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Hensel Interpolation. Lecture 9. Recall our typical mapping from GCD. GCD(F(x,y,z),G(x,y,z)). H(x,y,z). H p (x,y,z). GCD(F p (x,y,z),G p (x,y,z)). GCD(F p (x,0,0),G p (x,0,0)). H p (x,0,0). Start in the lower right corner of this diagram with some image of a factorization.

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hensel interpolation

Hensel Interpolation

Lecture 9

Richard Fateman CS 282 Lecture 9

recall our typical mapping from gcd
Recall our typical mapping from GCD

GCD(F(x,y,z),G(x,y,z))

H(x,y,z)

Hp(x,y,z)

GCD(Fp(x,y,z),Gp(x,y,z))

GCD(Fp(x,0,0),Gp(x,0,0))

Hp(x,0,0)

Richard Fateman CS 282 Lecture 9

start in the lower right corner of this diagram with some image of a factorization
Start in the lower right corner of this diagram with some image of a factorization
  • c(x)*h(x)=f(x) mod p
  • ... we have c mod p, h mod p. (and f mod p)
  • ... we want to know c and h over Z [x], in fact, over Z[x,y,z,...] as many variables as are necessary.
  • note that c(x) is the cofactor of f, (f/h), and h is the gcd(f,g).
  • ALSO we’ll sometimes need to fake these otherwise in general false assumptions..
  • f(x) and h(x) are monic [leading coeff 1]
  • f(x) and h(x) are “square-free”. That is, f(x)=k2(x)*... is forbidden.

Richard Fateman CS 282 Lecture 9

recall p adic notation
Recall p-adic notation

If p and N are integers, then the p-adic representation of N is

(a0, a1, ..., am) where for i = 1, ...,m = we have 0<ai<p

N = a0p0 + a1p1 + ... + ampm

For instance, the 3-adic representation of the integer 65 is (2,0,1,2).

We can extend this notion to polynomials. If p(x) and q(x) are polynomials, then the p-adic representation of q is (a1, a2, ..., am) where for i=1, ..., m-1, ai is in an integer, am is a nonzero integer and

q(x) = a0p(x)0 + a1p(x)1 + ... +amp(x)m

For instance the (x-1)-adic representation of x3 is (1,3,3,1).

Richard Fateman CS 282 Lecture 9

hensel algorithms
Hensel Algorithms

Start with a p-adic approximate and compute ever more accurate versions.

The original linear Hensel Construction lifts a factorization

from mod pi to mod pi+1 at the ith step

while the quadratic construction due to Zassenhaus lifts a factorization from mod (p2)i to mod (p2)i+1

That is, twice as many terms instead of one more term. Nevertheless, the linear version may require so much less computation at each step that it may be the algorithm of choice in lifting to a given modulus.

Richard Fateman CS 282 Lecture 9

starting up algorithm
Starting up.. Algorithm

Let u(x) be a monic polynomial . (This assumption is, in general, unwarranted, and overcoming it makes the algorithm messy.)

in Z[x] and assume v1(x) ¢ w1(x) = u(x) mod p and GCD (v1,w1) = 1 mod p where p is a prime.

The algorithm computes a sequence of pairs of polynomials

((vi,wi)) such that

vi(x) ¢ wi(x) = u(x) mod pi

Richard Fateman CS 282 Lecture 9

step i
Step I
  • Compute by means of the Extended Euclidean Algorithm
  • generalized to polynomials, two polynomials a(x) and b(x) in Zp[x] such that
  • (i) deg(a) < deg(w1),
  • deg(b) < deg(v1 )
  • (a(x)v1 + b(x)w1 = 1) mod p

Richard Fateman CS 282 Lecture 9

step ii
Step II

Now suppose we are given (vi, wi) and we wish to compute (vi+1,wi+1). Compute a polynomial ci such that

(pici(x) = vi(x)wi(x) - u(x)) mod pi+1

How hard is this? a multiply, subtraction, and division with remainder

Richard Fateman CS 282 Lecture 9

step iii
Step III

Compute (by polynomial division of a(x)ci(x) by w1(x)) the quotient qi(x) and remainder ai(x) such that

a(x)ci(x) = qi(x)w1(x) + ai(x) mod p

and set

bi(x) := (b(x)ci(x) + qi(x)v1(x)) mod p

Observe that deg(ai)<deg(w1), deg(bi)<deg(vi)

and mod p,

aiv1 + biw1 = (a ¢ ci - w1qi)v1+ (b ¢ ci + v1qi)w1

= ci(a ¢ v1 + b ¢ w1 )

= ci

Richard Fateman CS 282 Lecture 9

step iii observation
Step III- observation

{so in Z[x] , (or Zp2[x] )

aiv1 + biw1 = ci + p ¢ di(x) for some di(x) in Z[x]

Richard Fateman CS 282 Lecture 9

step iv
Step IV

Set

vi+1(x) := vi(x)-pibi(x) mod pi+1

wi+1(x) := wi(x)-piai(x) mod pi+1

Observe that, over the integers

vi+1(x)wi+1(x) = vi(x)wi(x)- pi[ai(x)vi(x)+bi(x)wi(x)] + p2i¢ ai(x)bi(x)

= u(x) + pi¢ ci(x) - pi[ci(x) +p ¢ di(x)] + p2i¢ ai(x)bi(x)

= u(x) - pi+1[di(x)-pi-1¢ ai(x)bi(x)

that is,

vi+1(x)wi+1(x) = u(x) mod pi+1 (QED)

Richard Fateman CS 282 Lecture 9