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Deciding Primality is in P

Deciding Primality is in P. M. Agrawal, N. Kayal, N. Saxena Speaker: Adi Akavia. Background. Sieve of Eratosthenes 240BC - (n) 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)

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Deciding Primality is in P

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  1. Deciding Primality is in P M. Agrawal, N. Kayal, N. SaxenaSpeaker: Adi Akavia

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

  3. This Paper unconditional, deterministic, polynomial • Def: r is specialwith respect to n if: • r is prime, • r-1 has a large prime factor q = (r2/3) , and • q|Or(n). • Tools: • simple algebra • High density Thm for numbers with properties (1) and (2). [Fou85, BH96] Def: ordern mod r, denoted Or(n), is the smallest power t s.t. nt 1 (mod r).h

  4. Proof: Develop (x-a)n using Newton-binomial. • Assume n is prime, then • Assume n is composite, then let q|n, let qk||n, then and , hence xq has non zero coefficient (mod n). Basic Idea • Fact: For anya s.t (a,n)=1: • n is prime  (x-a)nxn-a (mod n) • n is composite  (x-a)nxn-a (mod n) • Naive algo: Pick an arbitrarya, check if (x-a)nxn-a (mod n) • Problem: time complexity - (n).

  5. Basic Idea • Idea: Pick an arbitrarya, and some polynomial xr-1, with r = poly log n, check if (x-a)nxn-a (mod xr-1, n) • time complexity – poly(r) • n is prime  (x-a)nxn-a (mod xr-1, n) • n is composite ???? (x-a)nxn-a (mod xr-1, n) Not true for some (few) values of a,r !

  6. Improved Idea • Improved Idea: Pick many(poly log n)a’s, check for all of them if: (x-a)nxn-a (mod xr-1, n)Accept if equality holds for all a’s

  7. Some Algebra Reminders Def: Fp (p is prime) denotes the finite field of p elements {0,1,…,p-1}. Def: Fp[x] denotes the ring of polynomials over Fp. Def: Let f(x) be a k-degree polynomial. Def: Fp[x]/f(x) denotes the set of k-1-degree polynomials over Fp, with addition and multiplication modulo f(x). Thm: If f(x) is irreducible over Fp, then Fp[x]/f(x)  the unique field with pk elements.

  8. Fp[x]/f(x) - Addition • Let the polynomial f(x) over F2 be: • Represent polynomials as vectors (k-1 degree polynomial  vector of k coefficient): • Addition:

  9. Fp[x]/f(x) - Multiplication Multiplication: • First, multiply ‘modp’: • Next, apply ’modf(x)’:

  10. Fp[x]/f(x) - mod f(x) • Example: • In general for f(x) = xr-1:

  11. Irreducible Factors of (xr-1)/(x-1) • Fact: Consider the polynomial (xr-1)/(x-1) over Fp. All its irreducible factors are of degree d = deg(h(x))

  12. The Algorithm Input: integer n • Find r  O(log6n), s.t. r is special, • Let l = 2r1/2log n. • Small divisors test:For t=2,…,l, if t|n output COMPOSITE • Power test: If n is a power --n=pk, for k>1 output COMPOSITE . • Polynomials test:For a =1,…,l, if (x-a)n  xn-a (mod xr-1, n), output COMPOSITE . • Otherwise: output PRIME.

  13. Find r  O(log6n), s.t. r is special, • Let l = 2r1/2log n. • If exists a small ( < l+1) divisor, output COMPOSITE • If n is a power, output COMPOSITE . • For a = 1,…,l, if (x-a)n  xn-a (mod xr-1, n), output COMPOSITE . • Otherwise output PRIME. Saw: algorithm Yet to be seen: • Special r  O(log6n) exists (later) • If n is composite then one of the tests returns COMPOSITE.

  14. Find r  O(log6n), s.t. r is special, • Let l = 2r1/2log n. • If exists a small ( < l+1) divisor, output COMPOSITE • If n is a power, output COMPOSITE . • For a = 1,…,l, if (x-a)n  xn-a (mod xr-1, n), output COMPOSITE . • Otherwise output PRIME. Correctness Proof Lemma: n is composite algo returns ‘composite’. That is, • If n is composite, and • n has no divisor t  l, and • n is not a (prime) power • then a[1..l] s.t. (x-a)n  xn-a (mod xr-1, n)

  15. In the Proof - Using p and h(x) • Let p be a prime factor of n, and let h(x) be an irreducible factor of xr-1, • Suffices to show inequality (mod h(x), p) instead of: (mod xr-1, n), i.e. a[1..l] s.t. (x-a)n  xn-a (mod h(x), p) • Choose p and h(x) s.t. • q|Or(p), and • deg(h(x)) = Or(p) Such p exists: q|Or(n)and Or(n) = lcm{Or(pi)}, where n=p1p2…pk. Such h(x) exists: by previous fact.

  16. Proof • Assume by contradiction that n is composite, and passes all the tests, i.e. • n has no small factor, and • n is not a prime-power, and •  a[1..l](x-a)n  xn-a (mod h(x), p), • For any f(x), which is a multiple of polynomials (x-a) (where a[1..l]),f(x)n=f(xn). • Example: [(x-a1)(x-a2)]n = (xn-a1)(xn-a2)

  17. Proof • Therefore, consider the group generated by {(x-a)}a[1..l]: • Are there other integers m s.t. f(x)G, f(x)m  f(xm) ? • Yes! For example: p. • Any others? • Let I = { m | fG, f(x)m  f(xm) }. • Lemma: Iis multiplicative, i.e. u,vI uvI. • Hence, in particular {nipj : 0 ≤ i,j ≤ r1/2}  I. • Therefore,

  18. Proof – I[|G|] is large • Lemma: • Proof: Consider all polynomials of degree < d.They are all distinct in Fp[x]/h(x). Therefore • Hence, • However, we next show that dis big: q|Or(p)=d.

  19. Proof – I[|G|] is small • Lemma: Letm1, m2 I, thenm1  m2 (mod |G|)  m1  m2 (mod r) • Proof: Let g(x) be a generator of G. Let m2=m1+kr. • (*) m1m2 (mod r), then xm1xm2 (mod h(x)) (as xr  1 (mod h(x))) • Contradiction!

  20. Proof Summary • We saw that I[|G|] is small (unconditionally, using properties of xr-1), • However, if n is composite and not a prime power, then passing the polynomials test (i.e.nI) implies that I[|G|] is large.(using properties of the special r and of xr-1) • Therefore, the polynomials test must return ‘composite’.

  21. Back to Special Numbers • Recall: r is specialwith respect to n if: • r is prime, • r-1 has a large prime factor q = (r2/3) , and • q|Or(n). • We next show that Special r  O(log6n) exists.

  22. Find r  O(log6n), s.t. r is special, • Let l = 2r1/2log n. • If exists a small ( < l+1) divisor, output COMPOSITE • If n is a power, output COMPOSITE . • For a = 1,…,l, if (x-a)n  xn-a (mod xr-1, n), output COMPOSITE . • Otherwise output PRIME. Finding Special r Elaborating on step (1): • while r < c log6n • if r is prime • let q be the largest prime factor of r-1 • if (q4r1/2log n) and (n(r-1)/q  1 (mod r)) break; • rr+1 Complexity: O(log6n) iterations, each taking: O(r1/2 poly log r), hence total poly log n. • when ‘break’ is reached: r is prime, q is large, and q|Or(n)

  23. Recall: r is specialwith respect to n if: • r is prime, • q = (r2/3) prime factor of r-1, • q|Or(n). Special r  O(log6n) exists • Consider interval [..], ,=O(log6n). • Numbers with properties (1) and (2) are dense in [..] • immediate from density bounds for numbers with these properties and for primes. • For many primes r[..], property (3) holds. • For many r’s Or(n) > 1/3: Or(n) < 1/3 r | =(n-1)(n2-1)...(n^1/3-1). However, has no more than 2/3log n prime divisors. • Moreover, Or(n) > 1/3  q | Or(n): ifq doesn’t divide Or(n), then n(r-1)/q  1, therefore Or(n)  (r-1)/q. However(r-1)/q<1/3-- a contradiction. (here we utilize again the fact that q is large). • Hence, by counting argument, exists a special r[..].

  24. The End

  25. Proof - G is large, Cont. This is the reason for seeking a large q s.t. q|Or(n) Hence, Prop: d  2l Proof: Recall d=Or(p) and q|Or(p), hence d  q  2l(recall q4r1/2log n, l=2r1/2log n) Hence

  26. Algebraic Background – Extension Field Def: Consider fields F, E. E is an extension of F, if F is a subfield of E. Def: Galois fieldGF(pk) (p prime)is the unique (up to isomorphism) finite field containing pk 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)

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

  28. Fp[x]/f(x) - Example Let the irreducible polynomial f(x) be: Represent polynomials as vectors (k-1 degree polynomial  vector of k coefficient): Addition:

  29. Fp[x]/f(x) - Example Multiplication: • First, multiply ‘modp’: • Next, apply ’modf(x)’:

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