1. Computability and Complexity
27-1
Primes
Computability and Complexity
Andrei Bulatov

2. Computability and Complexity
27-2
The Problem
Instance: A positive integer k.
Question: Is k prime?
Primes
The complement of Primes, the Composite problem, belongs to NP. Therefore Primes is in coNP
Recently M.Agarwal et al. Proved that Primes can be solved in polynomial time
(see
However, the probabilistic algorithm we are going describe is far more efficient

3. Computability and Complexity
27-3
Residues
For a positive integer n, we denote
the set {0,1,2,…,n –1}
the set {1,2,…,n – 1}
addition, multiplication and exponentiation modulo n
together with these operations is called the set of residues modulo n
Every integer m, positive or negative, has a corresponding residue —
m mod n
For example,
17 mod 5 = 2
20 mod 5 = 0
-1 mod 5 = 4

4. Computability and Complexity
27-4
Complexity of Arithmetic
Given two integers, a and b, we can compute
a + b in O(max(log a, log b))
a  b in O(log a  log b)
cannot be computed in polynomial time, because the size of this number is blog a
It is possible modulo n
Let be the binary representation of b (k = log b)
Then that implies
First, we consecutively compute in
Then we compute the product again in

5. Computability and Complexity
27-5
Prime and Coprime
Integers a and b are called coprime if their greatest common divisor is 1
For example, 16 and 27 are coprime, and 15 and 18 are not
Theorem (Chinese Remainder Theorem)
If p and q are coprime then, for any a and b, there is x
such that
For example, if p = 5, q = 3, and a = 2, b = 1, then x can be chosen to be 7

6. Computability and Complexity
27-6
Fermat’s Theorem
Theorem (Fermat’s Little Theorem)
If p is prime then, for any we have
If the converse were true, we could use it for a probabilistic primality test:
Choose k residues modulo n;
Compute their n –1 powers;
Accept if all results are 1 (mod n), reject otherwise

7. Computability and Complexity
27-7
Carmichael Numbers
Unfortunately, the converse is true just “almost”
Definition
A number n passes Fermat’s test if for all a
coprime with n
A number that passes Fermat’s test is called pseudo-prime
One can straightforwardly check that, for any , coprime with 561,
561 is a Carmichael number
n is said to be a Carmichael number if, for any prime divisor p of n, p –1 | n – 1
Pseudo-prime = Prime + Carmichael

8. Computability and Complexity
27-8
Roots of 1
A square root of 1 modulo n is a number a such that
Clearly, 1 and -1 (that is n – 1) are always roots of 1, but if n is composite, then it may have more than two roots of 1
For example,
8 has four roots of 1: 1, -1, 3, and 5
561 has eight: 1, -1, 188, (find the remaining four)
Lemma
Any Carmichael number has at least 8 roots of 1

9. Computability and Complexity
27-9
Algorithm
On input n
if n is even, then if n = 2 accept, otherwise reject
select randomly
for i = 1 to k do
- if then reject
- let n – 1 = st where s is odd and is a power of 2
- compute the sequence modulo n
- if then
let j be the maximal with this property
if then reject
accept

10. Computability and Complexity
27-10
Analysis
First we show that the algorithm does not give false negatives, that is it accepts all prime numbers
If n = 2 then n is accepted. Let n be an odd prime number
Then n passes Fermat test
n cannot be rejected in the last line, because n has only two roots of 1

11. Computability and Complexity
27-11
Next we show that if n is composite, then Pr[n accepted]
A number such that a does not pass either Fermat test or the square root test, is called a witness
It is enough to prove that Pr[a is a witness]  1/2, or, in other words, that at least half of the elements of are witnesses
For every nonwitness d we find a witness d´ such that if then
For a nonwitness a the sequence either contains 1s only, or it contains -1 followed by 1s
Nonwitnesses of both types are present: 1 is a nonwitness of the first type, and -1 is a nonwitness of the second type

12. Computability and Complexity
27-12
Let d be a nonwitness of the second type such that the –1 appears in the largest position in the sequence
Let and
Since n is composite, n = qr for some coprime q and r
Note that
and
By the Chinese Reminder Theorem, there is t such that
therefore
Hence t is a witness, because but

13. Computability and Complexity
27-13
Now, for every nonwitness a we set a´ = a · t
a´ is a witness, because and
but
if then
Assume the contrary
Then, since we have
Finally, we have