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1 Deciding Primality is in P M. Agrawal, N. Kayal, N. Saxena Presentation by Adi Akavia

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2 Background Sieve of Eratosthenes 240BC - (n) Sieve of Eratosthenes 240BC - (n) Fermat’s Little Theorem (17 th century): Fermat’s Little Theorem (17 th century): p is prime, a 0 (mod p) a p-1 1 (mod p) (The converse does not hold – Carmichael numbers) Polynomial-time algorithms: Polynomial-time algorithms: [Miller 76] deterministic, assuming Extended Riemann Hypothesis. [Miller 76] deterministic, assuming Extended Riemann Hypothesis. [Solovay, Strassen 77; Rabin 80] unconditional, but randomized. [Solovay, Strassen 77; Rabin 80] unconditional, but randomized. [Goldwasser, Kilian 86] randomized produces certificate for primality! (for almost all numbers) [Goldwasser, Kilian 86] randomized produces certificate for primality! (for almost all numbers) [Atkin 86; Adelman Huang 92] primality certificate for all numbers. [Atkin 86; Adelman Huang 92] primality certificate for all numbers. [Adelman, Pomerance, Rumely 83] deterministic (log n) O(log log log n) -time. [Adelman, Pomerance, Rumely 83] deterministic (log n) O(log log log n) -time.

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3 This Paper unconditional, deterministic, polynomial Def (Sophie-Germain primes): primes (p-1)/2 s.t. p is also prime. Def (Sophie-Germain primes): primes (p-1)/2 s.t. p is also prime. Def: r is special with respect to n if: Def: r is special with respect to n if: r is prime, r is prime, r-1 has a large prime factor q = (r 2/3 ), and r-1 has a large prime factor q = (r 2/3 ), and q|O r (n). q|O r (n). Tools: Tools: simple algebra simple algebra High density conjecture for primes p s.t. (p-1)/2 is Sophie-Germain High density conjecture for primes p s.t. (p-1)/2 is Sophie-Germain High density Thm for primes p s.t. p-1 has a large ( (r 2/3 )prime factor. [Fou85, BH96] High density Thm for primes p s.t. p-1 has a large ( (r 2/3 )) prime factor. [Fou85, BH96] Def: order n mod r, denoted O r (n), is the smallest power t s.t. n t 1 (mod r).

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4 This Paper unconditional, deterministic, polynomial Def (Sophie-Germain primes): primes (p-1)/2 s.t. p is also prime. Def (Sophie-Germain primes): primes (p-1)/2 s.t. p is also prime. Def: r is “almost Sophie-Germain“ (ASG) if: Def: r is “almost Sophie-Germain“ (ASG) if: r is prime, r is prime, r-1 has a large prime factor q = (r 2/3 ) r-1 has a large prime factor q = (r 2/3 ) Tools: Tools: simple algebra simple algebra High density conjecture for primes p s.t. (p-1)/2 is Sophie-Germain High density conjecture for primes p s.t. (p-1)/2 is Sophie-Germain High density Thm for primes p that are ‘almost Sophie-Germain’. [Fou85, BH96] High density Thm for primes p that are ‘almost Sophie-Germain’. [Fou85, BH96]

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5 Basic Idea Fact: For any a s.t (a,n) =1: Fact: For any a s.t (a,n) =1: n is prime (x-a) n x n -a (mod n) n is prime (x-a) n x n -a (mod n) n is composite (x-a) n x n -a (mod n) n is composite (x-a) n x n -a (mod n) Naive algo: Pick an arbitrary a, check if (x-a) n x n -a (mod n) Naive algo: Pick an arbitrary a, check if (x-a) n x n -a (mod n) Problem: time complexity - (n). Problem: time complexity - (n). Proof: Develop (x-a) n using Newton-binomial. Assume n is prime, then Assume n is prime, then Assume n is composite, then let q|n, let q k ||n, then and, hence x q has non zero coefficient (mod n). Assume n is composite, then let q|n, let q k ||n, then and, hence x q has non zero coefficient (mod n).

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6 Basic Idea Idea: Pick an arbitrary a, and some polynomial x r -1, with r = poly log n, check if (x-a) n x n -a (mod x r -1, n) Idea: Pick an arbitrary a, and some polynomial x r -1, with r = poly log n, check if (x-a) n x n -a (mod x r -1, n) time complexity – poly(r) time complexity – poly(r) n is prime (x-a) n x n -a (mod x r -1, n) n is prime (x-a) n x n -a (mod x r -1, n) n is composite ?? ?? (x-a) n x n -a (mod x r -1, n) n is composite ?? ?? (x-a) n x n -a (mod x r -1, n) Not true for some (few) values of a,r !

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7 Improved Idea Improved Idea: Pick many (poly log n) a’s, check for all of them if: (x-a) n x n -a (mod x r -1, n) Accept if equality holds for all a’s Improved Idea: Pick many (poly log n) a’s, check for all of them if: (x-a) n x n -a (mod x r -1, n) Accept if equality holds for all a’s

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8 Algebraic Background – Extension Field Def: Consider fields F, E. E is an extension of F, if F is a subfield of E. Def: Galois field GF(p k ) (p prime) is the unique (up to isomorphism) finite field containing p k elements. (The cardinality of any finite fields is a prime-power.) Def: A polynomial f(x) is called irreducible in GF(p) if it does not factor over GF(p)

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9 Multiplicative Group Def: GF * (p k ) is the multiplicative group of the Galois Field GF(p k ), that is, GF * (p k ) = GF(p k )\{0}. Thm: GF * (p k ) is cyclic, thus it has a generator g:

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10 Constructing Galois Fields Def: F p denotes a finite field of p elements (p is prime). Def: Let f(x) be a k-degree polynomial. Def: Let F p [x]/f(x) be the set of k-1-degree polynomials over F p, with addition and multiplication modulo f(x). Thm: If f(x) is irreducible over GF(p), then GF(p k ) F p [x]/f(x).

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11 F p [x]/f(x) - Example Let the irreducible polynomial f(x) be: Represent polynomials as vectors (k-1 degree polynomial vector of k coefficient) : Addition:

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12 F p [x]/f(x) - Example Multiplication: First, multiply ‘mod p’: First, multiply ‘mod p’: Next, apply ’mod f(x)’: Next, apply ’mod f(x)’:

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13 The Algorithm Input: integer n 1. Find r O(log 6 n), s.t. r is special, 2. Let l = 2r 1/2 log n. 3. For t=2,…,l, if t|n output COMPOSITE 4. If n is (prime) power -- n=p k, for k>1 output COMPOSITE. 5. For a =1,…,l, if (x-a) n x n -a (mod x r -1, n), output COMPOSITE. 6. Otherwise: output PRIME. Def: r is special if: r is Almost Sophie-Germain, and q|O r (n) (where qthe large prime factor of r-1). q|O r (n) (where q is the large prime factor of r-1).

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14 Proof’s Structure Saw: primality test. We next show: Special r O(log 6 n) exists. Special r O(log 6 n) exists. For such r: if n is composite s.t. n passes steps (3) and (4), then a [1..l] s.t. (x-a) n x n -a (mod x r -1, n) (hence, returns COMPOSITE at step (5)) For such r: if n is composite s.t. n passes steps (3) and (4), then a [1..l] s.t. (x-a) n x n -a (mod x r -1, n) (hence, returns COMPOSITE at step (5)) 1. Find r O(log 6 n), s.t. r is special, 2. Let l = 2r 1/2 log n. 3. For t=2,…,l, if t|n output COMPOSITE 4. If n is a prime power, i.e. n=p k, for some prime p, output COMPOSITE. 5. For a =1,…,l, if (x-a) n x n -a (mod x r -1, n), output COMPOSITE. 6. Otherwise output PRIME.

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15 Finding Suitable r Elaborating on step (1): 1. while r < c log 6 n 1. if r is prime 2. let q be the largest prime factor of r-1 3. if (q 4r 1/2 log n) and (n (r-1)/q 1 (mod r)) break; 4. r r+1 Complexity: O(log 6 n) iterations, each taking: O(r 1/2 poly log r), hence total poly log n. when ‘break’ is reached: r is prime, q is large, and q|O r (n)when ‘break’ is reached: r is prime, q is large, and q|O r (n) 1. Find r O(log 6 n), s.t. r is special, 2. Let l = 2r 1/2 log n. 3. For t=2,…,l, if t|n output COMPOSITE 4. If n is a prime power, i.e. n=p k, for some prime p, output COMPOSITE. 5. For a =1,…,l, if (x-a) n x n -a (mod x r -1, n), output COMPOSITE. 6. Otherwise output PRIME.

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16 Lemma: Special r O(log 6 n) s.t. q|O r (n) exists. Proof: let , =O(log 6 n), consider the interval [ .. ]. let , =O(log 6 n), consider the interval [ .. ]. special numbers are dense in [ .. ] special numbers are dense in [ .. ] there are only few primes r [ .. ] s.t O r (n) < 1/3. there are only few primes r [ .. ] s.t O r (n) < 1/3. Hence, by counting argument, exists a special r [ .. ] s.t. O r (n) > 1/3. Hence, by counting argument, exists a special r [ .. ] s.t. O r (n) > 1/3. Moreover, O r (n) > 1/3 q | O r (n). Moreover, O r (n) > 1/3 q | O r (n). Therefore, exists a special r [ .. ] s.t. q|O r (n). Therefore, exists a special r [ .. ] s.t. q|O r (n). #special [ .. ] #special [1.. ] - #primes [1.. ] = (log 6 n / loglog n) (using density of special numbers, and lower bound on density of primes) O r (n) < 1/3 r | =(n-1)(n 2 -1)...(n^ 1/3 -1). However, has no more than 2/3 log n prime divisors assumeq O r (n), then n (r-1)/q 1, therefore O r (n) (r-1)/q. However(r-1)/q 1/3 -- a contradiction. assume q doesn’t divide O r (n), then n (r-1)/q 1, therefore O r (n) (r-1)/q. However (r-1)/q < 1/3 -- a contradiction.

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17 Lemma: Special r O(log 6 n) exists. Proof: let , =O(log 6 n), consider the interval [ .. ]. let , =O(log 6 n), consider the interval [ .. ]. ASG numbers are dense in [ .. ] ASG numbers are dense in [ .. ] there are only few primes r [ .. ] s.t O r (n) < 1/3. there are only few primes r [ .. ] s.t O r (n) < 1/3. Hence, by counting argument, exists a ASG r [ .. ] s.t. O r (n) > 1/3. Hence, by counting argument, exists a ASG r [ .. ] s.t. O r (n) > 1/3. Moreover, O r (n) > 1/3 q | O r (n). Moreover, O r (n) > 1/3 q | O r (n). Therefore, exists a special r [ .. ]. Therefore, exists a special r [ .. ]. #ASG [ .. ] #ASG [1.. ] - #primes [1.. ] = (log 6 n / loglog n) (using density of ASG numbers, and upper bound on density of primes) O r (n) < 1/3 r | =(n-1)(n 2 -1)...(n^ 1/3 -1). However, has no more than 2/3 log n prime divisors assumeq O r (n), then n (r-1)/q 1, therefore O r (n) (r-1)/q. However(r-1)/q 1/3 -- a contradiction. assume q doesn’t divide O r (n), then n (r-1)/q 1, therefore O r (n) (r-1)/q. However (r-1)/q < 1/3 -- a contradiction.

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18 Correctness Proof Lemma: n is composite step (5) returns ‘composite’. That is, If n is composite, and If n is composite, and n has no factor t l, and n has no factor t l, and n is not a prime-power n is not a prime-power then a [1..l] s.t. (x-a) n x n -a (mod x r -1, n) then a [1..l] s.t. (x-a) n x n -a (mod x r -1, n) 1. Find r O(log 6 n), s.t. r is special, 2. Let l = 2r 1/2 log n. 3. For t=2,…,l, if t|n output COMPOSITE 4. If n is a prime power, i.e. n=p k, for some prime p, output COMPOSITE. 5. For a =1,…,l, if (x-a) n x n -a (mod x r -1, n), output COMPOSITE. 6. Otherwise output PRIME.

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19 Proof Let p be a prime factor of n, and let h(x) be an irreducible factor of x r -1, Let p be a prime factor of n, and let h(x) be an irreducible factor of x r -1, It suffices to show inequality (mod h(x), p) instead of (mod x r -1, n), i.e. a [1..l] s.t. (x-a) n x n -a (mod h(x), p) It suffices to show inequality (mod h(x), p) instead of (mod x r -1, n), i.e. a [1..l] s.t. (x-a) n x n -a (mod h(x), p) Choose p and h(x) s.t. Choose p and h(x) s.t. q|O r (p), and q|O r (p), and deg(h(x)) = O r (p) deg(h(x)) = O r (p) Such p exists: Let n=p 1 p 2 …p k, then O r (n) = lcm{Or(p i )}. Therefore: q|O r (n) i q|O r (p i ) (as q is prime) Such h exists: by previous claim.

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20 Proof Assume by contradiction that n is composite, and passes all the tests, i.e. Assume by contradiction that n is composite, and passes all the tests, i.e. n has no small factor, and n has no small factor, and n is not a prime-power, and n is not a prime-power, and a [1..l] (x-a) n x n -a (mod h(x), p), a [1..l] (x-a) n x n -a (mod h(x), p),

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21 Proof Consider the group generated by {(x-a)} a [1..l] (mod h(x), p), i.e. Consider the group generated by {(x-a)} a [1..l] (mod h(x), p), i.e. Note: f(x) G, f(x) n f(x n ) Note: f(x) G, f(x) n f(x n ) Let I = { m | f G, f(x) m f(x m ) }. Let I = { m | f G, f(x) m f(x m ) }. Lemma: I is multiplicative, i.e. u,v I uv I. Lemma: I is multiplicative, i.e. u,v I uv I. Proof: x r -1|x vr -1, therefore Proof: x r -1|x vr -1, thereforehence

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22 Proof - n I I is large Prop: (i,j) (i’,j’) n i p j n i’ p j (since n p k ) Prop: (i,j) (i’,j’) n i p j n i’ p j (since n p k ) Lemma: , if u,v I s.t. (i,j) (i’,j’) u i v j u i’ v j’, then |I| [u v ] > 2. Lemma: , if u,v I s.t. (i,j) (i’,j’) u i v j u i’ v j’, then |I| [u v ] > 2. Corollary: , n I |I| [u v ] > 2. Proof: p I. Corollary: , n I |I| [u v ] > 2. Proof: p I. However, Lemma: However, Lemma: Corollary: n I |I| [|G|] > r. Corollary: n I |I| [|G|] > r. ( +1) 2 different pairs (i,j), each give a distinct value Consider all polynomials of degree bound <d. There are all distinct in F p [x]/h(x). Therefore

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23 Irreducible Factors of (x r -1)/(x-1) Def: Let h(x) denote any irreducible factor of (x r -1)/(x-1), and d = deg(h(x)) Def: Let h(x) denote any irreducible factor of (x r -1)/(x-1), and d = deg(h(x)) Claim: h(x), d=O r (p) Claim: h(x), d=O r (p) Proof: Denote k=O r (p). Note F p [x]/h(x) is of size p d, therefore F p [x]/h(x)* is cyclic of order p d -1. Proof: Denote k=O r (p). Note F p [x]/h(x) is of size p d, therefore F p [x]/h(x)* is cyclic of order p d -1. k|dx r 1 (mod h(x)), hence O h(x) (x) is r, therefore r|p d -1, i.e., p d 1 (mod r), and hence k|d (recall d=O r (p)). k|d: x r 1 (mod h(x)), hence O h(x) (x) is r, therefore r|p d -1, i.e., p d 1 (mod r), and hence k|d (recall d=O r (p)). d|kg be a generator, then hencep d -1 |p k -1therefore d|k. d|k: let g be a generator, then hence p d -1 | p k -1. and therefore d|k. Recall, if r is special with respect to n, then r-1 has a large prime factor q, s.t. q|O r (n). Choose p s.t. q|O r (p) (exists). Then d is large. exists

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24 Proof – I is small Lemma: Letm1, m2 I, then m1 m2 (mod |G|) m1 m2 (mod r) Lemma: Let m1, m2 I, then m1 m2 (mod |G|) m1 m2 (mod r) Lemma(I is small): |I| [|G|] r Lemma(I is small): |I| [|G|] r Proof: Proof: Each two elements in |I| [|G|] are different mod |G|. Each two elements in |I| [|G|] are different mod |G|. Therefore they are different mod r. Therefore they are different mod r. Hence |I| [|G|] r. Hence |I| [|G|] r. Contradiction! Contradiction! Proof: Let g(x) be a generator of G. Let m2=m1+kr. (*) m1 m2 (mod r), then x m1 x m2 (mod h(x)) (as x r 1 (mod h(x)))

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25 The End

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26 Proof - G is large, Cont. Hence, Prop: d 2l Proof: Recall d=O r (p) and q|O r (p), hence d q 2l (recall q 4r 1/2 log n, l=2r 1/2 log n) Hence This is the reason for seeking a large q s.t. q|O r (n)

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