microsoft excel 2013 barcode generator The symmetric key for the session is K in Software

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The symmetric key for the session is K
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(f mod p)Y mod p = (gY mod p)X mod p = fY mod p
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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
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The symmetric (shared) key in the Diffie-Hellman protocol is K=!tYmodp
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CRYPTOGRAPHY
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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:
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1 Alice chooses x = 3 and calculates R 1 = 7 3 mod 23 = 21
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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
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The value of K is the same for both Alice and Bob; EfY mod p = 7 18 mod 23 = 18
Idea ofDiffie-Hellman
The Diffie-Hellman 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 one-third 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 two-thirds completed key from Bob, she adds the last part, her x, to complete the key When Bob receives the two-thirds 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 g-y-x and the key in Bob's hand is g-x-y, these two keys are the same because EfY =gYx
Figure 3027 Diffie-Hellman idea
Alice fills up another one-third of the secret key using her random number
Alice
One-third of the key is public
bd !
Bob fills up another one-third 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
ASYMMETRIC-KEY 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 Man-in-the-Middle Attack Diffie-Hellman is a very sophisticated symmetric-key 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 Man-in-the-middle attack
p and g are public
Alice
Rz =gZmodp
R3 =gYmodp
(R1)t mod p
K z "" (R;F mod p
-----------1- ---------- -----------1---------K 1 =gXZmodp
Alice-Eve key
Eve-Bob 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 man-in-the-middle 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
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