Rational sieve

In mathematics, the rational sieve is a general algorithm for factoring integers into prime factors. It is a special case of the general number field sieve. While it is less efficient than the general algorithm, it is conceptually simpler. It serves as a helpful first step in understanding how the general number field sieve works.

Method

Suppose we are trying to factor the composite number n. We choose a bound B, and identify the factor base (which we will call P), the set of all primes less than or equal to B. Next, we search for positive integers z such that both z and z+n are B-smooth — i.e. all of their prime factors are in P. We can therefore write, for suitable exponents ${\displaystyle a_{i}}$,

${\displaystyle z=\prod _{p_{i}\in P}p_{i}^{a_{i}}}$

and likewise, for suitable ${\displaystyle b_{i}}$, we have

${\displaystyle z+n=\prod _{p_{i}\in P}p_{i}^{b_{i}}}$.

But ${\displaystyle z}$ and ${\displaystyle z+n}$ are congruent modulo ${\displaystyle n}$, and so each such integer z that we find yields a multiplicative relation (mod n) among the elements of P, i.e.

${\displaystyle \prod _{p_{i}\in P}p_{i}^{a_{i}}\equiv \prod _{p_{i}\in P}p_{i}^{b_{i}}{\pmod {n}}}$

(where the a_i and b_i are nonnegative integers.)

When we have generated enough of these relations (it's generally sufficient that the number of relations be a few more than the size of P), we can use the methods of linear algebra to multiply together these various relations in such a way that the exponents of the primes are all even. This will give us a congruence of squares of the form a²≡b² (mod n), which can be turned into a factorization of n = gcd(a-b,n)×gcd(a+b,n). This factorization might turn out to be trivial (i.e. n=n×1), in which case we have to try again with a different combination of relations; but with luck we will get a nontrivial pair of factors of n, and the algorithm will terminate.

Example

We will factor the integer n = 187 using the rational sieve. We'll arbitrarily try the value B=7, giving the factor base P = {2,3,5,7}. The first step is to test n for divisibility by each of the members of P; clearly if n is divisible by one of these primes, then we are finished already. However, 187 is not divisible by 2, 3, 5, or 7. Next, we search for suitable values of z; the first few are 2, 5, 9, and 56. The four suitable values of z give four multiplicative relations (mod 187):

${\displaystyle 2^{1}3^{0}5^{0}7^{0}=2\equiv 189=2^{0}3^{3}5^{0}7^{1}}$

(1)

${\displaystyle 2^{0}3^{0}5^{1}7^{0}=5\equiv 192=2^{6}3^{1}5^{0}7^{0}}$

(2)

${\displaystyle 2^{0}3^{2}5^{0}7^{0}=9\equiv 196=2^{2}3^{0}5^{0}7^{2}}$

(3)

${\displaystyle 2^{3}3^{0}5^{0}7^{1}=56\equiv 243=2^{0}3^{5}5^{0}7^{0}}$

(4)

There are now several essentially different ways to combine these and end up with even exponents. For example,

• (1)×(4): After multiplying these and canceling out the common factor of 7 (which we can do since 7, being a member of P, has already been determined to be coprime with n^[1]), this reduces to 2⁴ ≡ 3⁸ (mod n), or 4² ≡ 81² (mod n). The resulting factorization is 187 = gcd(81-4,187) × gcd(81+4,187) = 11×17.

Alternatively, equation (3) is in the proper form already:

• (3): This says 3² ≡ 14² (mod n), which gives the factorization 187 = gcd(14-3,187) × gcd(14+3,187) = 11×17.

Limitations of the algorithm

Like the general number field sieve, the rational sieve cannot factor numbers of the form p^m, where p is a prime and m is an integer. This is not a huge problem, though—such numbers are statistically rare, and moreover there is a simple and fast process to check whether a given number is of this form. Probably the most elegant method is to check whether ${\displaystyle \lfloor n^{1/b}\rfloor ^{b}=n}$ holds for any ${\displaystyle 1<b\leq \log _{2}(n)}$ using an integer version of Newton's method for the root extraction.^[2]

The biggest problem is finding a sufficient number of z such that both z and z+n are B-smooth. For any given B, the proportion of numbers that are B-smooth decreases rapidly with the size of the number. So if n is large (say, a hundred digits), it will be difficult or impossible to find enough z for the algorithm to work. The advantage of the general number field sieve is that one only needs to search for smooth numbers of order ${\displaystyle \exp(C(\log n)^{2/3}(\log \log n)^{1/3})}$ for some C > 0, rather than of order n as required here.^[3]

References

• A. K. Lenstra, H. W. Lenstra, Jr., M. S. Manasse, and J. M. Pollard, The Factorization of the Ninth Fermat Number, Math. Comp. 61 (1993), 319-349. Available at .
• A. K. Lenstra, H. W. Lenstra, Jr. (eds.) The Development of the Number Field Sieve, Lecture Notes in Mathematics 1554, Springer-Verlag, New York, 1993.

Footnotes

1. Note that common factors cannot in general be canceled in a congruence, but they can in this case, since the primes of the factor base are all required to be coprime to n, as mentioned above. See modular multiplicative inverse.
2. R. Crandall and J. Papadopoulos, On the implementation of AKS-class primality tests, available at
3. A. K. Lenstra, H. W. Lenstra, Jr., M. S. Manasse, and J. M. Pollard, The Factorization of the Ninth Fermat Number, Math. Comp. 61 (1993), p. 328