1. Selecting Forwarding Neighbors in Wireless Ad Hoc Networks
Yuchen Wang
Ying Luo
CS 6520 Final Project Apr. 27, 2001

2. Problem Background Broadcasting in ad hoc mobile networks
Flooding & Broadcast Storm
Flooding: Each node retransmits the message to its 1-hop neighbors
Broadcast Storm: redundant transmission, collisions
= Origin of message
= 1-hop adjacent to origin
= 3-hop adjacent to origin
= 2-hop adjacent to origin

3. Minimum Forwarding Set Problem
Given the origin of a message, there is a set of 1-hop neighbors and a set of 2-hop neighbors.
Find the Minimum Forwarding Set from the set of 1-hop neighbors such that every 2-hop neighbor is within the coverage of a Minimum Forwarding Set 1-hop neighbor
= 2-hop neighbors
= 1-hop neighbors
= origin of message
= Minimum Forwarding Set

4. Algorithm 1 Algorithm 1: Refine Disk Covering
Input: Unit-disk A, set of unit disks D centered inside A,
set of points P outside A such that P   {D  D}
Output: Subset F  D such that P   {D  F}
1. Partition the exterior of A into four quadrants Q1 - Q4 by two orthogonal lines through the center of A,
such that no point in P or center of disk in D belongs to any of the lines.
2. For q = 1, … , 4, do
(a) Find the set of disks Dq = {D1, … , D|Dq|} of D which have a non-empty intersection with the interior
of Qq. For each Dj  Dd find the two points of intersection, lj and rj, of the boundary circle of Dj with Jq, the boundary of Qq. We assume that lj < rj in a fixed orientation of Jq .
(b) Renumber the disks in Dq such that either lj < lj+1 or lj = lj+1 and rj < rj+1 for every j = 1, …, . |Dq| - 1.
Let Fq be the list of disks in Dq enumerated in this order.

5. (d) While there is a disk Dj  Fq whose points in Qq are covered by the disks, Dp and Ds, that precede, respectively succeed Dj in Fq, remove disk Dj from Fq (points of Dj in Fq are covered by Dp and Ds if Dj  P  Qq  Dp  Ds).
(c) Remove from the list Fq each disk Dj for which there is another disk Dk  Fq such that lk  lj < rj  rk.
3. Output F = F1  F2  F3  F4

6. 1. Partition in 4 quadrants by two orthogonal lines
Algorithm 1
1. Partition in 4 quadrants by two orthogonal lines Q1 Q2 Q4 Q3

7. Algorithm 1 2(a) find the intersection with the interior of Qq:
Disk 1’s two intersection points are L1,R1
Disk 2’s two intersection points are L2,R2
Disk 3’s two intersection points are L3,R4
Dish 4’s two intersection points are L4,R3
2(b) sort the disks in 1- Hop disk covering
the points are sorted as following:
L1< L2 <L3< L4 <R1 <R2 <R3 <R4
So the sorted disks’ re D1<D2<D3<D4
At present, Fq is the list of D1, D2, D3, D4 Q1 Q3 Q4 Q2 l2 R2 l3 R3 l4 R4 l1 R1 D1 D2 D3 D4

8. 2(c) remove fully covered disks
Algorithm 1 (cont.)
2(c) remove fully covered disks Q1 Q3 Q4 Q2 l2 R2 l3 R3 l4 R4 l1 R1 D1 D2 D3 D4 Q1 Q3 Q4 Q2 l3 R3 l1 R1 l2 R2 D1 D2 D3
Fq : D1 D2 D3 D4
Fq : D1 D2 D3
D4 is fully covered by D3, so remove it

9. Algorithm 1 2(d) Remove covered by two neighbors R3 l1 l2 R2 l3 R1 Q1Fq D1 D2 D3 D3 D2 D2 D3 D1 D1 D1 D1
D2 is covered by D1 and D3 , so remove it

10. 3 Output Q1 Q3 Q4 Q2 l3 R3 l1 R1 D1 D3
Output Fq : D1, D3

11. Runtime (using stack)
Refined disk covering algorithm can be implemented in O(n²) . n is the number of centers and the points. Obviously, the steps 1, 2(a), 2(c) 3 can be done in linear time. And the sorting step need O(nLogn) time. To implement step 2(d) , proceed with the first disk, keeping the surviving disks in a stack. At each iteration, check if the topmost disk, together with the disk two positions under the top, and repeat. If no, add the next disk on the top. Each iteration results in either one disk being discarded or a disk being aded to the stack. Thus, there are at most 2n iteration. Since each iteration can be implemented in O(n) time by traversing all points in P, this results in O(n²) runtime for step 2(d), as well as the entire algorithm.

12. Analysis algorithm
Theorem: Algorithm Refined Disk Covering finds at most 3 times more disks than the optimum
- Fact 1: in each quadrant the algorithm finds the optimum number of disks covering all points
- Fact 2: Each disk may cover
points in at most 3 quadrants.
Runtime: O(n2),
n = # of points + neighbors Q1 Q3 Q4 Q2