User:Florian Xaver/sandbox

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Gauss and Minkowski reduction (abstract)
A reduced version of a quadratic form is a canonical representative that has the same image as the original quadratic form with some conditions on its coefficients. In the 2D and 3D case of lattices, which are closely related to quadratic/ternary forms, finding the minimal basis vectors which are desired to be at most orthogonal yields the same result that is provided by the reduced form. This follows from the fact that the quadratic norm of a lattice is a special case of the quadratic/ternary form and hence links both theories together.

To this end, the Gauss-reduced two-dimensional or three-dimensional lattice basis is a special case of a reduced binary/ternary quadratic form. The Minkowski reduction is a generalization of the Gaussian reduction for higher dimensions, i.e. n-ary quadratic forms or n-dimensional basis.

Generally, a lattice in $$\R^n$$ is defined by

\Lambda = \left\{ a_1 b_1 + \ldots + a_n b_n \; | \; a_i \in\mathbb{Z} \right\} $$ where $$b_i \in \R^m, m \ge n$$, are the basis vectors of $$\Lambda$$. The rank of a lattice is defined by $$n$$ and $$d(\Lambda) := \sqrt{ \det B^\mathrm{T} B} $$ is the discriminant. It corresponds to the volume spanned by the basis vectors (fundamental parallelopiped).

Forms
=== Quadratic Forms  ===

A quadratic form is a homogeneous polynomial of degree two, ie.
 * $$q(x_1,\ldots,x_n) = \sum_{i,j=1}^{n} q_{ij}{x_i}{x_j} = x^\mathrm{T} Qx = (q_{11},\ldots,q_{nn}) $$

where $$q_{ij} \in \Z$$ are its coefficient, $$x \in \Z^n$$ and

Q=\begin{pmatrix} q_{11} & \ldots & q_{1n} \\ \vdots & \ddots & \vdots \\ q_{n1} & \ldots & q_{nn} \end{pmatrix} $$. For $$n=2$$ and $$n=3$$ it is termed a binary and a ternary quadratic form, respectively.

Two forms $$q_1$$ and $$q_2$$ are called equivalent, if the image sets of both are identical. Furthermore, the equivalence class
 * $$[q] = \{ q_1 \in Q_n \mid q_1 \sim q \}$$

where $$Q_n$$ is the set of all n-ary forms.

A reduced form is defined as a canonical form representative for such an equivalence class $$[q]$$. The Gauss-reduced form is a canoncial form of a binary or ternary quadratic form. Others exist for $$n > 2$$.

The minimum $$\lambda$$ of a form is defined by $$\lambda := \min\{\sqrt{|q(x_1,\ldots,x_n)|}: (x_1,\ldots,x_n)\in\Z^n, (x_1,\ldots,x_n)\neq(0,0,\ldots,0)\}\,$$.

=== Binary and Ternary Reduced Forms (Gauss Reduced Form) ===

For every positive binary quadratic form there exists an equivalent reduced form

q(x, y) = a x_1^2 + 2 bx_1x_2 + cx_2^2 $$ such that
 * $$ |b| \le a, \qquad 0 < a \le c $$.

For every positive ternary quadratic form $$ q(x, y) = \sum_{i=1}^3 q_{ii} x_i^2 + \sum_{i\neq j, i < j}  q_{ij} x_i x_j $$ there exists an equivalent, reduced form such that
 * $$ q_{11} \le q_{22} \le q_{33} $$ ,
 * $$ | q_{jk} | \le q_{jj}, \qquad j < k \le 3 $$

and

-q_{12} -q_{13} -q_{23} \le q_{11} + q_{22} $$,

-q_{12} +q_{13} +q_{23} \le q_{11} + q_{22} $$,

q_{12} -q_{13} +q_{23} \le q_{11} + q_{22} $$,

q_{12} +q_{13} -q_{23} \le q_{11} + q_{22} $$.

=== Successive Minima ===

The successive minimum $$\lambda_k$$ is the smallest real value $$\lambda'$$ for what there are $$k$$ linear independent lattice vectors with the norm $$ \le \lambda'$$. We have $$\lambda_n \ge \cdots \ge \lambda_1 $$.

Lattice vectors with minimal quadratic norm $$\lambda_k, k = 1, \cdots, 3$$ form a basis for $$n\le 3$$.

== Gaussian Reduced Lattices ==

Rank-2 Lattice
Assume a basis matrix $$B$$ of a rank-2 lattice. Then the length of any lattice vector $$v = B k, k \in \Z^2$$ is $$  {||  v  ||}_2^2  = k^\mathrm{T} Q k$$. We can write

Q = \begin{pmatrix} a & b \\ b & c \end{pmatrix}, k = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} $$ and thus
 * $$ {|| v  ||}_2^2 =  q ( x_1, x_2 ) = a x_1^2 +  2 b x_1  x_2 + c x_2^2$$.

If $$a,b,c \in \Z$$ then $$q(x_1,x_2)$$ is a binary positive quadratic form. If $$q(x_1,x_2)$$ satisfies the conditions of a Gauss reduced form (see above), the bounds of Gauss reduced forms lead

$$. The basis is termed Gauss reduced. The basis is not unique because the norm is the same for $$\pm b_i$$. Additionally we have $$|| b_i || = \lambda_i$$.
 * b_1 ||_2 \le || b_2 ||_2, |b_1 \cdot b_2 |\le \frac{1}{2} || b_1 ||_2^2

=== Rank-3 Lattice ===

For $$n = 3$$ the length of any lattice vector $$v = B k, k \in \Z^3$$ is $$  {||  v  ||}_2^2  = k^\mathrm{T} Q k$$ and

Q=\begin{pmatrix} q_{11} & q_{12} & q_{13} \\ q_{21} & q_{22} & q_{23} \\ q_{31} & q_{32} & q_{33} \end{pmatrix}, k = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} $$. Therefore
 * $$ {|| v  ||}_2^2 =  q ( x_1, x_2, x_3 ) $$.

This gives the definition of a Gauss reduced lattice of rank 3:

$$,
 * b_1 ||_2 \le || b_2 ||_2 \le || b_3 ||_2

$$,
 * b_1 \cdot b_2 | \le || b_1 ||_2, \qquad | b_1 \cdot b_3 | \le || b_1 ||_2, \qquad |  b_2 \cdot b_3 | \le || b_2 ||_2

-b_1 \cdot b_2 -b_1 \cdot b_3 -b_2 \cdot b_3 \le || b_1 ||_2^2 + || b_2 ||_2^2 $$,

-b_1 \cdot b_2 +b_1 \cdot b_3 +b_2 \cdot b_3 \le || b_1 ||_2^2 + || b_2 ||_2^2 $$,

b_1 \cdot b_2 -b_1 \cdot b_3 +b_2 \cdot b_3 \le || b_1 ||_2^2 + || b_2 ||_2^2 $$,

b_1 \cdot b_2 +b_1 \cdot b_3 -b_2 \cdot b_3 \le || b_1 ||_2^2 + || b_2 ||_2^2 $$.

An Algorithm for Gauss reduction
The algorithm to achieve Gaussian basis vectors in the 2-dimensional case is very simple:

begin Gaussian reduction repeat $$b_\text{tmp} := b_1$$ $$b_1 := b_2$$ $$b_2 := b_\text{tmp}$$ $$b_2 := b_2 - \left\lfloor \frac{b_1^\mathrm{T} b_2}{ || b_1 ||_2^2 }\right\rceil b_1$$ while $$\|b_1\|_2 > \|b_2\|_2$$ result: $$b_1,b_2$$ end

The main step in the algorithm subtracts multiples of the $$b_1$$ from the $$b_2$$ (the vector $$b_1$$ is the shorter vector of both). If this subtraction does not make $$b_2$$ smaller then $$b_1$$ the algorithm determines. The determination is for sure, because at one point the vectors cannot become smaller. The determination criterial ensures that the first condition of a Gaussian-reduced lattice is fulfilled and the modification step let the result satisfy the second condition.

Let us give an example. We choose the initial vectors as $$b_1=(3, 1)^\mathrm{T} $$ and $$b_2=(7, 3)^\mathrm{T} $$. In the first iteration we obtain that vector $$b_2$$ is modified to $$b_2 - 2 b_1$$ and hence $$b_2=(1, 1)^\mathrm{T} $$. Clearly this new vector is smaller than $$b_1$$ and therefore we have to do another iteration where we first swap the two vectors and obtain $$b_1=(1, 1)^\mathrm{T} $$ and  $$b_2=(3, 1)^\mathrm{T} $$. After the subtraction the new vector $$b_2$$ becomes $$(1, -1)^\mathrm{T} $$ since the subtraction factor is again 2. We see that both vectors are now equal in length and we are done. It is also easy to see that all the conditions of the Gaussian reduced basis are satisfied.

The proposed algorithm has complexity of $$\mathcal{O}(\log w)$$ where $$w=\frac{\|b_1\|_2 \|b_2\|_2}{d(\Lambda)} \ge 1$$ is the orthogonality defect of the input lattice. $$w=1$$ denotes orthogonality.

Minkowski Reduction
Gauss only considered reduction of forms up to 3rd order. For higher order lattices loser reduction conditions have to be used. Therefore Minkowski-reduced bases were introduced that have to fulfill that $$\|v\| \geq \|b_i\|$$ for all $$v \in \Lambda$$ for which $${b_1,b_2, \dots b_{i-1},v}$$ form a basis of $$\Lambda$$.

Minkowski-reduced bases are not unique, on the one hand the negative of the individual basis vectors form also a Minkovsky-reduced bases and on the other hand different ambiguities may occur that affect subsequent bases vectors.

For the 2-norm it is important to mention that Minkovsky-reduced bases are in many cases equal to the successive minima of a lattice, but this does not hold in general. Waerden's Theorem gives an upper bound on the possible difference.