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The following is the process of trying to generate a new detector when an old
detector is deleted in the filtration process listed above.
(1) Generate a new detector d' randomly.
(2) If d' is already included in the current set R , delete it and go back to (1).
(3) Perform partial matching between d' and elements in S one by one, get the
maximum matching length m' , if m' = l , go back to (1).
(4) Set the matching length threshold of d' as r' = m' +1, add d' to R .
Since this process could be time consuming, to limit the time cost for generating a
new detector, when an old detector is deleted in the updating process, step (1)-(3) in this
process will be performed for only a small constant number of cycles.
3.2 Strategy II: Stuffing Some Bits of Detector with a Special Symbol
Firstly, it is assumed that when a detector d (11100111) matches a self string
(00100110), the matching bits will be (10011), the maximum continuous matching
length m =5. If the indices of a detector's bits start from 1, the matching start point p =3.
The filtration process of a detector is given as follows.
If a detector d matches a self string, some selected matching bits of d will be stuffed
with a special symbol. For convenience, the special symbol is indicated by '#' in this
paper. Stuffing a matching bit with '#' makes this bit can not match either '0' or '1'.
Other matching bits can be kept unchanged. Therefore, the segment that consists of the
maximum continuous matching bits and represents some self patterns, is destroyed by
the stuffed '#'. And some of useful non-self patterns in this detector are reserved. The
key problem is how to choose the minimum number of bits to be stuffed with '#', and
these bits could destroy the self patterns and remain the largest number of non-self
patterns in this detector.
Fig. 2. Choose one bit to be stuffed with '#' when m<2r . '#' indicates a symbol other than '0'
and '1'. Stuffing a matching bit with '#' makes this bit can not match either '0' or '1'.
Firstly, we consider m<2r , then stuffing just one matching bit with '#' will be
enough. As shown in Fig. 2, if l =10, r =6, p =3 and m =7, replacing the 8th bit with '#' is
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