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5. 5. KHABBAZIAN AND BHARGAVA: EFFICIENT BROADCASTING IN MOBILE AD HOC NETWORKS 235 slice A0 P1 P2 . Since NA0 has at least one neighbor (i.e., NB0 ) 20: end for in this slice, there must be a selected node ND in the slice 21: if chk then that has forwarded the message. Using Lemma 1, we get 22: ind ind max B0 D R; hence, nodes ND and NB0 are neighbors. 23: else Therefore, ðND ; NB0 ; NC 0 Þ 2 Ã. However, this contradicts 24: ind ind min (1) because using Lemma 3, we have 25: end if 26: until S1 is in LBA ðListA ½indÞ OR ListA ½ind is in DC 0 A0 C 0 : LBA ðS1 Þ u t 27: if ind 6¼ 1 then 28: NSiþ1 ListA ½ind Algorithm 2 shows our proposed slice-based selection 29: end ifalgorithm. Suppose that node NA uses the proposedalgorithm to select the forwarding nodes from its neighbors. Lemma 4. Suppose the proposed algorithm selects m nodesLet us assume that NA stores all of its neighbors’ IDs and fNS1 ; NS2 ; . . . ; NSm g. For any 1 i m À 2, we havelocations in an array of length n, where n is the number of neighbors. The algorithm selects the first node NS1 ﬀA ðLBA ðSi Þ; LBA ðSiþ2 ÞÞ : 3randomly from the array. The first node can also be selecteddeterministically by, for example, selecting the node that is Proof. Based on (2), (3), and the algorithm terminationthe farthest away from NA . Let LBA ðP Þ and RBA ðP Þ denote condition, we can show thatthe left bulged slice and right bulged slice of P around A,respectively. Suppose that NSi is the last node selected by ﬀA ðLBA ðSi Þ; LBA ðSiþ2 ÞÞ ﬀA ðLBA ðSi Þ; LBA ðSiþ1 ÞÞthe algorithm. To select the next node, the algorithm iterates for any 1 i m À 2. By contradiction, assume thatthrough the array and selects the node NSiþ1 such that it isinside the slice LBA ðSi Þ, ﬀA ðLBA ðSi Þ; LBA ðSiþ1 ÞÞ 6¼ 0, and ﬀA ðLBA ðSi Þ; LBA ðSiþ2 ÞÞ : 3 8NB inside LBA ðSi Þ : ð2Þ Therefore, Siþ2 is inside LBA ðSi Þ. Thus, using (2), we ﬀA ðLBA ðSi Þ; LBA ðBÞÞ ﬀA ðLBA ðSi Þ; LBA ðSiþ1 ÞÞ: haveIf there is no such node, the algorithm selects NSiþ1 such that ﬀA ðLBA ðSi Þ; LBA ðSiþ2 ÞÞ ﬀA ðLBA ðSi Þ; LBA ðSiþ1 ÞÞ; 8NB inside CA;R : which is a contradiction. u t ð3Þ ﬀA ðLBA ðSi Þ; RBA ðSiþ1 ÞÞ ﬀA ðLBA ðSi Þ; RBA ðBÞÞ:The algorithm terminates by selecting the last node NSm if Theorem 2. The proposed slice-based selection algorithm willNSm is inside LBA ðS1 Þ or NS1 is inside LBA ðSm Þ or select at most 11 nodes.Smþ1 ¼ S1 . Proof. By contradiction, assume that the algorithm selects more than 11 nodes. Therefore, S11 is not in LBA ðS1 Þ.Algorithm 2 A slice-based selection algorithm Using Lemma 4, we getInput: ListA ½1 . . . n: List of all neighbors of NAOutput: A B-coverage set of NA : fNSi g ﬀA ðLBA ðS1 Þ; LBA ðS11 ÞÞ 1: ind 1; i 0 X 5 2: repeat ¼ ðﬀA ðLBA ðS2iÀ1 Þ; LBA ðS2iþ1 ÞÞÞ 5 Â : 3: ang max 0; ang min 2 i¼1 3 4: i iþ1 T h e r e f o r e , ﬀA ðLBA ðS11 Þ; LBA ðS1 ÞÞ ð2 À 5 Â Þ ¼ . 3 3 5: NSi ListA ½ind Consequently, S1 is inside LBA ðS11 Þ; thus, the proposed 6: chk false slice-based algorithm will terminate after selecting S11 .t u 7: for j ¼ 1; j lengthðListA Þ; j++ do 8: if ListA ½j is in LBA ðSi Þ then The above theorem gives an upper bound on the number of 9: if ﬀA ðLBA ðSi Þ; LBA ðListA ½jÞÞ ang max nodes selected by the proposed selection algorithm. In then Section 5, using simulation, we show that the average number of selected nodes (when the nodes are distributed10: chk true uniformly) is less than six.11: ind max j12: ang max ﬀA ðLBA ðSi Þ; LBA ðListA ½jÞÞ Theorem 3. Time complexity of the proposed slice-based selection13: end if algorithm is OðnÞ, where n is the number of neighbors.14: else Proof. The algorithm selects the first node in Oð1Þ. To select15: if ﬀA ðLBA ðSi Þ; RBA ðListA ½jÞÞ ang min each of the other nodes, the algorithm performs OðnÞ then operations by checking all the neighbors in the array.16: ind min j Therefore, the complexity of the algorithm is Oðm Â nÞ,17: ang min ﬀA ðLBA ðSi Þ; RBA ðListA ½jÞÞ where m is the number of selected nodes. Using18: end if Theorem 2, we have m 11; thus, the time complexity19: end if of algorithm is OðnÞ. u t
9. 9. KHABBAZIAN AND BHARGAVA: EFFICIENT BROADCASTING IN MOBILE AD HOC NETWORKS 239the square area. Corollary 2 shows that the probability thatNB broadcasts the message exponentially decreases whenthe distance AB decreases or when the node density increases. This result is further confirmed by simulation inSection 5.Example 2. Suppose R ¼ 250 m, L ¼ 1; 000 m, and L2 ¼ 300 (i.e., there are about 300 nodes in the network). Let P rbðBrdB Þ be the probability that NB broadcasts the message after receiving it from NA . Using Theorem 8, we g e t P rbðBrdB Þ 1:26 Â 10À2 , P rbðBrdB Þ 1:4 Â 10À3 , and P rbðBrdB Þ 10À4 when AB ¼ 100 m, AB ¼ 80 m, and AB ¼ 60 m, respectively. Let R1 ; R2 ; . . . Rk be k nonoverlapping regions inside thenetwork. Suppose that R is the event R ¼ fThe region R contains at least one nodeg:Since the nodes are placed by homogeneous planar Poisson Fig. 8. Finding a lower bound for ÁðIðDA;R ; DB;R ; DQ;QB ÞÞ.distribution, the events Ri are independent [20]. Conse-quently, we have Thus, using an induction hypothesis, we get À Á P rbðR1 ; R2 ; . . . ; Rk Þ ¼ P rbðR1 ÞP rbðR2 Þ . . . P rbðRk Þ: ð5Þ P rb R1 ; R2 ; . . . ; Rd ; Rdþ1 À Á ! P rb Rdþ1 P rbðR1 ; R2 ; . . . ; Rd ÞLemma 5 generalizes (5) to the case where the regions may À Áoverlap each other. ! P rbðR1 ÞP rbðR2 Þ . . . P rbðRd ÞP rb Rdþ1 :Lemma 5. Let R1 ; R2 ; . . . Rk be k regions inside the network. We u t have Lemma 6. Let DA;R and DB;R be two disks with radius R and P rbðR1 ; R2 ; . . . ; Rk Þ ! P rbðR1 ÞP rbðR2 Þ . . . P rbðRk Þ: centers A and B, respectively. Suppose AB R. As shown in Fig. 8, consider a point Q such that R QA and QB R.Proof. The proof is by induction on the number of regions. Let DQ;QB be a disk with radius QB. We have The lemma is true if the number of regions is one (i.e., ðR À ABÞ2 k ¼ 1). Suppose that the inequality holds for k ¼ d Á IðDA;R ; DB;R ; DQ;QB Þ ! ; regions. We have 3 À Á where I ðDA;R ; DB;R ; DQ;QB Þ is the intersection of disks DA;R , P rb R1 ; R2 ; . . . ; Rd jRdþ1 À Á ð6Þ DB;R , and DQ;QB , and ÁðRÞ is the area of region R. ¼ P rb R1 ÀRdþ1 ; R2 ÀRdþ1 ; . . . ; Rd ÀRdþ1 ; Proof. For any point P on the circle CA;R , we have where Ri is the complement of Ri , and Ri À Rj is the collection of all points inside Ri and outside Rj . Note that BP AP À AB ¼ R À AB: À Á P rb R1 ; R2 ; . . . ; Rd jR1 ÀRdþ1 ; R2 ÀRdþ1 ; . . . ; Rd ÀRdþ1 ¼ 1: Therefore, as shown in Fig. 8, the disk DB;ðRÀABÞ is inside I ðDA;R ; DB;R Þ. Consequently, we have Thus, P rbðR1 ; R2 ; . . . ; Rd Þ Á IðDA;R ; DB;R ; DQ;QB Þ ! Á IðDB;ðRÀABÞ ; DQ;QB Þ : À Á ð7Þ ! P rb R1 ÀRdþ1 ; R2 ÀRdþ1 ; . . . ; Rd ÀRdþ1 : Since QA ! R, using triangle inequality, we get It follows from (6) and (7) that QB ! QA À AB ! R À AB: À Á P rbðR1 ; R2 ; . . . ; Rd Þ ! P rb R1 ; R2 ; . . . ; Rd jRdþ1 : ð8Þ Therefore, we have We have ﬀQBC ! and ﬀQBD ! ; 3 3 P rbðR1 ; R2 ; . . . ; Rd Þ À Á À Á and hence, ﬀCBD ! 2 . Therefore, ¼ P rb Rdþ1 P rb R1 ; R2 ; . . . ; Rd jRdþ1 3 À Á À Á þ P rb Rdþ1 P rb R1 ; R2 ; . . . ; Rd jRdþ1 : Á DB;ðRÀABÞ Á IðDB;ðRÀABÞ ; DQ;QB Þ ! Therefore, using (8), we get 3 À Á À Á ðR À ABÞ2 P rb Rdþ1 P rb R1 ; R2 ; . . . ; Rd jRdþ1 ¼ : À Á 3 ! 1 À P rbðRdþ1 Þ P rbðR1 ; R2 ; . . . ; Rd Þ: u t