Diophantine Approximation and Basis Reduction

By Shu WangCAS 746 Presentation

6th, Feb, 2006

Overview Problem: Approximating real numbers by ra

tional numbers of low denominator and finding a so-called reduced basis in a lattice

Content The continued fraction method for approximati

ng one real number Lovász’s basis reduction method for lattices Applications

Notations

, , g.c.d, W.O.L.G

Dirichlet’s Theorem Let be a real number and let Then t

here exist two integers p and q such that

Example.

0 1

1 and 1p

qq q

0.2

, ?p q

Answer: 3 1p q

Proof of Dirichlet’s Theorem

Let we find two different integers i and j where

Consider the following series

Otherwise, according to pigeon-hole principle,

0 11

1M 2

1M 1

M

M …

1:M 1

0 , and {( ) }1

i j M i jM

{0 },{1 },{2 },{3 },...,{ }M 1

If ( | 0 :{ } ) then : , : 01

k k M k i k jM

1( , , | (0 , ) (1 ) : { },{ } )

1 1: max( , ), : min( , ) W.O.L.G Let : , :

1{( ) } { } { { } { }} { } { }

1

m mk l m k l M m M k l

M Mi k l j k l i l j k

i j l j l l k k l kM

Proof of Dirichlet’s Theorem - continued

Exercises

Let : , : . Then

{ } {( ) } 1

( 1)

1Since

1 11

1 ( 1)

So

q i j p q

qa qp qa p qa i j a

q q q q q M q

M

MM M q q

p

q q

Given a real number , we compute its rational approximation by following a series of steps as follows:

First we define

This sequence stops if becomes an integer We define an sequences called convergents that approximate to the abov

e

If becomes an integer then the last term of convergents equals to . We use to denote the term of the convergents of

The Continued Fraction Method

1

12 1 1

13 2 2

14 3 3

:

: ( )

: ( )

: ( )

1 1 12

23

1 1

1

1 1 12

23

1 1, , ,

1

i

i ( )kc -thk

The Continued Fraction Method (2)

We can determine a sequence where so that it corresponds to the convergent series

Suppose the first two terms are as follows:

What can we deduce from it?

If then . Contradiction exist.

31 2

1 2 3

, , , pp p

q q q

1 21

1 2

1, +

( )

p p

q q

,g.c.d( , ) 1i ii p q

1 21 1 2 1 1 2

1 2

=1, , , 1p p

q p p q p qq q

1 1q g.c.d( , ) 1i ip q

Proof

21 11 1

2

1

1 1 1+ + +

{ } {{ } }( ) { }

1+ +{ }

{ }

p

q a

aa

2 11 1

2 1

1 2 22 1 1 2 1 1

12

1 12 2 1

1 1

( ) ( )

( ) ( )

( )

1( ) ( ) ( ( )

( )

p p

q q

q q qp q p q

q k

p q k k k

2 1 1 2

11)

1

k

p q p q

The Continued Fraction Method (3)

Suppose we have found nonnegative integers such that

This implies why?

1 1, , ,k k k kp q p q

11 1

1

, 1. Where is even.k kk k k k

k k

p pp q p q k

q q

1 1g.c.d( , ) g.c.d( , ) 1k k k kp q p q

1 1

1 1

1 1

1 1

1

Suppose g.c.d.( , ) 1

Let g.c.d.( , ) 1, , , g.c.d.( , ) 1

1

1

( ) 1

( 1) ( 1)

Contradiction exist

Similarly, we can prove g.c.d.( ,

k k

k k k k

k k k k

k k

k k

k k

k

p q

p q k p ak q bk a b

p q p q

akq p bk

k aq bp

k aq bp

p

1) 1kq

The Continued Fraction Method (4)

We find the largest integer such that

We define

If then the sequence stop, otherwise we find the largest such that

We define and so on…… We can repeat the iteration and find the sequence

It turns out that this sequence is the same as the sequence of convergents of real number !

t 1

1

k k

k k

p tp

q tq

1 1 1 1: ; :k k k k k kp p tp q q tq

1 1/k kp q t

1

1

k k

k k

p up

q uq

1 1 1 1 1 1( ) ( ) 1

Which implies g.c.d( , )=1 !k k k k k k k k k k k k k k

k k

p q p q p q tq p tp q p q p q

p q

2 1 2 1: ; :k k k k k kp p up q q uq

31 2

1 2 3

, , , pp p

q q q

Proof We use to denote the term with respect to

First we prove when Prove by induction

Then we prove

Prove by induction

1

1

11 ( ) ( )i i

i i

p qi

q p

( )i

i

p

q -thi

( ) ( )kk

k

pc

q

0 1

Some Properties ofSequence

Denominators are monotonically increasing

For any real numbers and with , one of the convergents satisfy the Dirichlet’s theorem

Proof: Let be the last convergent for which holds. Then

The sequence converge to Proof by induction

/i ip q

1 1 1

1 1 1 1

1 1k k k k k k

k k k k k k k k

p p p q p q

q q q q q q q q

1 2 3, , ,...q q q

12

1 1

1 1k k k

k k k k k k

p p p

q q q q q q

0,0 1 /i ip q

1 and 1p

qq q

/k kp q 1kq

1

1k

k k k k

p

q q q q

/i ip q

Algorithm of Continued Fraction Method

Initially . Suppose then we compute by

using the following rule: If k is even and , subtract times the second column of

from the first column; If k is odd and , subtract times the first column of

from the second column; The matrices is in the following form:

The found in this way are the same as in the convergents

Proved by induction

0

1

: 1 0

0 1

A

:k k

k k k

k k

A

1kA

0k /k k 1kA

0k /k k 1kA

1 1 2 3 2 3 4

1 1 2 3 2 3 4

1 1 2 3 2 3 4

1

1 0 , 0 , , ,

0 1 1

q q q q q q q

p p p p p p p

0 1 2, , ,...A A A

,k kp q ,k kp q

Time complexity of Continued Fraction Method Corollary. Given rational number , the continued fraction

method finds integers and as described in Dirichelet’s theorem in time polynomially bounded by the size of

Proved similar to Euclidean algorithm Theorem. Let be a real number, and let and be natural

numbers with . Then occurs as convergent for

Corollary. There exist a polynomial algorithm which, for given rational number and natural number M, tests if there exists a rational number with . If so, finds this rational number.

p q

0 p q2/ 1/ 2p q q /p q

/p q

2/ 1/ 2p q M

Summary Given a real number , there exist

a rational number with small that is close enough to

Continued fraction method compute a rational number that equals to if is a rational number. Otherwise converge to

The algorithm for continued fraction method is a polynomial Euclidean-like algorithm

/p q

q

/p q /p q

Basis Reduction in Lattices - Overview

Problem: Given a lattice (represented by its basis), finds a reduced “short” (nearly orthogonal) basis.

Applications: Finding a short nonzero vector in a lattice Simultaneous Diophantine approximation Finding the Hermite normal form Basis reduction has numerous applications in

cryptanalysis of public-key encryption schemes: knapsack cryptosystems, RSA with particular settings, and so forth

Basic Concepts Review Lattice. Given a sequence of vectors

, and a group we say generate if . We call a lattice and the basis of . In other words, a lattice can be seen as an integer linear combinations of its basis. It is a subset of the subspace generated by its basis.

A matrix can be seen as a sequence of column (row) vectors, therefore a lattice can be generated by columns (rows) of a matrix

1 2, ,..., ma a a1 2, ,..., ma a a

1 1 2 2 1... | ,...,m m ma a a 1 2, ,..., ma a a

Basic Concepts Review - 2 Let A and B both be a nonsingular matrix of order n, and whose column

both generate the same lattice , then and this is called the det of lattice . In other words, det is independent to chose of basis

Proof: Lemma 1: If B is obtained by interchanging two columns (rows) of A, th

en det B = -det A. Proof: Complicated (component-wise) proof by induction

Lemma 2: If A has two identical columns (rows), then det A = 0. Proof: Let A be a matrix with two identical rows, let B be a matrix constructe

d from A by interchanging these two column (rows). Then det B = det A because these two matrices are equal. However, from Lemma 1 we know that det B = -det A. So det B = det A = 0

Lemma 3: The determinant of an nxn matrix can be computed by expansion of any row or column.

Also called Laplace Expansion Theorem, component-wisely proved by Laplace.

Lemma 4: If B is obtained by multiplying a column (row) of A by k, then det B = k det A.

Proof. We can calculate det B by expanding the same column (row) of B as that of A, which yields det B = k det A.

det det A B

Basic Concepts Review - 3 Lemma 5: If A, B and C are identical except that the i-th colu

mn (row) of C is the sum of the i-th columns (rows) of A and B, then det C = det A + det B.

Proof. We can calculate det B by expanding the i-th column of C, then we can prove det C = det A + det B by using the distributivity of multiplication of matrices

Lemma 6: If B is obtained by adding a multiple of one column (row) i of A to another column (row) j, then det B = det A.

Proof. Let A’ be the matrix that constructed by replacing column (row) i of A to j, then det A’ = 0 because A’ has two identical columns. Matrix A, A’ and B satisfy Lemma 5 so that det B = det A + det A’ = det A

Lemma 7: If If B is obtained by elementary column operations from A, then |det B| = |det A|.

Proof. Directly from Lemma 1, 4 and 6.

From chapter 4, we know that if matrix A and B generate the same lattice then they have the same Hermite Normal Form by elementary column operations, therefore from Lemma 7 we have |det B| = |det A|.

Geometric Meaning of Determinant

The determinant of corresponds to the volume of the parallelepiped

Where is any basis for

Hadamard Inequality theorem:

When are orthogonal to each other, the equality holds.

We now have the lower bound of , what about the upper bound?

Hermite showed that Minkowski showed that

Schnorr proved that for each fixedthen there exist a polynomial algorithm finding a basis satisfying

1 1 2 2... | 0 1 for 1,...,n n ib b b i n

1,..., nb b

1 2det , where denotes the Euclidean norm Tnb b b x x x

1 2 nb b b

( 1) / 41 2 (4 / 3) det n n

nb b b

/ 21 2 (2 / ) det (2 / ) det n

n nb b b n V n e

1,..., nb b

0

( 1)1 2 (1 ) det n n

nb b b

Basis Reduction Theorem A matrix is called positive definite if

There exist a polynomial algorithm which, for given positive definite rational matrix D, finds a basis

for the lattice satisfying ‖b1‖ ‖b2‖…‖bn‖≤ where ‖x‖

We prove this theorem by showing the LLL algorithm

for all 0, 0Tx x Ax

1 2, ,..., nb b b n( 1) / 42 det n n D

: Tx Dx

The Lenstra, Lenstra and Lovász Algorithm We construct a series of basis for as follows: The first basis is the unit basis. We construct the next basis inductively using the following

steps: 1. Denote as the matrix with columns , we

calculate

2.

3. Choose, if possible, an index i such that ‖b2*‖2>2‖b*

i+1‖2. Exchange bi and bi+1, and start with step 1 again. If no such i exists, the algorithm stops.

n

iB 1 2, ,..., nb b b* 1

1 1 1 1( )T Ti i i i i i ib b B B DB B Db

1 1

* * *1 1

for 2,...,

for 1, 2,...,1

1/ 2

i i i i

i i j j

i n

j i i

b b b b

b b b

The Lenstra, Lenstra and Lovász Algorithm - Continued The LLL algorithm is an approximation of the Gra

m-Schmidt orthogonalization process which finds a orthogonal basis in a subspace of

The LLL algorithm terminates in polynomial time, with intermediate numbers polynomially bounded by the size of D

Complicated proof see p.68 – p.71

n

Finding a Short Nonzero Vector in a Lattice In 1891, Minkowski proved a classical result: any n-dimensional latti

ce contains a nonzero vector b with

where denotes the volume of the n-dimensional unit ball. However, no polynomial algorithm finding such a vector b is known.

With the basis reduction method, by taking the shortest vector one can find a “longer short vector” in a lattice, which satisfy

However, this vector is generally not the shortest one in the lattice

The CVP (Closest Vector Problem): “Given a lattice and vector a, find b with (any kind of) norm of b-a as small as possible” is proven to be NP-complete

The SVP (Shortest Nonzero Vector Problem): “Given a lattice, finding a vector in the lattice as small as possible” is even proven to be NP-hard to approximate within some constant [Dan 2001]

1/det 2( ) n

n

bV

nV

( 1) / 4 1/2 (det )n n nb D

Simultaneous Diophantine Approximation Dirichlet showed that Let be real numbers with Then

there exist two integers and q such that

No polynomial method is known for this problem, unless when n=1, where we can use the continued fraction method

However, we can use basis reduction method to find a weaker approximation of the problem in polynomial time

0 1 1 2, ,..., ,n 1 2, ,..., np p p

for 1,..., and 1 nii

pi n q

q q

Finding the Hermite Normal Form Given a matrix A, we can use basis reduction method to calculate ve

ctor and record it in such a way that it can be transform to Hermite Normal Form by elementary column operations

Some of the other applications Lenstra’s Integer Linear Programming algorithm Factoring polynomials (over rationals) in polynomial time Breaking cryptographic codes Disproving Mertens’ conjecture Solving low density subset sum problems

1 2, ,..., nb b b

Summary The continued fraction method for ap

proximating one real number by rational numbers

Lovász’s basis reduction method for finding a short basis in a lattice

Applications

Thank you