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Theoretical analysis in Java Development qr barcode in Java Theoretical analysis




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17.4 Theoretical analysis using barcode encoder for java control to generate, create qr-codes image in java applications. .NET Framework 4.5 Beta Lemma 17.4.3 Java QR Let S be a circular area with center o and radius R.

Assume that node x lies within S and P(A S1 . A S, S1 S) = S1 /S. Let d(x) denote the random variable of the distance from x to o, then P(d(x) = r . x S) = P ROOF. For any 0 < r R, we have P(d(x) = r x S) = lim r 2 /( R 2 ) (r )2 /( R 2 ). 2r/R 2 0 0 r R, r > R. (17.6). 2r . R2 (17.7). For any r > R, we have x S, which implies P(d(x) = r x S) = 0. / QR for Java Lemma 17.4.

4 Let S be a circular area with center o and radius R. Given two nodes a and b independently deployed within S according to the 2D uniform distribution, we have 3 3 P(. ab > R a S, b S) = , (17.8) 4 where ab denotes the distance between a and b. P ROOF. We use Figure 17.

3 to help illustrate the proof. Let r denote the distance from a to o, let Co denote the circle with center o and radius R, and let Ca denote the circle with center a and radius R. Let c and d be the intersecting points between the two circles Co and Ca , and let = coa = doa.

Let SI (r ) denote the area of the intersection of the two circles Co and Ca with . oa = r , and le t SII (r ) denote the area of S minus SI (r ). Then we have P(. ab > R a S, b S) =. 2r SII (r ) dr, R2 S (17.9). where (17.9) Java QR comes from Lemma 17.4.

4. We rst calculate SI (r ): 2 r r r , R2 SI (r ) = 2 R 2 arccos 2R 2 2. (17.10). b R c a SII SI r/2 e Illustration for proving Lemma 17.4.4. Defense against traf c-injection attacks where = arc cos[r/(2R)]. Then SII (r ) can be calculated as r r r SII (r ) = R 2 2 arccos 2 R2 2R 2 R. . (17.11).

On integrat ing (17.11) into (17.9), we have P(.

ab > R a S , b S) = 3 3/(4 ). Lemma 17.4.

5 Assume that n nodes A = {a1 , . . .

, an } are independently deployed inside a circular area S according to the 2D uniform distribution with R being the radius, then we have. n P(. ai a j > R : ai , a j A) P(. a1 a2 > R)(2) .. (17.12). P ROOF. P(. ai a j > R : ai , a j A) = P(. a1 a2 > R, . . .

, . a1 an > R, . . .

, . an 1 an > R) = P(. a1 a2 > R a1 a3 > R, . . .

, . an 1 an > R) P(. a1 a3 > R, . . .

, . an 1 an > R) = P(. a1 a2 > R a1 ai > R, . a2 ai > R : 3 i n) P(. a1 a3 > R, . . .

, . an 1 an > R). Given a1 ai > R and a2 ai > R for a qr bidimensional barcode for Java ny 3 i n, we can draw a circle with center ai and radius R. To conform to the statement that ai , a j A, . ai a j > R, both QR Code 2d barcode for Java a1 and a2 cannot lie inside the area of intersection of this circle and the circle having o as its center. That is, a1 and a2 are now restricted to lying within an area of S S, which is smaller than S. So the probability that .

a1 a2 is larger th QR Code for Java an R under such restrictions will become smaller than it would be without such restrictions. That is, P(. a1 a2 > R a1 ai > R, . a2 ai > R) P(. a1 a2 > R : 3 Java QR Code i n). (17.13) Following the same arguments, we can have P(.

ai a j > R : ai , a j A) .
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