9 sub-clusters as follows. Therefore, nk(a+28) Pcids(n,a,B,Y,r0p）≥ F(Kj) 会28.nss的*， nk(a+23) k-1 (6-9) 合刚-g8.n nk(a+28)nk(a+28) =日(n-). P(Sn∩Sjx） (6-5) Substituting Egn.(6-6),Egn.(6-7)and Egn.(6-9)into Eqn. (6-5),and let n-oo,the result follows. ■ nk(a+ 2 会名s-22sns C.Sufficient Condition of Proposition 3 -(君6 In this part,we also do segmentation for each cluster.As shown in Figure 6.1,we first cut the cluster into disjoint rings with radii ofk标，k=l,2,·,n号，where手=n-tR. And then cut the k-th ring equally into 2k-1 sub-areas.This segmentation has two features:(1)the whole region of the With Eqn.(6-1)and Lemma 7,we have cluster is covered and(2)each sub-area can be covered by a n(+28) circle with a radius rk.As shown in Figure 6.2,we can bound 日王6小≥mma6-+P)r≥e5 Tk by (6-6) (ksin nk(a+2 会三ss会三-w-y ≤7V+(P (6-10) ≤V1+π2 Sm2n2k(a+28)e-2kzne(r-F)2 ≤4. =e-2.4vot座-头 ≌(1+6)e-2 Cluster Region (6-7) where u6 >0 and can be arbitrary close to 0. Sub-area Then we evaluate the term P(Sixnj).Let k= (a1,a2,...,ak)and=(b1,b2,...,b).Suppose there are exactly t (0<t <k-1)identical elements between the sequences k and k,i.e.,={s(1),s(2),...,s(t)} (1,2,...,k}that as(i)=bs(i),i <t.P(Six n Sj)is determined by three items.i)For eacht,therearechoices on 3.ii)For the t timeslots that Si and Si overlap.the /na+2\ probability is not larger than Fig.6.1:Segmentation illustration iii)For the other k-t timeslots,since the sub-areas don't overlap with each other,the probability is not larger than After this segmentation,we get n+2 sub-areas,each with na+28 ·（1-26-)-0 Putting all these an area of Similar to previous subsection,we can obtain 2 nk(+28)sub-clusters.The average number of nodes in each together,we can prove that (o可)=n(t-a+2）,which shows sub-cluster is·（,az） P()≤()n, (6-8) that when n is sufficiently large,each sub-cluster will not be empty. for sufficient large n. The proof is carried out from the perspective of sub-clusters.9 sub-clusters as follows. Pcids(n, α, β, γ, rw.p. c ) ≥ Xm j=1 n kX (α+2β) κ=1 P(K s jκ) ≥ Xm j=1 n kX (α+2β) κ=1 P(Sjκ) − Xm j=1 X κ̸=κ′ P(Sjκ ∩ Sjκ′ ) − Xm i=1 X j̸=i n kX (α+2β) κ=1 n kX (α+2β) κ′=1 P(Siκ ∩ Sjκ′ ) ≥ Xm j=1 n kX (α+2β) κ=1 P(Sjκ) − Xm j=1 X κ̸=κ′ P(Sjκ ∩ Sjκ′ ) − Xm j=1 n kX (α+2β) κ=1 P(Sjκ) 2 . (6-5) With Eqn. (6-1) and Lemma 7, we have Xm j=1 n kX (α+2β) κ=1 P(Sjκ) ≥ mnk(α+2β) € 1−π(r+ ¯r) 2 Šknα ≥ θe−ξ . (6-6) Xm j=1 n kX (α+2β) κ=1 P(Sjκ) 2 ≤ Xm j=1 n kX (α+2β) κ=1 € 1 − π(r − r¯) 2 Šknα 2 ≤m2n 2k(α+2β) e −2kπnα(r−r¯) 2 =e −2ξ e 4 √ kπ· √[k(α+2β)+γ] log n+ξ log n − 2kπ log2 n ,(1 + µ6)e −2ξ , (6-7) where µ6 > 0 and can be arbitrary close to 0. Then we evaluate the term P(Sjκ ∩ Sjκ′ ). Let κ = (a1, a2, ..., ak) and κ ′ = (b1, b2, ..., bk). Suppose there are exactly t (0 ≤ t ≤ k − 1) identical elements between the sequences κ and κ ′ , i.e., ∃ −→s = {s(1), s(2), ..., s(t)} ⊆ {1, 2, ..., k} that as(i) = bs(i) , i ≤ t. P(Sjκ ∩ Sjκ′ ) is determined by three items. i) For each t, there are  k t  choices on −→s . ii) For the t timeslots that Sjκ and Sjκ′ overlap, the probability is not larger than  n α+2β 1 t ·  1−π(r −r¯) 2 tnα . iii) For the other k − t timeslots, since the sub-areas don’t overlap with each other, the probability is not larger than  n α+2β 2 k−t ·  1 − 2π(r − r¯) 2 (k−t)n α . Putting all these together, we can prove that P(Sjκ ∩ Sjκ′ ) ≤  k t  n t−2k k γ (6-8) for sufficient large n. Therefore, Xm j=1 X κ̸=κ′ P(Sjκ ∩ Sjκ′ ) ≤ m k X−1 t=0  k t  n t−2k k γ = n γ k X−1 t=0  k t  n t−2k k γ = Θ(n − γ k ). (6-9) Substituting Eqn. (6-6), Eqn. (6-7) and Eqn. (6-9) into Eqn. (6-5), and let n → ∞, the result follows. C. Sufficient Condition of Proposition 3 In this part, we also do segmentation for each cluster. As shown in Figure 6.1, we first cut the cluster into disjoint rings with radii of kr, k ¯¯ = 1, 2, · · · , n α+2β 2 , where r¯¯ = n − α+2β 2 R. And then cut the k-th ring equally into 2k −1 sub-areas. This segmentation has two features: (1) the whole region of the cluster is covered and (2) each sub-area can be covered by a circle with a radius rk. As shown in Figure 6.2, we can bound rk by rk = r¯¯ É cos2 π 2k − 1 + (k sin π 2k − 1 ) 2 ≤ r¯¯ r 1 + ( kπ 2k − 1 ) 2 ≤ r¯¯ p 1 + π 2 ≤ 4r.¯¯ (6-10) r 2 R n r 2 α β + = Cluster Region Sub-area Fig. 6.1: Segmentation illustration After this segmentation, we get n α+2β sub-areas, each with an area of πr¯¯ 2 . Similar to previous subsection, we can obtain n k(α+2β) sub-clusters. The average number of nodes in each sub-cluster is ϖ ·  1 nα+2β) k = n k € ϵ k −(α+2β) Š , which shows that when n is sufficiently large, each sub-cluster will not be empty. The proof is carried out from the perspective of sub-clusters