# rough bound on length closest vector in a lattice

I'm reading "Complexity of Lattice Problems" by Micciancio and Goldwasser. Currently I am reading how the decision CVP problem is equivalent to the search CVP problem. Let $$B = [b_1,...,b_n]$$ be a set of linearly independent vectors in $$\mathbb R^m$$ and consider the lattice $$\mathcal{L}(B)$$. Furthermore, consider a target vector $$t$$. I struggle to see why $$R = \sum_{i = 1}^{n}||b_i||^2$$ is an upper bound on the squared distance from $$t$$ to the lattice (ie the squared distance from $$t$$ to the closest vector to $$t$$ in the lattice). I think this should be kind of obvious but I'm unfortunately not seeing it. Thanks!

This is true if $$t$$ lies in the subspace spanned by $$B$$. For example, $$B = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$ and $$t = \begin{pmatrix} 0 \\ 10 \end{pmatrix}$$ is a counter example. So assuming that $$t\in span(B)$$, $$t$$ lies in a fundamental region defined by the basis $$B$$ so that the distance of $$t$$ to the lattice is bounded by the maximum distance of any two points of this fundamental region. For any two points $$x$$ and $$y$$ in a fundamental region, we have $$x-y = \sum b_iz_i \ , \ \text{ with } |z_i| \leq 1 \ .$$ Thus $$\|x-y\|^2 = \|\sum b_iz_i\|^2 \leq \sum z_i^2\|b_i\|^2 \leq \sum \|b_i\|^2 \ .$$