from the interval $[\hat{x}_{\ell }, \hat{x}_{r}]$ is at most $4 \pi \rho / \tau$, that is, $\min \left\{ |x_{mid} - z| : z \in [\hat{x}_{r}, \hat{x}_{\ell }] \right\} \le 4 \pi \rho / \tau$. Then, we set $x_{\ell } = x_{mid} - 20 \pi \rho / \tau$ and $x_{r} = x_{mid} + 20 \pi \rho / \tau$, thus ensuring that $[\hat{x}_{\ell }(r_1, r_2, r_3), \hat{x}_{r}(r_1, r_2, r_3)] \subseteq [x_{\ell }(r_1, r_3), x_r(r_1, r_3)]$ which implies, in particular, that $x(s) \in [x_{\ell }, x_{r}]$ for every $s \in \tilde{\mathcal{F}}_2 - V$ due to the weaker version of the lemma established in “Weaker version of the Lemma” section. Lemma 3 itself follows as our choice of $x_{\ell }$ and $x_{r}$ implies that $x_{\ell } \le x_r \le x_{\ell } + 40 \pi \rho / \tau$.

1.1 Disk notation and definitions

First, we establish some technical geometric properties of disks in the Euclidean plane. Let $D_i$ be a disk of radius $\varrho _i$ centered at $r_i$, where $1 \le \varrho _i \le 1 + \rho$, and let $\mathcal{D}_i$ be the intersection of $D_i$ with the upper halfplane $y > \bar{y}$ for every $i = 1, 2, 3$.

Let $p_{i}^L$ (respectively, $p_{i}^R$, $p_{i}^U$, $p_{i}^D$) be the leftmost (resp., rightmost, uppermost, lowermost) point in $D_i$, as illustrated in Fig. 6. Let $x_{i}^K = x \left( p_{i}^K\right)$ and $y_{i}^K = y \left( p_{i}^K\right)$ for every $K \in \{L,R,U,D\}$ and $i \in \{1,2,3\}$. Note that, $y_{i}^L=y_{i}^R= y \left( r_i\right)$, $x_{i}^U=x_{i}^D= x \left( r_i\right)$, $x_{i}^{L}= x \left( r_i\right) -\varrho _i$, $x_{i}^R= x \left( r_i\right) +\varrho _i$ and $y_{i}^U= y \left( r_i\right) +\varrho _i$, $y_{i}^D= x \left( r_i\right) -\varrho _i$ for $i=1,2,3$.

Given some non-trivial set S of points in $\mathbb {R}^2$, we define the boundary of S, denoted by $\partial {S}$, to be the intersection of the closure of S and the closure of its complement, where closures in this context take on their standard meaning in point-set topology. In particular, for a non-empty subset $I \subseteq \{1,2,3\}$, let $\partial {\left( \bigcup _{i \in I} D_i \right) }$ be the boundary of $\bigcup _{i\in I}D_i$. Given two points $p, q \in \partial {D_i}$, let $\partial D_{i}[p,q]$ be the curve going from p to q along $\partial {D_i}$ in the clockwise direction. For $i=1,2,3$, denote the upper boundary of $D_i$ by $\partial D_{i}^U = \partial D_{i}[p_{i}^L, p_{i}^R]$ and denote the upper right boundary and upper left boundary) of $D_i$ by $\partial D_{i}^{UR} = \partial D_{i}[p_{i}^U, p_{i}^R]$ and $\partial D_{i}^{UL} = \partial D_{i}[p_{i}^L, p_{i}^U]$, respectively (refer to Fig. 6 for illustration). Let

$$\begin{aligned} g_{i}(x) \, = \, \left\{ \begin{array}{ll} y \left( p\right) \text{ s.t. } p \in \partial D_{i}^U \text{ and } x \left( p\right) = x, &{} x \in [x_{i}^{L}, x_{i}^R] \\ 0, &{} \hbox {otherwise} \end{array} \right. \end{aligned}$$

be the function that maps each $x \in [x_{i}^{L}, x_{i}^R]$ to the y coordinate of its (unique) projection on the upper boundary of $D_{i}$.

Given some point $p \in \mathbb {R}^2$, let $\mathcal{H}_L(p)$ and $\mathcal{H}_R(p)$) denote the halfplanes $x \le x \left( p\right)$ and $x \ge x \left( p\right)$, respectively. Given two points $p, q \in \mathbb {R}^2$, let $L( {p, q} )$ be the straight line that passes through the points p and q and let ${\overline{p, q}}$ be the segment connecting p and q.

Fig. 6

a the emphasized curve is the upper boundary. b the upper emphasized curve is the upper right quarter; the lower emphasized curve is the curve from p to q in the clockwise direction

1.2 Two disk toolbox

Consider two indices $1 \le i < j \le 3$.

Fact 1

The leftmost and rightmost points in $D_i\cup D_j$ are $p_{i}^L$ and $p_{j}^R$, respectively. Moreover, $p_{i}^L \not \in D_j$ and $p_{j}^R \not \in D_i$.

Proof

Recall that $x \left( r_i\right) \le x \left( r_j\right) -\tau$, $\rho <\tau$, and $1\le \varrho _k\le 1+\rho$ for $k=1,2,3$, which implies that

$$\begin{aligned} x_{i}^{L}= & {} x \left( r_i\right) -\varrho _i ~\le ~ x \left( r_i\right) -1 ~<~ x \left( r_i\right) -1+\tau -\rho \\\le & {} x \left( r_j\right) -1-\rho ~\le ~ x \left( r_j\right) -\varrho _j ~=~ x_{j}^{L}, \end{aligned}$$

and

$$\begin{aligned} x_{j}^R= & {} x \left( r_j\right) +\varrho _j ~\ge ~ x \left( r_j\right) +1 ~>~ x \left( r_j\right) +1-\tau +\rho \\\ge & {} x \left( r_i\right) +1+\rho ~\ge ~ x \left( r_i\right) +\varrho _i ~=~ x_{i}^R. \end{aligned}$$

Hence, $x_{i}^{L}\not \in [x_{j}^{L},x_{j}^R]$ and $x_{j}^R\not \in [x_{i}^{L},x_{i}^R]$, and therefore $p_{i}^L\not \in D_j$ and $p_{j}^R\not \in D_i$. $\square$

It follows that there are two intersection points between $\partial {D_i}$ and $\partial {D_j}$—denote them by ${u_{ij}}$ and ${u'_{ij}}$, where ${u_{ij}} \in \partial {(D_{i} \cup D_{j})} [p_{i}^L, p_{j}^R]$ and ${u'_{ij}} \in \partial {(D_{i} \cup D_{j})} [p_{j}^R, p_{i}^L]$. By definition, we have $\partial D_{j}[{u_{ij}}, {u'_{ij}}] \cap D_i = \{{u_{ij}}, {u'_{ij}}\}$ and $\partial D_{j}[{u'_{ij}}, {u_{ij}}] \subseteq D_i$, whereas Fact 1 guarantees that $p_{j}^R \not \in D_i$, thus ${u'_{ij}} \not \in \partial D_{j}[{u_{ij}}, p_{j}^R]$, or equivalently, ${u_{ij}} \in \partial D_{j}[{u'_{ij}}, p_{j}^R]$. Define the upper boundary of $D_i\cup D_j$ to be $\partial D_{ij}^U= \partial D_{i}[p_{i}^L, {u_{ij}}] \cup \partial D_{j}[{u_{ij}}, p_{j}^R]$. Thus,

$$\begin{aligned} {u_{ij}}\in \partial D_{ij}^U \text{ and } {u'_{ij}}\not \in \partial D_{ij}^U\end{aligned}$$

as illustrated in Fig. 7.

Fig. 7

The emphasized curve is $\partial D_{ij}^U$, the upper boundary of $D_i\cup D_j$

Claim

(“${u_{ij}}$ is the upper intersection point”). $y \left( {u_{ij}}\right) > y \left( {u'_{ij}}\right)$.

Proof

Observe that the quadrilateral whose vertices are $r_i$, ${u_{ij}}$, $r_j$, and ${u'_{ij}}$ is a kite. As such, the ${\overline{r_i, r_j}}$ and ${\overline{{u_{ij}}, {u'_{ij}}}}$ segments are orthogonal, and hence $y \left( {u_{ij}}\right) \le y \left( {u'_{ij}}\right)$ implies that $x \left( r_i\right) \ge x \left( r_j\right)$, in contradiction to the assumption on $r_i$ and $r_j$. The assertion follows. $\square$

Following Claim A.2, we refer to ${u_{ij}}$ (respectively, ${u'_{ij}}$) as the upper (resp., lower) intersection point between $\partial {D} _i$ and $\partial {D} _j$. Next, we turn to proving a series of claims concerning the inclusion status of the upper intersection point in the upper boundaries of $D_i$ and/or $D_j$, starting with the case that ${u_{ij}} \notin \partial D_{i}^U$ (refer to Fig. 8 for illustration).

Fig. 8

The case where ${u_{ij}}$ is not on the upper boundary of $D_i$. The filled region $\mathcal{D}_j$ is contained in $\mathcal{D}_i$ and $\mathcal{D}_j\cap \mathcal{H}_R({u_{ij}})=\emptyset$

Claim

If ${u_{ij}} \notin \partial D_{i}^U$, then $g_{i}(x) \ge g_{j}(x)$ for every $x \in [x_{i}^{L}, x_{i}^R]$.

Proof

Assume, towards contradiction, that there exists a point $q \in \partial D_{j}^U$ such that $x \left( q\right) \in [x_{i}^{L},x_{i}^R]$ and $y \left( q\right) > g_{i}( x \left( q\right) )$. Recall that $g_{i}( x \left( r_j\right) )>0> y \left( p_{j}^D\right) =y_{j}^D$. Therefore, the fact that $y \left( q\right) >g_{i}( x \left( q\right) )$ implies that q is above the upper boundary of $D_i$, and the fact that $g_{i}( x \left( r_j\right) )>y_{j}^D$ implies that $p_{j}^D$ is below the upper boundary of $D_i$. On the other hand, the constraints on $r_{i}$ and $r_{j}$ ensure that $x \left( p_{j}^L\right) \in [x_{i}^{L},x_{i}^R]$, hence $x \left( q^{\prime }\right) \in [x_{i}^{L},x_{i}^R]$ for every $q^{\prime }\in \partial D_{j}[p_{j}^D,q]$. This implies that the curves $\partial D_{i}^U$ and $\partial D_{j}[p_{j}^D,q]$ intersect—let p be the intersection point. But since $p \in \partial D_{i}^U$ and ${u_{ij}} \in \partial {D} _i - \partial D_{i}^U$, it follows that $y \left( p\right) > y \left( {u_{ij}}\right)$, in contradiction to Claim A.2 stating that ${u_{ij}}$ is the upper intersection point of $\partial {D} _i$ and $\partial {D} _j$. The assertion follows. $\square$

Claim

If ${u_{ij}} \notin \partial D_{i}^U$, then ${u_{ij}}\in \partial D_{j}^{UR}$.

Proof

Claim (A.2) implies that $g_{i}( x \left( r_j\right) ) \ge g_{j}( x \left( r_j\right) )$, thus there exists a point $p \in D_i$ such that $x \left( p\right) = x \left( r_j\right)$ and $y \left( p\right) \ge y_{j}^U > \bar{y} \ge y \left( r_i\right)$. This implies that $\Vert {r_i, p_{j}^U} \Vert \le \Vert {r_i, p} \Vert$, hence $p_{j}^U \in D_i$. It is also clear that $p_{j}^R \notin D_i$ by the constraints on $r_{i}$ and $r_{j}$ Combining these two facts, we conclude that $\partial {D_i}$ and $\partial {D_j}$ intersects along $\partial D_{j}^{UR}$—let p be the intersection point. Since ${u'_{ij}} \in \partial {(D_{i} \cup D_{j})} [p_{j}^R, p_{i}^L]$, we conclude that p cannot be ${u'_{ij}}$, thus $p = {u_{ij}}$. $\square$

Fig. 9

a the filled region $\mathcal{H}_L({u_{ij}})\cap \mathcal{D}_j$ is contained in $\mathcal{D}_i$. b $x \left( q\right) = x \left( q^{\prime }\right) = x \left( p^{\prime }\right)$, $q^{\prime }\in \partial D_{D_j}[p_{j}^L,{u_{ij}}]$ and $p^{\prime }\in \partial D_{D_i}[p_{i}^L,{u_{ij}}]$

Claim

If ${u_{ij}} \notin \partial D_{i}^U$, then $\mathcal{D}_j \subseteq \mathcal{D}_i$.

Proof

The assertion is established by showing that $\mathcal{D}_j \cap \mathcal{H}_R({u_{ij}}) = \emptyset$ and $\mathcal{D}_j \cap \mathcal{H}_L({u_{ij}}) \subseteq \mathcal{D}_i$, starting with the former. Claim A.2 guarantees that ${u_{ij}} \in \partial D_{j}^{UR}$, hence $y \left( {u_{ij}}\right) = \max \{ y \left( q\right) \mid q \in D_j \cap \mathcal{H}_R({u_{ij}}) \}$. Moreover, since ${u_{ij}} \notin \partial D_{i}^U$, we know that $y \left( {u_{ij}}\right) < y \left( r_i\right) \le \bar{y}$. Therefore, $\bar{y} > y \left( {u_{ij}}\right) \ge y \left( q\right)$ for every $q \in D_j \cap \mathcal{H}_R({u_{ij}})$, which implies that $\mathcal{D}_j \cap \mathcal{H}_R({u_{ij}}) = \emptyset$, as promised.

To see that $\mathcal{D}_j \cap \mathcal{H}_L({u_{ij}}) \subseteq \mathcal{D}_i$, consider some point $q \in \mathcal{D}_j \cap \mathcal{H}_L({u_{ij}})$. As $x \left( {u_{ij}}\right) \le x_{i}^R$, it must be that $x \left( q\right) \in [x_{j}^{L},x_{i}^R]\subseteq [x_{i}^{L},x_{i}^R]$. Therefore, by Claim (A.2), $g_{i}( x \left( q\right) ) \ge g_{j}( x \left( q\right) )$, thus there exists a point $p = p(q) \in D_i$ such that $x \left( p\right) = x \left( q\right)$ and $y \left( p\right) \ge y \left( q\right)$. Recall that $y \left( q\right) > \bar{y} \ge y \left( r_i\right)$, which implies that $| y \left( q\right) - y \left( r_i\right) | \le | y \left( p\right) - y \left( r_i\right) |$. Hence $\Vert {r_i,q} \Vert \le \Vert {r_i,p} \Vert \le \varrho _i$, and thus $q \in D_i$. Recalling (again) that $y \left( q\right) > \bar{y}$ implies that $q \in \mathcal{D}_i$ which completes the proof. $\square$

Fig. 10

In sub-figures a, b $x \left( p^{\prime }\right) = x \left( q\right)$. a $p^{\prime }\in \partial D_{i}[t,{u_{ij}}]$ implying that $\Vert {r_j,p^{\prime }} \Vert \ge \Vert {r_j,{u_{ij}}} \Vert =\varrho _j$. b $p^{\prime }\in \partial D_{i}[p_{i}^L,t]$ implying that $\Vert {r_j,p^{\prime }} \Vert \ge \Vert {r_j,p_{i}^L} \Vert >\varrho _j$

Combining Claims A.2 and A.2 yields Corollary 1. Corollary 2 is derived from a symmetric line of arguments.

Corollary 1

If ${u_{ij}} \notin \partial D_{i}^U$, then ${u_{ij}} \in \partial D_{j}^{UR}$ and $\mathcal{D}_j \subseteq \mathcal{D}_i$.

Corollary 2

If ${u_{ij}} \notin \partial D_{j}^U$, then ${u_{ij}} \in \partial D_{i}^{UL}$ and $\mathcal{D}_i \subseteq \mathcal{D}_j$.

We now turn to investigate the case that complements Corollaries 1 and 2, i.e., the case that the upper intersection point is included in both upper boundaries. ${u_{ij}} \in \partial D_{i}^U \cap \partial D_{j}^U$ as illustrated in Fig. 9.

Claim

If ${u_{ij}} \in \partial D_{i}^U \cap \partial D_{j}^U$, then $\mathcal{D}_j \cap \mathcal{H}_L({u_{ij}}) \subseteq \mathcal{D}_i$ (refer to Fig. 9a for illustration).

Proof

Consider some point $q \in \mathcal{D}_j \cap \mathcal{H}_L({u_{ij}})$ (recall that $x \left( q\right) \le x \left( {u_{ij}}\right)$). Let $p^{\prime }\in \partial {D} _i$ (respectively, $q^{\prime }\in \partial {D} _j$) be the point on the upper boundary of $D_i$ (resp., $D_j$) such that $x \left( p^{\prime }\right) = x \left( q\right)$, (resp., $x \left( q^{\prime }\right) = x \left( q\right)$) ($p^{\prime }$ is well defined since $x \left( q\right) \in [x_{j}^{L}, x \left( {u_{ij}}\right) ]\subseteq [x_{i}^{L},x_{i}^R]$). Let $t_i$ (resp., $t_j$) be the left intersection point between $\partial {D} _i$ (resp., $\partial {D} _j$) and the straight line $L( {r_i,r_j} )$. Since $q\in \mathcal{H}_L({u_{ij}})$, we have that $p^{\prime }\in \partial D_{i}[p_{i}^L,{u_{ij}}]$ and $q^{\prime }\in \partial D_{j}[p_{j}^L,{u_{ij}}]$, as illustrated in Fig. 9b. To prove that $q\in \mathcal{D}_i$, we consider two cases.

Case 1: $y \left( r_i\right) \ge y \left( r_j\right)$. We first show that

$$\begin{aligned} \Vert {r_j,p^{\prime }} \Vert \ge \varrho _j. \end{aligned}$$

(1)

To prove this, we consider two subcases.

Subcase 1a: $p^{\prime } \in \partial D_{i}[t_i,{u_{ij}}]$ (see Fig. 10a). Let $\alpha =\angle ({u_{ij}},r_i,r_j)$ and let $\beta =\angle (p^{\prime },r_i,r_j)$. It is clear that $\pi \ge \beta \ge \alpha$, thus by the law of cosines, $\Vert {r_j,p^{\prime }} \Vert \ge \Vert {r_j,{u_{ij}}} \Vert =\varrho _j$, as required.

Subcase 1b: $p^{\prime } \in \partial D_{i}[f_{\{i\}}(1),t_i]$ (see Fig. 10b). Let $\alpha =\angle (f_{\{i\}}(1),r_i,r_j)$ and let $\beta =\angle (p^{\prime },r_i,r_j)$. It is clear that $\pi \ge \beta \ge \alpha$, thus by the law of cosines, $\Vert {r_j,p^{\prime }} \Vert \ge \Vert {r_j,f_{\{i\}}(1)} \Vert >\varrho _j$, where the last inequality holds since $p_{i}^L\not \in D_j$. Also, the fact that $q\in \mathcal{D}_j$ implies that $\Vert {r_j,q} \Vert \le \varrho _j$. Combining this together with (1), we get that $\Vert {r_j,p^{\prime }} \Vert \ge \Vert {r_j,q} \Vert$. Recall that $x \left( q\right) = x \left( p^{\prime }\right)$, thus

$$\begin{aligned}&\Vert {r_j,p^{\prime }} \Vert ^2 - \Vert {r_j,q} \Vert ^2 \\ = \,&\left[ \left( x \left( p^{\prime }\right) - x \left( r_j\right) \right) ^2 + \left( y \left( p^{\prime }\right) - y \left( r_j\right) \right) ^2 \right] \\&- \left[ \left( x \left( q\right) - x \left( r_j\right) \right) ^2 + \left( y \left( q\right) - y \left( r_j\right) \right) ^2 \right] \\ =\,&\left( y \left( p^{\prime }\right) - y \left( r_j\right) \right) ^2 - \left( y \left( q\right) - y \left( r_j\right) \right) ^2 \, \ge \, 0. \end{aligned}$$

Combining this together with the inequalities $y \left( p^{\prime }\right) \ge y \left( r_i\right) \ge y \left( r_j\right)$ and $y \left( q\right) > \bar{y} \ge \max \{ y \left( r_i\right) , y \left( r_j\right) \}$, we get that $y \left( p^{\prime }\right) \ge y \left( q\right)$ and $( y \left( q\right) - y \left( r_i\right) )^2 \le ( y \left( p^{\prime }\right) - y \left( r_i\right) )^2$. Hence, $\Vert {r_i, q} \Vert \le \Vert {r_i, p^{\prime }} \Vert$ and therefore, $q \in \mathcal{D}_i$.

Case 2: $y \left( r_i\right) < y \left( r_j\right)$ (see Fig. 11). The slope of $L( {r_i,r_j} )$ is positive, and therefore, $t_j$ is not on the upper boundary of $D_j$, and in particular, $t_j \notin \partial D_{j}[p_{i}^L, {u_{ij}}]$.

Let $\alpha =\angle (q^{\prime },r_j,r_i)$ and $\beta =\angle ({u_{ij}},r_j,r_i)$. Recall that $x \left( q^{\prime }\right) \le x \left( {u_{ij}}\right)$. Together with the fact that $q^{\prime }$ and ${u_{ij}}$ are on the upper boundary of $D_j$, this implies that $q^{\prime }\in \partial D_{j}[t_j,{u_{ij}}]$, thus it is clear that $\pi \ge \beta \ge \alpha$. By the law of cosines, we get that $\Vert {r_i,q^{\prime }} \Vert \le \Vert {r_i,{u_{ij}}} \Vert = \varrho _i$. Combining this with the fact that $x \left( q^{\prime }\right) = x \left( q\right)$ and $y \left( q^{\prime }\right) \ge y \left( q\right) >\bar{y}\ge y \left( r_i\right)$, we conclude that $\Vert {r_i,q} \Vert \le \Vert {r_i,q^{\prime }} \Vert = \varrho _i$. Hence, $q \in D_i$ which establishes the assertion. $\square$

Claim A.2 can be extended using a symmetric line of arguments to yield the following corollary.

Corollary 3

If ${u_{ij}} \in \partial D_{i}^U \cap \partial D_{j}^U$, then $\mathcal{D}_j \cap \mathcal{H}_L({u_{ij}}) \subseteq \mathcal{D}_i$ and $\mathcal{D}_i \cap \mathcal{H}_R({u_{ij}}) \subseteq \mathcal{D}_j$.

Observation 2

The upper intersection point satisfies $y \left( {u_{ij}}\right) \ge \min \{ y \left( r_i\right) , y \left( r_j\right) \}\ge 0$ and $y \left( {u'_{ij}}\right) \le \max \{ y \left( r_i\right) , y \left( r_j\right) \}\le \bar{y}$. Thus, also (1) ${u_{ij}}\in \partial D_{i}^U$ or ${u_{ij}}\in \partial D_{j}^U$; and (2) ${u'_{ij}}\not \in \partial D_{i}^U$ or ${u'_{ij}}\not \in \partial D_{j}^U$.

Proof

By Corollaries 1 and 2, ${u_{ij}} \in \partial D_{i}^U$ or ${u_{ij}} \in \partial D_{j}^U$. Thus, $y \left( {u_{ij}}\right) \ge \min \{ y \left( r_i\right) , y \left( r_j\right) \}\ge 0$. A symmetric argument proves that ${u'_{ij}}\not \in \partial D_{i}^U$ or ${u'_{ij}}\not \in \partial D_{j}^U$, which implies that $y \left( {u'_{ij}}\right) \le \max \{ y \left( r_i\right) , y \left( r_j\right) \}\le \bar{y}$. $\square$

Fig. 11

$q^{\prime }\in \partial D_{j}[t_j,{u_{ij}}]$, implying that $\Vert {r_i,q^{\prime }} \Vert \le \Vert {r_i,{u_{ij}}} \Vert =\varrho _i$

Fig. 12

In both sub-figures the filled region is $\mathcal{D}_2\setminus (\mathcal{D}_1\cup \mathcal{D}_3)$. a The region is limited by $x \left( {u_{12}}\right)$ and $x \left( {u_{23}}\right)$. b The region is limited by $x \left( {u_{12}}\right)$ and $x \left( p_{2}^R\right)$

1.3 Weaker version of the Lemma

In this subsection we prove the weaker version of the lemma. We begin by defining some notations and then define the aforementioned two reals $\hat{x}_\ell (r_1,r_2,r_3)$ and $\hat{x}_r(r_1,r_2,r_3)$ of the lemma. Recall that $F _i$ (respectively, $\tilde{ F }_i$) is the same as $D_i$ with radius 1 (resp., radius $1+\rho$) and $\mathcal{F}_i$ (resp., $\tilde{\mathcal{F}}_i$) is the same as $\mathcal{D}_i$ with radius 1 (resp., radius $1+\rho$), for $i=1,2,3$. For two given disks $D,D^{\prime }$ such that $\partial {D} \cap \partial {D} ^{\prime }\not =\emptyset$, let ${I({D,D^{\prime }})}$ be the upper (in terms of maximizing the $y$-coordinate) intersection point of $\partial {D}$ and $\partial {D} ^{\prime }$. (Note that ${u_{ij}}={I({D_i,D_j})}$ for $1\le i<j\le 3$). Let

$$\begin{aligned} \hat{x}_\ell = \left\{ \begin{array}{ll} x \left( {I({ F _1,\tilde{ F }_2})}\right) , &{} \quad \text {if } {I({ F _1,\tilde{ F }_2})}\in \partial \tilde{ F }_{2}^U, \\ x \left( r_2\right) -1-\rho , &{} \quad \text {otherwise}, \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \hat{x}_r= \left\{ \begin{array}{ll} x \left( {I({\tilde{ F }_2, F _3})}\right) , &{} \quad \text {if } {I({\tilde{ F }_2, F _3})}\in \partial \tilde{ F }_{2}^U, \\ x \left( r_2\right) +1+\rho , &{} \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

Claim

Suppose that $\tilde{\mathcal{F}}_2 - V \ne \emptyset$. Then $x \left( s\right) \in [\hat{x}_\ell ,\hat{x}_r]$, for all points $\tilde{\mathcal{F}}_2 - V \ne \emptyset$ (see Fig. 12).

Proof

Suppose that $\tilde{\mathcal{F}}_2 - V \ne \emptyset$. Then, obviously $\tilde{\mathcal{F}}_2 - (\mathcal{F}_1 \cup \mathcal{F}_3) \ne \emptyset$ and, in particular, $\tilde{\mathcal{F}}_2 \not \subseteq \mathcal{F}_1$. Therefore, by Corollary 1 with $\mathcal{D}_i = \mathcal{F}_1$ and $\mathcal{D}_j = \tilde{\mathcal{F}}_2$, it follows that ${u_{ij}}={I({ F _1,\tilde{ F }_2})}$ is on the upper boundary of $D_i = F _1$, that is,

$$\begin{aligned} {I({ F _1,\tilde{ F }_2})} \in \partial F _{1}^U. \end{aligned}$$

(2)

Consider a point $s\in \tilde{ F }_2\setminus F _1$. We consider two cases. If ${I({ F _1,\tilde{ F }_2})}\in \partial \tilde{ F }_{2}^U$, then by (2), with $\mathcal{D}_i=\mathcal{F}_1$ and $\mathcal{D}_j=\tilde{\mathcal{F}}_2$, we assume that the conditions of Corollary 3 hold, which implies that $\mathcal{H}_L({u_{12}})\cap \mathcal{D}_2\setminus \mathcal{D}_1=\emptyset$ and equivalently $\mathcal{H}_L({I({ F _1,\tilde{ F }_2})})\cap \tilde{ F }_2\setminus F _1=\emptyset$. Thus, $\hat{x}_\ell = x \left( {I({ F _1,\tilde{ F }_2})}\right) \le x \left( s\right)$, as required. Otherwise, ${I({ F _1,\tilde{ F }_2})}\not \in \partial \tilde{ F }_{2}^U$, and then obviously $\hat{x}_\ell \le x \left( s\right)$, since $\hat{x}_\ell = x \left( r_2\right) -1-\rho$.

A symmetric arguments also proves that $x \left( s\right) \le \hat{x}_r$. The claim follows. $\square$

1.3.1 Bounding $\hat{x}_r-\hat{x}_\ell$ from above

To complete the weaker lemma version proofs, it remain to prove that

$$\begin{aligned} \hat{x}_r-\hat{x}_\ell \le 16\pi \rho / \tau . \end{aligned}$$

(3)

We first prove that

$$\begin{aligned} \hat{x}_r- x \left( {I({ F _2, \tilde{ F }_3})}\right) \, \le \, 8 \pi \rho / \tau \end{aligned}$$

(4)

and symmetrically that

$$\begin{aligned} x \left( {I({\tilde{ F }_1, F _2})}\right) - \hat{x}_\ell \, \le \, 8 \pi \rho / \tau . \end{aligned}$$

(5)

Then, we prove that

$$\begin{aligned} x \left( {I({\tilde{ F }_1, F _2})}\right) > x \left( {I({ F _2,\tilde{ F }_3})}\right) , \end{aligned}$$

(6)

which implies together with (4) and (5) that (3) indeed holds, and hence the weaker lemma version holds too.

Now, we turn our attention to proving that (4) holds; the line of arguments that establish (5) is analogous. For an index $i \in \{1, 2, 3\}$ and for two points $p, q \in \partial { F _i}$, denote by $\ell _{\partial F _i}\left[ p,q\right]$ the length of the curve $\partial F _{i}[p,q]$ going along $\partial { F _i}$ from p to q in the clockwise direction. It is clear that in a Euclidean geometry $\Vert {p,q} \Vert \le \ell _{\partial F _i}\left[ p,q\right]$. The following claim provides important upper bounds on certain curve lengths.

Claim

Consider some two indices $1\le i < j\le 3$. Then,

$$\begin{aligned} \ell _{\partial F _i}\left[ {I({ F _i, \tilde{ F }_j})} , {I({ F _i, F _j})}\right] \, \le \, 4 \pi \rho / \tau \end{aligned}$$

and

$$\begin{aligned} \ell _{\partial F _j}\left[ {I({ F _i, F _j})}, {I({\tilde{ F }_i, F _j})}\right] \, \le \, 4 \pi \rho / \tau . \end{aligned}$$

In other words, the (clockwise) distance along $\partial { F _i}$ (resp., $\partial { F _j}$) from its intersection point with $\tilde{ F }_j$ (resp., $F _i$) to its intersection point with $F _j$ (resp., $\tilde{ F }_i$) is at most $4 \pi \rho / \tau$.

Fig. 13

The clockwise distance along $F _i$ from ${I({ F _i, \tilde{ F }_j})}$ to ${I({ F _i, F _j})}$ is equal to the clockwise distance along $F _j$ from ${I({ F _i, F _j})}$ to ${I({\tilde{ F }_i, F _j})}$, i.e., $\ell _{\partial F _i}\left[ {I({ F _i, \tilde{ F }_j})}, {I({ F _i, F _j})}\right] = \ell _{\partial F _j}\left[ {I({ F _i, F _j})}, {I({\tilde{ F }_i, F _j})}\right]$ and both are equal to $2 \pi \gamma$, where $\gamma = \beta - \alpha$

Proof

Let

$$\begin{aligned} \gamma \, = \, \angle ({I({ F _i, \tilde{ F }_j})} ,r_{i}, {I({ F _i, F _j})}) \end{aligned}$$

and

$$\begin{aligned} \gamma ^{\prime } \, = \, \angle ({I({ F _i, F _j})}, r_{j}, {I({\tilde{ F }_i, F _j})}), \end{aligned}$$

where the angles in this proof are measured in turns. (Fig. 13 provides helpful illustrations throughout this proof.) The (clockwise) distance along $F _{i}$ from ${I({ F _i,\tilde{ F }_j})}$ to ${I({ F _i, F _j})}$ and the (clockwise) distance along $F _{j}$ from ${I({ F _i, F _j})}$ to ${I({\tilde{ F }_i, F _j})}$ satisfy

$$\begin{aligned} \ell _{\partial F _i}\left[ {I({ F _i,\tilde{ F }_j})},{I({ F _i, F _j})}\right] \, = \, 2 \pi \gamma \end{aligned}$$

and

$$\begin{aligned} \ell _{\partial F _j}\left[ {I({ F _i, F _j})},{I({\tilde{ F }_i, F _j})}\right] \, = \, 2 \pi \gamma ^{\prime }. \end{aligned}$$

By symmetry considerations, we conclude that $\gamma ^{\prime } = \gamma$, so it suffices to devise an upper bound on $\gamma$ which is our goal in the remainder of this proof.

Consider the angles

$$\begin{aligned} \alpha \, = \, \angle (r_j, r_i, {I({ F _i, F _j})}) \quad \text {and}\quad \beta \,= & {} \, \angle (r_j, r_i, {I({ F _i, \tilde{ F }_j})}). \end{aligned}$$

It is clear that $\gamma = \beta - \alpha$ as illustrated in Fig. 13. Fixing $a = \Vert {r_i, r_j} \Vert$, we apply the law of cosines to the triangles $\Delta (r_j, r_i, {I({ F _i, F _j})})$ and $\Delta (r_j, r_i, {I({ F _i, \tilde{ F }_j})})$, concluding that

$$\begin{aligned} \alpha \, = \, \arccos \left( \frac{a^{2} + 1 - 1}{2 a} \right) \, = \, \arccos \left( \frac{a}{2} \right) \end{aligned}$$

and

$$\begin{aligned} \beta \, = \, \arccos \left( \frac{a^{2} + 1 - (1 + \rho )^{2}}{2 a} \right) \, = \, \arccos \left( \frac{a}{2} - \frac{2 \rho + \rho ^{2}}{2 a} \right) , \end{aligned}$$

hence

$$\begin{aligned} \gamma \, = \, \beta - \alpha \, = \, \arccos \left( \frac{a}{2} - \frac{2 \rho + \rho ^{2}}{2 a} \right) - \arccos \left( \frac{a}{2} \right) . \end{aligned}$$

Recall that $\tau \le a \le 1$ and $\rho \le \tau / 4$ which implies that

$$\begin{aligned} \frac{a}{2} \, \le \, 1 / 2 \quad \text {and}\quad \frac{a}{2} - \frac{2 \rho + \rho ^{2}}{2 a} \, \ge \, \frac{a}{2} - \frac{(a / 2) + (a / 4)^{2}}{2 a} \, > \, - 1 / 4. \end{aligned}$$

Since $\frac{d}{dx} \arccos x = \frac{-1}{\sqrt{1 - x^{2}}}$ and since $\frac{-1}{\sqrt{1 - x^{2}}} \ge -\sqrt{4 / 3}$ for every $-1 / 4 \le x \le 1 / 2$, we can bound $\gamma = \arccos \left( \frac{a}{2} - \frac{2 \rho + \rho ^{2}}{2 a} \right) - \arccos \left( \frac{a}{2} \right)$ from above by $\sqrt{4 / 3} \cdot \left| \frac{a}{2} - \left( \frac{a}{2} - \frac{2 \rho + \rho ^{2}}{2 a} \right) \right|$, hence

$$\begin{aligned} \gamma \, \le \,&\sqrt{4 / 3} \cdot \frac{2 \rho + \rho ^{2}}{2 a} \, = \, \sqrt{4 / 3} \cdot \frac{\rho }{a} (1 + \rho / 2) \\ \le \,&\sqrt{4 / 3} \cdot \frac{\rho }{a} (1 + 1 / 8) \, < \, 2 \rho / \tau . \end{aligned}$$

The assertion follows. $\square$

Hereafter, let $p_{i}^L$ (respectively, $p_{i}^R$) be the leftmost (resp., rightmost) point in $F _i$ and let $\tilde{p}_{i}^L$ (respectively, $\tilde{p}_{i}^R$) be the leftmost (resp., rightmost) point in $\tilde{ F }_i$. Note that, if ${I({\tilde{ F }_2, F _3})}\not \in \partial \tilde{ F }_{2}^U$ (respectively, ${I({ F _1,\tilde{ F }_2})}\not \in \partial \tilde{ F }_{2}^U$), then $\hat{x}_r= x \left( \tilde{p}_{2}^R\right)$ (resp., $\hat{x}_\ell = x \left( \tilde{p}_{2}^L\right)$).

Claim

Our choice of parameters guarantees that $\hat{x}_r-{I({ F _2, \tilde{ F }_3})} \le \ell _{\partial F _2}\left[ {I({ F _2, \tilde{ F }_3})}, {I({ F _2, F _3})}\right] + \max \Big \{ \rho , \ell _{\partial F _3}\left[ {I({ F _2, F _3})}, {I({\tilde{ F }_2, F _3})}\right] \Big \}$.

Proof

We consider two cases.

Case 1: ${I({\tilde{ F }_2, F _3})}\in \partial \tilde{ F }_{2}^U$ (see Fig. 14). Then $\hat{x}_r= x \left( {I({\tilde{ F }_2, F _3})}\right)$. Thus,

$$\begin{aligned}&\hat{x}_r- x \left( {I({ F _2, \tilde{ F }_3})}\right) \\ \le \,&\Vert {{I({ F _2, \tilde{ F }_3})}, {I({\tilde{ F }_2, F _3})}} \Vert \\ \le \,&\Vert {{I({ F _2, \tilde{ F }_3})}, {I({ F _2, F _3})}} \Vert + \Vert {{I({ F _2, F _3})}, {I({\tilde{ F }_2, F _3})}} \Vert \\ \le \,&\ell _{\partial F _2}\left[ {I({ F _2, \tilde{ F }_3})}, {I({ F _2, F _3})}\right] {+}\ell _{\partial F _3}\left[ {I({ F _2, F _3})}, {I({\tilde{ F }_2, F _3})}\right] . \end{aligned}$$

Fig. 14

Case 1 ${I({\tilde{ F }_2, F _3})}\in \partial \tilde{ F }_{2}^U$. Thus, $\hat{x}_r- x \left( {I({ F _2,\tilde{ F }_3})}\right) \le \ell _{\partial F _2}\left[ {I({ F _2,\tilde{ F }_3})},{I({ F _2, F _3})}\right] {+}\ell _{\partial F _3}\left[ {I({ F _2, F _3})},{I({\tilde{ F }_2, F _3})}\right]$

Case 2 ${I({\tilde{ F }_2, F _3})}\not \in \partial \tilde{ F }_{2}^U$. Then $\hat{x}_r= x \left( \tilde{p}_{2}^R\right)$. We show that $\Vert {{I({ F _2,\tilde{ F }_3})},\tilde{p}_{2}^R} \Vert \le \ell _{\partial F _2}\left[ {I({ F _2,\tilde{ F }_3})},{I({ F _2, F _3})}\right] +\max \left\{ \rho ~,~\ell _{\partial F _3}\left[ {I({ F _2, F _3})},{I({\tilde{ F }_2, F _3})}\right] \right\}$. To prove this, we consider two subcases.

Subcase 2.a  ${I({ F _2, F _3})}\not \in \partial F _{2}^U$ (see Fig. 15). Then $p_{2}^R\in \partial F _{2}[{I({ F _2,\tilde{ F }_3})},{I({ F _2, F _3})}]$, recalling that $\Vert {\tilde{p}_{2}^R,p_{2}^R} \Vert =\rho$, hence

$$\begin{aligned} \hat{x}_r- x \left( {I({ F _2, \tilde{ F }_3})}\right) \, \le \,&\Vert {{I({ F _2, \tilde{ F }_3})}, \tilde{p}_{2}^R} \Vert \\ \le \,&\ell _{\partial F _2}\left[ {I({ F _2, \tilde{ F }_3})}, {I({ F _2, F _3})}\right] + \rho \, . \end{aligned}$$

Fig. 15

Case 2.a ${I({\tilde{ F }_2, F _3})}\not \in \partial \tilde{ F }_{2}^U$ and ${I({ F _2, F _3})}\not \in \partial F _{2}^U$. Thus, $p_{2}^R\in \partial F _{2}[{I({ F _2,\tilde{ F }_3})},{I({ F _2, F _3})}]$. Hence, $\hat{x}_r- x \left( {I({ F _2,\tilde{ F }_3})}\right) \le \Vert {p_{2}^R,\tilde{p}_{2}^R} \Vert + \ell _{\partial F _2}\left[ {I({ F _2,\tilde{ F }_3})},{I({ F _2, F _3})}\right]$

Subcase 2.b  ${I({ F _2, F _3})}\in \partial F _{2}^U$ (see Fig. 16). Let p be the intersection point between the horizontal line $y= y \left( r_2\right)$ and the curve $\partial F _{3}[{I({ F _2, F _3})},{I({\tilde{ F }_2, F _3})}]$, i.e., $p\in \partial F _{3}[{I({ F _2, F _3})},{I({\tilde{ F }_2, F _3})}]$ and $y \left( p\right) = y \left( r_2\right)$. Thus,

$$\begin{aligned} \hat{x}_r- x \left( {I({ F _2, \tilde{ F }_3})}\right) \, \le \,&\Vert {p_{2}^R, p} \Vert + \ell _{\partial F _3}\left[ p, {I({ F _2, F _3})}\right] \nonumber \\&+ \ell _{\partial F _3}\left[ {I({ F _2, F _3})}, {I({ F _2, \tilde{ F }_3})}\right] . \end{aligned}$$

(7)

In addition,

$$\begin{aligned} \Vert {\tilde{p}_{2}^R,p} \Vert \le \Vert {{I({\tilde{ F }_2, F _3})},p} \Vert \le \ell _{\partial F _3}\left[ {I({\tilde{ F }_2, F _3})},p\right] , \end{aligned}$$

(8)

where the left inequality holds, since (1) $\Vert {{I({\tilde{ F }_2, F _3})},p} \Vert \ge 1+\rho -\Vert {r_2,p} \Vert$ by the triangle inequality; and (2) $\Vert {\tilde{p}_{2}^R,p} \Vert =1+\rho -\Vert {r_2,p} \Vert$. Combining (7) and (8), we get that

$$\begin{aligned} \hat{x}_r- x \left( {I({ F _2, \tilde{ F }_3})}\right) \, \le \,&\ell _{\partial F _3}\left[ {I({\tilde{ F }_2, F _3})}, {I({ F _2, F _3})}\right] \\&+ \ell _{\partial F _2}\left[ {I({ F _2, F _3})}, {I({ F _2, \tilde{ F }_3})}\right] , \end{aligned}$$

as required. $\square$

Fig. 16

Case 2.b ${I({\tilde{ F }_2, F _3})}\not \in \partial \tilde{ F }_{2}^U$ and ${I({ F _2, F _3})}\in \partial F _{2}^U$. The point p is on $\partial F _{3}[{I({ F _2, F _3})},{I({\tilde{ F }_2, F _3})}]$. Hence, $\hat{x}_r- x \left( {I({ F _2,\tilde{ F }_3})}\right) \le \ell _{\partial F _2}\left[ {I({ F _2,\tilde{ F }_3})},{I({ F _2, F _3})}\right] + \ell _{\partial F _3}\left[ {I({ F _2, F _3})},p\right] + \Vert {p,\tilde{p}_{2}^R} \Vert$

Inequality (4) holds by Claim A.3.1, Claim A.3.1, and since $\rho \le 4\pi \rho / \tau$ (since $\tau \le 1$). A symmetric arguments proves that (5) holds too.

1.3.2 Establishing inequality (6)

Claim

Suppose that $(\tilde{\mathcal{F}}_1\cap \tilde{\mathcal{F}}_3)\setminus \mathcal{F}_2{\not =}\emptyset$. Then, $x \left( {I({\tilde{ F }_1, F _2})}\right) > x \left( {I({ F _2,\tilde{ F }_3})}\right)$ (See Fig. 17a).

Fig. 17

a the case where $x \left( {I({ F _2,\tilde{ F }_3})}\right) < x \left( {I({\tilde{ F }_1, F _2})}\right)$. The filled region is $(\tilde{\mathcal{F}}_1\cap \tilde{\mathcal{F}}_3)\setminus \mathcal{F}_2\not =\emptyset$. bf b the case where $x \left( {I({\tilde{ F }_1, F _2})}\right) < x \left( {I({ F _2,\tilde{ F }_3})}\right)$. The filled region is $\tilde{\mathcal{F}}_1\cap \tilde{\mathcal{F}}_3$, contained in $\mathcal{F}_2$, thus $(\tilde{\mathcal{F}}_1\cap \tilde{\mathcal{F}}_3)\setminus \mathcal{F}_2=\emptyset$

Proof

For simplicity, let $\mathcal{D}_1=\tilde{\mathcal{F}}_1$, $\mathcal{D}_2=\mathcal{F}_2$ and $\mathcal{D}_3=\tilde{\mathcal{F}}_3$. Note that, ${u_{12}}={I({\tilde{ F }_1, F _2})}$ and ${u_{23}}={I({ F _2,\tilde{ F }_3})}$. We prove an equivalent assertion. That is, see Fig. 17a,

$$\begin{aligned} \text{ if } (\mathcal{D}_1\cap \mathcal{D}_3)\setminus \mathcal{D}_2\not =\emptyset , \text{ then } x \left( {u_{12}}\right) > x \left( {u_{23}}\right) . \end{aligned}$$

(9)

Suppose that $(\mathcal{D}_1\cap \mathcal{D}_3)\setminus \mathcal{D}_2\not =\emptyset$. We begin by proving that

$$\begin{aligned} {u_{12}},{u_{23}}\in \partial D_{2}^U. \end{aligned}$$

(10)

To see this, note that if ${u_{12}}\not \in \partial D_{2}^U$, then by Corollary 2 with $i=1$ and $j=2$ we get that $\mathcal{D}_1\setminus \mathcal{D}_2=\emptyset$, and if ${u_{23}}\not \in \partial D_{2}^U$, then by Corollary 1 with $i=2$ and $j=3$ we get that $\mathcal{D}_3\setminus \mathcal{D}_2=\emptyset$. In both cases, $(\mathcal{D}_1\cap \mathcal{D}_3)\setminus \mathcal{D}_2=\emptyset$, contradicting the premise of the claim.

Next, we prove the claim itself by contradiction. Suppose that $x \left( {u_{12}}\right) \le x \left( {u_{23}}\right)$. Then by (10),

$$\begin{aligned} {u_{23}}\in \partial D_{2}[{u_{12}},p_{2}^R] \text{ and } {u_{12}}\in \partial D_{2}[p_{2}^L,{u_{23}}]. \end{aligned}$$

(11)

We examine two cases. The first case is when ${u_{12}}\not \in \partial D_{1}^U$ or ${u_{23}}\not \in \partial D_{3}^U$. If ${u_{12}}\not \in \partial D_{1}^U$ (see Fig. 18), then $y \left( {u_{12}}\right) < y \left( r_i\right)$, and by Corollary 1 with $i=1$ and $i=2$, it follows that ${u_{12}}\in \partial D_{2}^{UR}$. This implies that $x \left( {u_{12}}\right) \ge x \left( r_2\right)$, which together with the left side of (11) implies that ${u_{23}}\in \partial D_{2}^{UR}$ and that $y \left( {u_{23}}\right) \le y \left( {u_{12}}\right)$, and thus

$$\begin{aligned} y \left( {u_{23}}\right) < y \left( r_1\right) \le \bar{y} ~~~~ \text{ and } ~~~~ x \left( {u_{23}}\right) \ge x \left( r_2\right) >0. \end{aligned}$$

(12)

In addition, the fact that ${u_{23}}\in \partial D_{2}^{UR}$ implies that ${u_{23}}\in \partial D_{3}^U$ (by the negative direction of Corollary 2 with $i=2$ and $j=3$). Moreover, $[0,\bar{y}]^2\subset D_3$, hence $(0,\bar{y})^2\cap \partial {D} _3=\emptyset$. Combining this together with (12), it follows that $x \left( {u_{23}}\right) \ge \bar{y}\ge x \left( r_3\right)$, hence ${u_{23}}\in \partial D_{3}^{UR}$, as illustrated in Fig. 18. Therefore, $\mathcal{H}_L({u_{23}})\cap \mathcal{D}_3\setminus \mathcal{D}_2=\emptyset$, by Corollary 3, with $i=2$ and $j=3$, and $\bar{y}> y \left( {u_{23}}\right) =\max \{ y \left( q\right) \mid p\in D_3\cap \mathcal{H}_R({u_{23}})\}$, hence $\mathcal{H}_R({u_{23}})\cap \mathcal{D}_3=\emptyset$, thus $(\mathcal{D}_1\cap \mathcal{D}_3)\setminus \mathcal{D}_2=\emptyset$, contradiction. (A symmetric argument derives a contradiction when ${u_{23}}\not \in \partial D_{3}^U$.)

Fig. 18

The case where ${u_{12}}\not \in \partial D_{1}^U$ and $x \left( {u_{12}}\right) < x \left( {u_{23}}\right)$. Here $y \left( {u_{23}}\right) \le \bar{y}$ and ${u_{23}}\in \partial D_{3}^{UR}$, thus $\mathcal{D}_3\cap \mathcal{H}_R({u_{23}})=\emptyset$. The filled region $\mathcal{D}_3$ is contained in $\mathcal{D}_2$, i.e., $\mathcal{D}_3\setminus \mathcal{D}_2=\emptyset$

The second case is when ${u_{12}}\in \partial D_{1}^U$ and ${u_{23}}\in \partial D_{3}^U$. Then $\mathcal{H}_R({u_{12}})\cap \mathcal{D}_1\setminus \mathcal{D}_2=\emptyset$, by Corollary 3 with $i=1$ and $j=2$, and $\mathcal{H}_L({u_{23}})\cap \mathcal{D}_3\setminus \mathcal{D}_2=\emptyset$, by Corollary 3 with $i=2$ and $j=3$ (recall that $x \left( {u_{12}}\right) \le x \left( {u_{23}}\right)$). Hence $(\mathcal{D}_1\cap \mathcal{D}_3)\setminus \mathcal{D}_2=\emptyset$, as illustrated in Fig. 17b, which is contradiction. Thus, $x \left( {u_{12}}\right) > x \left( {u_{23}}\right)$ and (9) does holds which also implies the claim. $\square$

We now have

$$\begin{aligned} \hat{x}_r-\hat{x}_\ell\le & {} \hat{x}_r- x \left( {I({ F _2,\tilde{ F }_3})}\right) \nonumber \\&+ x \left( {I({\tilde{ F }_1, F _2})}\right) -\hat{x}_\ell ~\le ~ 16\pi \rho / \tau , \end{aligned}$$

(13)

where the first inequality holds by inequality Claim A.3.2 and the last inequality holds by (4) and (5). The weaker version of the lemma follows.

1.4 Full version of the Lemma

Let

$$\begin{aligned} x_{mid} \equiv x \left( {I({ F _1, F _3})}\right) \end{aligned}$$

and let

$$\begin{aligned} x_\ell \equiv x_{mid}-20\pi \rho / \tau \quad \text {and}\quad x_r\equiv x_{mid}+20\pi \rho / \tau . \end{aligned}$$

(14)

Notice that $x_\ell$ and $x_r$ are both determined independently of $r_2$ since $x_{mid}$ is determined independently of $r_2$. Lemma 3 is established by proving that $[\hat{x}_\ell , \hat{x}_r] \subseteq [x_\ell , x_r]$ (recall that the definitions of $\hat{x}_\ell = \hat{x}_\ell (r_1, r_2, r_3)$ and $\hat{x}_r= \hat{x}_r(r_1, r_2, r_3)$ depend on $r_1$, $r_3$, and also on $r_2$). Let

$$\begin{aligned} p_\ell \, = \, \left\{ \begin{array}{ll} x_{mid}, &{} \quad {I({ F _1, F _3})}\in \partial F _{3}^U \\ x \left( r_3\right) -1, &{} \quad \hbox {otherwise} \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} p_r\, = \, \left\{ \begin{array}{ll} x_{mid}, &{} \quad {I({ F _1, F _3})}\in \partial F _{1}^U \\ x \left( r_1\right) +1, &{} \quad \hbox {otherwise.} \end{array} \right. \end{aligned}$$

We establish the following properties regarding $p_\ell$ and $p_r$:

• (P1) $x_{mid}=p_\ell$ or $x_{mid}=p_r$;

• (P2) $p_{\ell } \le p_r \le p_\ell + 4 \pi \rho / \tau$; and

• (P3) $[\hat{x}_\ell ,\hat{x}_r] \cap [p_\ell ,p_r] \ne \emptyset$.

Combined with (14), these three properties ensure that $[\hat{x}_\ell ,\hat{x}_r] \subseteq [x_\ell ,x_r]$, as desired. We first show that $\hat{x}_\ell \ge x_\ell$. By Property (P1) and Property (P2), it follows that

$$\begin{aligned} p_\ell \ge x_{mid}-4\pi \rho / \tau . \end{aligned}$$

(15)

Let $x^{\prime }\in [\hat{x}_\ell ,\hat{x}_r]\cap [p_\ell ,p_r]$ (by Property (P3) such a number does exists). We have

$$\begin{aligned} \hat{x}_\ell \ge x^{\prime }-|\hat{x}_r-\hat{x}_\ell |\ge p_\ell - 16\pi \rho / \tau \ge x_{mid}-20\pi \rho / \tau = x_\ell , \end{aligned}$$

where the second inequality hold since (1) $x^{\prime }\ge p_\ell$ and (2) $\hat{x}_r-\hat{x}_\ell \le 16\pi \rho /\tau$, by (13) (weaker version of the lemma); and the third inequality holds by (15). A symmetric argument also proves that $\hat{x}_r\le x_r$. Therefore, $[\hat{x}_\ell ,\hat{x}_r]\subseteq [x_\ell ,x_r]$, which yields Lemma 3. It remains to prove these three properties. The following observation and the following claim proves Property (P1) and Property (P2), respectively. Property (P3) is more involve and its proof appears in “Proof of property” subsection.

Observation 3

(“Property (P1)”). $p_\ell =x_{mid} ~~~ \text{ or } ~~~ p_r=x_{mid}.$

Proof

By Observation 2 (with $D_1= F _1$, $D_3= F _3$ and ${u_{13}}={I({ F _1, F _3})}$), we have that ${I({ F _1, F _3})}\in \partial F _{1}^U$ or ${I({ F _1, F _3})}\in \partial F _{3}^U$. Thus,

$$\begin{aligned} p_\ell = x \left( {I({ F _1, F _3})}\right) ~~~ \text{ or } ~~~ p_r= x \left( {I({ F _1, F _3})}\right) . \end{aligned}$$

The assertion follows. $\square$

Claim

(“Property (P2)”). $p_r-p_\ell \le 4\pi \rho / \tau$.

Proof

If ${I({ F _1, F _3})}\in \partial F _{1}^U$ and ${I({ F _1, F _3})}\in \partial F _{3}^U$, then $p_\ell =p_r$ and the claim follows. Otherwise (${I({ F _1, F _3})}\not \in \partial F _{1}^U$), we have ${I({ F _1, F _3})}\in \partial F _{3}^U$, by Corollary 1 with $D_1=\mathcal{F}_1$, $D_3=\mathcal{F}_3$ and ${u_{13}}={I({ F _1, F _3})}$. Thus, $p_\ell = x \left( {I({ F _1, F _3})}\right)$. Next, we prove that ${I({ F _1,\tilde{ F }_3})}\in \partial F _{1}^U$. Assume, toward contradiction, that ${I({ F _1,\tilde{ F }_3})}\not \in \partial F _{1}^U$. Then, by Corollary 1 with $\mathcal{D}_1=\mathcal{F}_1$, $\mathcal{D}_3=\tilde{\mathcal{F}}_3$ and ${u_{13}}={I({ F _1,\tilde{ F }_3})}$, it follows that $\tilde{\mathcal{F}}_3\setminus \mathcal{F}_1=\emptyset$, hence $(\tilde{\mathcal{F}}_3\cap \tilde{\mathcal{F}}_1)- V =\emptyset$, contradiction.

We have ${I({ F _1, F _3})}\not \in \partial F _{1}^U$ and ${I({ F _1,\tilde{ F }_3})}\in \partial F _{1}^U$, which implies that $p_{1}^R\in \partial F _{1}[{I({ F _1,\tilde{ F }_3})},{I({ F _1, F _3})}]$. Therefore,

$$\begin{aligned} \Vert {p_{1}^R,{I({ F _1, F _3})}} \Vert\le & {} \ell _{\partial F _1}\left[ {I({ F _1,\tilde{ F }_3})},{I({ F _1, F _3})}\right] ~\le ~ 4\pi \rho / \tau , \end{aligned}$$

where the last inequality holds by Claim A.3.1. Recall that $p_r= x \left( r_1\right) +1= x \left( p_{1}^R\right)$, and $p_\ell = x \left( {I({ F _1, F _3})}\right)$, thus $p_r-p_\ell \le 4\pi \rho / \tau$ as well. A symmetric argument also proves the case where ${I({ F _1, F _3})}\not \in \partial F _{3}^U$. The claim follows.      $\square$

1.4.1 Proof of property (P3)

We first prove a claim concerning three disks.

Claim

If ${u_{13}}\in \partial D_{2}^U$, then $\mathcal{D}_2\setminus (\mathcal{D}_1\cup \mathcal{D}_3)=\emptyset$ (See Fig. 19).

Fig. 19

The case where ${u_{13}}\in \partial D_{2}^U$. It follows that $\mathcal{D}_2\setminus (\mathcal{D}_1\cup \mathcal{D}_3)=\emptyset$

Proof

Suppose that ${u_{13}}\in \partial D_{2}^U$. If ${u_{13}}={u_{12}}$ and ${u_{13}}={u_{23}}$, then by Corollary 1 or Corollary 3, with $i=1$ and $j=2$, it follows that $\mathcal{H}_L({u_{12}})\cap \mathcal{D}_2\setminus \mathcal{D}_1=\emptyset$, and by symmetric arguments, $\mathcal{H}_R({u_{23}})\cap \mathcal{D}_2\setminus \mathcal{D}_3=\emptyset$. Thus, $\mathcal{D}_2\setminus (\mathcal{D}_1\cup \mathcal{D}_3)=\emptyset$, since $x \left( {u_{12}}\right) = x \left( {u_{23}}\right) = x \left( {u_{13}}\right)$.

If ${u_{13}}\not ={u_{12}}$, then ${u_{13}}={u'_{12}}\in \partial D_{2}^U$, which implies together with Claim A.2 that $y \left( {u_{12}}\right) > y \left( {u'_{12}}\right) \ge y \left( r_2\right)$, hence

$$\begin{aligned} {u_{12}}\in \partial D_{2}^U. \end{aligned}$$

(16)

In addition,

$$\begin{aligned} x \left( {u'_{12}}\right) < x \left( {u_{12}}\right) , \end{aligned}$$

(17)

since ${u_{12}}\in \partial D_{2}[{u'_{12}},p_{2}^R]$ and ${u'_{12}}\in \partial D_{2}^U$.

Therefore, by Observation 2, ${u_{13}}={u'_{12}}\not \in \partial D_{1}^U$, which implies together with Claim A.2 that $y \left( {u'_{13}}\right)< y \left( {u_{13}}\right) < y \left( r_1\right)$. Thus,

$$\begin{aligned} {u_{13}}\not \in \partial D_{1}^U \text{ and } {u'_{13}}\not \in \partial D_{1}^U. \end{aligned}$$

(18)

A symmetric argument proves that, if ${u_{13}}\not ={u_{23}}$, then

$$\begin{aligned} {u_{13}}\not \in \partial D_{3}^U \text{ and } {u'_{13}}\not \in \partial D_{3}^U. \end{aligned}$$

(19)

Therefore, the right relation of (18) and the right relation of (19) imply that the case where ${u_{13}}\not ={u_{12}}$ and ${u_{13}}\not ={u_{23}}$ leads to contradiction with Observation 2.

Suppose that ${u_{13}}\not ={u_{12}}$ and ${u_{13}}={u_{23}}$. By Corollary 1 or Corollary 3 with $i=2$, $j=3$ and ${u'_{12}}={u_{13}}={u_{23}}$, we have

$$\begin{aligned} \mathcal{H}_R({u'_{12}})\cap \mathcal{D}_2\setminus \mathcal{D}_3=\emptyset , \end{aligned}$$

since ${u_{23}}\in \partial D_{2}^U$. By (16) and Corollary 1 or Corollary 3 with $i=1$ and $j=2$, we have

$$\begin{aligned} \mathcal{H}_L({u_{12}})\cap \mathcal{D}_2\setminus \mathcal{D}_1=\emptyset . \end{aligned}$$

Combining the above two equalities with (17), we get that $\mathcal{D}_2\setminus (\mathcal{D}_1\cup \mathcal{D}_3)=\emptyset$. A symmetric argument proves that, if ${u_{13}}\not ={u_{23}}$, then $\mathcal{D}_2\setminus (\mathcal{D}_1\cup \mathcal{D}_3)=\emptyset$. The claim follows. $\square$

Claim

Suppose that $\tilde{\mathcal{F}}_2\setminus (\mathcal{F}_1\cup \mathcal{F}_3)\not =\emptyset$. Then, there exists a point $q\in \tilde{ F }_2$ such that $x \left( q\right) = x \left( {I({ F _1, F _3})}\right)$ and $y \left( q\right) > y \left( {I({ F _1, F _3})}\right)$. (See Fig. 20)

Proof

Let $\mathcal{D}_1=\mathcal{F}_1$, $\mathcal{D}_2=\tilde{\mathcal{F}}_2$ and $\mathcal{D}_3=\mathcal{F}_3$ and consequently, $D_1= F _1$, $D_2=\tilde{ F }_2$, $D_3= F _3$, $\varrho _1=\varrho _3=1$, $\varrho _2=1+\rho$ and ${u_{13}}={I({ F _1, F _3})}$.

Thus, $\mathcal{D}_2\setminus (\mathcal{D}_1\cup \mathcal{D}_3)\not =\emptyset$. Suppose that $y \left( q\right) \le y \left( {u_{13}}\right)$, for every $q\in D_2$ such that $x \left( q\right) = x \left( {u_{13}}\right)$. Let $r^{\prime }_2$ be a point such that $x \left( r^{\prime }_2\right) = x \left( r_2\right)$ and $q\in \partial D_{2}^U$. (There exists such a point since $x \left( r_1\right)< x \left( r_2\right) < x \left( r_3\right)$ and $\varrho _2\ge \varrho _1=\varrho _3=1$.) Thus, ${u_{13}}\in \partial D_{2^{\prime }}^U$. In addition, $\mathcal{D}_{2}\subseteq \mathcal{D}_{2^{\prime }}$, since $y \left( r^{\prime }_2\right) \ge y \left( r_2\right)$. Combining this together with Claim A.4.1, we get that $\mathcal{D}_{2^{\prime }}\setminus (\mathcal{D}_1\cup \mathcal{D}_3)=\emptyset$, hence also $\mathcal{D}_{2}\setminus (\mathcal{D}_1\cup \mathcal{D}_3)=\emptyset$. The claim follows. $\square$

Fig. 20

The case where $\tilde{\mathcal{F}}_2 \setminus (\mathcal{F}_1 \cup \mathcal{F}_3) \not = \emptyset$. The point $q \in \tilde{ F }_2$ has the following two properties: $x \left( q\right) = x \left( {I({ F _1, F _3})}\right)$; and $y \left( q\right) > y \left( {I({ F _1, F _3})}\right)$

We now ready to prove Property (P3).

Claim

(“Property (P3)”). $[\hat{x}_\ell ,\hat{x}_r]\cap [p_\ell ,p_r]\not =\emptyset$.

Proof

We first prove that

$$\begin{aligned} {I({ F _1,\tilde{ F }_2})}\in \partial F _{1}[p_{1}^L,{I({ F _1, F _3})}] \end{aligned}$$

(20)

and

$$\begin{aligned} {I({\tilde{ F }_2, F _3})}\in \partial F _{3}[{I({ F _1, F _3})},p_{3}^R] \, . \end{aligned}$$

(21)

If ${I({ F _1, F _3})}\in \tilde{ F }_2$, then the above assertions hold trivially.

Next, we prove that, indeed, ${I({ F _1, F _3})}\in \tilde{ F }_2$. Assume toward contradiction that ${I({ F _1, F _3})}\not \in \tilde{ F }_2$. This implies, in particular, that

$$\begin{aligned} {I({ F _1, F _3})}\not \in [0,\bar{y}]^2, \end{aligned}$$

(22)

since $[0,\bar{y}]^2\subseteq \tilde{ F }_2$.

In addition, by Claim A.4.1, there exists q such that x(q)=x(I(1,3)) and y(q)=y(I(1,3)), there exists a point $q\in \tilde{ F }_2$ such that $x \left( q\right) = x \left( {I({ F _1, F _3})}\right)$ and $y \left( q\right) \ge y \left( {I({ F _1, F _3})}\right)$.

This implies that $y \left( {I({ F _1, F _3})}\right) < y \left( r_2\right)$, since $\Vert {{I({ F _1, F _3})},r_2} \Vert >1+\rho$; and $\Vert {q^{\prime },r_2} \Vert \le \Vert {q,r_2} \Vert \le 1+\rho$, for every point $q^{\prime }$ such that $x \left( q^{\prime }\right) = x \left( q\right)$ and $y \left( r_2\right) \le y \left( q^{\prime }\right) \le y \left( q\right)$. In addition, by Observation 2 (with $D_1= F _1$, $D_3= F _3$ and ${u_{13}}={I({ F _1, F _3})}$), we have $y \left( {I({ F _1, F _3})}\right) \ge 0$. Thus, also

$$\begin{aligned} y \left( {I({ F _1, F _3})}\right) \in [0,\bar{y}]. \end{aligned}$$

This implies, together with (22), that

$$\begin{aligned} x \left( {I({ F _1, F _3})}\right) \not \in [0,\bar{y}]. \end{aligned}$$

Thus, if $y \left( r_1\right) \ge y \left( r_3\right)$, then trivially $x \left( {I({ F _1, F _3})}\right) >0$, thus by the above inequality, also $x \left( {I({ F _1, F _3})}\right) >\bar{y}$. This implies that

$$\begin{aligned} | x \left( {I({ F _1, F _3})}\right) - x \left( r_2\right) |<| x \left( {I({ F _1, F _3})}\right) - x \left( r_1\right) | \end{aligned}$$

(23)

(recalling that $\bar{y}> x \left( r_2\right) > x \left( r_1\right)$). Thus ,

$$\begin{aligned} ( y \left( {I({ F _1, F _3})}\right) - y \left( r_2\right) )^2 \,> \,&(1 + \rho )^2 - ( x \left( {I({ F _1, F _3})}\right) - x \left( r_2\right) )^2 \\ > \,&1 + 2 \rho + \rho ^2 \\&- ( x \left( {I({ F _1, F _3})}\right) - x \left( r_1\right) )^2 \\ = \,&( y \left( {I({ F _1, F _3})}\right) - y \left( r_1\right) )^2 \\&+ 2 \rho + \rho ^2, \end{aligned}$$

where the first inequality holds since ${I({ F _1, F _3})}\not \in \tilde{ F }_2$; the second inequality holds by (23); and the last equality holds since $\Vert {{I({ F _1, F _3})},r_1} \Vert =1$. Hence,

$$\begin{aligned} y \left( r_2\right) - y \left( {I({ F _1, F _3})}\right) >| y \left( r_1\right) - y \left( {I({ F _1, F _3})}\right) |+\rho , \end{aligned}$$

since $y \left( r_2\right) > y \left( {I({ F _1, F _3})}\right)$ and $( y \left( {I({ F _1, F _3})}\right) - y \left( r_1\right) )^2\le 1$. Therefore, $y \left( r_2\right) > y \left( r_1\right) +\rho$. A symmetric argument proves that if $y \left( r_3\right) \ge y \left( r_1\right)$, then $y \left( r_2\right) > y \left( r_3\right) +\rho$. Thus,

$$\begin{aligned} y \left( r_2\right) >\max \{ y \left( r_1\right) , y \left( r_3\right) \}+\rho . \end{aligned}$$

This implies that $p_{2}^U\not \in \tilde{ F }_1$ and $p_{2}^U\not \in \tilde{ F }_3$. Therefore, $x \left( {I({\tilde{ F }_1, F _2})}\right) < x \left( r_2\right)$ and $x \left( {I({ F _2,\tilde{ F }_3})}\right) > x \left( r_2\right)$. Hence,

$$\begin{aligned} x \left( {I({\tilde{ F }_1, F _2})}\right) < x \left( {I({ F _2,\tilde{ F }_3})}\right) . \end{aligned}$$

Thus, by Claim A.3.2 (with $\mathcal{D}_1=\tilde{\mathcal{F}}_1$, $\mathcal{D}_2=\mathcal{F}_2$, $\mathcal{D}_3=\tilde{\mathcal{F}}_3$, ${u_{12}}={I({\tilde{ F }_1, F _2})}$ and ${u_{23}}={I({ F _2,\tilde{ F }_3})}$), we have that $(\tilde{\mathcal{F}}_1 \cap \tilde{\mathcal{F}}_3) - \mathcal{F}_2 = \emptyset$, which leads to contradiction. Relations (20) and (21) follow.

Now, we prove that $\hat{x}_\ell \le p_r$. If ${I({ F _1, F _3})}\in \partial F _{1}^U$, then $p_r= x \left( {I({ F _1, F _3})}\right)$ and $x \left( {I({ F _1,\tilde{ F }_2})}\right) \le x \left( {I({ F _1, F _3})}\right)$ by (20). Thus, $x \left( {I({ F _1,\tilde{ F }_2})}\right) \le p_r$. Otherwise, ${I({ F _1, F _3})}\not \in \partial F _{1}^U$, and then $p_r= x \left( r_1\right) +1\ge x \left( {I({ F _1,\tilde{ F }_2})}\right)$. Recall that $\hat{x}_\ell \le x \left( {I({ F _1,\tilde{ F }_2})}\right)$, thus $\hat{x}_\ell \le p_r$. A symmetric argument also proves that $p_\ell \le \hat{x}_r$. The claim follows. $\square$

Lemma 3 follows.