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how to make barcode in c#.net An application of number theory in Software
An application of number theory Generating Quick Response Code In None Using Barcode encoder for Software Control to generate, create QR Code JIS X 0510 image in Software applications. QR Recognizer In None Using Barcode scanner for Software Control to read, scan read, scan image in Software applications. The renowned mathematician G H Hardy once declared of his work: I have never done anything useful Hardy was an expert in the theory of numbers, which has long been regarded as one of the purest areas of mathematics, untarnished by material motivation and consequence Yet the work of thousands of number theorists over the centuries, Hardy s included, is now crucial to the operation of Web browsers and cell phones and to the security of nancial transactions worldwide Make QR Code In Visual C# Using Barcode creation for VS .NET Control to generate, create Denso QR Bar Code image in .NET applications. Quick Response Code Generator In VS .NET Using Barcode drawer for ASP.NET Control to generate, create QR Code 2d barcode image in ASP.NET applications. Privatekey schemes: onetime pad and AES
Encoding QR In VS .NET Using Barcode printer for Visual Studio .NET Control to generate, create QR image in .NET applications. Creating QR Code JIS X 0510 In Visual Basic .NET Using Barcode maker for Visual Studio .NET Control to generate, create QR Code image in Visual Studio .NET applications. If Alice wants to transmit an important private message to Bob, it would be wise of her to scramble it with an encryption function, e : messages encoded messages Of course, this function must be invertible for decoding to be possible and is therefore a bijection Its inverse is the decryption function d( ) In the onetime pad, Alice and Bob meet beforehand and secretly choose a binary string r of the same length say, n bits as the important message x that Alice will later send Alice s encryption function is then a bitwise exclusiveor, e r (x) = x r: each position in the encoded message is the exclusiveor of the corresponding positions in x and r For instance, if r = 01110010, then the message 11110000 is scrambled thus: er (11110000) = 11110000 01110010 = 10000010 This function er is a bijection from nbit strings to nbit strings, as evidenced by the fact that 37 ECC200 Encoder In None Using Barcode creation for Software Control to generate, create Data Matrix ECC200 image in Software applications. Encode Code 39 Full ASCII In None Using Barcode creator for Software Control to generate, create Code 39 Full ASCII image in Software applications. it is its own inverse! er (er (x)) = (x r) r = x (r r) = x 0 = x, where 0 is the string of all zeros Thus Bob can decode Alice s transmission by applying the same encryption function a second time: d r (y) = y r How should Alice and Bob choose r for this scheme to be secure Simple: they should pick r at random, ipping a coin for each bit, so that the resulting string is equally likely to be any element of {0, 1}n This will ensure that if Eve intercepts the encoded message y = e r (x), she gets no information about x Suppose, for example, that Eve nds out y = 10; what can she deduce She doesn t know r, and the possible values it can take all correspond to different original messages x: GS1 128 Printer In None Using Barcode generation for Software Control to generate, create GTIN  128 image in Software applications. GS1  13 Maker In None Using Barcode creator for Software Control to generate, create EAN13 image in Software applications. 00 01 10 11 x Barcode Generation In None Using Barcode maker for Software Control to generate, create barcode image in Software applications. Printing Barcode In None Using Barcode printer for Software Control to generate, create bar code image in Software applications. e10 e11 e00 e01
Encoding Code 2 Of 5 In None Using Barcode generation for Software Control to generate, create 2/5 Industrial image in Software applications. Encoding Bar Code In None Using Barcode encoder for Font Control to generate, create barcode image in Font applications. So given what Eve knows, all possibilities for x are equally likely! The downside of the onetime pad is that it has to be discarded after use, hence the name A second message encoded with the same pad would not be secure, because if Eve knew x r and z r for two messages x and z, then she could take the exclusiveor to get x z, which might be important information for example, (1) it reveals whether the two messages begin or end the same, and (2) if one message contains a long sequence of zeros (as could easily be the case if the message is an image), then the corresponding part of the other message will be exposed Therefore the random string that Alice and Bob share has to be the combined length of all the messages they will need to exchange The onetime pad is a toy cryptographic scheme whose behavior and theoretical properties are completely clear At the other end of the spectrum lies the advanced encryption standard (AES), a very widely used cryptographic protocol that was approved by the US National Institute of Standards and Technologies in 2001 AES is once again privatekey: Alice and Bob have to agree on a shared random string r But this time the string is of a small xed size, 128 to be precise (variants with 192 or 256 bits also exist), and speci es a bijection e r from 128bit strings to 128bit strings The crucial difference is that this function can be used repeatedly, so for instance a long message can be encoded by splitting it into segments of 128 bits and applying er to each segment The security of AES has not been rigorously established, but certainly at present the general public does not know how to break the code to recover x from e r (x) except using techniques that are not very much better than the bruteforce approach of trying all possibilities for the shared string r 38 EAN 13 Recognizer In C#.NET Using Barcode recognizer for Visual Studio .NET Control to read, scan read, scan image in .NET framework applications. 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Unlike the previous two protocols, the RSA scheme is an example of publickey cryptography: anybody can send a message to anybody else using publicly available information, rather like addresses or phone numbers Each person has a public key known to the whole world and a secret key known only to him or herself When Alice wants to send message x to Bob, she encodes it using his public key He decrypts it using his secret key, to retrieve x Eve is welcome to see as many encrypted messages for Bob as she likes, but she will not be able to decode them, under certain simple assumptions The RSA scheme is based heavily upon number theory Think of messages from Alice to Bob as numbers modulo N ; messages larger than N can be broken into smaller pieces The encryption function will then be a bijection on {0, 1, , N 1}, and the decryption function will be its inverse What values of N are appropriate, and what bijection should be used Property Pick any two primes p and q and let N = pq For any e relatively prime to (p 1)(q 1): 1 The mapping x xe mod N is a bijection on {0, 1, , N 1} 2 Moreover, the inverse mapping is easily realized: let d be the inverse of e modulo (p 1)(q 1) Then for all x {0, , N 1}, (xe )d x mod N The rst property tells us that the mapping x x e mod N is a reasonable way to encode messages x; no information is lost So, if Bob publishes (N, e) as his public key, everyone else can use it to send him encrypted messages The second property then tells us how decryption can be achieved Bob should retain the value d as his secret key, with which he can decode all messages that come to him by simply raising them to the dth power modulo N Example Let N = 55 = 5 11 Choose encryption exponent e = 3, which satis es the condition gcd(e, (p 1)(q 1)) = gcd(3, 40) = 1 The decryption exponent is then d = 3 1 mod 40 = 27 Now for any message x mod 55, the encryption of x is y = x 3 mod 55, and the decryption of y is x = y 27 mod 55 So, for example, if x = 13, then y = 13 3 = 52 mod 55 and 13 = 5227 mod 55 Let s prove the assertion above and then examine the security of the scheme Proof If the mapping x xe mod N is invertible, it must be a bijection; hence statement 2 implies statement 1 To prove statement 2, we start by observing that e is invertible modulo (p 1)(q 1) because it is relatively prime to this number To see that (x e )d x mod N , we examine the exponent: since ed 1 mod (p 1)(q 1), we can write ed in the form 1 + k(p 1)(q 1) for some k Now we need to show that the difference xed x = x1+k(p 1)(q 1) x is always 0 modulo N The second form of the expression is convenient because it can be simpli ed using Fermat s little theorem It is divisible by p (since x p 1 1 mod p) and likewise 39 Barcode Maker In .NET Framework Using Barcode printer for ASP.NET Control to generate, create bar code image in ASP.NET applications. Generate EAN13 In Java Using Barcode printer for BIRT Control to generate, create GS1  13 image in Eclipse BIRT applications. Figure 19 RSA Bob chooses his public and secret keys He starts by picking two large (nbit) random primes p and q Print Barcode In Java Using Barcode encoder for Java Control to generate, create barcode image in Java applications. Code39 Printer In Java Using Barcode encoder for Android Control to generate, create Code 3/9 image in Android applications. His secret key is d, the inverse of e modulo (p 1)(q 1), computed using the extended Euclid algorithm Alice wishes to send message x to Bob His public key is (N, e) where N = pq and e is a 2nbit number relatively prime to (p 1)(q 1) A common choice is e = 3 because it permits fast encoding She looks up his public key (N, e) and sends him y = (x e mod N ), computed using an ef cient modular exponentiation algorithm He decodes the message by computing y d mod N by q Since p and q are primes, this expression must also be divisible by their product N Hence xed x = x1+k(p 1)(q 1) x 0 (mod N ), exactly as we need The RSA protocol is summarized in Figure 19 It is certainly convenient: the computations it requires of Alice and Bob are elementary But how secure is it against Eve The security of RSA hinges upon a simple assumption: Given N, e, and y = xe mod N , it is computationally intractable to determine x This assumption is quite plausible How might Eve try to guess x She could experiment with all possible values of x, each time checking whether x e y mod N , but this would take exponential time Or she could try to factor N to retrieve p and q, and then gure out d by inverting e modulo (p 1)(q 1), but we believe factoring to be hard Intractability is normally a source of dismay; the insight of RSA lies in using it to advantage

