Pollard's p − 1 algorithm

Pollard's p − 1 algorithm is a number theoretic integer factorization algorithm, invented by John Pollard in 1974. It is a special-purpose algorithm, meaning that it is only suitable for integers with specific types of factors; it is the simplest example of an algebraic-group factorisation algorithm.

The factors it finds are ones for which the number preceding the factor, p − 1, is powersmooth; the essential observation is that, by working in the multiplicative group modulo a composite number N, we are also working in the multiplicative groups modulo all of N's factors.

The existence of this algorithm leads to the concept of strong primes, being primes for which p − 1 has at least one large prime factor. Almost all sufficiently large primes are strong; if a prime used for cryptographic purposes turns out to be non-strong, it is much more likely to be through malice than through an accident of random number generation.

Contents

Base concepts

Let n be a composite integer with prime factor p. By Fermat's little theorem, we know that for all integers a coprime to p and for all positive integers K:

a^{K(p-1)} \equiv 1\pmod{p}

If a number x is congruent to 1 modulo a factor of n, then the gcd (x − 1, n) will be divisible by that factor.

The idea is to make the exponent a large multiple of p − 1 by making it a number with very many prime factors; generally, we take the product of all prime powers less than some limit B. Start with a random x, and repeatedly replace it by x^w \mod n as w runs through those prime powers. Check at each stage, or once at the end if you prefer, whether gcd (x − 1, n) is not equal to 1.

Multiple factors

It is possible that for all the prime factors p of n, p − 1 is divisible by small primes, at which point the Pollard p − 1 algorithm gives you n again.

Algorithm and running time

The basic algorithm can be written as follows:

Inputs: n: a composite integer
Output: a non-trivial factor of n or failure
  1. select a smoothness bound B
  2. M \gets \prod_{\text{primes}~q \le B} q^{ \lfloor \log_q{B} \rfloor }
  3. randomly pick a coprime to n (note: we can actually fix a, random selection here is not imperative)
  4. g \gets \gcd(a^M - 1, n) (note: the powering can be done mod n)
  5. if 1 < g < n then return g
  6. if g = 1 then select a higher B and go to step 2 or return failure
  7. if g = n then go to step 2 or return failure

If g = 1 in step 6, this indicates that for all p − 1 that none were B-powersmooth. If g = n in step 7, this usually indicates that all factors were B-powersmooth, but in rare cases it could indicate that a had a small order modulo n.

The running time of this algorithm is O(B × log B × log2n); larger values of B make it run more slowly, but are more likely to produce a factor.

How to choose B?

Since the algorithm is incremental, it can just keep running with the bound constantly increasing.

Assume that p − 1, where p is the smallest prime factor of n, can be modelled as a random number of size less than √n. By Dixon's theorem, the probability that the largest factor of such a number is less than (p − 1)ε is roughly εε; so there is a probability of about 3−3 = 1/27 that a B value of n1/6 will yield a factorisation.

In practice, the elliptic curve method is faster than the Pollard p − 1 method once the factors are at all large; running the p − 1 method up to B = 106 will find a quarter of all twelve-digit factors and 1/27 of all eighteen-digit factors, before proceeding to another method.

Large prime variant

A variant of the basic algorithm is sometimes used; instead of requiring that p − 1 has all its factors less than B, we can require it to have all but one of its factors less than some B1, and the remaining factor less than some B2. Let p1 be the smallest prime greater than B1, p2 the next-largest, and so on; let dn = pn − pn−1. The distribution of prime numbers is such that the dn will all be fairly small.

Having computed c = a^M \mod n, we can easily compute once and for all E_r = c^r \mod n for all r which appear as a value of d_n. Compute t_1 = c^{p_1} \mod n. We can then stop doing exponentiation, and compute

t_2 (= c^{p_2} \mod n) = t_1 E_{d_2} \mod n, t_3 = t_2 E_{d_3} \mod n, \dots

with one multiplication rather than one exponentiation at each step; this is quicker by roughly a factor log B than doing the exponentiations. It can also be accelerated significantly using Fast Fourier transforms.

Implementations

References

See also

External links