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On the Number of Maximal Cliques in
Two-Dimensional Random Geometric Graphs:
Euclidean and Hyperbolic

Hodaka Yamaji

Graduate School of Information Science and Technology

The University of Tokyo

Tokyo 113-8656, Japan [email protected]

Abstract: Maximal clique enumeration appears in various real-world networks, such as social networks and protein-protein interaction networks for different applications. For general graph inputs, the number of maximal cliques can be up to 3|V|/33^{|V|/3}. However, many previous works suggest that the number is much smaller than that on real-world networks, and polynomial-delay algorithms enable us to enumerate them in a realistic-time span. To bridge the gap between the worst case and practice, we consider the number of maximal cliques in two popular models of real-world networks: Euclidean random geometric graphs and hyperbolic random graphs. We show that the number of maximal cliques on Euclidean random geometric graphs is lower and upper bounded by exp(Ω(|V|1/3))\exp(\Omega(|V|^{1/3})) and exp(O(|V|1/3+ϵ))\exp(O(|V|^{1/3+\epsilon})) with high probability for any ϵ>0\epsilon>0. For a hyperbolic random graph, we give the bounds of exp(Ω(|V|(3γ)/6))\exp(\Omega(|V|^{(3-\gamma)/6})) and exp(O(|V|(3γ+ϵ)/6)))\exp(O(|V|^{(3-\gamma+\epsilon)/6)})) where γ\gamma is the power-law degree exponent between 2 and 3.

Keywords: Maximal Cliques, Random Geometric Graphs, Real-World Networks

1 Introduction

1.1 Background

Detecting all maximal cliques in a graph is a crucial analysis tool for real-world networks from various fields: social networks, protein-protein interaction networks, and web graphs because cliques correspond to meaningful components in the networks [22, 21, 3]. Not only does it have many direct applications, but its algorithms and techniques are used in other clique-related methods such as clique percolation [19] and kk-clique counting [17]. This is because we can detect all cliques by enumerating only the maximal ones.

For general graph inputs, the number of maximal cliques \mathcal{M} can be up to 3|V|/33^{|V|/3} [20]. Therefore, enumerating all of them is NP-hard. However, many studies report that in real-world networks, \mathcal{M} is much smaller than that. Thus, polynomial-delay algorithms, the running time of which is bounded by poly(|V|)poly(|V|)\cdot\mathcal{M}, can enumerate all maximal cliques in realistic-time span even for networks with millions of vertices [8]. Also, classic Bron-Kerbosch algorithm [5] (plus graph orientation [7]) is known to be efficient in many instances [9], although its worst running time is O(3|V|/3)O^{*}(3^{|V|/3}) and not bounded in terms of \mathcal{M}. Here we strike upon the question: why is the number of maximal cliques small on real-world networks?

In the study of real-world networks, networks that appear naturally in various fields are considered. In terms of the graph structure, it seems that networks from different domains are entirely different from each other. However, it is known that they share specific common properties. For example, they have a power-law degree distribution: the number of nodes with a vertex degree of kk is proportional to kγk^{-\gamma}. In many cases, γ\gamma is between two and three, and these networks are called scale-free. Additionally, they have the triadic closure property, meaning that if two vertices have common neighbors, they are likely to be connected. The property is often described with a measure called the clustering coefficient, and real-world graphs often have a high clustering coefficient. Other common properties include tree-like structures, small diameter, and small clique number.

One of the most combinatorially studied models of real-world networks is hyperbolic random graphs [18]. The graph is generated by independently placing vertices according to a particular distribution in a two-dimensional space with negative curvature and connecting two vertices within a certain distance. It is thus classified as a random geometric graph. It is known that a hyperbolic plane naturally induces power-law degree distribution [18]. Also, a hyperbolic random graph satisfies a high clustering coefficient with high probability [14], which distinguishes itself from other power-law models such as Barabási-Albert [2] and Chung-Lu random graphs. Parameters studied in this model include: the number of kk-cliques, clique number [12], treewidth [4], modularity [6], and diameter [13].

1.2 Our Main Results and Contributions

In this paper, we consider the number of maximal cliques in hyperbolic random graphs. We also consider the number on two-dimensional Euclidean random geometric graphs, which are the Euclidean counterpart to hyperbolic random graphs. Euclidean random geometric graphs are also thought to be good representations of some types of real-world graphs [15, 16], although they do not possess power-law degree distribution. Our findings are as follows:

Theorem 1.1 (Main 1)

Let r<1r<1 be a constant. Let \mathcal{M} be the number of maximal cliques in a two-dimensional Euclidean random geometric graph whose connection distance is rr. There exists positive constants C1C_{1} and C2C_{2} such that for all ϵ>0\epsilon>0,

Pr[exp(C1|V|1/3)exp(C2|V|1/3+ϵ)]1\displaystyle\Pr[\exp(C_{1}|V|^{1/3})\leq\mathcal{M}\leq\exp(C_{2}|V|^{1/3+\epsilon})]\rightarrow 1

as |V||V|\rightarrow\infty.

Theorem 1.2 (Main 2)

Let γ(2,3)\gamma\in(2,3). Let \mathcal{M} be the number of maximal cliques in a hyperbolic random graph whose power-law degree exponent is γ\gamma. There exist positive constants C1C_{1} and C2C_{2} such that for all ϵ>0\epsilon>0,

Pr[exp(C1|V|(3γ)/6)exp(C2|V|(3γ)/6+ϵ)]1\displaystyle\Pr[\exp(C_{1}|V|^{(3-\gamma)/6})\leq\mathcal{M}\leq\exp(C_{2}|V|^{(3-\gamma)/6+\epsilon})]\rightarrow 1

as |V||V|\rightarrow\infty.

For hyperbolic random graphs, we consider the case when γ(2,3)\gamma\in(2,3) for hyperbolic random graphs. In this case, the graphs are scale-free and have exp(Ω(|V|(3γ)/2))\exp(\Omega(|V|^{(3-\gamma)/2})) cliques with high probability [12]. In general graphs, the number of maximal cWe and the bound we obtained is much smaller than that.

To prove the main theorem, we consider what is called an octahedral graph OtO_{t}. The definition of the graph is the following. Let tK2=(V,E)tK_{2}=(V,E) where V={1,2,,2t}V=\{1,2,...,2t\} and E={(i,i+t):1it}E=\{(i,i+t):1\leq i\leq t\}. Therefore, tK2tK_{2} is a graph with tt pairwise disjoint edges. An octahedral graph OtO_{t} is the complement of tK2tK_{2}. We have the following theorems from the previous study.

Theorem 1.3 (Forklore)

If the graph has OtO_{t} as a vertex-induced subgraph, then the number of maximal cliques is lower-bounded by 2t2^{t}.

Theorem 1.4 (M. Farber, M. Hujter, and Z. Tuza [10])

If |V|4t|V|\geq 4t and there exists no Ot+1O_{t+1} as a vertex-induced subgraph, then the number of maximal cliques is upper-bounded by (|V|/t)2t(|V|/t)^{2t}.

Let τ(G)\tau(G) be the maximum tt such that a graph GG has OtO_{t} as its vertex-induced subgraph. With these theorems, all that remains is to bound τ\tau. If τ\tau is a constant, then the number of maximal cliques is polynomial. Unfortunately, this is not the case for the two random geometric graphs. However, our bounds on τ\tau are much smaller than the obvious O(|V|)O(|V|) bound.

Intuitively, τ\tau is small because any pair of unconnected vertices have 2t22t-2 common neighbors, which is against the triadic closure property, i.e. two vertices with common neighbors are likely to be connected. As tt gets larger, the more severe the violation becomes. This explanation can be mathematically justified on Euclidean and hyperbolic random geometric graphs. The arguments on the two different random graphs are basically the same even though the definitions of distance are different, and the parallel postulate does not hold on a hyperbolic plane.

Our contributions are briefly summarized as follows. Firstly, to the best of our knowledge, this is the first work that assesses the number of maximal cliques on random geometric graphs. We shed light on the importance of OtO_{t} and develop geometric and probabilistic techniques to determine its size. Those techniques apply to both Euclidean and hyperbolic planes. Secondly, what we have found is yet another result followed by cc-closed graphs [11] supporting that the triadic closure property plays an essential role in maximal cliques. Lastly, we give an upper bound of τ\tau on real-world graph datasets. It turns out that τ\tau is often at most 5-20, even on networks with hundreds of thousands of vertices (See Table 1 at the end of this paper). The upper bound of \mathcal{M} given by τ\tau is still far from the actual value. However, it is still surprising how small τ\tau is in practice.

1.3 Proof Sketch

Here we explain how we bound τ\tau. For the lower bound, we construct regions so that vertices on them would form OtO_{t}. Here we use a technique called Poissonization introduced in [23].

To derive the upper bound, We define a set of segments 𝒮:={vw¯:v,wV(G),\mathcal{S}:=\{\overline{vw}:v,w\in V(G), the distance between vv and ww is greater than the connection distance}\}. Since 𝒮\mathcal{S} is somewhat like the complement of the graph, a vertex-induced subgraph that is isomorphic to OtO_{t} corresponds to tt segments that are pairwise disjoint (or we call “independent” in this paper). We soon find out that any two independent segments must intersect. Therefore, we can define an “angle” between them. By the pigeonhole principle, if tt is large enough, then there exists a set of segments with decent cardinality whose “angles” are small (Lemma 2.10 and Lemma 3.14). We also prove that if two intersecting segments have small “angles”, the geometry is very restricted. If we fix one segment, then endpoints of the other segments can only exist in small regions (Corollary 2.8 and Corollary 3.12). By combining the two facts, we derive that tt cannot get too large because the number of vertices in those small regions cannot be too large with high probability.

Unlike Euclidean random geometric graphs, vertices in hyperbolic random graphs are placed by non-uniform distribution. To deal with this, we classify the segments by the density around their endpoints. We treat the segments in dense regions slightly differently from those in sparse regions. However, the overall strategies of the proof are the same as Euclidean random geometric graphs.

Refer to caption
Refer to caption
Figure 1: Illustrations of the proof sketch. The left one illustrates tt independent segments, corresponding to OtO_{t} and the claim of Lemma 2.10 and Lemma 3.14. The right one illustrations Corollary 2.8 and Corollary 3.12.

1.4 Organization of the Paper

In Section 2, we define Euclidean random geometric graphs and prove the main theorem. In Section 3.1-3.3, we define hyperbolic random graphs and discuss how the proof of the random geometric graphs on a Euclidean plane can be extended to a hyperbolic plane. For the upper bound, we prove the weaker version of the main theorem. In Section 3.4, we discuss how to treat the non-uniformity of hyperbolic random graphs and prove the complete version of the main theorem. Finally, Section 4 is the conclusion.

2 Euclidean Random Geometric Graphs

2.1 Definitions

Let n+n\in\mathbb{N}^{+}, and r(0,1)r\in(0,1). A two-dimensional Euclidean random geometric graph Gn,rG_{n,r} is obtained as below:

  • The vertex set is V={1,2,,n}V=\{1,2,...,n\}.

  • The vertices are identically and independently distributed on [0,1]2[0,1]^{2} according to a probability density function f(x,y)=1f(x,y)=1.

  • The edge set EE is given by {(u,v):dist(u,v)r}\{(u,v):dist(u,v)\leq r\}

Here, dist(u,v)=(uxvx)2+(uyvy)2dist(u,v)=\sqrt{(u_{x}-v_{x})^{2}+(u_{y}-v_{y})^{2}} where (ux,uy)(u_{x},u_{y}) and (vx,vy)(v_{x},v_{y}) are the xyxy-coordinates of uu and vv respectively. From here, we often identify a vertex with its position.

Given a region UU on [0,1]2[0,1]^{2} (where we can perform integration), define F(U):=Uf(x,y)𝑑x𝑑yF(U):=\int_{U}f(x,y)dxdy. F(U)F(U) is equal to the probability that a vertex lies on UU. Note that we can assume no three vertices lie on a single line since the probability is 0. This avoids some corner case arguments.

2.1.1 Poissonization

This is introduced in [23]. Let NnN_{n} be a Poisson random variable with mean nn. Consider GNn,rG_{N_{n},r} where the number of vertices is chosen by the Poisson distribution NnN_{n}. Then, on such a random graph, the number of vertices on a region UU follows a Poisson distribution with mean nF(U)nF(U). In particular, the probability that UU has no vertex on itself is enF(U)e^{-nF(U)}. Moreover, for disjoint regions U1U_{1} and U2U_{2}, the number of vertices on each region is an independent Poisson distribution with mean nF(U1)nF(U_{1}) and nF(U2)nF(U_{2}), respectively. Therefore, on Poissonized random geometric graphs, arguments of multiple disjoint regions are easier. Let AA be some property of a graph. Then

Pr[GNn,r has A]\displaystyle\Pr[G_{N_{n},r}\text{ has }A] Pr[GNn,r has A|Nn=n]Pr[Nn=n]\displaystyle\geq\Pr[G_{N_{n},r}\text{ has }A|N_{n}=n]\cdot\Pr[N_{n}=n]
=Pr[Gn,r has A]ennnn!\displaystyle=\Pr[G_{n,r}\text{ has }A]\cdot\frac{e^{-n}n^{n}}{n!}

Here, ennn/n!=Θ(n1/2)e^{-n}n^{n}/n!=\Theta(n^{-1/2}) by the Stirling’s approximation. Thus, we get

Pr[Gn,r has A]Θ(n1/2)Pr[GNn,r has A]\displaystyle\Pr[G_{n,r}\text{ has }A]\leq\Theta(n^{1/2})\cdot\Pr[G_{N_{n},r}\text{ has }A] (2.1)

Although our results are on the normal random geometric graph Gn,rG_{n,r}, we use this technique to derive the lower bound of the number of maximal cliques.

2.2 Lower Bound of the Number of Maximal Cliques

Let k4k\geq 4 be an integer. Let o=(1/2,1/2)o^{\prime}=(1/2,1/2). Consider taking a polar coordinate system whose origin is oo^{\prime}. Let θ0:=π/(3k)\theta_{0}:=\pi/(3k). For 1i2k1\leq i\leq 2k, Define Ui:={(r,ϕ):r1rr2, 3(i1)θ0ϕ(3(i1)+1)θ0}U_{i}:=\{(r^{\prime},\phi):r_{1}\leq r^{\prime}\leq r_{2},\ 3(i-1)\theta_{0}\leq\phi\leq(3(i-1)+1)\theta_{0}\} where r1:=r/2+2cos(θ0/2)r_{1}:=r/\sqrt{2+2\cos(\theta_{0}/2)} and r2:=r/2+2cos(2θ0)r_{2}:=r/\sqrt{2+2\cos(2\theta_{0})}. With some calculations, we can confirm the following.

Refer to caption
Figure 2: Definition of r1r_{1}, r2r_{2}, θ0\theta_{0} and UiU_{i} in the lower bound construction
Proposition 2.1

Let 1ik1\leq i\leq k and 1j2k1\leq j\leq 2k. For a vertex vv on UiU_{i} and a vertex ww on UjU_{j},

dist(v,w)\displaystyle dist(v,w) >r(j=i+k)\displaystyle>r\ (j=i+k)
dist(v,w)\displaystyle dist(v,w) r(ji+k)\displaystyle\leq r\ (j\neq i+k)

Proof: Let rvr_{v} and rwr_{w} be radial coordinates of vv and ww, respectively. If j=i+kj=i+k, then

{dist(v,w)}2\displaystyle\{dist(v,w)\}^{2} =rv2+rw22rvrwcoswov\displaystyle=r_{v}^{2}+r_{w}^{2}-2r_{v}r_{w}\cos\angle wo^{\prime}v
rv2+rw22rvrwcos(πθ0)\displaystyle\geq r_{v}^{2}+r_{w}^{2}-2r_{v}r_{w}\cos(\pi-\theta_{0})
2r12+2r12cosθ0\displaystyle\geq 2r_{1}^{2}+2r_{1}^{2}\cos\theta_{0}
>r2\displaystyle>r^{2}

Otherwise, {dist(v,w)}22r22+2r22cos2θ0r2\{dist(v,w)\}^{2}\leq 2r_{2}^{2}+2r_{2}^{2}\cos 2\theta_{0}\leq r^{2}\Box

For 1ik1\leq i\leq k, let tit_{i} be 0-1 random variables which are equal to 1 if and only if both UiU_{i} and Ui+kU_{i+k} have at least one vertex on themselves. Let t=i=1ktit=\sum_{i=1}^{k}t_{i}. Then, there exists OtO_{t} as a vertex-induced subgraph. We are left to lower bound tt.

Proposition 2.2

F(U1)=Ω(θ03)F(U_{1})=\Omega(\theta_{0}^{3}) as θ00\theta_{0}\rightarrow 0

Proof:

F(U1)\displaystyle F(U_{1}) =12(r22θ0r12θ0)\displaystyle=\frac{1}{2}\left(r_{2}^{2}\theta_{0}-r_{1}^{2}\theta_{0}\right)
=r2θ04(11+cos(2θ0)11+cos(θ0/2))\displaystyle=\frac{r^{2}\theta_{0}}{4}\left(\frac{1}{1+\cos(2\theta_{0})}-\frac{1}{1+\cos(\theta_{0}/2)}\right)
=r2θ04(12+θ022+O(θ04)(12+θ0232+O(θ04)))\displaystyle=\frac{r^{2}\theta_{0}}{4}\left(\frac{1}{2}+\frac{\theta_{0}^{2}}{2}+O(\theta_{0}^{4})-\left(\frac{1}{2}+\frac{\theta_{0}^{2}}{32}+O(\theta_{0}^{4})\right)\right)
=15128r2θ03+O(θ05)\displaystyle=\frac{15}{128}r^{2}\theta_{0}^{3}+O(\theta_{0}^{5})
=Ω(θ03)\displaystyle=\Omega(\theta_{0}^{3})

On the third line, we used the Taylor expansion.  \Box

Proposition 2.3

If k=n1/3k=n^{1/3}, then there exists a positive constant CC such that Pr[tCn1/3]0\Pr[t\leq Cn^{1/3}]\rightarrow 0 as nn\rightarrow\infty

Proof: Consider the Poissonization GNn,rG_{N_{n},r}. For each ii, the number of vertices on UiU_{i} follows an independent and identical Poisson distribution with mean nF(U1)nF(U_{1}). Therefore, t1,,tkt_{1},...,t_{k} are independent and identical random variables. Let p:=1enF(U1)p:=1-e^{-nF(U_{1})}. This is equal to the probability that UiU_{i} has at least one vertex on itself. By the Proposition 2.2, p=Ω(1)p=\Omega(1). Also, we have 𝔼[t]=kp2=Ω(n1/3)\mathbb{E}[t]=kp^{2}=\Omega(n^{1/3}). Since t1,,tnt_{1},...,t_{n} are independent, we can apply the Chernoff bound and obtain Pr[tkp2/2]exp(Ω(n1/3))\Pr[t\leq kp^{2}/2]\leq\exp(-\Omega(n^{1/3})). Now we are back to the normal model Gn,rG_{n,r}. Using (2.1), it follows that Pr[tkp2/2]Ω(n1/2)exp(Ω(n1/3))\Pr[t\leq kp^{2}/2]\leq\Omega(n^{1/2})\cdot\exp(-\Omega(n^{1/3})). This goes to 0 as nn\rightarrow\infty.  \Box

By combining the proposition with the Theorem 1.3, we obtain the main theorem regarding (Gn,r)\mathcal{M}(G_{n,r}).

Corollary 2.4

There exists a positive constant CC such that Pr[exp(Cn1/3)(Gn,r)]1\Pr[\exp(Cn^{1/3})\leq\mathcal{M}(G_{n,r})]\rightarrow 1 as nn\rightarrow\infty.

2.3 Upper Bound of the Number of Maximal Cliques

Let vw¯\overline{vw} denote the segment between vertices vv and ww on a Euclidean plane. Let 𝒮:={vw¯:v,wV(G),dist(v,w)>r}\mathcal{S}:=\{\overline{vw}:v,w\in V(G),dist(v,w)>r\}. Note that 𝒮\mathcal{S} is like a complement of the random graph. Two segments v1v2¯\overline{v_{1}v_{2}} and w1w2¯\overline{w_{1}w_{2}} in 𝒮\mathcal{S} are called independent if v1w1¯\overline{v_{1}w_{1}}, v1w2¯\overline{v_{1}w_{2}}, v2w1¯\overline{v_{2}w_{1}}, and v2w2¯\overline{v_{2}w_{2}} are not in 𝒮\mathcal{S}. Two or more segments in 𝒮\mathcal{S} are called independent if any pair of the segments are independent. From the definition, it is obvious that the following proposition holds.

Proposition 2.5

There is no Ot+1O_{t+1} as a vertex-induced subgraph \Leftrightarrow There is no set of independent segments S𝒮S\subseteq\mathcal{S} whose cardinality is t+1t+1.

Therefore, we are left to bound such tt on Gn,rG_{n,r}. With elementary geometry, we can confirm the following.

Proposition 2.6

On a Euclidean plane, two independent segments intersect.

We defer the proof to the next section. Thanks to the proposition, we can define an angle between two independent segments thanks to the proposition.

Let s:=v1v2¯s:=\overline{v_{1}v_{2}} and s:=w1w2¯s^{\prime}:=\overline{w_{1}w_{2}} be independent segments. Suppose that the counter-clockwise order of four endpoints is v1w1w2v2v_{1}w_{1}w_{2}v_{2}. Let qq be the intersection of two segments. Define a directed angle (s,s):=w1qv1\angle(s,s^{\prime}):=\angle w_{1}qv_{1}. For convinience, define (s,s):=0\angle(s,s):=0. It holds that (s,s)=π(s,s)\angle(s,s^{\prime})=\pi-\angle(s^{\prime},s). Therefore, the order of the two segments matters.

From here, we consider conditions that two independent segments mWe angle between them is known. Again, s=v1v2¯,s=w1w2¯𝒮s=\overline{v_{1}v_{2}},s^{\prime}=\overline{w_{1}w_{2}}\in\mathcal{S} be independent segments. Suppose v1w1v2w2v_{1}w_{1}v_{2}w_{2} is the counter-clockwise order of four endpoints. Let mm be the midpoint of ss. Consider taking a polar coordinate system whose origin and polar axis are mm and mv1\overrightarrow{mv_{1}}. Let (r0,ϕ)(r_{0},\phi) be the polar coordinate of w1w_{1}. Note that r0=dist(m,w1)r_{0}=dist(m,w_{1}) and ϕ=w1mv1\phi=\angle w_{1}mv_{1} (refer to Figure 3). The next proposition states that if the directed angle (s,s)\angle(s,s^{\prime}) is small, then r0r_{0} is bounded tight.

Proposition 2.7

Let 0θ0π/30\leq\theta_{0}\leq\pi/3. If (s,s)θ0\angle(s,s^{\prime})\leq\theta_{0}, then r1r0r2r_{1}\leq r_{0}\leq r_{2} where

r1\displaystyle r_{1} :=(1/2+cosθ0)cosθ0r\displaystyle:=(-1/2+\cos\theta_{0})\sqrt{\cos\theta_{0}}\cdot r
r2\displaystyle r_{2} :=(3/2cosθ0)r\displaystyle:=(3/2-\cos\theta_{0})r

Proof: Let θ:=(s,s)\theta:=\angle(s,s^{\prime}). Let qq be the intersection of ss and ss^{\prime}. Let a:=dist(v1,v2),b:=dist(w1,w2),c=dist(v1,q),d=dist(w1,q)a:=dist(v_{1},v_{2}),b:=dist(w_{1},w_{2}),c=dist(v_{1},q),d=dist(w_{1},q). Note that aa is the length of ss and a>ra>r holds. The same thing stands for bb. From the definition of independence of segments,

dist(v1,w2)\displaystyle dist(v_{1},w_{2}) rc2+(bd)22c(bd)cos(πθ)r2\displaystyle\leq r\Rightarrow c^{2}+(b-d)^{2}-2c(b-d)\cos(\pi-\theta)\leq r^{2}
dist(v2,w1)\displaystyle dist(v_{2},w_{1}) r(ac)2+d22(ac)dcos(πθ)r2\displaystyle\leq r\Rightarrow(a-c)^{2}+d^{2}-2(a-c)d\cos(\pi-\theta)\leq r^{2}

must hold. From the first inequality, we get

c2+(bd)22c(bd)cos(πθ)r2\displaystyle c^{2}+(b-d)^{2}-2c(b-d)\cos(\pi-\theta)\leq r^{2}
(c+(bd)cos(θ))2+((bd)sin(θ))2r2\displaystyle\Leftrightarrow(c+(b-d)\cos(\theta))^{2}+((b-d)\sin(\theta))^{2}\leq r^{2}
c+(bd)cos(θ)r\displaystyle\Rightarrow c+(b-d)\cos(\theta)\leq r
cdrbcos(θ)\displaystyle\Rightarrow c-d\leq r-b\cos(\theta)

With the same argument, the second inequality yields (racos(θ))cd-(r-a\cos(\theta))\leq c-d. Regardless of how v1,m,q,v2v_{1},m,q,v_{2} are lined up on ss, we have r02=(a/2c)2+d2+2(a/2c)dcos(θ)r_{0}^{2}=(a/2-c)^{2}+d^{2}+2(a/2-c)d\cos(\theta). We can upper and lower bound r0r_{0} as below.

r02\displaystyle r_{0}^{2} =(a/2c)2+d2+2(a/2c)dcos(θ)\displaystyle=(a/2-c)^{2}+d^{2}+2(a/2-c)d\cos(\theta)
(a/2c+d)2\displaystyle\leq(a/2-c+d)^{2}
(a/2+racos(θ))2\displaystyle\leq(a/2+r-a\cos(\theta))^{2}
r2(3/2cos(θ0))2\displaystyle\leq r^{2}(3/2-\cos(\theta_{0}))^{2}
r02\displaystyle r_{0}^{2} =(a/2c)2+d2+2(a/2c)dcos(θ)\displaystyle=(a/2-c)^{2}+d^{2}+2(a/2-c)d\cos(\theta)
(a/2c+d)2cos(θ)\displaystyle\geq(a/2-c+d)^{2}\cos(\theta)
(a/2r+bcos(θ))2cos(θ)\displaystyle\geq(a/2-r+b\cos(\theta))^{2}\cos(\theta)
r2(1/2+cos(θ0))2cos(θ0)\displaystyle\geq r^{2}(-1/2+\cos(\theta_{0}))^{2}\cos(\theta_{0})

\Box

Refer to caption
Figure 3: Definitions of points, length, and angles of two independent segments in the Proposition 2.7

If v1,q,m,v2v_{1},q,m,v_{2} are lined up in this order, we can bound ϕ\phi as 0ϕθθ00\leq\phi\leq\theta\leq\theta_{0}. By considering the other case (v1,m,q,v2v_{1},m,q,v_{2} is lined up in this order) in the same way, we have the following.

Corollary 2.8

Let 0θ0π/30\leq\theta_{0}\leq\pi/3. If (s,s)θ0\angle(s,s^{\prime})\leq\theta_{0}, then either w1w_{1} or w2w_{2} must lie on a region U1U_{1} or a region U2U_{2} where

U1\displaystyle U_{1} :={(r0,ϕ):r1r0r2,0ϕθ0}\displaystyle:=\{(r_{0},\phi):r_{1}\leq r_{0}\leq r_{2},0\leq\phi\leq\theta_{0}\}
U2\displaystyle U_{2} :={(r0,ϕ):r1r0r2,πϕπ+θ0}\displaystyle:=\{(r_{0},\phi):r_{1}\leq r_{0}\leq r_{2},\pi\leq\phi\leq\pi+\theta_{0}\}
Refer to caption
Figure 4: U1U_{1} and U2U_{2} in the Corollary 2.8

Then we obtain the next lemma: if the directed angle is bounded small, the expected number of independent segments is also small.

Lemma 2.9

Let s𝒮s\in\mathcal{S} be a segment. For a constant 0θ0π/30\leq\theta_{0}\leq\pi/3, let XX be a number of segments ss^{\prime} where ss and ss^{\prime} are independent and (s,s)θ0\angle(s,s^{\prime})\leq\theta_{0}. Then 𝔼[X]=nO(θ03)\mathbb{E}[X]=nO(\theta_{0}^{3}) as θ00\theta_{0}\rightarrow 0.

Proof: By the Corollary 2.8, XX is at most the number of vertices in U:=U1U2U:=U_{1}\cup U_{2}. Therefore, 𝔼[X]nF(U)\mathbb{E}[X]\leq nF(U). We are left to prove F(U1)=F(U2)=O(θ03)F(U_{1})=F(U_{2})=O(\theta_{0}^{3}) as θ00\theta_{0}\rightarrow 0, and

F(U1)\displaystyle F(U_{1}) 12r22θ012r12θ0\displaystyle\leq\frac{1}{2}r_{2}^{2}\theta_{0}-\frac{1}{2}r_{1}^{2}\theta_{0}
=12r2θ0((3/2cos(θ0))2(1/2+cos(θ0))2cos(θ0))\displaystyle=\frac{1}{2}r^{2}\theta_{0}\left((3/2-\cos(\theta_{0}))^{2}-(-1/2+\cos(\theta_{0}))^{2}\cos(\theta_{0})\right)
=12r2θ0((22cosθ0)+(1/2+cosθ0)2(1cosθ0))\displaystyle=\frac{1}{2}r^{2}\theta_{0}\left((2-2\cos\theta_{0})+(-1/2+\cos\theta_{0})^{2}(1-\cos\theta_{0})\right)
=916r2θ03+O(θ05)\displaystyle=\frac{9}{16}r^{2}\theta_{0}^{3}+O(\theta_{0}^{5})

To derive the fourth line, we use the Taylor expansion.  \Box

To fully use the Lemma 2.9, we prove another lemma, which states that if there exists large OtO_{t}, we can always take a set of segments so that its size is unneglectable, and the directed angles between them are small.

Lemma 2.10

Let kk be a positive integer. If there exists a set of tt independent segments S𝒮S\subseteq\mathcal{S}, then there exists a segment ss^{\prime} and a set of independent segments S𝒮S^{\prime}\subseteq\mathcal{S} such that

{|S|t/ksSs′′S.(s,s′′)π/k\displaystyle\left\{\begin{array}[]{l}|S^{\prime}|\geq\left\lceil t/k\right\rceil\\ s^{\prime}\in S^{\prime}\\ \forall s^{\prime\prime}\in S^{\prime}.\angle(s^{\prime},s^{\prime\prime})\leq\pi/k\\ \end{array}\right. (2.5)

Proof: Let s0=v1v2¯s_{0}=\overline{v_{1}v_{2}} be some segment in SS. For an integer 1ik1\leq i\leq k, define
Si:={sS:i1kπ(s0,s)<ikπ}S_{i}:=\left\{s\in S:\frac{i-1}{k}\pi\leq\angle(s_{0},s)<\frac{i}{k}\pi\right\}. Since 1ikSi=S\bigcup_{1\leq i\leq k}S_{i}=S, we have i=1k|Si|=t\sum_{i=1}^{k}|S_{i}|=t. By the pigeonhole principle, there exists an index ii such that |Si|t/k|S_{i}|\geq\lceil t/k\rceil. Consider taking the segment ss^{\prime} in SiS_{i} so that (s0,s)=mins′′Si(s0,s′′)\angle(s_{0},s^{\prime})=\min_{s^{\prime\prime}\in S_{i}}\angle(s_{0},s^{\prime\prime}). If we can prove (s,s′′)(s′′,s0)(s,s0)\angle(s^{\prime},s^{\prime\prime})\leq\angle(s^{\prime\prime},s_{0})-\angle(s^{\prime},s_{0}), we are done. Let aa be the intersection of s0s_{0} and ss^{\prime}, bb be that of s0s_{0} and s′′s^{\prime\prime}, and cc be that of ss^{\prime} and s′′s^{\prime\prime}. If a=ba=b, then it obviously holds. We have two other cases: v1,b,a,v2v_{1},b,a,v_{2} are lined up on s0s_{0} in this order, and v1,a,b,v2v_{1},a,b,v_{2} are lined up on s0s_{0} in this order. For the former case, a=(s0,s)\angle a=\angle(s_{0},s^{\prime}), b=π(s0,s′′)\angle b=\pi-\angle(s_{0},s^{\prime\prime}) and c=(s,s′′)\angle c=\angle(s^{\prime},s^{\prime\prime}) where a\angle a, b\angle b, c\angle c are the inner angle of abc\triangle abc. Therefore,

a+b+c=π\displaystyle\angle a+\angle b+\angle c=\pi (2.6)
(s,s′′)=π((s0,s))(π(s0,s′′))=(s0,s′′)(s0,s)\displaystyle\Leftrightarrow\angle(s^{\prime},s^{\prime\prime})=\pi-(\angle(s_{0},s^{\prime}))-(\pi-\angle(s_{0},s^{\prime\prime}))=\angle(s_{0},s^{\prime\prime})-\angle(s_{0},s^{\prime})

Here, (2.6) follows from the fact that the sum of the inner angles of a triangle is equal to π\pi.
The latter case can be shown similarly.  \Box

We finally combine the two lemmas to upper bound τ\tau. After that, we apply Theorem 1.4 to get the upper bound of the number of maximal cliques.

Theorem 2.11

Let ϵ\epsilon be a positive constant. Let t=n1/3+ϵt=n^{1/3+\epsilon}. Then,

Pr[Gn,r has Ot as its vertex-induced subgraph]0\displaystyle\Pr[G_{n,r}\text{ has $O_{t}$ as its vertex-induced subgraph}]\rightarrow 0

as nn\rightarrow\infty.

Proof: Let k:=n1/3k:=n^{1/3} and θ0:=π/k\theta_{0}:=\pi/k. Suppose Gn,rG_{n,r} has OtO_{t} as its vertex-induced subgraph. Apply Lemma 2.10 for tt and kk. Then there exists a segment ss^{\prime} and a set of independent segments SS^{\prime} such that (2.5) holds. We claim that for a certain segment ss^{\prime}, the probability that there exists a set of independent segments SS^{\prime} such that (2.5) holds is at most exp(Ω(nϵ))\exp(-\Omega(n^{\epsilon})). By Lemma 2.9, the number of segments that satisfy the third condition of (2.5) is at most the number of vertices on UU, and that expected number is bounded by O(nθ03)=O(1)O(n\theta_{0}^{3})=O(1). However, to satisfy (2.5)(\ref{three}), The cardinarity of SS^{\prime} needs be at least nϵn^{\epsilon}. This means that the number of vertices on UU also needs to be at least nϵn^{\epsilon}. Since the vertices are placed independently, we can apply the Chernoff bound to get the probability bound exp(Ω(nϵ))\exp(-\Omega(n^{\epsilon})).

Finally, we apply a union bound over all segments of which there are at most n(n1)/2n(n-1)/2. We have

Pr[Gn,r has Ot as its vertex-induced subgraph]n(n1)2exp(Ω(nϵ))\displaystyle\Pr[G_{n,r}\text{ has $O_{t}$ as its vertex-induced subgraph}]\leq\frac{n(n-1)}{2}\exp(-\Omega(n^{\epsilon}))

This goes to 0 as nn\rightarrow\infty.  \Box

Corollary 2.12

There exist a positive constant CC such that for all ϵ>0\epsilon>0, Pr[(Gn,r)exp(Cn1/3+ϵ)]1\Pr[\mathcal{M}(G_{n,r})\leq\exp(Cn^{1/3+\epsilon})]\rightarrow 1 as nn\rightarrow\infty

3 Hyperbolic Random Graphs

3.1 Definitions

Given n+n\in\mathbb{N^{+}}, γ(2,)\gamma\in(2,\infty), and CC\in\mathbb{R}, let R:=2logn+CR:=2\log n+C, and α=(γ1)/2\alpha=(\gamma-1)/2. A hyperbolic random graph Gn,γ,CG_{n,\gamma,C} is obtained as below:

  • The vertex set is V={1,2,,n}V=\{1,2,...,n\}.

  • The vertices are identically and independently distributed on a hyperbolic plane. The probability density function by a polar coordinate is:

    f(r,θ)={12παsinh(αr)cosh(αR)1(rR)0(r>R)\displaystyle f(r,\theta)=\begin{cases}\frac{1}{2\pi}\cdot\frac{\alpha\sinh(\alpha r)}{\cosh(\alpha R)-1}&(r\leq R)\\ 0&(r>R)\end{cases}
  • The edge set EE is given by {(u,v):dist(u,v)R}\{(u,v):dist(u,v)\leq R\}

Refer to caption
Figure 5: A hyperbolic random graph (n=500n=500, γ=2.1\gamma=2.1)

In our work, we are interested in the case 2<γ<3(1/2<α<1)2<\gamma<3\ (1/2<\alpha<1) where the generated graph is “scale-free” with high probability. Let (ru,ϕu)(r_{u},\phi_{u}) and (rv,ϕv)(r_{v},\phi_{v}) be polar coordinates of uu and vv respectively. Let θ:=π|π|ϕuϕv||\theta:=\pi-|\pi-|\phi_{u}-\phi_{v}||. Then, the hyperbolic cosine formula suggests that

cosh(dist(u,v))=coshrucoshrvsinhrusinhrvcosθ\displaystyle\cosh(dist(u,v))=\cosh r_{u}\cosh r_{v}-\sinh r_{u}\sinh r_{v}\cos\theta

Again, define F(U):=Uf(r,θ)𝑑r𝑑θF(U):=\int_{U}f(r,\theta)drd\theta. On a hyperbolic plane, the area of UU is equal to sinhrdrdθ=:μ(U)\int\sinh rdrd\theta=:\mu(U). Let ρ(r,θ):=dFdμ=f(r,θ)sinhr\rho(r,\theta):=\frac{dF}{d\mu}=\frac{f(r,\theta)}{\sinh r}. Since the value of ρ(r,θ)\rho(r,\theta) does not depend on θ\theta, we sometimes denote it as ρ(r)\rho(r). Intuitively, ρ\rho is the the probability that a vertex lies on a single unit square around (r,θ)(r,\theta). By differentiating ρ\rho, we can confirm that it is monotonically decreasing with respect to rr. Thus, the graph is dense near the origin of the plane.

Let oo be the origin of the plane. We define B0(d)B_{0}(d) as the set of points pp such that dist(p,o)<ddist(p,o)<d. In other words, B0(d):={(r,ϕ):r<d}B_{0}(d):=\{(r,\phi):r<d\}.

3.2 Lower Bound of the Number of Maximal Cliques

To obtain the lower bound of (Gn,r,C)\mathcal{M}(G_{n,r,C}), we do the same thing as the Euclidean random geometric graphs i.e. We take regions so that vertices on them form OtO_{t}.

Let k4k\geq 4 be an integer. Let θ0:=π/(3k)\theta_{0}:=\pi/(3k). For 1i2k1\leq i\leq 2k, Define Ui:={(r,ϕ):r1rr2, 3(i1)θ0ϕ(3(i1)+1)θ0}U_{i}:=\{(r,\phi):r_{1}\leq r\leq r_{2},\ 3(i-1)\theta_{0}\leq\phi\leq(3(i-1)+1)\theta_{0}\} where r1r_{1} and r2r_{2} are determined as

cosh(2r1)\displaystyle\cosh(2r_{1}) =cosh(R)cos2(θ0/2)\displaystyle=\frac{\cosh(R)}{\cos^{2}(\theta_{0}/2)}
cosh(2r2)\displaystyle\cosh(2r_{2}) =cosh(R)1cos2(θ0)\displaystyle=\frac{\cosh(R)-1}{\cos^{2}(\theta_{0})}

We now assume that r1<r2Rr_{1}<r_{2}\leq R. We later confirm this holds under proper θ0\theta_{0} and nn.

Proposition 3.1

Let 1ik1\leq i\leq k and 1j2k1\leq j\leq 2k. For a vertex vv on UiU_{i} and a vertex ww on UjU_{j},

dist(v,w)\displaystyle dist(v,w) >R(j=i+k)\displaystyle>R\ (j=i+k)
dist(v,w)\displaystyle dist(v,w) R(ji+k)\displaystyle\leq R\ (j\neq i+k)

Proof: Let rvr_{v} and rwr_{w} be radial coordinates of vv and ww, respectively. If j=i+kj=i+k, then

cosh(dist(v,w))\displaystyle\cosh(dist(v,w)) =coshrvcoshrwsinhrvsinhrwcoswov\displaystyle=\cosh r_{v}\cosh r_{w}-\sinh r_{v}\sinh r_{w}\cos\angle wov
coshrvcoshrwsinhrvsinhrwcos(πθ0)\displaystyle\geq\cosh r_{v}\cosh r_{w}-\sinh r_{v}\sinh r_{w}\cos(\pi-\theta_{0})
cosh2r1+sinh2r1cosθ0\displaystyle\geq\cosh^{2}r_{1}+\sinh^{2}r_{1}\cos\theta_{0}
cos2(θ0/2)cosh(2r1)+sin2(θ0/2)\displaystyle\geq\cos^{2}(\theta_{0}/2)\cosh(2r_{1})+\sin^{2}(\theta_{0}/2)
>cosh(R)\displaystyle>\cosh(R)

Otherwise,

cosh(dist(v,w))\displaystyle\cosh(dist(v,w)) cos2(θ0)cosh(2r2)+sin2(θ0)\displaystyle\leq\cos^{2}(\theta_{0})\cosh(2r_{2})+\sin^{2}(\theta_{0})
cosh(R)1+sin2(θ0)cosh(R)\displaystyle\leq\cosh(R)-1+\sin^{2}(\theta_{0})\leq\cosh(R)

\Box

For 1ik1\leq i\leq k, let tit_{i} be 0-1 random variables which are equal to 1 if and only if both UiU_{i} and Ui+kU_{i+k} have at least one vertex on themselves. Let t=i=1kXit=\sum_{i=1}^{k}X_{i}. Then, there exists OtO_{t} as a vertex-induced subgraph. Again, we are left to lower bound tt.

Proposition 3.2

If θ0>1/n\theta_{0}>1/\sqrt{n}, then F(U1)=Ω(nαθ03)F(U_{1})=\Omega(n^{-\alpha}\theta_{0}^{3}) as nn\rightarrow\infty and θ00\theta_{0}\rightarrow 0

Proof: We first claim that r2=logn+Θ(1)r_{2}=\log n+\Theta(1). It is sufficient to prove cosh(2r2)=Θ(n2)\cosh(2r_{2})=\Theta(n^{2}), and this follows since R=2logn+Θ(1)R=2\log n+\Theta(1) and cos(θ0)>12\cos(\theta_{0})>\frac{1}{2}. With the same argument, we have r1=logn+Θ(1)r_{1}=\log n+\Theta(1). Then, F(U1)F(U_{1}) can be calculated as below.

F(U1)\displaystyle F(U_{1}) =r1r20θ0f(r,θ)𝑑r𝑑θ\displaystyle=\int_{r_{1}}^{r_{2}}\int_{0}^{\theta_{0}}f(r,\theta)drd\theta
=r1r20θ0ρ(r,θ)sinh(r)𝑑r𝑑θ\displaystyle=\int_{r_{1}}^{r_{2}}\int_{0}^{\theta_{0}}\rho(r,\theta)\sinh(r)drd\theta
ρ(r2)r1r20θ0sinh(r)𝑑r\displaystyle\geq\rho(r_{2})\int_{r_{1}}^{r_{2}}\int_{0}^{\theta_{0}}\sinh(r)dr
=θ02παcosh(αR)1sinh(αr2)sinh(r2)(cosh(r2)cosh(r1))\displaystyle=\frac{\theta_{0}}{2\pi}\frac{\alpha}{\cosh(\alpha R)-1}\frac{\sinh(\alpha r_{2})}{\sinh(r_{2})}(\cosh(r_{2})-\cosh(r_{1}))
=Θ(θ0n1α)(cosh(r2)cosh(r1))\displaystyle=\Theta(\theta_{0}n^{-1-\alpha})(\cosh(r_{2})-\cosh(r_{1}))

On the third line, we use the fact that ρ(r)\rho(r) is a monotonically decreasing function. Here,

cosh(r2)cosh(r1)\displaystyle\cosh(r_{2})-\cosh(r_{1})
=1cosh(r2)+cosh(r1)(cosh2(r2)cosh2(r1))\displaystyle=\frac{1}{\cosh(r_{2})+\cosh(r_{1})}(\cosh^{2}(r_{2})-\cosh^{2}(r_{1}))
=Θ(n1)(cosh(2r2)cosh(2r1))\displaystyle=\Theta(n^{-1})(\cosh(2r_{2})-\cosh(2r_{1}))
=Θ(n1)cosh(R)(1cos2(θ0)1cos2(θ0/2)+1cosh(R)cos2(θ0/2))\displaystyle=\Theta(n^{-1})\cosh(R)\left(\frac{1}{\cos^{2}(\theta_{0})}-\frac{1}{\cos^{2}(\theta_{0}/2)}+\frac{1}{\cosh(R)\cos^{2}(\theta_{0}/2)}\right)
=Θ(n)(34θ02+O(θ04)+Θ(n2))\displaystyle=\Theta(n)\left(\frac{3}{4}\theta_{0}^{2}+O(\theta_{0}^{4})+\Theta(n^{-2})\right)
=Θ(nθ02)\displaystyle=\Theta(n\theta_{0}^{2})

The last equality holds since θ02>n1n2\theta_{0}^{2}>n^{-1}\gg n^{-2}.  \Box

As a byproduct of the Proposition 3.2, we obtain the condition which r1<r2Rr_{1}<r_{2}\leq R holds.

Corollary 3.3

If θ0>1/n\theta_{0}>1/\sqrt{n}, then with sufficiently large nn and sufficiently small θ0\theta_{0}, r1<r2Rr_{1}<r_{2}\leq R.

Proof: Recall that r2=logn+Θ(1)r_{2}=\log n+\Theta(1) and R=2logn+Θ(1)R=2\log n+\Theta(1). Therefore, we have r2Rr_{2}\leq R with sufficiently large nn. Also, recall that cosh(r2)cosh(r1)=Θ(nθ02)>0\cosh(r_{2})-\cosh(r_{1})=\Theta(n\theta_{0}^{2})>0 as nn\rightarrow\infty and θ00\theta_{0}\rightarrow 0, proving r1<r2r_{1}<r_{2} with sufficiently large nn and sufficiently small θ0\theta_{0}.  \Box

Proposition 3.4

If k=n(1α)/3k=n^{(1-\alpha)/3}, then there exists a positive constant CC such that Pr[tCn(1α)/3]0\Pr[t\leq Cn^{(1-\alpha)/3}]\rightarrow 0 as nn\rightarrow\infty

Proof: As in the Proposition 2.3, consider the Poissonization GNn,γ,CG_{N_{n},\gamma,C}. The proof is the same. We only need to comfirm that θ0=π/(3n(1α)/3)>1/n\theta_{0}=\pi/(3n^{(1-\alpha)/3})>1/\sqrt{n} for sufficiently large nn and nF(U1)=Ω(1)nF(U_{1})=\Omega(1).  \Box

Again, by combining the proposition with the Theorem 1.3, we obtain the main theorem regarding (Gn,γ,C)\mathcal{M}(G_{n,\gamma,C}).

Corollary 3.5

There exists a positive constant CC^{\prime} such that Pr[exp(Cn(1α)/3)(Gn,γ,C)]1\Pr[\exp(C^{\prime}n^{(1-\alpha)/3})\leq\mathcal{M}(G_{n,\gamma,C})]\rightarrow 1 as nn\rightarrow\infty.

3.3 Upper Bound of the Number of Maximal Cliques

To upper bound (Gn,γ,C)\mathcal{M}(G_{n,\gamma,C}), we prove the hyperbolic versions of the Lemma 2.9 and 2.10. Again, we start with defining segments. Let vw¯\overline{vw} denote the segment (i.e., the shortest geodesic) between vertices vv and ww on a hyperbolic plane. Let 𝒮:={vw¯:v,wG(V),dist(v,w)>R}\mathcal{S}:=\{\overline{vw}:v,w\in G(V),dist(v,w)>R\}. We define the independence of segments in the same way as for the Euclidean case. We first confirm that two independent segments intersect. To do so, we use the following lemma from [12].

Lemma 3.6 ([12])

Let aa and bb be points on a hyperbolic plane where dist(a,b)=Rdist(a,b)=R. Let ll be a line that passes through aa and bb. Let cc and dd be points on the same halfplane induced by ll. Suppose dist(a,c)Rdist(a,c)\leq R, dist(b,c)Rdist(b,c)\leq R, dist(a,d)Rdist(a,d)\leq R, and dist(b,d)Rdist(b,d)\leq R. Then, dist(c,d)Rdist(c,d)\leq R.

Also, we use the following basic facts of non-Euclidean geometry.

Proposition 3.7 (Euclid’s Proposition I-13)

If a line stood on another line makes angles, it will make two right angles, or the sum of the angles is equal to two right angles.

Proposition 3.8 (Euclid’s Proposition I-19)

In any given triangle, the side for a larger angle is larger.

Proposition 3.9 (Saccheri–Legendre Theorem [24])

The sum of all three inner angles in a triangle is less than or equal to two right angles.

Note that all of those Lemma and Propositions do not rely on the parallel postulate. Therefore, they are valid on both a Euclidean and a hyperbolic plane.

Proposition 3.10

On a hyperbolic plane, Two independent segments intersect.

Proof: Let s:=v1v2¯s:=\overline{v_{1}v_{2}} and s:=w1w2¯s^{\prime}:=\overline{w_{1}w_{2}} be two independent segments. Let ll be a line that passes through v1v_{1} and v2v_{2}. We claim that w1w_{1} and w2w_{2} are located on the different halfplanes induced by ll. We prove this by contradiction. Suppose w1w_{1} and w2w_{2} are on the same halfplane. Take points pp and qq on the segment ss so that dist(v1,p)=(aR)/2dist(v_{1},p)=(a-R)/2, dist(p,q)=Rdist(p,q)=R, and dist(q,v2)=(aR)/2dist(q,v_{2})=(a-R)/2, where a:=dist(v1,v2)a:=dist(v_{1},v_{2}). If we can prove dist(p,w1)Rdist(p,w_{1})\leq R, then we also have dist(q,w1)Rdist(q,w_{1})\leq R, dist(p,w2)Rdist(p,w_{2})\leq R, and dist(q,w2)Rdist(q,w_{2})\leq R by symmetricity. By Lemma 3.5, it follows that dist(w1,w2)Rdist(w_{1},w_{2})\leq R. However, this is a contradiction since s𝒮s^{\prime}\in\mathcal{S}.

Now the goal is to prove dist(p,w1)R.dist(p,w_{1})\leq R. Focus on v1w1v2\triangle v_{1}w_{1}v_{2}. By the definition of independent segments, w1v2¯\overline{w_{1}v_{2}} is shorter than v1v2¯\overline{v_{1}v_{2}}. Therefore, by the Proposition 3.9, w1v1v2<v1w1v2\angle w_{1}v_{1}v_{2}<\angle v_{1}w_{1}v_{2}. We have w1v1v2<π/2\angle w_{1}v_{1}v_{2}<\pi/2. Otherwise, w1v1v2π/2\angle w_{1}v_{1}v_{2}\geq\pi/2 and v1w1v2>π/2\angle v_{1}w_{1}v_{2}>\pi/2, and these contradict with the Proposition 3.8. With the same argument, w1pv2<π/2\angle w_{1}pv_{2}<\pi/2. By the Proposition 3.7, w1pv1=πw1pv2π/2w_{1}pv_{1}=\pi-\angle w_{1}pv_{2}\geq\pi/2. Therefore, on w1pv1\triangle w_{1}pv_{1}, w1pv1>w1v1p\angle w_{1}pv_{1}>\angle w_{1}v_{1}p. Again, by the Proposition 3.8, the length of v1w1¯\overline{v_{1}w_{1}} is greater than pw1¯\overline{pw_{1}}. Recall that dist(v1,w1)Rdist(v_{1},w_{1})\leq R. Thus, We can conclude that dist(p,w1)Rdist(p,w_{1})\leq R.  \Box

Notice that the proof of the Proposition 3.10 does not rely on the parallel postulate. Therefore, this proof is also valid on a Euclidean plane, proving the Proposition 2.6.

We can also define the angle of segments on a hyperbolic plane. Again, s=v1v2¯,s=w1w2¯𝒮s=\overline{v_{1}v_{2}},s^{\prime}=\overline{w_{1}w_{2}}\in\mathcal{S} be independent segments. Suppose v1w1v2w2v_{1}w_{1}v_{2}w_{2} is the counter-clockwise order of four endpoints. Let qq be the intersection of the two segments. Define (s,s):=w1qv1\angle(s,s^{\prime}):=\angle w_{1}qv_{1}. Let mm be the midpoint of ss. Consider taking a polar coordinate system whose origin and polar axis are mm and mv1\overrightarrow{mv_{1}}. Let (r0,ϕ)(r_{0},\phi) be the polar coordinate of w1w_{1}. Also, define a:=dist(v1,v2),b:=dist(w1,w2),c=dist(v1,q),d=dist(w1,q)a:=dist(v_{1},v_{2}),b:=dist(w_{1},w_{2}),c=dist(v_{1},q),d=dist(w_{1},q). The definitions are summarized in Figure 3.

Lemma 3.11

Let θ:=(s,s)\theta:=\angle(s,s^{\prime}). For all 0<θ<π0<\theta<\pi,

R<a,bR+2Δ(θ)\displaystyle R<a,b\leq R+2\Delta(\theta) (3.1)
ec,edsinθ211+e2ReR/2\displaystyle e^{c},e^{d}\geq\frac{\sin\theta}{2}\frac{1}{1+e^{-2R}}e^{R/2} (3.2)
cosh(R/2Δ(θ))1+cos(θ)2cosh(r0)cosh(R/2+2Δ(θ))\displaystyle\cosh(R/2-\Delta(\theta))\frac{1+\cos(\theta)}{2}\leq\cosh(r_{0})\leq\cosh(R/2+2\Delta(\theta)) (3.3)

where Δ(θ):=log(2(1+e2R)/(1+cos(θ)))\Delta(\theta):=\log\left(2(1+e^{-2R})/(1+\cos(\theta))\right)

Proof: By the definition of independent segments, dist(v1,w2)dist(v_{1},w_{2}), dist(v2,w1)dist(v_{2},w_{1}), dist(v1,w1)dist(v_{1},w_{1}), and dist(v2,w2)dist(v_{2},w_{2}) are all less than or equal to RR. Therefore,

cosh(c)cosh(bd)+sinh(c)sinh(bd)cos(θ)\displaystyle\cosh(c)\cosh(b-d)+\sinh(c)\sinh(b-d)\cos(\theta) cosh(R)\displaystyle\leq\cosh(R) (3.4)
cosh(ac)cosh(d)+sinh(ac)sinh(d)cos(θ)\displaystyle\cosh(a-c)\cosh(d)+\sinh(a-c)\sinh(d)\cos(\theta) cosh(R)\displaystyle\leq\cosh(R) (3.5)
cosh(c)cosh(d)sinh(c)sinh(d)cos(θ)\displaystyle\cosh(c)\cosh(d)-\sinh(c)\sinh(d)\cos(\theta) cosh(R)\displaystyle\leq\cosh(R) (3.6)
cosh(ac)cosh(bd)sinh(ac)sinh(bd)cos(θ)\displaystyle\cosh(a-c)\cosh(b-d)-\sinh(a-c)\sinh(b-d)\cos(\theta) cosh(R)\displaystyle\leq\cosh(R) (3.7)

From (3.4), we have

(3.4)cosh(bd+c)\displaystyle(\ref{hcf1})\Leftrightarrow\cosh(b-d+c) 1+cos(θ)2+cosh(bdc)1cos(θ)2cosh(R)\displaystyle\frac{1+\cos(\theta)}{2}+\cosh(b-d-c)\frac{1-\cos(\theta)}{2}\leq\cosh(R)
\displaystyle\Rightarrow 1+cos(θ)2ebd+ceR+eR\displaystyle\frac{1+\cos(\theta)}{2}e^{b-d+c}\leq e^{R}+e^{-R}
\displaystyle\Rightarrow bd+cR+Δ(θ)\displaystyle b-d+c\leq R+\Delta(\theta) (3.8)

By doing the same thing on (3.5), we obtain

(Ra)Δ(θ)\displaystyle-(R-a)-\Delta(\theta) cdRb+Δ(θ)\displaystyle\leq c-d\leq R-b+\Delta(\theta)
Δ(θ)\displaystyle\Rightarrow-\Delta(\theta) cdΔ(θ)\displaystyle\leq c-d\leq\Delta(\theta) (3.9)

The last inequalities hold since a,b>Ra,b>R. Plugging this in to (3.9) yields (3.1).
From (3.7), we have

(4.6)cosh(ac+bd)\displaystyle(4.6)\Leftrightarrow\cosh(a-c+b-d) 1cos(θ)2+cosh(acb+d)1+cos(θ)2eR+eR2\displaystyle\frac{1-\cos(\theta)}{2}+\cosh(a-c-b+d)\frac{1+\cos(\theta)}{2}\leq\frac{e^{R}+e^{-R}}{2}
\displaystyle\Rightarrow 1cos(θ)2e2RcdeR+eR\displaystyle\frac{1-\cos(\theta)}{2}e^{2R-c-d}\leq e^{R}+e^{-R}
\displaystyle\Rightarrow 1cos(θ)2eR1+e2Rec+d\displaystyle\frac{1-\cos(\theta)}{2}\frac{e^{R}}{1+e^{-2R}}\leq e^{c+d}

Multiplying this with (3.9) gives us

ec,edsinθ211+e2ReR/2\displaystyle e^{c},e^{d}\geq\frac{\sin\theta}{2}\frac{1}{1+e^{-2R}}e^{R/2}

For r0r_{0},

cosh(a/2c)cosh(d)+sinh(a/2c)sinh(d)cos(θ)=cosh(r)\displaystyle\cosh(a/2-c)\cosh(d)+\sinh(a/2-c)\sinh(d)\cos(\theta)=\cosh(r)

Note that this holds no matter how mm and qq are lined up on ss. We have

cosh(a/2c+d)1+cos(θ)2cosh(r)cosh(a/2c+d)\displaystyle\cosh(a/2-c+d)\frac{1+\cos(\theta)}{2}\leq\cosh(r)\leq\cosh(a/2-c+d)

Finally, use (3.9) to derive (3.3).  \Box

Corollary 3.12

Let θ0(0,π)\theta_{0}\in(0,\pi). If (s,s)θ0\angle(s,s^{\prime})\leq\theta_{0}, then either w1w_{1} or w2w_{2} must lie on U1U_{1} or U2U_{2} where

U1\displaystyle U_{1} :={(r0,ϕ):r1r0r2,0ϕθ0}\displaystyle:=\{(r_{0},\phi):r_{1}\leq r_{0}\leq r_{2},0\leq\phi\leq\theta_{0}\}
U2\displaystyle U_{2} :={(r0,ϕ):r1r0r2,πϕπ+θ0}\displaystyle:=\{(r_{0},\phi):r_{1}\leq r_{0}\leq r_{2},\pi\leq\phi\leq\pi+\theta_{0}\}

Here,

r1\displaystyle r_{1} :=cosh1(cosh(R/2Δ(θ0))1+cos(θ0)2)\displaystyle:=\cosh^{-1}\left(\cosh(R/2-\Delta(\theta_{0}))\frac{1+\cos(\theta_{0})}{2}\right)
r2\displaystyle r_{2} :=R/2+2Δ(θ0)\displaystyle:=R/2+2\Delta(\theta_{0})
Corollary 3.13

Let ss be a segment. Let θ0>1/n\theta_{0}>1/\sqrt{n}. Let XX be a number of segments ss^{\prime} which is independent from ss and (s,s)θ0\angle(s,s^{\prime})\leq\theta_{0}, then

𝔼[X]=sup(r,θ)U{ρ(r,θ)}O(n2θ03)\displaystyle\mathbb{E}[X]=\sup_{(r,\theta)\in U}\{\rho(r,\theta)\}O(n^{2}\theta_{0}^{3})

as nn\rightarrow\infty and θ00\theta_{0}\rightarrow 0.

Proof: In this proof, we simply denote Δ(θ0)\Delta(\theta_{0}) as Δ\Delta. It is sufficient to prove that
F(U1)=sup(r,θ)U1{ρ(r,θ)}O(nθ03)F(U_{1})=\sup_{(r,\theta)\in U_{1}}\{\rho(r,\theta)\}O(n\theta_{0}^{3}). Here,

F(U1)\displaystyle F(U_{1}) =U1f(r,θ)𝑑r𝑑θ\displaystyle=\int_{U_{1}}f(r,\theta)drd\theta
=U1ρ(r,θ)𝑑μ\displaystyle=\int_{U_{1}}\rho(r,\theta)d\mu
supU1ρ(r,θ)U1𝑑μ\displaystyle\leq\sup_{U_{1}}\rho(r,\theta)\int_{U_{1}}d\mu
=2θ0supU1ρ(r,θ)(cosh(r2)cosh(r1))\displaystyle=2\theta_{0}\sup_{U_{1}}\rho(r,\theta)(\cosh(r_{2})-\cosh(r_{1}))
=2θ0supUρ(r,θ)(cosh(R/2+2Δ)cosh(R/2Δ)\displaystyle=2\theta_{0}\sup_{U}\rho(r,\theta)(\cosh(R/2+2\Delta)-\cosh(R/2-\Delta)
+1cos(θ0)2cosh(R/2Δ))\displaystyle+\frac{1-\cos(\theta_{0})}{2}\cosh(R/2-\Delta))

Here, we can bound

cosh(R/2+2Δ)cosh(R/2Δ)\displaystyle\cosh(R/2+2\Delta)-\cosh(R/2-\Delta)
=cosh(R/2+2Δ)cosh(R/2Δ)3Δ3Δ\displaystyle=\frac{\cosh(R/2+2\Delta)-\cosh(R/2-\Delta)}{3\Delta}\cdot 3\Delta
3sinh(R/2+2Δ)Δ\displaystyle\leq 3\sinh(R/2+2\Delta)\cdot\Delta
3sinh(R/2+2Δ)(2(1+e2R)1+cosθ01)\displaystyle\leq 3\sinh(R/2+2\Delta)\cdot\left(\frac{2(1+e^{-2R})}{1+\cos\theta_{0}}-1\right)
=O(n)(1cosθ0+2e2R1+cosθ0)\displaystyle=O(n)\cdot\left(\frac{1-\cos\theta_{0}+2e^{-2R}}{1+\cos\theta_{0}}\right)
=O(n)(O(θ02)+O(n4))=O(nθ02)\displaystyle=O(n)\cdot\left(O(\theta_{0}^{2})+O(n^{-4})\right)=O(n\theta_{0}^{2})

The inequality follows because cosh(x)\cosh(x) is a convex function. The last equality follows by the assumption θ02>n1n4\theta_{0}^{2}>n^{-1}\gg n^{-4}. Also,

1cos(θ0)2cosh(R/2Δ)=O(θ02)O(n)=O(nθ02)\displaystyle\frac{1-\cos(\theta_{0})}{2}\cosh(R/2-\Delta)=O(\theta_{0}^{2})\cdot O(n)=O(n\theta_{0}^{2})

\Box

To fully use the Lemma 3.13, we prove the hyperbolic version of the Lemma 2.10. The statement is identical.

Lemma 3.14

Let kk be a positive integer. If there exists a set of tt independent segments S𝒮S\subseteq\mathcal{S}, then there exists a segment ss^{\prime} and a set of independent segments S𝒮S^{\prime}\subseteq\mathcal{S} such that

{|S|t/ksSs′′S.(s,s′′)π/k\displaystyle\left\{\begin{array}[]{l}|S^{\prime}|\geq\left\lceil t/k\right\rceil\\ s^{\prime}\in S^{\prime}\\ \forall s^{\prime\prime}\in S^{\prime}.\angle(s^{\prime},s^{\prime\prime})\leq\pi/k\\ \end{array}\right. (3.13)

Proof: The proof of Lemma 2.10 is also valid on a hyperbolic plane, except for the part (2.6) where we used the fact that the sum of inner angles of a triangle is equal to π\pi. This can be replaced with the Proposition 3.8. Then we get

a+b+cπ\displaystyle\angle a+\angle b+\angle c\leq\pi (3.14)
(s,s′′)π((s0,s))(π(s0,s′′))=(s0,s′′)(s0,s)\displaystyle\Leftrightarrow\angle(s^{\prime},s^{\prime\prime})\leq\pi-(\angle(s_{0},s^{\prime}))-(\pi-\angle(s_{0},s^{\prime\prime}))=\angle(s_{0},s^{\prime\prime})-\angle(s_{0},s^{\prime})

Thus, we can prove the same statement.  \Box

It is not so hard from here to prove the limited version of the main theorem. The next proposition states that if vertices of OtO_{t} do not fall on a region near the origin, we can bound its size.

Proposition 3.15

Let V:={vV(Gn,γ,C):the radial coordinate of v is greater than R/2}V^{\prime}:=\{v\in V(G_{n,\gamma,C}):\text{the radial coordinate of $v$ is greater than $R/2$}\}. Let Gn,γ,CG_{n,\gamma,C}^{\prime} be a subgraph of Gn,γ,CG_{n,\gamma,C} induced by VV^{\prime}. Let t=n(1α)/3+ϵt=n^{(1-\alpha)/3+\epsilon} where ϵ\epsilon is arbitrary positive constant. Then,

Pr[Gn,γ,Chas Ot as its vertex-induced subgraph]0\displaystyle\Pr[G_{n,\gamma,C}^{\prime}\ \text{has $O_{t}$ as its vertex-induced subgraph}]\rightarrow 0

as nn\rightarrow\infty.

Proof: We will not go into detail here. However, the statement can be proved in the same way as the Theorem 2.11. We end up claiming ρ(R/2)O(n2θ03)=O(1)\rho(R/2)\cdot O(n^{2}\theta_{0}^{3})=O(1) when θ0:=π/n(1α)/3\theta_{0}:=\pi/n^{(1-\alpha)/3}. Since ρ(R/2)=O(n1α)\rho(R/2)=O(n^{-1-\alpha}), it follows.  \Box

3.4 Dealing With Non-Uniformity

We prove the additional facts about two independent segments. Again, we will use the definitions of points, angles, and length in Figure 3.

Corollary 3.16

For all 0<(s,s)<π0<\angle(s,s^{\prime})<\pi, 12cosh(R/2Δ(π/2))cosh(r0)cosh(R/2+2Δ(π/2))\frac{1}{2}\cosh(R/2-\Delta(\pi/2))\leq\cosh(r_{0})\leq\cosh(R/2+2\Delta(\pi/2)).

Proof: Let θ:=(s,s)\theta:=\angle(s,s^{\prime}). Due to symmetricity, (3.3) in the Lemma 3.11 also holds for πθ\pi-\theta. Especially, when θ=πθ\theta=\pi-\theta^{\prime} for some 0<θπ/20<\theta^{\prime}\leq\pi/2, we have

cosh(R/2Δ(θ))1+cos(θ)2cosh(r0)cosh(R/2+2Δ(θ))\displaystyle\cosh(R/2-\Delta(\theta^{\prime}))\frac{1+\cos(\theta^{\prime})}{2}\leq\cosh(r_{0})\leq\cosh(R/2+2\Delta(\theta^{\prime}))

Since cosh(R/2Δ(θ))1+cosh(θ)2\cosh(R/2-\Delta(\theta))\frac{1+\cosh(\theta)}{2} and cosh(R/2+2Δ(θ))\cosh(R/2+2\Delta(\theta)) are monotonic functions on 0<θπ/20<\theta\leq\pi/2, we have the bound on cosh(r0)\cosh(r_{0}).  \Box

Corollary 3.17

With sufficiently large nn, 18cosh(R/2)<cosh(r0)\frac{1}{8}\cosh(R/2)<\cosh(r_{0}) and R/23log2<r0R/2-3\log 2<r_{0}.

Proof: It is sufficient to prove that we have 18cosh(R/2)<cosh(R/23log2)<12cosh(R/2Δ(π/2))\frac{1}{8}\cosh(R/2)<\cosh(R/2-3\log 2)<\frac{1}{2}\cosh(R/2-\Delta(\pi/2)) if nn is large enough. We omit the detail here since it is a simple calculation task.  \Box

The following proposition states that ϕ\phi can be lower bounded in terms of θ:=(s,s)\theta:=\angle(s,s^{\prime}).

Proposition 3.18

if θ>1/n\theta>1/\sqrt{n}, then with sufficiently large nn,

sinϕsin2θ9\displaystyle\sin\phi\geq\frac{\sin^{2}\theta}{9}

Proof: If θ=π/2\theta=\pi/2, it clearly holds. Otherwise, Let θ:=(s,s)\theta:=\angle(s,s^{\prime}). Let qq be the intersection of ss and ss^{\prime}. Let a:=dist(v1,v2),b:=dist(w1,w2),c=dist(v1,q),d=dist(w1,q)a:=dist(v_{1},v_{2}),b:=dist(w_{1},w_{2}),c=dist(v_{1},q),d=dist(w_{1},q). By the sine formula on a hyperbolic plane, it holds (no matter how mm and qq are lined up on ss) that

sinθsinhr0=sinϕsinhd\displaystyle\frac{\sin\theta}{\sinh r_{0}}=\frac{\sin\phi}{\sinh d}

Therefore,

sin\displaystyle\sin ϕsinhdsinhr0sinθ\displaystyle\phi\geq\frac{\sinh d}{\sinh r_{0}}\sin\theta
ededer0sinθ\displaystyle\geq\frac{e^{d}-e^{-d}}{e^{r_{0}}}\sin\theta
sinθ2eR/21+e2R1eR/2e2Δ(π/2)sinθ\displaystyle\geq\frac{\frac{\sin\theta}{2}\frac{e^{R/2}}{1+e^{-2R}}-1}{e^{R/2}e^{2\Delta(\pi/2)}}\sin\theta
sin2θ8(1eR(1+e2R)38eR/2e2Δ(π/2)sinθ)\displaystyle\geq\frac{\sin^{2}\theta}{8}\left(\frac{1-e^{-R}}{(1+e^{-2R})^{3}}-\frac{8}{e^{R/2}e^{2\Delta(\pi/2)}\sin\theta}\right)
sin2θ9\displaystyle\geq\frac{\sin^{2}\theta}{9}

For sufficiently large nn since eR0e^{-R}\rightarrow 0 and eR/2sinθ=Ω(n)+e^{R/2}\sin\theta=\Omega(\sqrt{n})\rightarrow+\infty\Box

Lemma 3.19

Let ss and ss^{\prime} be independent segments. Let hh be the distance between the midpoint mm of ss and the origin. Suppose one of ss’s endpoints is in B0(R/2)B_{0}(R/2), and sin((s,s))12eh/4\sin(\angle(s,s^{\prime}))\geq 12e^{-h/4}, then with sufficiently large nn, both endpoints of ss^{\prime} are outside of B0(R/2)B_{0}(R/2).

Proof: Note that we would never have the case where both endpoints of a segment in 𝒮\mathcal{S} are in B0(R/2)B_{0}(R/2). We prove the proposition by contradiction. By symmetricity and the fact that sin(π(s,s))=sin((s,s))\sin(\pi-\angle(s,s^{\prime}))=\sin(\angle(s,s^{\prime})), We can assume that w1w_{1} is in B0(R/2)B_{0}(R/2) without loss of generality.

Recall that ϕ=v1mw1\phi=\angle v_{1}mw_{1}. We have ϕ=v1mo+w1mo\phi=\angle v_{1}mo+\angle w_{1}mo. Therefore, either v1mw1\angle v_{1}mw_{1} or w1mo\angle w_{1}mo must be greater than or equal to ϕ/2\phi/2. First, suppose v1moϕ/2\angle v_{1}mo\geq\phi/2. By the Lemma 3.18,

sinv1mo2sinϕ4{sin((s,s))}2364eh/2\displaystyle\sin\frac{\angle v_{1}mo}{2}\geq\frac{\sin\phi}{4}\geq\frac{\{\sin(\angle(s,s^{\prime}))\}^{2}}{36}\geq 4e^{-h/2}

Therefore,

cosh{dist(v1,o)}\displaystyle\cosh\left\{dist(v_{1},o)\right\} =cosh(a/2)cos(h)sinh(a/2)sinh(h)cos(v1mo)\displaystyle=\cosh(a/2)\cos(h)-\sinh(a/2)\sinh(h)\cos(\angle v_{1}mo)
>cosh(a/2)cosh(h)sin2(v1mo2)\displaystyle>\cosh(a/2)\cosh(h)\sin^{2}\left(\frac{\angle v_{1}mo}{2}\right)
cosh(R/2)12eh16eh\displaystyle\geq\cosh(R/2)\frac{1}{2}e^{h}16e^{-h}
8cosh(R/2)\displaystyle\geq 8\cosh(R/2)

This is a contradiction since dist(v1,o)dist(v_{1},o) must be less than or equal to R/2R/2. On the other hand, suppose w1moϕ/2\angle w_{1}mo\geq\phi/2. Then,

cosh{dist(w1,o)}\displaystyle\cosh\left\{dist(w_{1},o)\right\} cosh(r0)cosh(h)sin2(w1mo2)\displaystyle\geq\cosh(r_{0})\cosh(h)\sin^{2}\left(\frac{\angle w_{1}mo}{2}\right)
>18cosh(R/2)12eh16eh\displaystyle>\frac{1}{8}\cosh(R/2)\frac{1}{2}e^{h}16e^{-h}
cosh(R/2)\displaystyle\geq\cosh(R/2)

Again, this is a contradiction.  \Box

Refer to caption
Figure 6: The situation of the Lemma 3.19. The yellow region is B0(R/2)B_{0}(R/2). If (s,s)\angle(s,s^{\prime}) is large, then by the Proposition 3.18, ϕ\phi must be large. Thus, either length of v1o¯\overline{v_{1}o} or w1o¯\overline{w_{1}o} exceeds R/2R/2.

Using the Lemma 3.19, we divide independent segments into some sets so that segments in each set satisfy either condition.

  • Angles between them are not so small but are located at the sparse area (See (3.19))

  • Angles between them are small but might be located at the dense area (See (3.24))

Lemma 3.20

Let kk be a positive integer. If there exists a set of tt independent segments S𝒮S\subseteq\mathcal{S}, then there exists a segment ss^{\prime}, a set of independent segments S𝒮S^{\prime}\subseteq\mathcal{S} such that,

{|S|t3ksSs′′S.(s,s′′)π/ks′′S.Both endpoints of s′′ are outside of B0(R/2)\displaystyle\left\{\begin{array}[]{c}|S^{\prime}|\geq\left\lceil\frac{t}{3k}\right\rceil\\ s^{\prime}\in S^{\prime}\\ \forall s^{\prime\prime}\in S^{\prime}.\angle(s^{\prime},s^{\prime\prime})\leq\pi/k\\ \forall s^{\prime\prime}\in S^{\prime}.\text{Both endpoints of $s^{\prime\prime}$ are outside of $B_{0}(R/2)$}\end{array}\right. (3.19)

or

{|S|t3ksSs′′S.(s,s′′)12eh/4π/ks′′S.Both endpoints of s′′ are outside of B0(max{R/23log2h,0})\displaystyle\left\{\begin{array}[]{c}|S^{\prime}|\geq\left\lceil\frac{t}{3k}\right\rceil\\ s^{\prime}\in S^{\prime}\\ \forall s^{\prime\prime}\in S^{\prime}.\angle(s^{\prime},s^{\prime\prime})\leq 12e^{-h/4}\pi/k\\ \forall s^{\prime\prime}\in S^{\prime}.\text{Both endpoints of $s^{\prime\prime}$ are outside of $B_{0}(\max\{R/2-3\log 2-h,0\}$})\\ \end{array}\right. (3.24)

for some h[0,R]h\in[0,R].

Proof: Take a segment s0Ss_{0}\in S whose endpoint is in B0(R/2)B_{0}(R/2). If there is none, then we are done with the Lemma 3.14. Let θs0:=arcsin(min{12eh/4,1})\theta_{s_{0}}:=\arcsin(\min\{12e^{-h/4},1\}). For each index 1ik1\leq i\leq k, we define

Si\displaystyle S_{i} :={sS:i1kθs0(s0,s)<ikθs0}\displaystyle:=\left\{s\in S:\frac{i-1}{k}\theta_{s_{0}}\leq\angle(s_{0},s)<\frac{i}{k}\theta_{s_{0}}\right\}
Ti\displaystyle T_{i} :={sS:θs0+i1k(π2θs0)(s0,s)<θs0+ik(π2θs0)}\displaystyle:=\left\{s\in S:\theta_{s_{0}}+\frac{i-1}{k}(\pi-2\theta_{s_{0}})\leq\angle(s_{0},s)<\theta_{s_{0}}+\frac{i}{k}(\pi-2\theta_{s_{0}})\right\}
Ui\displaystyle U_{i} :={sS:πθs0+i1kθs0(s0,s)<πθs0+ikθs0}\displaystyle:=\left\{s\in S:\pi-\theta_{s_{0}}+\frac{i-1}{k}\theta_{s_{0}}\leq\angle(s_{0},s)<\pi-\theta_{s_{0}}+\frac{i}{k}\theta_{s_{0}}\right\}

Since SiTiUi=SS_{i}\cup T_{i}\cup U_{i}=S, i=0k1|Si|+|Ti|+|Ui|t\sum_{i=0}^{k-1}|S_{i}|+|T_{i}|+|U_{i}|\geq t. By the pigeonhole principle, there exists an index ii such that either SiS_{i}, UiU_{i}, or TiT_{i} has the cardinality greater or equal to t/(3k)\lceil t/(3k)\rceil. Let SS^{\prime} be such a set. Consider taking the segment ss^{\prime} in SS^{\prime} so that (s0,s)=mins′′S(s0,s′′)\angle(s_{0},s^{\prime})=\min_{s^{\prime\prime}\in S^{\prime}}\angle(s_{0},s^{\prime\prime}). If S=SiS^{\prime}=S_{i} or S=UiS^{\prime}=U_{i}, (s,s′′)(s0,s′′)(s0,s)θs0/k12eh/4π/k\angle(s^{\prime},s^{\prime\prime})\leq\angle(s_{0},s^{\prime\prime})-\angle(s_{0},s^{\prime})\leq\theta_{s_{0}}/k\leq 12e^{-h/4}\pi/k. The third condition of (3.24) is satisfied. Also, by the Corollary 3.17, endpoints of s′′s^{\prime\prime} must be outside of a circle whose radius and center are R/23log2R/2-3\log 2 and mm. Therefore, the fourth condition of (3.19) is satisfied. We have proved that ss^{\prime} and SS^{\prime} fulfill (3.24). If S=TiS^{\prime}=T_{i}, then we must have 12eh/4<112e^{-h/4}<1 so that S(=Ti)S^{\prime}(=T_{i}) is not empty. Therefore, sin((s0,s))sin(θs0)=12eh/4\sin(\angle(s_{0},s))\geq\sin(\theta_{s_{0}})=12e^{-h/4}. By Lemma 3.19, both endpoints of s′′Ss^{\prime\prime}\in S^{\prime} must be outside of B0(R/2)B_{0}(R/2), and (3.19) holds.  \Box

Theorem 3.21

Let ϵ\epsilon be a positive constant. Let t=n(1α)/3+ϵt=n^{(1-\alpha)/3+\epsilon}. Then,

Pr[Gn,γ,C has Ot as its vertex-induced subgraph]0\displaystyle\Pr[G_{n,\gamma,C}\text{ has $O_{t}$ as its vertex-induced subgraph}]\rightarrow 0

as nn\rightarrow\infty.

Proof: Let k:=n(1α)/3k:=n^{(1-\alpha)/3} and θ0:=π/k\theta_{0}:=\pi/k. Suppose Gn,rG_{n,r} has OtO_{t} as its vertex-induced subgraph. Apply Lemma 3.20 for tt and kk. As in the proof of 2.11, Then there exists a segment ss^{\prime} and a set of independent segments SS^{\prime} such that either (3.19) or (3.24) holds. We first claim that for a certain segment ss^{\prime}, the probability that there exists a set of independent segments SS^{\prime} such that (3.19) holds is at most exp(Ω(nϵ))\exp(-\Omega(n^{\epsilon})). By the Corollary 3.13, the expected number of segments which is possibly in SS^{\prime} can be bounded by ρ(R/2)O(n2θ03)=O(n1α)O(n2n(1α))=O(1)\rho(R/2)\cdot O(n^{2}\theta_{0}^{3})=O(n^{-1-\alpha})\cdot O(n^{2}\cdot n^{-(1-\alpha)})=O(1). However, the size of SS^{\prime} must be nϵn^{\epsilon}. By the Chernoff bound, we have the probability bound. We next claim that for a certain segment ss^{\prime}, the probability that there exists a set of independent segments SS^{\prime} such that (3.24) holds is at most exp(Ω(nϵ))\exp(-\Omega(n^{\epsilon})). Again, it is sufficient to prove that the expected number of segments that satisfy the third and the fourth condition of (3.24) is O(1)O(1). Let θs0:=12eh/4π/k=12eh/4θ0\theta_{s_{0}}:=12e^{-h/4}\pi/k=12e^{-h/4}\theta_{0}. Here,

(The expected value)\displaystyle(\text{The expected value}) nρ(max{R3log2h,0})O(nθs03)\displaystyle\leq n\rho(\max\{R-3\log 2-h,0\})\cdot O(n\theta_{s_{0}}^{3})
=nα2π(cosh(αR)1)1e(1α)(R/23log2h)O(ne3h/4θ03)\displaystyle=n\cdot\frac{\alpha}{2\pi}\cdot(\cosh(\alpha R)-1)^{-1}e^{-(1-\alpha)(R/2-3\log 2-h)}\cdot O(ne^{-3h/4}\theta_{0}^{3})
=nO(n2αn1+αe(1α)h)O(ne3h/4θ03)\displaystyle=nO(n^{-2\alpha}n^{-1+\alpha}e^{(1-\alpha)h})\cdot O(ne^{-3h/4}\theta_{0}^{3})
=O(n1αeh(3/4(1α))θ03)\displaystyle=O(n^{1-\alpha}e^{-h(3/4-(1-\alpha))}\theta_{0}^{3})
=O(n1αθ03)(1/2<α<1)\displaystyle=O(n^{1-\alpha}\theta_{0}^{3})\ (\because 1/2<\alpha<1)
=O(1)\displaystyle=O(1)

Finally, we apply a union bound over all segments and obtain

Pr[Gn,γ,C has Ot as its vertex-induced subgraph]n(n1)22exp(Ω(nϵ))\displaystyle\Pr[G_{n,\gamma,C}\text{ has $O_{t}$ as its vertex-induced subgraph}]\leq\frac{n(n-1)}{2}\cdot 2\exp(-\Omega(n^{\epsilon}))

which goes to 0 as nn\rightarrow\infty.  \Box

Theorem 3.22

There exist a positive constant CC^{\prime} such that for all ϵ>0\epsilon>0,

Pr[(Gn,γ,C)exp(Cn(1α)/3+ϵ)]1\displaystyle\Pr[\mathcal{M}(G_{n,\gamma,C})\leq\exp(C^{\prime}n^{(1-\alpha)/3+\epsilon})]\rightarrow 1

as nn\rightarrow\infty

4 Conclusions

In this paper, we started with the question: why is the number of maximal cliques small on real-world networks? To give an explanation to it, we consider the number of maximal cliques in two models of real-world networks: Euclidean random geometric graphs and hyperbolic random graphs. To bound the number, we focus on vertex-induced subgraphs, which are isomorphic to OtO_{t}. Similar geometric and probabilistic techniques apply to both random graphs. For future works, we are interested in whether the property of OtO_{t} has positive effects on clique enumeration algorithms on real-world graphs. Also, the generalization to a higher dimensional space is an open question.

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Graph |V||V| |E||E| τ\tau
amazon0601 403394 2443408 5
as-skitter 1696415 11095298 15
ca-AstroPh 18772 198050 7
com-dblp 317080 1049866 4
com-youtube 1134890 2987624 7
email-Enron 36692 183831 7
email-EuAll 265214 364481 8
facebook_combined 4039 88234 19
hrg 180705 968205 4
loc-gowalla_edges 196591 950327 8
soc-buzznet 101164 2763067 14
soc-digg 770800 5907133 17
soc-Epinions1 75879 405740 10
web-Google 875713 4322051 5
web-NotreDame 325729 1090108 5
web-Stanford 281903 1992636 5
wiki-Talk 2394385 4659565 13
wiki-topcats 1791489 25444207 10
wiki-Vote 7115 100762 8
Table 1: Upper bound of τ\tau (i.e. maximum tt such that a graph contains OtO_{t} as its vertex-induced subgraph) on SNAP graph datasets