0" (mod

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arid Bob computes

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p) =

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(mod p )

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= gab

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(mod p )

a' (mod p ) = (g")b (mod p ) = g a b (mod p ) .

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Alice and Bob now share gab (mod p ) , which can be used as a symmetric key.

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Figure 6.2: Difie Hellman key exchange. Trudy sees ga (mod p ) and g6 (mod p ) and she breaks the key exchange protocol if she can find gab (mod p ) . To do 50, it appears that she must find a from ga (mod p ) or b from gb (mod p ) . Therefore, the strength of the DiffieHcllinan key exchange protocol is believed to depend on the computational

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6.3 DIFFIE-HELLMAN K E Y EXCHANGE

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complexity of solving the discrete logarithm problem. That is, there is no efficient solution to the problem of finding y given ICY (mod p ) , the base z and the modulus p . Suppose, for example, we want to solve the equation 2" = 9 (mod 11). Since the numbers are so small, this is easy to solve by exhaustively searching through all of the possible exponents. We see that

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2'= 1 (mod 11) 2l

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2 (mod 11)

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22 = 4 (mod 11) 23 = 8 (mod 11) 24 = 16 = 5 (mod 11) 25 = 5 . 2

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10 (mod 11)

1 0 . 2 = 9 (mod 11)

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and, therefore, II: = 6 is the desired solution. However, for large p , an exhaustive search is not feasible. Although there are some efficient methods for solving certain classes of discrete logarithm problems, there is no known efficient algorithm for solving g z = t (mod p ) for z in general, where g, t and p are given. Some of the current discrete log algorithms are analyzed in Section 7.3.

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Man-in-the-Middle At tack

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The Diffie-Hellrnan key exchange is subject to a man-in-the-middle attack if there is no procedure to authenticate the participants during the key exchange. Suppose that Trudy wants to read messages that are being sent between Alice and Bob, where Alice and Bob use the Diffie-Hellman key exchange. First, Trudy chooses an exponent t. She then intercepts ga (mod p ) and gb (mod p ) and sends g t (mod p ) to Alice and Bob. At this point, Alice believes gt (mod p ) came from Bob, and Bob believes gt (mod p ) came from Alice. Now Trudy computes K A = ( g a ) t (mod p ) and K B = (gb)t (mod p ) . Alice, not realizing that Trudy is in the middle, follows the Diffie-Hellman protocol and computes K A . Similarly, Bob computes K g . Then when Alice sends a message to Bob (encrypted with K A ) , Trudy can intercept it, decrypt it and re-encrypt it (or encrypt a different message) with K B before sending it on to Bob. In this manner, Trudy can read (and alter, if she so desires) all messages between Alice and Bob, and neither Alice nor Bob will suspect that there is any problem. Figure 6.3 illustrates this man-in-the-middle attack. The man-in-the-middle attack on the Diffic-Hellman key exchange can be prevented provided the parties are properly authenticated. For example, an authentication protocol that uses digital signatures would assure Alice and

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PUBLlC K E Y SYSTEMS

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Figure 6.3: Man-in-the-middle attack on Diffie-Hellniari

Bob that the received messages originated from the correct person. Since Trudy cannot forge Alice s or Bob s signatures, her man-in-the-middle attack would be thwarted. In addition, such a protocol will prevent a replay attack. There are several ways to prevent the man-in-the-middle attack on DiffieHellrnan. For example, the Station-to-Statzon protocol, devised by Diffie, van Oorschot and Wiener [39], could be used for authentication purposes. Problem 3 gives a simple example illustrating a technique that prevents the man-in-t he-middle attack.