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By Raphael Pass, Abhi Shelat

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Something is not quite right here, since f mult is conjectured to be hard to invert on some, but not all, inputs2 . The strong definition of a one-way function is too restrictive to capture this notion, so we now determine whether the function satisfies the weak notion of one-wayness. In order to do so, we must first introduce an assumption and some basic facts from number theory. 1 The Factoring Assumption Denote the (finite) set of primes that are smaller than 2n as Πn = {q | q < 2n and q is prime} Consider the following assumption, which we shall assume for the remainder of this course: that by the way we have defined f mult , (1, xy) will never be a pre-image of xy.

Before showing this, let us consider some examples. 3 Multiplication, Primes, and Factoring In this section, we consider examples of one-way functions. A first candidate is the function f mult : N2 → N defined by f mult ( x, y) = 1 if x = 1 ∨ y = 1 x · y otherwise Is this a one-way function? Clearly, by the multiplication algorithm, f mult is easy to compute. But f mult is not always hard to invert. If at least one of x and y is even, then their product will be even as well. This happens with probability 34 if the input ( x, y) is picked uniformly at random from N2 .

Kleinjung were able to factor a 663-bit challenge number (of the form described above). 2 GHz Opteron CPU. See [bbfk05] for details. In January 2010, Kleinjung and 12 colleagues [kaf+ 10] announced the factorization of the RSA-768 challenge modulus. They describe the amount of work required for this task as follows: We spent half a year on 80 processors on polynomial selection. This was about 3% of the main task, the sieving, which was done on many hundreds of machines and took almost two years.

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A Course in Cryptography by Raphael Pass, Abhi Shelat

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