The polynomial strategy is a strategy to obtain a good algorithm for the finding primes project by means of computing certain polynomials in various rings that detect primes. The initial target is to obtain an elementary, deterministic O(N1 / 2 + o(1)) for obtaining a prime of size N, with the hope to beat the square root barrier later.
The basic ideas are as follows.
I. First, observe that to find a prime quickly, it would suffice to have an algorithm which, when given an interval [a,b] as input, determines whether that interval contains at least one prime. Indeed, once one has that algorithm, one can find a prime by starting with a large interval (e.g. ) that is already known to contain a prime (by Bertrand's postulate), and then reduce that interval to a single prime by a binary search using only O(k) steps.
II. Now fix [a,b], with a, b both of size O(N) (say), and consider the polynomial defined by
where p ranges over primes. Clearly, this polynomial is non-zero iff [a,b] is a prime. So it will suffice to have an efficient algorithm (the benchmark time to beat here is O(N1 / 2 + o(1))) which would detect whether f is non-zero quickly. (The reason for working modulo 2 will be made clearer in the next step.)
III. The reason we work modulo 2 is that f modulo 2 can be expressed in an appealingly "low-complexity" fashion (from the perspective of arithmetic circuit complexity). In particular, note that for a given number n, the number of divisors between 1 and n is odd if n is prime, and even if n is composite and non-square. So, we morally have
except for an error supported on square monomials which hopefully can be compensated for later (and which can be computed in O(N1 / 2 + o(1)) time in any event).
The expression (1) can be rearranged as
This expression has an arithmetic circuit complexity of O(N1 / 2 + o(1)), thanks to the Pascal's triangle modulo 2 identity
and binary decomposition of the inner sum in (2). Unfortunately, this arithmetic circuit requires much more than O(N1 / 2 + o(1)) time to compute, due to the large amount of memory needed to store f (about O(N) bits), so arithmetic operations are very expensive.
IV. The next trick is to observe that in order to show that f mod 2 is nonzero, it suffices to show that f(x) mod (2,g(x)) is nonzero for some low-degree g (e.g. degree No(1))), or more generally that f(xt) mod (2,g(x)) is non-zero. The point is that because of the low arithmetic complexity of f, this criterion can be verified in just O(N1 / 2 + o(1)) time when g is low-degree.
For instance, if g(x)=x-1, then we are basically just evaluating f(1) mod 2, i.e. the parity of pi(x), which is known to be computable in time o(N1 / 2), see prime counting function.
To finish the job, we need a result of the following form:
- Goal 1 If f(xt) = 0 mod (2,g(x)) for many t, then f is zero.
Heuristics (see here) suggest that one needs to take t to be about N1 / 2 + o(1) in size. Using an FFT, it seems that one can compute all the f(xt) mod 2,g(x) simultaneously in time N1 / 2 + o(1) (how??), so the main job would be to establish Goal 1.