Talk:Lenstra–Lenstra–Lovász lattice basis reduction algorithm

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Requested move

I attempted to clarify text related to some issues:

There is not 'the' shortest vector, since there are always at least two of them
One has to be very careful about stuff like "2nd shortest vectors", as such talk is often incorrect for dimension >= 5. For example, consider the parity lattice defined by $\{(x_{i})\mid x_{i}\sim x_{j}\mod 2\forall i,j\}$ . It has $n$ linearly independent vectors of length 1, yet its last successive minimum $\lambda _{n}={\sqrt {n}}>1$ (once the dimension n is >= 5). 131.234.72.252 (talk) 13:52, 16 June 2008 (UTC)[reply]
I apologize, what I wrote yesterday in a hurry is obviously incorrect. The parity lattice has a full set of linearly independent vectors of length 2 (not 1), and there are no shorter nonzero vectors in the lattice (for dimension >= 5). However, those vectors do not form a lattice basis, because the vectors of odd parity cannot be written as integer linear combinations of vectors of even parity. 89.245.87.65 (talk) 06:24, 17 June 2008 (UTC)[reply]