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HARD PROBLEMS IN E in .NET Generate 39 barcode in .NET HARD PROBLEMS IN E




How to generate, print barcode using .NET, Java sdk library control with example project source code free download:
7.2. HARD PROBLEMS IN E using barcode creator for visual .net control to generate, create code 39 image in visual .net applications. ISO QR Code standard 1 ( ), we obtain a circu Code 3/9 for .NET it C (y) = C(y, z)1 that contradicts the lemma s hypoth2 esis concerning F1 . Using Claim 7.

15.1, we show how to obtain a circuit that violates the lemma s hypothesis concerning F2 , and in doing so we complete the proof of the lemma..

Claim 7.15.2: There exists a circuit C of size s2 (n ) such that Proof: Otherwise, using any z {0, 1}n that satis es Pr[C(Y, z)1 = F1 (Y )] > Pr[C (Z ) = F2 (Z )] . Pr[C(Y, Z ) = F(Y, Z ) Visual Studio .NET USS Code 39 Z G] 1 ( ) 2. > 2 (n ). Proof: The second inequal Visual Studio .NET barcode 3/9 ity is due to Eq. (7.

11), and thus we focus on establishing the rst inequality. We construct the circuit C as suggested in the foregoing outline. Speci cally, we take a poly(n/ )-large sample, denoted S, from the distribution def (Y, F1 (Y )) and let C (z) = C(y, z)2 , where (y, v) is uniformly selected among the elements of S for which C(y, z)1 = v holds.

Details follow. Let m be a suf ciently large number that is upper-bounded by a polynomial in n/ , and consider a random sequence of m pairs, generated by taking m independent samples from the distribution (Y, F1 (Y )). We stress that we do not assume here that such a sample, denoted S, can be produced by an ef cient (uniform) algorithm (but, jumping ahead, we remark that such a sequence can be xed non-uniformly).

For each z G {0, 1}n , we denote by Sz the set of pairs (y, v) S for which C(y, z)1 = v. Note that Sz is a random sample of the residual probability space de ned by (Y, F1 (Y )) conditioned on C(Y, z)1 = F1 (Y ). Also, with overwhelmingly high probability, .

Sz = (n/ 2 ), because z G visual .net Code 39 implies Pr[C(Y, z)1 = F1 (Y )] /2 and m = (n/ 3 ).16 Thus, for each z G, with overwhelming probability (taken over the choices of S), the sample Sz provides a good approximation of the conditional probability space.

17 In particular, with probability greater than 1 2 n , it holds that. {(y, v) Sz : C(y, z)2 = F2 (z)}. Pr[C(Y, z)2 = F2 (z) . C(Y, z)1 = F1 (Y )] . . Sz 2 (7.12) Thus, with posi tive probability, Eq. (7.

12) holds for all z G {0, 1}n . The circuit C computing F2 is now de ned as follows. The circuit will contain a set S = {(yi , vi ) : i = 1, .

. . , m} (i.

e., S is hard-wired into the circuit C ) such that the following two conditions hold: 1. For every i [m] it holds that vi = F1 (yi ).

2. For each good z the set Sz = {(y, v) S : C(y, z)1 = v} satis es Eq. (7.

12). (In particular, Sz is not empty for any good z.).

Note that the expected si 3 of 9 for .NET ze of Sz is m /2 = (n/ 2 ). Using the Chernoff Bound, we get Pr S [.

Sz < m /4] = exp( (n/ visual .net ANSI/AIM Code 39 2 )) < 2 n . 17 For Tz = {y : C(y, z)1 = F1 (y)}, we are interested in a sample S of Tz such that .

{y S : C(y, z)2 = F2 (z)}. /. S . approximates Pr[C(Y, z)2 Code 3/9 for .NET = F2 (z) . Y Tz ] up to an additi ve term of /2. Using the Chernoff Bound again, we note that a random S Tz of size (n/ 2 ) provides such an approximation with probability greater than 1 2 n ..

THE BRIGHT SIDE OF HARDNESS On input z, the circuit C rst determines the set Sz , by running C for m times and checking, for each i = 1, . . .

, m, whether or not C(yi , z) = vi . In case Sz is empty, the circuit returns an arbitrary value. Otherwise, the circuit selects uniformly a pair (y, v) Sz and outputs C(y, z)2 .

(The latter random choice can be eliminated by an averaging argument; see Exercise 7.16.) Using the de nition of C and Eq.

(7.12), we have:. Pr[C (Z ) = F2 (Z )] . Pr[Z = z] Pr[C (z) = F2 (z)] Pr[Z = z] .
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