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Table 6.2: Gram-Schmidt, Process
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// Gram--Schmidt M = (bo, b l , . . . , b,) GS(Rf) 20 = bo for j = I t o n 2; = b . .I 3 for i =0 t o j - 1 yl/ij = ( x i . bj)/\lx:l/il12
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end GS Now, suppose that Trudy wants to recover the plaintext that corresponds to ciphertext C = 548. Since Trudy knows the public key T and ciphertext C = 548, she needs to firid a set of u l , for i = 0 , 1 , . . . , 7 , with the rcstrictiori that each u,E (0, l}, and
This can be written as the matrix equation
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where T is Alicc's public knapsack and U = ( 2 1 0 , u 1 , . . . , u 7 ) , arid the a, are unknown. but earh must be cither 0 or 1. This is of the form AU = B (as discussed above), so Trudy rewrites the matrix equation as AJV = W and applies the LLL Algorithm to 1 . . this case, Trudy firids 2 1In - 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 - 8 2 123 287
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0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 83 248 373
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0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 10 471 -548
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The LLL Algorithm outputs a matrix h consisting of short vectors in the 1.
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6.3 DIFFIE-HELLMAN KEY EXCHANGE
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--1-1 0 1 0 1 0 0 1 0 - 1 1 0 1-1 0 0 0 0 1-1 0 0 0 - 1 1 2 1 -1 -1 1 0 -1 0 -1 0 M = 0 0 1 0 - 2 - 1 0 1 0 0 0 0 1 1 1 1 - 1 1 0 0 0 1 0 0 - 1 0 - 1 0 0 0 0 0 0 1 1 - 1 - 1 - 1 1 0 0 1-1 2 0 The entries in the fourth column of M have the correct form to be a solution to this knapsack problem. Therefore, Trudy obtains the putative solution
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u = (1,0,0,1,0,1,1,0)
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Using the public key and ciphertext C = 548, she can easily verify that U is indeed the original plaintext sent by Bob.
Knapsack Conclusion
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Much research has been done on the knapsack problem since the MerkleHellman cryptosystem was broken. Several different knapsack variants have been created and some of these appear to yield secure cryptosystems. However, people have been reluctant to use these systems, since knapsack continues to be equated with broken, even to this day. For more information on knapsack cryptosystems, see [37, 89, 1091.
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Diffie-Hellman Key Exchange
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[If] you look right under the center of a streetlight,
you don t find anything that wasn t known before. I f you look out into the darkness, you don t discover anything, cause you can t see anything. So you re always working at the edge o f the streetlight, trying to find your keys. - Whitfield Diffie
In symmetric key cryptography, both parties use a common key to encrypt and decrypt messages. However, when using such a system, there is a critical issue that needs to be dealt with, that is, how can Alice and Bob agree upon a key Can this be accomplished in a secure manner over a public channel
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These and related questions were on the minds of Diffie, Hellman and Merkle during the 1970s. At the time, there was no solution in sight for this vexing kxy distribution problem. In 1976, Diffie arid Hellman published their seminal paper [38] which made the case that public key cryptography should be possible and proposed the Diffie~-Hellman key exchange protocol as a solution to the key distribution problem; see also [l20]. Ironically, the first to discover DiffieeHellman was actually Malcolm Williamson of GCHQ (roughly, the British equivalent of NSA) [93]. However, this does nothing to diminish the accomplishment of Diffie and Hellman, since Williamson's work was classified. The Diffie-Hellman key exchange, as illustrated in Figure 6.2; opemtes in the following way. Let p be a large prime and g an integer, where 2 5 g 5 p-2. Both the prime p and generator g are publicly known. Alice chooses a random niiniher a , where 1 5 a 5 p - 2 and calculates a = ga (mod p ) . She then serids Q to Bob. The number a is private, that is, a is known only to Alice. Bob also chooses a random number b, where 1 5 b 5 p - 2 and calculates fi = 9' (mod p ) and sends /3 to Alice. The number b is private, known only to Bob. Alice t,lien calculates