is n a prime number
TRANSCRIPT
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Is n a Prime Number?
Manindra Agrawal
IIT Kanpur
March 27, 2006, Delft
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Overview
1 The Problem
2 Two Simple, and Slow, Methods
3
Modern Methods
4 Algorithms Based on Factorization of Group Size
5 Algorithms Based on Fermat’s Little Theorem
6 An Algorithm Outside the Two Themes
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Outline
1 The Problem
2 Two Simple, and Slow, Methods
3 Modern Methods
4 Algorithms Based on Factorization of Group Size
5 Algorithms Based on Fermat’s Little Theorem
6 An Algorithm Outside the Two Themes
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The Problem
Given a number n, decide if it is prime.
Easy: try dividing by all numbers less than n.
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The Problem
Given a number n, decide if it is prime.
Easy: try dividing by all numbers less than n.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 4 / 47
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The Problem
Given a number n, decide if it is prime efficiently.
Not so easy: several non-obvious methods have been found.
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The Problem
Given a number n, decide if it is prime efficiently.
Not so easy: several non-obvious methods have been found.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 5 / 47
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Efficiently Solving a Problem
There should exist an algorithm for solving the problem taking apolynomial in input size number of steps.
For our problem, this means an algorithm taking logO (1) n steps.
Caveat: An algorithm taking log12
n steps would be slower than analgorithm taking loglog log log n n steps for all practical valuesof n!
Notation:
log is logarithm base 2.O (logc n) stands for O (logc n log logO (1) n).
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Efficiently Solving a Problem
There should exist an algorithm for solving the problem taking apolynomial in input size number of steps.
For our problem, this means an algorithm taking logO (1) n steps.
Caveat: An algorithm taking log12
n steps would be slower than analgorithm taking loglog log log n n steps for all practical valuesof n!
Notation:
log is logarithm base 2.O (logc n) stands for O (logc n log logO (1) n).
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Efficiently Solving a Problem
There should exist an algorithm for solving the problem taking apolynomial in input size number of steps.
For our problem, this means an algorithm taking logO (1) n steps.
Caveat: An algorithm taking log12
n steps would be slower than analgorithm taking loglog log log n n steps for all practical valuesof n!
Notation:
log is logarithm base 2.O (logc n) stands for O (logc n log logO (1) n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 6 / 47
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Efficiently Solving a Problem
There should exist an algorithm for solving the problem taking apolynomial in input size number of steps.
For our problem, this means an algorithm taking logO (1) n steps.
Caveat: An algorithm taking log
12
n steps would be slower than analgorithm taking loglog log log n n steps for all practical valuesof n!
Notation:
log is logarithm base 2.O (logc n) stands for O (logc n log logO (1) n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 6 / 47
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Outline
1
The Problem
2 Two Simple, and Slow, Methods
3 Modern Methods
4 Algorithms Based on Factorization of Group Size
5 Algorithms Based on Fermat’s Little Theorem
6 An Algorithm Outside the Two Themes
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The Sieve of Eratosthenes
Proposed by Eratosthenes (ca. 300 BCE).
1 List all numbers from 2 to n in a sequence.
2 Take the smallest uncrossed number from the sequence and cross outall its multiples.
3 If n is uncrossed when the smallest uncrossed number is greater than√n then n is prime otherwise composite.
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Time Complexity
If n is prime, algorithm crosses out all the first√
n numbers before
giving the answer.So the number of steps needed is Ω(
√n).
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Time Complexity
If n is prime, algorithm crosses out all the first√
n numbers before
giving the answer.So the number of steps needed is Ω(
√n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 9 / 47
’
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Wilson’s Theorem
Based on Wilson’s theorem (1770).
Theorem
n is prime iff (n − 1)! = −1 (mod n).
Computing (n − 1)! (mod n) naıvely requires Ω(n) steps.
No significantly better method is known!
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W ’ T
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Wilson’s Theorem
Based on Wilson’s theorem (1770).
Theorem
n is prime iff (n − 1)! = −1 (mod n).
Computing (n − 1)! (mod n) naıvely requires Ω(n) steps.
No significantly better method is known!
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 10 / 47
O
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Outline
1
The Problem
2 Two Simple, and Slow, Methods
3 Modern Methods
4 Algorithms Based on Factorization of Group Size
5 Algorithms Based on Fermat’s Little Theorem
6 An Algorithm Outside the Two Themes
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F a a I a
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Fundamental Idea
Nearly all the efficient algorithms for the problem use the following idea.
Identify a finite group G related to number n.
Design an efficienty testable property P (·) of the elements G suchthat P (e ) has different values depending on whether n is prime.
The element e is either from a small set (in deterministic algorithms)or a random element of G (in randomized algorithms).
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Fundamental Idea
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Fundamental Idea
Nearly all the efficient algorithms for the problem use the following idea.
Identify a finite group G related to number n.
Design an efficienty testable property P (·) of the elements G suchthat P (e ) has different values depending on whether n is prime.
The element e is either from a small set (in deterministic algorithms)or a random element of G (in randomized algorithms).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 12 / 47
Fundamental Idea
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Fundamental Idea
Nearly all the efficient algorithms for the problem use the following idea.
Identify a finite group G related to number n.
Design an efficienty testable property P (·) of the elements G suchthat P (e ) has different values depending on whether n is prime.
The element e is either from a small set (in deterministic algorithms)or a random element of G (in randomized algorithms).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 12 / 47
Fundamental Idea
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Fundamental Idea
Nearly all the efficient algorithms for the problem use the following idea.
Identify a finite group G related to number n.
Design an efficienty testable property P (·) of the elements G suchthat P (e ) has different values depending on whether n is prime.
The element e is either from a small set (in deterministic algorithms)or a random element of G (in randomized algorithms).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 12 / 47
Groups and Properties
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Groups and Properties
The group G is often: A subgroup of Z ∗n or Z ∗n [ζ ] for an extension ring Z n[ζ ].
A subgroup of E (Z n), the set of points on an elliptic curve modulo n.The properties vary, but are from two broad themes.
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Groups and Properties
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Groups and Properties
The group G is often: A subgroup of Z ∗n or Z ∗n [ζ ] for an extension ring Z n[ζ ].
A subgroup of E (Z n), the set of points on an elliptic curve modulo n.The properties vary, but are from two broad themes.
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Groups and Properties
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Groups and Properties
The group G is often: A subgroup of Z ∗n or Z ∗n [ζ ] for an extension ring Z n[ζ ].
A subgroup of E (Z n), the set of points on an elliptic curve modulo n.The properties vary, but are from two broad themes.
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Theme I: Factorization of Group Size
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Theme I: Factorization of Group Size
Compute a complete, or partial, factorization of the size of G
assuming that n is prime.
Use the knowledge of this factorization to design a suitable property.
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Theme I: Factorization of Group Size
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Theme I: Factorization of Group Size
Compute a complete, or partial, factorization of the size of G
assuming that n is prime.
Use the knowledge of this factorization to design a suitable property.
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Theme II: Fermat’s Little Theorem
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Theme II: Fermat s Little Theorem
Theorem (Fermat, 1660s)
If n is prime then for every e , e n = e (mod n).
Group G = Z ∗n and property P is P (e ) ≡ e n = e in G .
This property of Z n is not a sufficient test for primality of n.
So try to extend this property to a neccessary and sufficient condition.
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Theme II: Fermat’s Little Theorem
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Theme II: Fermat s Little Theorem
Theorem (Fermat, 1660s)
If n is prime then for every e , e n = e (mod n).
Group G = Z ∗n and property P is P (e ) ≡ e n = e in G .
This property of Z n is not a sufficient test for primality of n.
So try to extend this property to a neccessary and sufficient condition.
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Outline
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Outline
1 The Problem
2 Two Simple, and Slow, Methods
3 Modern Methods
4 Algorithms Based on Factorization of Group Size
5 Algorithms Based on Fermat’s Little Theorem
6 An Algorithm Outside the Two Themes
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Lucas Theorem
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Theorem (E. Lucas, 1891)
Let n − 1 =t
i =1 p d i i where p i ’s are distinct primes. n is prime iff there is
an e ∈ Z n such that e n−1 = 1 and gcd(e n−1p i − 1, n) = 1 for every
1≤
i ≤
t .
The theorem also holds for a random choice of e .
We can choose G = Z ∗n and P to be the property above.
The test will be efficient only for numbers n such that n−1 is smooth.
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Lucas Theorem
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Theorem (E. Lucas, 1891)
Let n − 1 =t
i =1 p d i i where p i ’s are distinct primes. n is prime iff there is
an e ∈ Z n such that e n−1 = 1 and gcd(e n−1p i − 1, n) = 1 for every
1≤
i ≤
t .
The theorem also holds for a random choice of e .
We can choose G = Z ∗n and P to be the property above.
The test will be efficient only for numbers n such that n−1 is smooth.
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Lucas Theorem
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Theorem (E. Lucas, 1891)
Let n − 1 =t
i =1 p d i i where p i ’s are distinct primes. n is prime iff there is
an e ∈ Z n such that e n−1 = 1 and gcd(e n−1p i − 1, n) = 1 for every
1≤
i ≤
t .
The theorem also holds for a random choice of e .
We can choose G = Z ∗n and P to be the property above.
The test will be efficient only for numbers n such that n−1 is smooth.
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Lucas-Lehmer Test
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G is a subgroup of Z ∗n [√
3] containing elements of order n + 1.
The property P is: P (e ) ≡ e n+1
2 = −1 in Z n[√
3].
Works only for special Mersenne primes of the form n = 2p
−1, p
prime.
For such n’s, n + 1 = 2p .
The property needs to be tested only for e = 2 +√
3.
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Lucas-Lehmer Test
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G is a subgroup of Z ∗n [√
3] containing elements of order n + 1.
The property P is: P (e ) ≡ e n+1
2 = −1 in Z n[√
3].
Works only for special Mersenne primes of the form n = 2p
−1, p
prime.
For such n’s, n + 1 = 2p .
The property needs to be tested only for e = 2 +√
3.
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Lucas-Lehmer Test
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G is a subgroup of Z ∗n [√
3] containing elements of order n + 1.
The property P is: P (e ) ≡ e n+1
2 = −1 in Z n[√
3].
Works only for special Mersenne primes of the form n = 2p
−1, p
prime.
For such n’s, n + 1 = 2p .
The property needs to be tested only for e = 2 +√
3.
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Time Complexity
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Raising 2 +√
3 to n+12 th power requires O (log n) multiplication
operations in Z n.
Overall time complexity is O (log2 n).
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Time Complexity
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Raising 2 +√
3 to n+12 th power requires O (log n) multiplication
operations in Z n.
Overall time complexity is O (log2 n).
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Pocklington-Lehmer Test
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Theorem (Pocklington, 1914)If there exists a e such that e n−1 = 1 (mod n) and gcd(e
n−1p j − 1, n) = 1
for distinct primes p 1, p 2, . . . , p t dividing n− 1 then every prime factor of n
has the form k ·
t j =1 p j + 1.
Similar to Lucas’s theorem.
Let G = Z ∗n and property P precisely as in the theorem.
The property is tested for a random e .
For the test to work, we needt
j =1 ≥ √n.
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Pocklington-Lehmer Test
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Theorem (Pocklington, 1914)If there exists a e such that e n−1 = 1 (mod n) and gcd(e
n−1p j − 1, n) = 1
for distinct primes p 1, p 2, . . . , p t dividing n− 1 then every prime factor of n
has the form k ·
t j =1 p j + 1.
Similar to Lucas’s theorem.
Let G = Z ∗n and property P precisely as in the theorem.
The property is tested for a random e .
For the test to work, we needt
j =1 ≥ √n.
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Pocklington-Lehmer Test
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Theorem (Pocklington, 1914)If there exists a e such that e n−1 = 1 (mod n) and gcd(e
n−1p j − 1, n) = 1
for distinct primes p 1, p 2, . . . , p t dividing n− 1 then every prime factor of n
has the form k ·
t j =1 p j + 1.
Similar to Lucas’s theorem.
Let G = Z ∗n and property P precisely as in the theorem.
The property is tested for a random e .
For the test to work, we needt
j =1 ≥ √n.
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Time Complexity
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Depends on the difficulty of finding prime factorizations of n − 1whose product is at least
√n.
Other operations can be carried out efficiently.
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Time Complexity
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Depends on the difficulty of finding prime factorizations of n − 1whose product is at least
√n.
Other operations can be carried out efficiently.
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Elliptic Curves Based Tests
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Elliptic curves give rise to groups of different sizes associated with thegiven number.
With good probability, some of these groups have sizes that can beeasily factored.
This motivated primality testing based on elliptic curves.
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Elliptic Curves Based Tests
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Elliptic curves give rise to groups of different sizes associated with thegiven number.
With good probability, some of these groups have sizes that can beeasily factored.
This motivated primality testing based on elliptic curves.
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Elliptic Curves Based Tests
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Elliptic curves give rise to groups of different sizes associated with thegiven number.
With good probability, some of these groups have sizes that can beeasily factored.
This motivated primality testing based on elliptic curves.
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Goldwasser-Kilian Test
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This is a randomized primality proving algorithm.
Under a reasonable hypothesis, it is polynomial time on all inputs.
Unconditionally, it is polynomial time on all but negligible fraction of numbers.
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Goldwasser-Kilian Test
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This is a randomized primality proving algorithm.
Under a reasonable hypothesis, it is polynomial time on all inputs.
Unconditionally, it is polynomial time on all but negligible fraction of numbers.
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Goldwasser-Kilian Test
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This is a randomized primality proving algorithm.
Under a reasonable hypothesis, it is polynomial time on all inputs.
Unconditionally, it is polynomial time on all but negligible fraction of numbers.
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Goldwasser-Kilian Test
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Consider a random elliptic curve over Z n.
By a theorem of Lenstra (1987), the number of points of the curve isnearly uniformly distributed in the interval [n + 1− 2
√n, n + 1 + 2
√n]
for prime n.Assuming a conjecture about the density of primes in small intervals,it follows that there are curves with 2q points, for q prime, withreasonable probability.
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Goldwasser-Kilian Test
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Consider a random elliptic curve over Z n.
By a theorem of Lenstra (1987), the number of points of the curve isnearly uniformly distributed in the interval [n + 1− 2
√n, n + 1 + 2
√n]
for prime n.Assuming a conjecture about the density of primes in small intervals,it follows that there are curves with 2q points, for q prime, withreasonable probability.
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Goldwasser-Kilian Test
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Consider a random elliptic curve over Z n.
By a theorem of Lenstra (1987), the number of points of the curve isnearly uniformly distributed in the interval [n + 1− 2
√n, n + 1 + 2
√n]
for prime n.Assuming a conjecture about the density of primes in small intervals,it follows that there are curves with 2q points, for q prime, withreasonable probability.
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Goldwasser-Kilian Test
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Theorem (Goldwasser-Kilian)
Suppose E (Z n) is an elliptic curve with 2q points. If q is prime and there
exists A ∈ E (Z n) = O such that q · A = O then either n is provably prime
or provably composite.
Proof.
Let p be a prime factor of n with p ≤ √n.We have q · A = O in E (Z p ) as well.
If A = O in E (Z p ) then n can be factored.
Otherwise, since q is prime,
|E (Z p )
| ≥q .
If 2q < n + 1 − 2√n then n must be composite.
Otherwise, p + 1 + 2√
p > n2 −
√n which is not possible.
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Goldwasser-Kilian Test
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Theorem (Goldwasser-Kilian)
Suppose E (Z n) is an elliptic curve with 2q points. If q is prime and there
exists A ∈ E (Z n) = O such that q · A = O then either n is provably prime
or provably composite.
Proof.
Let p be a prime factor of n with p ≤ √n.We have q · A = O in E (Z p ) as well.
If A = O in E (Z p ) then n can be factored.
Otherwise, since q is prime, |E (Z p )| ≥ q .
If 2q < n + 1 − 2√n then n must be composite.
Otherwise, p + 1 + 2√
p > n2 −
√n which is not possible.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 25 / 47
Goldwasser-Kilian Test
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Theorem (Goldwasser-Kilian)
Suppose E (Z n) is an elliptic curve with 2q points. If q is prime and there
exists A ∈ E (Z n) = O such that q · A = O then either n is provably prime
or provably composite.
Proof.
Let p be a prime factor of n with p ≤ √n.We have q · A = O in E (Z p ) as well.
If A = O in E (Z p ) then n can be factored.
Otherwise, since q is prime, |E (Z p )| ≥ q .
If 2q < n + 1 − 2√n then n must be composite.
Otherwise, p + 1 + 2√
p > n2 −
√n which is not possible.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 25 / 47
Goldwasser-Kilian Test
8/8/2019 Is n a prime number
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Theorem (Goldwasser-Kilian)
Suppose E (Z n) is an elliptic curve with 2q points. If q is prime and there
exists A ∈ E (Z n) = O such that q · A = O then either n is provably prime
or provably composite.
Proof.
Let p be a prime factor of n with p ≤ √n.We have q · A = O in E (Z p ) as well.
If A = O in E (Z p ) then n can be factored.
Otherwise, since q is prime, |E (Z p )| ≥ q .
If 2q < n + 1 − 2√n then n must be composite.
Otherwise, p + 1 + 2√
p > n2 −
√n which is not possible.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 25 / 47
Goldwasser-Kilian Test
8/8/2019 Is n a prime number
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1 Find a random elliptic curve over Z n with 2q points.
2 Prove primality of q recursively.
3 Randomly select an A such that q · A = O .4 Infer n to be prime or composite.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 26 / 47
Goldwasser-Kilian Test
8/8/2019 Is n a prime number
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1 Find a random elliptic curve over Z n with 2q points.
2 Prove primality of q recursively.
3 Randomly select an A such that q · A = O .4 Infer n to be prime or composite.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 26 / 47
Goldwasser-Kilian Test
8/8/2019 Is n a prime number
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1 Find a random elliptic curve over Z n with 2q points.
2 Prove primality of q recursively.
3 Randomly select an A such that q · A = O .4 Infer n to be prime or composite.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 26 / 47
Goldwasser-Kilian Test
8/8/2019 Is n a prime number
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1 Find a random elliptic curve over Z n with 2q points.
2 Prove primality of q recursively.
3 Randomly select an A such that q · A = O .4 Infer n to be prime or composite.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 26 / 47
Analysis
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The algorithm never incorrectly classifies a composite number.
With high probability it correctly classifies prime numbers.
The running time is O (log11 n).
Improvements by Atkin and others result in a conjectured runningtime of O (log4 n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 27 / 47
Analysis
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The algorithm never incorrectly classifies a composite number.
With high probability it correctly classifies prime numbers.
The running time is O (log11 n).
Improvements by Atkin and others result in a conjectured runningtime of O (log4 n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 27 / 47
Analysis
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The algorithm never incorrectly classifies a composite number.
With high probability it correctly classifies prime numbers.
The running time is O (log11 n).
Improvements by Atkin and others result in a conjectured runningtime of O (log4 n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 27 / 47
Analysis
8/8/2019 Is n a prime number
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The algorithm never incorrectly classifies a composite number.
With high probability it correctly classifies prime numbers.
The running time is O (log11 n).
Improvements by Atkin and others result in a conjectured runningtime of O (log4 n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 27 / 47
Adleman-Huang Test
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The previous test is not unconditionally polynomial time on a smallfraction of numbers.
Adleman-Huang (1992) removed this drawback.
They first used hyperelliptic curves to reduce the problem of testingfor n to that of a nearly random integer of similar size.
Then the previous test works with high probability.
The time complexity becomes O (logc n) for c > 30!
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 28 / 47
Adleman-Huang Test
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The previous test is not unconditionally polynomial time on a smallfraction of numbers.
Adleman-Huang (1992) removed this drawback.
They first used hyperelliptic curves to reduce the problem of testingfor n to that of a nearly random integer of similar size.
Then the previous test works with high probability.
The time complexity becomes O (logc n) for c > 30!
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 28 / 47
Adleman-Huang Test
8/8/2019 Is n a prime number
http://slidepdf.com/reader/full/is-n-a-prime-number 67/136
The previous test is not unconditionally polynomial time on a smallfraction of numbers.
Adleman-Huang (1992) removed this drawback.
They first used hyperelliptic curves to reduce the problem of testingfor n to that of a nearly random integer of similar size.
Then the previous test works with high probability.
The time complexity becomes O (logc n) for c > 30!
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 28 / 47
Outline
8/8/2019 Is n a prime number
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1 The Problem
2 Two Simple, and Slow, Methods
3 Modern Methods
4 Algorithms Based on Factorization of Group Size
5 Algorithms Based on Fermat’s Little Theorem
6 An Algorithm Outside the Two Themes
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 29 / 47
Solovay-Strassen Test
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A Restatement of FLTIf n is odd prime then for every e , 1 ≤ e < n, e
n−12 = ±1 (mod n).
When n is prime, e is a quadratic residue in Z n iff e n−1
2 = 1 (mod n).
Therefore, if n is prime then
e
n
= e
n−12 (mod n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 30 / 47
Solovay-Strassen Test
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A Restatement of FLTIf n is odd prime then for every e , 1 ≤ e < n, e
n−12 = ±1 (mod n).
When n is prime, e is a quadratic residue in Z n iff e n−1
2 = 1 (mod n).
Therefore, if n is prime then
e
n
= e
n−12 (mod n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 30 / 47
Solovay-Strassen Test
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A Restatement of FLTIf n is odd prime then for every e , 1 ≤ e < n, e
n−12 = ±1 (mod n).
When n is prime, e is a quadratic residue in Z n iff e n−1
2 = 1 (mod n).
Therefore, if n is prime then
e
n
= e
n−12 (mod n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 30 / 47
Solovay-Strassen Test
8/8/2019 Is n a prime number
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Proposed by Solovay and Strassen (1973).
A randomized algorithm based on above property.
Never incorrectly classifies primes and correctly classifies compositeswith probability at least 1
2 .
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 31 / 47
Solovay-Strassen Test
8/8/2019 Is n a prime number
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Proposed by Solovay and Strassen (1973).
A randomized algorithm based on above property.
Never incorrectly classifies primes and correctly classifies compositeswith probability at least 1
2 .
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 31 / 47
Solovay-Strassen Test
8/8/2019 Is n a prime number
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Proposed by Solovay and Strassen (1973).
A randomized algorithm based on above property.
Never incorrectly classifies primes and correctly classifies compositeswith probability at least 1
2 .
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 31 / 47
Solovay-Strassen Test
8/8/2019 Is n a prime number
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1 If n is an exact power, it is composite.2 For a random e in Z n, test if
e
n
= e
n−12 (mod n).
3 If yes, classify n as prime otherwise it is proven composite.
The time complexity is O (log2 n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 32 / 47
Solovay-Strassen Test
8/8/2019 Is n a prime number
http://slidepdf.com/reader/full/is-n-a-prime-number 76/136
1 If n is an exact power, it is composite.2 For a random e in Z n, test if
e
n
= e
n−12 (mod n).
3 If yes, classify n as prime otherwise it is proven composite.
The time complexity is O (log2 n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 32 / 47
Solovay-Strassen Test
8/8/2019 Is n a prime number
http://slidepdf.com/reader/full/is-n-a-prime-number 77/136
1 If n is an exact power, it is composite.2 For a random e in Z n, test if
e
n
= e
n−12 (mod n).
3 If yes, classify n as prime otherwise it is proven composite.
The time complexity is O (log2 n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 32 / 47
Solovay-Strassen Test
8/8/2019 Is n a prime number
http://slidepdf.com/reader/full/is-n-a-prime-number 78/136
1 If n is an exact power, it is composite.2 For a random e in Z n, test if
e
n
= e
n−12 (mod n).
3 If yes, classify n as prime otherwise it is proven composite.
The time complexity is O (log2 n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 32 / 47
Analysis
Consider the case when n is a product of two primes p and q
8/8/2019 Is n a prime number
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Consider the case when n is a product of two primes p and q .
Let a, b ∈ Z p , c ∈ Z q with a residue and b non-residue in Z p .Clearly, < a, c >
n−12 =< b , c >
n−12 (mod q ).
If < a, c >n−1
2 =< b , c >n−1
2 (mod n) then one of them is not in1,−1 and so compositeness of n is proven.
Otherwise, either< a, c >
n
=< a, c >
n−12 (mod n),
or < b , c >
n
=< b , c >n−1
2 (mod n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 33 / 47
Analysis
Consider the case when n is a product of two primes p and q
8/8/2019 Is n a prime number
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Consider the case when n is a product of two primes p and q .
Let a, b ∈ Z p , c ∈ Z q with a residue and b non-residue in Z p .Clearly, < a, c >
n−12 =< b , c >
n−12 (mod q ).
If < a, c >n−1
2 =< b , c >n−1
2 (mod n) then one of them is not in1,−1 and so compositeness of n is proven.
Otherwise, either< a, c >
n
=< a, c >
n−12 (mod n),
or < b , c >
n =< b , c >
n−12 (mod n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 33 / 47
Analysis
Consider the case when n is a product of two primes p and q
8/8/2019 Is n a prime number
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Consider the case when n is a product of two primes p and q .
Let a, b ∈ Z p , c ∈ Z q with a residue and b non-residue in Z p .Clearly, < a, c >
n−12 =< b , c >
n−12 (mod q ).
If < a, c >n−1
2 =< b , c >n−1
2 (mod n) then one of them is not in1,−1 and so compositeness of n is proven.
Otherwise, either< a, c >
n
=< a, c >
n−12 (mod n),
or < b , c >n
=< b , c >n−1
2 (mod n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 33 / 47
Analysis
Consider the case when n is a product of two primes p and q
8/8/2019 Is n a prime number
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Consider the case when n is a product of two primes p and q .
Let a, b ∈ Z p , c ∈ Z q with a residue and b non-residue in Z p .Clearly, < a, c >
n−12 =< b , c >
n−12 (mod q ).
If < a, c >n−1
2 =< b , c >n−1
2 (mod n) then one of them is not in1,−1 and so compositeness of n is proven.
Otherwise, either< a, c >
n
=< a, c >
n−12 (mod n),
or < b , c >n
=< b , c >n−1
2 (mod n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 33 / 47
Miller’s Test
8/8/2019 Is n a prime number
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Theorem (Another Restatement of FLT)
If n is odd prime and n = 1 + 2s · t , t odd, then for every e , 1 ≤ e < n,
the sequence e
2s −1·t
(mod n), e
2s −2·t
(mod n), . . ., e
t
(mod n) has either all 1’s or a −1 somewhere.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 34 / 47
Miller’s Test
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This theorem is the basis for Miller’s test (1973).
It is a deterministic polynomial time test.
It is correct under Extended Riemann Hypothesis.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 35 / 47
Miller’s Test
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This theorem is the basis for Miller’s test (1973).
It is a deterministic polynomial time test.
It is correct under Extended Riemann Hypothesis.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 35 / 47
Miller’s Test
8/8/2019 Is n a prime number
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This theorem is the basis for Miller’s test (1973).
It is a deterministic polynomial time test.
It is correct under Extended Riemann Hypothesis.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 35 / 47
Miller’s Test
8/8/2019 Is n a prime number
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1 If n is an exact power, it is composite.
2 For each e , 1 < e ≤ 4log2 n, check if the sequence e 2s −1
·t (mod n),e 2
s −2·t (mod n), . . ., e t (mod n) has either all 1’s or a −1 somewhere.
3
If yes, classify n as prime otherwise composite.
The time complexity of the test is O (log4 n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 36 / 47
Miller’s Test
8/8/2019 Is n a prime number
http://slidepdf.com/reader/full/is-n-a-prime-number 88/136
1 If n is an exact power, it is composite.
2 For each e , 1 < e ≤ 4log2 n, check if the sequence e 2s −1
·t (mod n),e 2
s −2·t (mod n), . . ., e t (mod n) has either all 1’s or a −1 somewhere.
3
If yes, classify n as prime otherwise composite.
The time complexity of the test is O (log4 n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 36 / 47
Miller’s Test
8/8/2019 Is n a prime number
http://slidepdf.com/reader/full/is-n-a-prime-number 89/136
1 If n is an exact power, it is composite.
2 For each e , 1 < e ≤ 4log2 n, check if the sequence e 2s −1
·t (mod n),e 2
s −2·t (mod n), . . ., e t (mod n) has either all 1’s or a −1 somewhere.
3
If yes, classify n as prime otherwise composite.
The time complexity of the test is O (log4 n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 36 / 47
Miller’s Test
8/8/2019 Is n a prime number
http://slidepdf.com/reader/full/is-n-a-prime-number 90/136
1 If n is an exact power, it is composite.
2 For each e , 1 < e ≤ 4log2 n, check if the sequence e 2s −1
·t (mod n),e 2
s −2·t (mod n), . . ., e t (mod n) has either all 1’s or a −1 somewhere.
3
If yes, classify n as prime otherwise composite.
The time complexity of the test is O (log4 n).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 36 / 47
Rabin’s Test
8/8/2019 Is n a prime number
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A modification of Miller’s algorithm proposed soon after (1974).
Selects e randomly instead of trying all e in the range [2, 4log2 n].
Randomized algorithm that never classfies primes incorrectly and
correctly classifies composites with probabilty at least
3
4 .Time complexity is O (log2 n).
The most popular primality testing algorithm.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 37 / 47
Rabin’s Test
8/8/2019 Is n a prime number
http://slidepdf.com/reader/full/is-n-a-prime-number 92/136
A modification of Miller’s algorithm proposed soon after (1974).
Selects e randomly instead of trying all e in the range [2, 4log2 n].
Randomized algorithm that never classfies primes incorrectly and
correctly classifies composites with probabilty at least
3
4 .Time complexity is O (log2 n).
The most popular primality testing algorithm.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 37 / 47
Rabin’s Test
8/8/2019 Is n a prime number
http://slidepdf.com/reader/full/is-n-a-prime-number 93/136
A modification of Miller’s algorithm proposed soon after (1974).
Selects e randomly instead of trying all e in the range [2, 4log2 n].
Randomized algorithm that never classfies primes incorrectly and
correctly classifies composites with probabilty at least
3
4 .Time complexity is O (log2 n).
The most popular primality testing algorithm.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 37 / 47
Rabin’s Test
8/8/2019 Is n a prime number
http://slidepdf.com/reader/full/is-n-a-prime-number 94/136
A modification of Miller’s algorithm proposed soon after (1974).
Selects e randomly instead of trying all e in the range [2, 4log2 n].
Randomized algorithm that never classfies primes incorrectly and
correctly classifies composites with probabilty at least
3
4 .Time complexity is O (log2 n).
The most popular primality testing algorithm.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 37 / 47
AKS Test
8/8/2019 Is n a prime number
http://slidepdf.com/reader/full/is-n-a-prime-number 95/136
Theorem (A Generalization of FLT)
If n is prime then for every e , 1 ≤ e < n, (ζ + e )n = ζ n + e in Z n[ζ ],
ζ r = 1.
A test proposed in 2002 based on this generalization.
The only known deterministic, unconditionally correct, polynomialtime algorithm.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 38 / 47
AKS Test
8/8/2019 Is n a prime number
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Theorem (A Generalization of FLT)
If n is prime then for every e , 1 ≤ e < n, (ζ + e )n = ζ n + e in Z n[ζ ],
ζ r = 1.
A test proposed in 2002 based on this generalization.
The only known deterministic, unconditionally correct, polynomialtime algorithm.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 38 / 47
AKS Test
8/8/2019 Is n a prime number
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Theorem (A Generalization of FLT)
If n is prime then for every e , 1 ≤ e < n, (ζ + e )n = ζ n + e in Z n[ζ ],
ζ r = 1.
A test proposed in 2002 based on this generalization.
The only known deterministic, unconditionally correct, polynomialtime algorithm.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 38 / 47
AKS Test
8/8/2019 Is n a prime number
http://slidepdf.com/reader/full/is-n-a-prime-number 98/136
1 If n is an exact power or has a small divisor, it is composite.
2 Select a small number r carefully, let ζ r = 1 and consider Z n[ζ ].
3 For each e , 1≤
e ≤
2√
r log n, check if (ζ + e )n = ζ n + e in Z n[ζ ].
4 If yes, n is prime otherwise composite.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27 2006 Delft 39 / 47
AKS Test
8/8/2019 Is n a prime number
http://slidepdf.com/reader/full/is-n-a-prime-number 99/136
1 If n is an exact power or has a small divisor, it is composite.
2 Select a small number r carefully, let ζ r = 1 and consider Z n[ζ ].
3 For each e , 1≤
e ≤
2√
r log n, check if (ζ + e )n = ζ n + e in Z n[ζ ].
4 If yes, n is prime otherwise composite.
Manindra Agrawal (IIT Kanpur) Is a Prime Number? March 27 2006 Delft 39 / 47
AKS Test
8/8/2019 Is n a prime number
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1 If n is an exact power or has a small divisor, it is composite.
2 Select a small number r carefully, let ζ r = 1 and consider Z n[ζ ].
3 For each e , 1≤
e ≤
2√
r log n, check if (ζ + e )n = ζ n + e in Z n[ζ ].
4 If yes, n is prime otherwise composite.
M A (IIT K ) Is P N ? M 27 2006 D 39 / 47
AKS Test
8/8/2019 Is n a prime number
http://slidepdf.com/reader/full/is-n-a-prime-number 101/136
1 If n is an exact power or has a small divisor, it is composite.
2 Select a small number r carefully, let ζ r = 1 and consider Z n[ζ ].
3 For each e , 1≤
e
≤2√
r log n, check if (ζ + e )n = ζ n + e in Z n[ζ ].
4 If yes, n is prime otherwise composite.
M A (IIT K ) In
P N ? M 27 2006 D 39 / 47
Analysis
Suppose n has at least two prime factors and let p be one of them
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Suppose n has at least two prime factors and let p be one of them.
Let S ⊆ Z p [ζ ] such that for every element f (ζ ) ∈ S , f ζ )n = f (ζ n) inZ p [ζ ].
It follows that for every f (ζ ) ∈ S , f (ζ )m = f (ζ m) for any m of theform ni · p j .
Since n is not a power of p , this places an upper bound on the size of S .
If ζ + e ∈ S for every 1 ≤ e ≤ 2√
r log n, then all their products arealso in S .
This makes the size of S bigger than the upper bound above.
M A (IIT K ) In
P N ? M 27 2006 D 40 / 47
Analysis
Suppose n has at least two prime factors and let p be one of them
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Suppose n has at least two prime factors and let p be one of them.
Let S ⊆ Z p [ζ ] such that for every element f (ζ ) ∈ S , f ζ )n = f (ζ n) inZ p [ζ ].
It follows that for every f (ζ ) ∈ S , f (ζ )m = f (ζ m) for any m of theform ni · p j .
Since n is not a power of p , this places an upper bound on the size of S .
If ζ + e ∈ S for every 1 ≤ e ≤ 2√
r log n, then all their products arealso in S .
This makes the size of S bigger than the upper bound above.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 40 / 47
Analysis
Suppose n has at least two prime factors and let p be one of them.
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Suppose n has at least two prime factors and let p be one of them.
Let S ⊆ Z p [ζ ] such that for every element f (ζ ) ∈ S , f ζ )n = f (ζ n) inZ p [ζ ].
It follows that for every f (ζ ) ∈ S , f (ζ )m = f (ζ m) for any m of theform ni · p j .
Since n is not a power of p , this places an upper bound on the size of S .
If ζ + e ∈ S for every 1 ≤ e ≤ 2√
r log n, then all their products arealso in S .
This makes the size of S bigger than the upper bound above.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 40 / 47
Analysis
Suppose n has at least two prime factors and let p be one of them.
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Suppose as at east t o p e acto s a d et p be o e o t e
Let S ⊆ Z p [ζ ] such that for every element f (ζ ) ∈ S , f ζ )n = f (ζ n) inZ p [ζ ].
It follows that for every f (ζ ) ∈ S , f (ζ )m = f (ζ m) for any m of theform ni · p j .
Since n is not a power of p , this places an upper bound on the size of S .
If ζ + e ∈ S for every 1 ≤ e ≤ 2√
r log n, then all their products arealso in S .
This makes the size of S bigger than the upper bound above.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 40 / 47
Analysis
Suppose n has at least two prime factors and let p be one of them.
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pp p p
Let S ⊆ Z p [ζ ] such that for every element f (ζ ) ∈ S , f ζ )n = f (ζ n) inZ p [ζ ].
It follows that for every f (ζ ) ∈ S , f (ζ )m = f (ζ m) for any m of theform ni · p j .
Since n is not a power of p , this places an upper bound on the size of S .
If ζ + e ∈ S for every 1 ≤ e ≤ 2√
r log n, then all their products arealso in S .
This makes the size of S bigger than the upper bound above.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 40 / 47
Time Complexity
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Number r is O (log5
n).Time complexity of the algorithm is O (log12 n).
An improvement by Hendrik Lenstra (2002) reduces the timecomplexity to O (log15/2 n).
Lenstra and Pomerance (2003) further reduce the time complexity toO (log6 n).
Bernstein (2003) reduced the time complexity to O (log4 n) at thecost of making it randomized.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 41 / 47
Time Complexity
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Number r is O (log5
n).Time complexity of the algorithm is O (log12 n).
An improvement by Hendrik Lenstra (2002) reduces the timecomplexity to O (log15/2 n).
Lenstra and Pomerance (2003) further reduce the time complexity toO (log6 n).
Bernstein (2003) reduced the time complexity to O (log4 n) at thecost of making it randomized.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 41 / 47
Time Complexity
5
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Number r is O (log5
n).Time complexity of the algorithm is O (log12 n).
An improvement by Hendrik Lenstra (2002) reduces the timecomplexity to O (log15/2 n).
Lenstra and Pomerance (2003) further reduce the time complexity toO (log6 n).
Bernstein (2003) reduced the time complexity to O (log4 n) at thecost of making it randomized.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 41 / 47
Time Complexity
5
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Number r is O (log5
n).Time complexity of the algorithm is O (log12 n).
An improvement by Hendrik Lenstra (2002) reduces the timecomplexity to O (log15/2 n).
Lenstra and Pomerance (2003) further reduce the time complexity toO (log6 n).
Bernstein (2003) reduced the time complexity to O (log4 n) at thecost of making it randomized.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 41 / 47
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Outline
1 The Problem
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2 Two Simple, and Slow, Methods
3 Modern Methods
4 Algorithms Based on Factorization of Group Size
5 Algorithms Based on Fermat’s Little Theorem
6 An Algorithm Outside the Two Themes
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Adleman-Pomerance-Rumeli Test
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Proposed in 1980.
Is conceptually the most complex algorithm of them all.
Uses multiple groups, ideas derived from both themes, plus new ones!
It is a deterministic algorithm with time complexity logO (log log log n) n.Was speeded up by Cohen and Lenstra (1981).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 43 / 47
Adleman-Pomerance-Rumeli Test
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Proposed in 1980.
Is conceptually the most complex algorithm of them all.
Uses multiple groups, ideas derived from both themes, plus new ones!
It is a deterministic algorithm with time complexity logO (log log log n) n.Was speeded up by Cohen and Lenstra (1981).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 43 / 47
Adleman-Pomerance-Rumeli Test
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Proposed in 1980.
Is conceptually the most complex algorithm of them all.
Uses multiple groups, ideas derived from both themes, plus new ones!
It is a deterministic algorithm with time complexity logO (log log log n) n.Was speeded up by Cohen and Lenstra (1981).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 43 / 47
Adleman-Pomerance-Rumeli Test
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Proposed in 1980.
Is conceptually the most complex algorithm of them all.
Uses multiple groups, ideas derived from both themes, plus new ones!
It is a deterministic algorithm with time complexity logO
(log log logn
) n.Was speeded up by Cohen and Lenstra (1981).
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 43 / 47
Overview of the Algorithm
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Tries to compute a factor of n.
Let p be a factor of n, p ≤ √n.
Find two sets of primes q 1, q 2, . . . , q t and r 1, r 2, . . . , r u satisfying:
t i =1 q i = logO (log log log n) n. For each j ≤ u , r j − 1 is square-free and has only q i ’s as prime divisors.
u j =1 r j >
√n.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 44 / 47
Overview of the Algorithm
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Tries to compute a factor of n.
Let p be a factor of n, p ≤ √n.
Find two sets of primes q 1, q 2, . . . , q t and r 1, r 2, . . . , r u satisfying:
t
i =1 q i = logO (log log log n) n. For each j ≤ u , r j − 1 is square-free and has only q i ’s as prime divisors.
u j =1 r j >
√n.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 44 / 47
Overview of the Algorithm
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Tries to compute a factor of n.
Let p be a factor of n, p ≤ √n.
Find two sets of primes q 1, q 2, . . . , q t and r 1, r 2, . . . , r u satisfying:
t
i =1 q i = logO (log log log n) n. For each j ≤ u , r j − 1 is square-free and has only q i ’s as prime divisors.
u j =1 r j >
√n.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 44 / 47
Overview of the Algorithm
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Tries to compute a factor of n.
Let p be a factor of n, p ≤ √n.
Find two sets of primes q 1, q 2, . . . , q t and r 1, r 2, . . . , r u satisfying:
t
i =1 q i = logO (log log log n) n. For each j ≤ u , r j − 1 is square-free and has only q i ’s as prime divisors.
u j =1 r j >
√n.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 44 / 47
Overview of the Algorithm
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Tries to compute a factor of n.
Let p be a factor of n, p ≤ √n.
Find two sets of primes q 1, q 2, . . . , q t and r 1, r 2, . . . , r u satisfying:
t
i =1 q i = logO
(log log logn
) n. For each j ≤ u , r j − 1 is square-free and has only q i ’s as prime divisors.
u j =1 r j >
√n.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 44 / 47
Overview of the Algorithm
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Tries to compute a factor of n.
Let p be a factor of n, p ≤ √n.
Find two sets of primes q 1, q 2, . . . , q t and r 1, r 2, . . . , r u satisfying:
t
i =1 q i = logO
(log log logn
) n. For each j ≤ u , r j − 1 is square-free and has only q i ’s as prime divisors.
u j =1 r j >
√n.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 44 / 47
Overview of the Algorithm
∗
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Let g j be a generator for the group F r j .Let p = g
γ j j (mod r j ) and γ j = δi , j (mod q i ) for every q i | r j − 1.
Compute ‘associated’ primes r j i ∈ r 1, r 2, . . . , r u for each q i .
Cycle through all tuples (α1, α2, . . . , αt ) with 0 ≤ αi < q i .
From a given tuple (α1, α2, . . . , αt ), derive numbers β i , j for1 ≤ j ≤ u , 1 ≤ i ≤ t and q i | r j − 1 such that If p = g αi
j i (mod r j ) for every j then δi , j = β i , j for every j and for every
i with q i | r j − 1.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 45 / 47
Overview of the Algorithm
L j b f h F∗
j
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Let g j be a generator for the group F r j .Let p = g
γ j j (mod r j ) and γ j = δi , j (mod q i ) for every q i | r j − 1.
Compute ‘associated’ primes r j i ∈ r 1, r 2, . . . , r u for each q i .
Cycle through all tuples (α1, α2, . . . , αt ) with 0 ≤ αi < q i .
From a given tuple (α1, α2, . . . , αt ), derive numbers β i , j for1 ≤ j ≤ u , 1 ≤ i ≤ t and q i | r j − 1 such that If p = g αi
j i (mod r j ) for every j then δi , j = β i , j for every j and for every
i with q i | r j − 1.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 45 / 47
Overview of the Algorithm
L j b f h F∗
j
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Let g j be a generator for the group F r j .Let p = g
γ j j (mod r j ) and γ j = δi , j (mod q i ) for every q i | r j − 1.
Compute ‘associated’ primes r j i ∈ r 1, r 2, . . . , r u for each q i .
Cycle through all tuples (α1, α2, . . . , αt ) with 0 ≤ αi < q i .
From a given tuple (α1, α2, . . . , αt ), derive numbers β i , j for1 ≤ j ≤ u , 1 ≤ i ≤ t and q i | r j − 1 such that If p = g αi
j i (mod r j ) for every j then δi , j = β i , j for every j and for every
i with q i | r j − 1.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 45 / 47
Overview of the Algorithm
L j b f h F∗
j
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Let g j be a generator for the group F r j .Let p = g
γ j j (mod r j ) and γ j = δi , j (mod q i ) for every q i | r j − 1.
Compute ‘associated’ primes r j i ∈ r 1, r 2, . . . , r u for each q i .
Cycle through all tuples (α1, α2, . . . , αt ) with 0 ≤ αi < q i .
From a given tuple (α1, α2, . . . , αt ), derive numbers β i , j for1 ≤ j ≤ u , 1 ≤ i ≤ t and q i | r j − 1 such that If p = g αi
j i (mod r j ) for every j then δi , j = β i , j for every j and for every
i with q i | r j − 1.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 45 / 47
Overview of the Algorithm
L t j b t f th F∗
rj
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Let g j be a generator for the group F r j .Let p = g
γ j j (mod r j ) and γ j = δi , j (mod q i ) for every q i | r j − 1.
Compute ‘associated’ primes r j i ∈ r 1, r 2, . . . , r u for each q i .
Cycle through all tuples (α1, α2, . . . , αt ) with 0 ≤ αi < q i .
From a given tuple (α1, α2, . . . , αt ), derive numbers β i , j for1 ≤ j ≤ u , 1 ≤ i ≤ t and q i | r j − 1 such that If p = g αi
j i (mod r j ) for every j then δi , j = β i , j for every j and for every
i with q i | r j − 1.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 45 / 47
Overview of the Algorithm
F δi j ’ b t t d il
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From δi , j ’s, p can be constructed easily: Use Chinese remaindering to compute γ j ’s from δi , j ’s. Use Chinese remaindering to compute p (mod
u
j =1 r j ) from g γ j
j ’s. Since
u
j =1 r j >√
n ≥ p , the residue equals p .
β i , j ’s are computed using higher reciprocity laws in extension rings
Z n[ζ i ], ζ q i i = 1.
For most of the composite numbers, the algorithm will fail duringcomputation of β i , j ’s.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 46 / 47
Overview of the Algorithm
From δi j ’s p can be constr cted easil
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From δi , j s, p can be constructed easily: Use Chinese remaindering to compute γ j ’s from δi , j ’s. Use Chinese remaindering to compute p (mod
u
j =1 r j ) from g γ j
j ’s. Since
u
j =1 r j >√
n ≥ p , the residue equals p .
β i , j ’s are computed using higher reciprocity laws in extension rings
Z n[ζ i ], ζ q i i = 1.
For most of the composite numbers, the algorithm will fail duringcomputation of β i , j ’s.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 46 / 47
Overview of the Algorithm
From δi j ’s p can be constructed easily:
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From δi , j s, p can be constructed easily: Use Chinese remaindering to compute γ j ’s from δi , j ’s. Use Chinese remaindering to compute p (mod
u
j =1 r j ) from g γ j
j ’s. Since
u
j =1 r j >√
n ≥ p , the residue equals p .
β i , j ’s are computed using higher reciprocity laws in extension rings
Z n[ζ i ], ζ q i i = 1.
For most of the composite numbers, the algorithm will fail duringcomputation of β i , j ’s.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 46 / 47
Overview of the Algorithm
From δi j ’s p can be constructed easily:
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From δi , j s, p can be constructed easily: Use Chinese remaindering to compute γ j ’s from δi , j ’s. Use Chinese remaindering to compute p (mod
u
j =1 r j ) from g γ j
j ’s. Since
u
j =1 r j >√
n ≥ p , the residue equals p .
β i , j ’s are computed using higher reciprocity laws in extension rings
Z n[ζ i ], ζ q i i = 1.
For most of the composite numbers, the algorithm will fail duringcomputation of β i , j ’s.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 46 / 47
Overview of the Algorithm
From δi j ’s p can be constructed easily:
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From δi , j s, p can be constructed easily: Use Chinese remaindering to compute γ j ’s from δi , j ’s. Use Chinese remaindering to compute p (mod
u
j =1 r j ) from g γ j
j ’s. Since
u
j =1 r j >√
n ≥ p , the residue equals p .
β i , j ’s are computed using higher reciprocity laws in extension rings
Z n[ζ i ], ζ q i i = 1.
For most of the composite numbers, the algorithm will fail duringcomputation of β i , j ’s.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 46 / 47
Overview of the Algorithm
From δi ,j ’s p can be constructed easily:
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From δi , j s, p can be constructed easily: Use Chinese remaindering to compute γ j ’s from δi , j ’s. Use Chinese remaindering to compute p (mod
u
j =1 r j ) from g γ j
j ’s. Since
u
j =1 r j >√
n ≥ p , the residue equals p .
β i , j ’s are computed using higher reciprocity laws in extension rings
Z n[ζ i ], ζ q i i = 1.For most of the composite numbers, the algorithm will fail duringcomputation of β i , j ’s.
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 46 / 47
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Outstanding Questions
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Is there a ‘practical’, polynomial time deterministic primality test?
Is there a ‘practical’, provably polynomial time, primality proving test?
Are there radically different ways of testing primality efficiently?
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 47 / 47
Outstanding Questions
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Is there a ‘practical’, polynomial time deterministic primality test?
Is there a ‘practical’, provably polynomial time, primality proving test?
Are there radically different ways of testing primality efficiently?
Manindra Agrawal (IIT Kanpur) Is n a Prime Number? March 27, 2006, Delft 47 / 47