By Neal Koblitz

ISBN-10: 3540634460

ISBN-13: 9783540634461

From the experiences: "This is a textbook in cryptography with emphasis on algebraic tools. it really is supported by means of many routines (with solutions) making it applicable for a direction in arithmetic or computing device technology. [...] total, this is often a superb expository textual content, and may be very priceless to either the coed and researcher." Mathematical stories

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**Additional resources for Algebraic Aspects of Cryptography (Algorithms and Computation in Mathematics)**

**Example text**

An efficient algorithm for this promise problem would suffice to break RSA. Of course, in the unlikely event that an efficient algorithm was found that could factor a product of two primes but not a product of three primes, it would be easy to modify RSA to use moduli that are products of three primes. §6. 2 is of a different sort from Example 5 . 1 , although it is also related to the problem of cryptanalyzing a public key encryption system. 2 is concerned with solving the underlying mathematical problem upon which RSA is based, whereas the cracking problem is the narrowest possible description of the cryptanalyst's task.

It is very unlikely that this could be done. ) On the other hand, one can use the binary search method to reduce the Exact Traveling Salesrep decision problem to the Traveling Salesrep decision problem. It is also possible (but a little harder - see [Papadimitriou 1 994], p. 41 1-4 1 2) to show the converse: that Traveling Salesrep reduces to Exact Traveling Salesrep. In such a case we say that the two problems are polynomial time equivalent (sometimes the term "NP-equivalent" is used). We conclude this section with a definition that applies to problems that are not necessarily in NP.

1 , except dividing now by the l-bit integer b, we find that each division takes longer than before (if l is large), namely, O(kl) bit operations. How many times do we have to divide? Here notice that the number of base-b digits in n is O(k j l). Thus, the total number of bit operations required to do all of the necessary divisions is O(k / l) O(kl) = 0(k 2 ). This turns out to be the same answer as in Example 3. 1 . That is, our estimate for the conversion time does not depend upon the base to which we're converting (no matter how large it may be).

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