Rabin Cryptosystem Conclusion

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Although the Rabiri cryptosystem is effective and it was developed shortly after RSA, it has never enjoyed anything like the popularity of RSA. Perhaps, this is because only one of the four possible decrypts of a ciphertext corresponds to the plaintext. However, this issue is easily resolved.3 It is easily verified that the security of the Rabin cryptosystem is equivalent to factoring, while this is not known to be the case for RSA. Consequently, it is conceivable that R S A could be compromised without solving the factoring problem. Viewed in this light, it is not unreasonable to argue that the Rabin cryptosystem rests on a somewhat more secure foundation than RSA. Of course, to date, thc most significant general attack on RSA is to factor thc modulus. So, in a practical sense, thc security of RSA and Rabin are indistinguishable today. Without an overwhelming reason to abandon the gold st,aridard o f RSA, it is unlikely that the Rabin cryptosysteni will gain a 1110rcsignificant following in the public kcy arena.

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"HSA was patented (the patents have now expired) and promoted by RSA Security, Inc.

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Jri contrast, thc Rahin cipher had no comparable corporate hacking.

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6.7 NTR U CIPHER

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NTRU Cipher

That s a great deal to make one word mean, Alice said in a thoughtful tone. When I make a word do a lot of work like that, said Humpty Durnpty, (I always pay it extra. - Through the Looking Glass

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Relative to many other public key cryptosystems, the NTRU cipher, rumored to stand for Nth-degree TRUncated polynomial ring or Number Theorists aRe Us, is young. Invented in 1995 by Hoffstein, Pipher, and Silverman [68], it is somewhat more complicated than RSA or the Rabin cryptosystem. The security of NTRU derives from the difficulty of a certain factoring problem in a polynomial ring. We have more to say about this below, but first we describe the NTRU system and give an example. The NTRU cipher depends on three positive integer parameters ( N , , Q) p and four sets of polynomials of degree N - 1 with integer coefficients. The sets if polynomials are denoted L f ,L,, L,, and L,. The parameters p and g are chosen so that gcd(p,q) = 1 and q > p , where q must be much larger than p . All of the NTRU polynomials are in the set of truncated polynomials of degree N - 1 having integer coefficients. That is, an NTRU polynomial is of the form

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u(x) = uo + a12 + a2x2

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+ . . + uN-2xN-2 + U N - l X N-1 ,

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where the ai are integers (taken modulo p or g, depending on the specific polynomial). Polynomials are added in the usual way. Multiplication is perfomed modulo xN - 1, meaning that polynomials are multiplied in the usual way, but xN is replaced by 1, x N f l is replaced by x, xN+2is replaced by x2 and so on. We use the symbol LL* denote this type of polynomial multo tiplication. In mathematical terms, all of the NTRU polynomials are in the quotient ring

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- 1) The message space L , consists of all polynomials in R modulo p . Assuming that p is odd, we define the message space as

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{M(x) E R

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all coefficients of M lie in [ - ( p

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1 ) / 2 , ( p - 1)/2]}.

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As a notational convenience, let L ( d 0 ,d l ) be the set of polynomials in R with do coefficients that are +1 and dl coefficents that are -1, and all remaining coefficients are 0. For example,

+ x2 + x3 - x5 + x9 E L ( 3 , 2 ) ,

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PUBLIC KE Y SYSTEn;%S

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since the nonzero coefficients consists of three +Is and two -1s. Given the NTRU parameters ( N ,p , (I), we must select three additional parameters, denoted d f , d,: arid d, which should be selected from the recommended NTRU parameters (the current set of recommended paranieters can be found at [108]). These additional parameters are used t o define the sets of polynomials

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L f = L ( d f df ,

L, = L(d,, d,),

and L,, = L ( d , d ) .

Now Alice generates her NTRU key pair as follows. She first chooses two polynomials f ( z ) and g z ! where f ( z ) E L f and f ( x )is invertible modulo p () and modulo q , and g(z) E L,. She can find a suitable f ( z )using the algorithm in Table 6.3 [137]. Table 6.3: Algorithm t o Find Inverse Polynomial

// // //

Input: polynomial u ( J ) , primc p Output: b ( r ) = u ( z ) p l in ( 2 / p 2 ) [ z ] / ( z N- 1) Initialization k = 0, b(z)= 1, .(z) = 0, f ( z ) = u ( z ) , g(z) = zN - 1 // find inverse