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Lattice reduction

# Overview

Having introduced the LLL reduction, we now provide a more general notions of a reduced basis for a lattice as well as provide bounds for the basis vectors. The key idea behind introducing these definitions is that once we know some basis vector is []-reduced, we can bound the sizes of the basis, which is important when algorithms require short vectors in a lattice. For fast algorithms, LLL-reduction is typically the most important notion as it can be computed quickly. Two main definitions appear often when discussing lattice reductions, which we will provide here.

# Definitions

A basis
$\left\{b_i\right\}_{i=1}^d$
is size-reduced if
$\left|\mu_{i,j}\right|\leq\frac12$
. Intuitively this captures the idea that a reduced basis being "almost orthogonal".
Let
$L$
be a lattice,
$1\leq i\leq\dim L=d$
, we define the
$i^\text{th}$
successive minima
$\lambda_i(L)$
as
$\lambda_i(L)=\min\left(\max_{1\leq j\leq i}\left(\left\lVert v_j\right\rVert\right):v_j\in L\text{ are linearly independent}\right)$
Intuitively,
$\lambda_i(L)$
is the length of the "
$i^\text{th}$
shortest lattice vector". This intuition is illustrated by the definition of
$\lambda_1$
:
$\lambda_1(L)=\min\left(\left\lVert v\right\rVert:v\in L\right)$
However this is not precise as if
$v$
is the shortest lattice vector, then
$-v$
is also the shortest lattice vector.
Unfortunately, a basis
$b_i$
for
$L$
where
$\lambda_i(L)=\left\lVert b_i\right\rVert$
for dimensions
$5$
and above. This tells us that we can't actually define "the most reduced basis" in contrast to the 2D case (see Lagrange's algorithm) and we would need some other definition to convey this intuition.
An alternate definition of
$\lambda_i(L)$
that will be helpful is the radius of the smallest ball centered at the origin such that the ball contains at least
$i$
linearly independent vectors in
$L$
.

# Exercises

1) Show that both definitions of
$\lambda_i$
are equivalent
2) Consider the lattice
$L=\begin{pmatrix}2&0&0&0&0\\0&2&0&0&0\\0&0&2&0&0\\0&0&0&2&0\\1&1&1&1&1\end{pmatrix}$
. Show that the successive minima are all
$2$
but no basis
$b_i$
can satisfy
$\left\lVert b_i\right\rVert=\lambda_i$
.