Miller-Rabin primality test

From Academic Kids

The Miller-Rabin primality test is a primality test: an algorithm which determines whether a given number is prime, similar to the Fermat primality test and the Solovay-Strassen primality test. Its original version, due to G. L. Miller, is deterministic, but it relies on the unproven generalized Riemann hypothesis; M. O. Rabin modified it to obtain an unconditional probabilistic algorithm.

Contents

1 Concepts

2 Algorithm and running time

3 Additional information

4 External links

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Concepts

Just like with the Fermat and Solovay-Strassen tests, with the Miller-Rabin test we will rely on an equality or set of equalities that hold true for prime values, and then see whether or not they hold for a number that we want to test for primality.

Let n be an odd prime, then we can write n − 1 as 2^s × d, where s is an integer and d is odd -- this is the same as factoring out 2 from n repeatedly. Then, one of the following must be true for some <math>a\in \left(\mathbb{Z}/n\mathbb{Z}\right)^*<math>:

<math>

a^{d} \equiv 1\pmod{n} <math> or

<math>

a^{2^r\cdot d} \equiv -1\pmod{n}<math> for some <math>0 \le r \le s-1 <math>

To show that one of these must be true, recall Fermat's little theorem:

<math>

a^{n-1} \equiv 1\pmod{n} <math>

So, if we keep taking square roots of a^{n − 1}, we will either get 1 or −1. If we get −1 then the second equality holds and we are done.

In the case when we've taken out every power of 2 and the second equality never held, we are left with the first equality which also must be equal to 1 or −1, as it too is a square root. However, if the second equality never held, then it couldn't have held for r = 0, meaning that

<math>

a^{2^0\cdot d} = a^d \not\equiv -1\pmod{n} <math>

Thus in the case the second equality doesn't hold, the first equality must.

The Miller-Rabin primality test is based on the above equalities. If we want to test to see if n is prime, then if

<math>

a^{d} \not\equiv 1\pmod{n} <math> and

<math>

a^{2^rd} \not\equiv -1\pmod{n}<math> for all <math>0 \le r \le s-1 <math>

then a is called a strong witness for the compositeness of n. Otherwise a is called a strong liar and n is a probable prime.

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Algorithm and running time

The algorithm can be written as follows:

Inputs: n: a value to test for primality; k: a parameter that determines the number of times to test for primality

Output: composite if n is composite, otherwise probably prime

write n − 1 as 2^s × d by factoring powers of 2 from n − 1

repeat k times:

pick a randomly in the range [1, n − 1]

if a^d mod n ≠ 1 and <math>a^{2^rd}<math> mod n ≠ −1 for all r in the range [0, s − 1] then return composite

return probably prime

Using modular exponentiation by repeated squaring, the running time of this algorithm is O(k × log³ n), where k is the number of different values of a we test, furthermore fast FFT multiplication can push the running time down to Õ(k × log² n), thus this algorithm is polynomial time and efficient.

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Additional information

Like all probabilistic primality tests, there are values of n that will repeatedly produce liars, thus claiming that n is prime when it is actually composite -- these values are known as pseudoprimes.

The more values of a we test, the better the accuracy of the test. It can be shown that there always exists a strong witness for any odd composite n, and that at least 3/4 of the values for a are strong witnesses for the compositeness of n. Thus, the set of strong liars is smaller than the set of Euler liars (used in the Solovay-Strassen primality test).

In general usage the actual number of witnesses is much greater than the lower bound. For instance if we test a 1024-bit odd integer n using the lower bound we would need to test 44 different values of a to guarantee that the chance a given n was declared prime when it was actually composite is less than 2^-80, which means that the value of n could be used safely in cryptographic purposes. However, in practice we generally need to test only 6 different values of a to guarantee this probability. Compare this to 90 iterations for Solovay-Strassen.

Assuming the truth of the generalized Riemann hypothesis, one can prove that, if all the values of a up to 2(log n)² have been tested and n is still pronounced a "probable prime", then it is in fact guaranteed to be prime. This leads to a deterministic primality test that has runtime Õ((log n)⁴).

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