As illustrated in Figure 6, consider a non-faulty sensor
node Si 2 (S??’C1) which sits close to the event boundary but outside the event
region. Assume sensor nodes are uniformly distributed. Since only a small portion
of the disk is covered by the event region, the median of x(i)
1 , x(i)
2 , ..., x(i)
k
will be obtained from a sensor node outside the event region as well. Thus, Si
won??™t be detected by simply applying the faulty sensor detection algorithm.
In order to solve the problem, the faulty sensor detection algorithm should
be modified. Now, if we randomly draw a line through Si, it will equally divide
any closed disk centered at Si into two half disks. As illustrated in Figure 6,
the line randomly chosen intersects the boundary of the disk at points P1 and
P2, and the boundary of events intersects with the same disk at points A and
B. Based on equation (18), one can calculate di for each half disk. Again,
assume sensors are deployed uniformly, the half disk yielding the largest |di|,
denoted as NN(Si), would be the one containing P1, P2,B and Si. To make
the detection of sensor nodes near the event boundary more e?®ective, the
random bisection approach replaces the old di for Si with the di resulting
from NN(Si). Then perform calculation in equation (21). If |yi| ?? ?µ, the
random bisection approach claims Si as an edge sensor.
Fig. 6. An illustration of random bisection. The half disk yielding the largest |di|
is the one containing P1, P2, and Si.
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