Complexity

1
Complexity
18-1
Primes
Complexity Andrei Bulatov
2
Complexity
18-2
The Problem
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
http//www.cse.iitk.ac.in/news/primality.html)
However, the probabilistic algorithm we are going
describe is far more efficient
3
Complexity
18-3
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
Complexity
18-4
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
Complexity
18-5
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
Complexity
18-6
Fermats Theorem
Theorem (Fermats 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
Complexity
18-7
Carmichael Numbers
Unfortunately, the converse is true just almost
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
Complexity
18-8
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, 373 (find the
remaining four)
Lemma Any
Carmichael number has at least 8 roots of 1
9
Complexity
18-9
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
Complexity
18-10
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
Complexity
18-11
Next we show that if n is composite, then
Prn 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 Pra 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
Complexity
18-12
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
Complexity
18-13
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