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Finding Reduced Basis for Lattices
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For given positive

Do there exist such that:

(is s a subset sum of the miís)?

Slide 15

Sums of squares

Sums of squares

Every prime that is 1mod4 can be

written as sum of two squares.

Those squares are found using LLL

Slide 16

abc Conjecture

abc Conjecture

For define the radical

(Thatís the product of distinct prime factors of a,b,c). suppose gcd(a,b,c)=1.

abc conjecture: For every x>1 there exists only finitely many a,b,c with gcd(a,b,c) = 1 and a + b = c such that

The search for examples uses LLL

Slide 17

Proposition:

Proposition:

B1,bn are reduced basis for a lattice L in b1*, bn* defined as before. Then:

1.

2.

3.

4.

(i.e. the 1st vector is ďreasonablyĒ short).

Reduced basis, what is it good for?

Slide 18

Algorith.doc

Algorith.doc

Example.doc

Slide 19

Algorith terminates:

Algorith terminates:

so each is a pos. real number

D changes only if some bi* is changed, which only occurs at case 1 of the algorith. The number is reduced by a factor of ¾ since is, while the other

diís are unchanged. Hence D reduced by factor of ¾ .

Slide 20

diís are bounded from below which bounds D from below.

diís are bounded from below which bounds D from below.

So thereís an upper bound for # of times we pass through case 1.

Slide 21

In end of case 1, k = k-1

In end of case 1, k = k-1

End of case 2, k = k+1

Start with k = 2, and

So # of times we pass through case 2

Is at most n-1 more than the # of times we pass through case 1,

Hence the algorith terminates.

Slide 22

Complexity:

Complexity:

Initialization step with rationales:

# of times pass through case 1:

# of times pass through case 2:

Case 1 requires operations

Case 2 we have values of p

Each requires operations

Slide 23

Hence we get a total of

Hence we get a total of

Operations.

Polynomial Time.

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