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microsoft excel 2013 barcode generator The symmetric key for the session is K in Software
The symmetric key for the session is K Encode Code 3 Of 9 In None Using Barcode creator for Software Control to generate, create Code39 image in Software applications. Scanning ANSI/AIM Code 39 In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. (f mod p)Y mod p = (gY mod p)X mod p = fY mod p
USS Code 39 Creation In C#.NET Using Barcode drawer for Visual Studio .NET Control to generate, create ANSI/AIM Code 39 image in .NET applications. Paint Code 39 Extended In .NET Framework Using Barcode generator for ASP.NET Control to generate, create ANSI/AIM Code 39 image in ASP.NET applications. Bob has calculated K = (R 1 mod p = (If mod p mod p = lfY mod p Alice has calculated K = (R 2)X mod p = (gY mod p)X mod = lfY mod p Both have reached the same value without Bob knowing the value of x and without Alice knowing the value of y Encoding Code 3 Of 9 In .NET Framework Using Barcode creator for Visual Studio .NET Control to generate, create Code 39 Full ASCII image in VS .NET applications. ANSI/AIM Code 39 Encoder In VB.NET Using Barcode printer for Visual Studio .NET Control to generate, create Code39 image in VS .NET applications. The symmetric (shared) key in the DiffieHellman protocol is K=!tYmodp
Code 128C Maker In None Using Barcode generator for Software Control to generate, create Code 128 Code Set C image in Software applications. Encode Barcode In None Using Barcode generator for Software Control to generate, create barcode image in Software applications. CRYPTOGRAPHY
Print UCC  12 In None Using Barcode maker for Software Control to generate, create GS1  12 image in Software applications. ANSI/AIM Code 39 Printer In None Using Barcode generation for Software Control to generate, create Code 39 Full ASCII image in Software applications. Example 3010
DataMatrix Generation In None Using Barcode encoder for Software Control to generate, create Data Matrix 2d barcode image in Software applications. Generating UCC128 In None Using Barcode generator for Software Control to generate, create GS1 128 image in Software applications. Let us give a trivial example to make the procedure clear Our example uses small numbers, but note that in a real situation, the numbers are very large Assume g = 7 and p = 23 The steps are as follows: Paint Postnet 3 Of 5 In None Using Barcode generator for Software Control to generate, create USPS POSTNET Barcode image in Software applications. Data Matrix 2d Barcode Decoder In None Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications. 1 Alice chooses x = 3 and calculates R 1 = 7 3 mod 23 = 21
Read UPCA Supplement 2 In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. Bar Code Printer In None Using Barcode encoder for Office Word Control to generate, create barcode image in Microsoft Word applications. 2 3 4 5 6 Code 128 Code Set B Encoder In Visual C#.NET Using Barcode generator for Visual Studio .NET Control to generate, create Code 128 Code Set B image in .NET framework applications. GTIN  128 Creation In VB.NET Using Barcode creation for .NET framework Control to generate, create GTIN  128 image in VS .NET applications. Bob chooses y = 6 and calculates R 2 = 7 6 mod 23 = 4 Alice sends the number 21 to Bob Bob sends the number 4 to Alice Alice calculates the symmetric key K = 4 3 mod 23 = 18 Bob calculates the symmetric key K = 21 6 mod 23 = 18 Linear 1D Barcode Encoder In VB.NET Using Barcode creation for VS .NET Control to generate, create Linear image in Visual Studio .NET applications. EAN / UCC  13 Creation In Java Using Barcode generation for Java Control to generate, create USS128 image in Java applications. The value of K is the same for both Alice and Bob; EfY mod p = 7 18 mod 23 = 18
Idea ofDiffieHellman
The DiffieHellman concept, shown in Figure 3027, is simple but elegant We can think of the secret key between Alice and Bob as made of three parts: g, x, and y The first part is public Everyone knows onethird of the key; g is a public value The other two parts must be added by Alice and Bob Each adds one part Alice adds x as the second part for Bob; Bob adds y as the second part for Alice When Alice receives the twothirds completed key from Bob, she adds the last part, her x, to complete the key When Bob receives the twothirds completed key from Alice, he adds the last part, his y, to complete the key Note that although the key in Alice's hand consists of gyx and the key in Bob's hand is gxy, these two keys are the same because EfY =gYx Figure 3027 DiffieHellman idea
Alice fills up another onethird of the secret key using her random number
Alice
Onethird of the key is public
bd ! Bob fills up another onethird of the secret key using his random number
The two keys are the same because it does not matter if x is filled first or y
SECTION 303
ASYMMETRICKEY CRYPTOGRAPHY
Note also that although the two keys are the same, Alice cannot find the value y used by Bob because the calculation is done in modulo p; Alice receives gY mod p from Bob, not gY ManintheMiddle Attack DiffieHellman is a very sophisticated symmetrickey creation algorithm If x and y are very large numbers, it is extremely difficult for Eve to find the key, knowing only p and g An intruder needs to determine x and y if R 1 and R2 are intercepted But finding x from R 1 and y from R 2 are two difficult tasks Even a sophisticated computer would need perhaps years to find the key by trying different numbers In addition, Alice and Bob will change the key the next time they need to communicate However, the protocol does have a weakness Eve does not have to find the value of x and y to attack the protocol She can fool Alice and Bob by creating two keys: one between herself and Alice and another between herself and Bob Figure 3028 shows the situation Figure 3028 Maninthemiddle attack p and g are public
Alice
Rz =gZmodp
R3 =gYmodp
(R1)t mod p
K z "" (R;F mod p
1  1K 1 =gXZmodp
AliceEve key
EveBob key
Kz = gZY modp
The following can happen: 1 Alice chooses x, calculates R1 = gX mod p, and sends R1 to Bob 2 Eve, the intruder, intercepts R l' She chooses z, calculates R2 = Ef mod p, and sends R 2 to both Alice and Bob CRYPTOGRAPHY
3 Bob chooses y, calculates R 3 = gY mod p, and sends R 3 to Alice; R 3 is intercepted by Eve and never reaches Alice 4 Alice and Eve calculate K I = Etz mod p, which becomes a shared key between Alice and Eve Alice, however, thinks that it is a key shared between Bob and herself 5 Eve and Bob calculate K 2 = gZY mod p, which becomes a shared key between Eve and Bob Bob, however, thinks that it is a key shared between Alice and himself In other words, two keys, instead of one, are created: one between Alice and Eve and one between Eve and Bob When Alice sends data to Bob encrypted with K 1 (shared by Alice and Eve), it can be deciphered and read by Eve Eve can send the message to Bob encrypted by K 2 (shared key between Eve and Bob); or she can even change the message or send a totally new message Bob is fooled into believing that the message has come from Alice A similar scenario can happen to Alice in the other direction This situation is called a maninthemiddle attack because Eve comes in between and intercepts R I , sent by Alice to Bob, and R 3, sent by Bob to Alice It is also known as a bucket brigade attack because it resembles a short line of volunteers passing a bucket of water from person to person

