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3 small grammer check

Excellent question. I don't know the answer and perhaps what I am suggesting is obvious. Nevertheless, I think it's on the right track.

Let $B$ be a basis for $\Lambda$ and let $P$ be the corresponding (origin centred) fundamental parallelepiped. That is, $P$ is the region given by $Bu$ where $u \in [-0.5,0.5]^n$. Clearly $2P \cap \Lambda$ contains representatives of $\Lambda/2\Lambda$. However, the elements in $2P \cap \Lambda$ are generally not the minimal length representatives for $\Lambda/2\Lambda$, those are contained in $2\mathcal{V} \cap \Lambda$ where $\mathcal{V}$ is the (closed) Voronoi cell of $\Lambda$.

Let $d$ be the smallest real number such that $\mathcal{V} \subseteq dP$. Clearly $2dP \cap \Lambda$ is a superset of $2\mathcal{V} \cap \Lambda$ and a suitable value of $c$ is therefore $\lfloor d \rfloor$. We can refine this a little by asking for the diagonal matrix $D$ such that $\mathcal{V} \subseteq DP$. Then we have $n$ different $c$'s, specifically $c_i = \lfloor d_i \rfloor$ where $d_i$ are the diagonal elements of $D$.

The problem is now to find the value $d$ (or matrix $D$). LooselyThat is, how much do we need to scale the fundamental parallelepiped so that it completely contains the Voronoi cell? Perhaps if $R$ is suitably reduced, say LLL reduced or Korkine–Zolotareff reduced, then bounds on $d$ can be found?

2 Added a sentence to aid intuitive understanding

Excellent question. I don't know the answer and perhaps what I am suggesting is obvious. Nevertheless, I think it's on the right track.

Let $B$ be a basis for $\Lambda$ and let $P$ be the corresponding (origin centred) fundamental parallelepiped. That is, $P$ is the region given by $Bu$ where $u \in [-0.5,0.5]^n$. Clearly $2P \cap \Lambda$ contains representatives of $\Lambda/2\Lambda$. However, the elements in $2P \cap \Lambda$ are generally not the minimal length representatives for $\Lambda/2\Lambda$, those are contained in $2\mathcal{V} \cap \Lambda$ where $\mathcal{V}$ is the (closed) Voronoi cell of $\Lambda$.

Let $d$ be the smallest real number such that $\mathcal{V} \subseteq dP$. Clearly $2dP \cap \Lambda$ is a superset of $2\mathcal{V} \cap \Lambda$ and a suitable value of $c$ is therefore $\lfloor d \rfloor$. We can refine this a little by asking for the diagonal matrix $D$ such that $\mathcal{V} \subseteq DP$. Then we have $n$ different $c$'s, specifically $c_i = \lfloor d_i \rfloor$ where $d_i$ are the diagonal elements of $D$.

The problem is now to find the value $d$ (or matrix $D$). Loosely, how much do we need to scale the fundamental parallelepiped so that it completely contains the Voronoi cell? Perhaps if $R$ is suitably reduced, say LLL reduced or Korkine–Zolotareff reduced, then bounds on $d$ can be found?

1

Excellent question. I don't know the answer and perhaps what I am suggesting is obvious. Nevertheless, I think it's on the right track.

Let $B$ be a basis for $\Lambda$ and let $P$ be the corresponding (origin centred) fundamental parallelepiped. That is, $P$ is the region given by $Bu$ where $u \in [-0.5,0.5]^n$. Clearly $2P \cap \Lambda$ contains representatives of $\Lambda/2\Lambda$. However, the elements in $2P \cap \Lambda$ are generally not the minimal length representatives for $\Lambda/2\Lambda$, those are contained in $2\mathcal{V} \cap \Lambda$ where $\mathcal{V}$ is the (closed) Voronoi cell of $\Lambda$.

Let $d$ be the smallest real number such that $\mathcal{V} \subseteq dP$. Clearly $2dP \cap \Lambda$ is a superset of $2\mathcal{V} \cap \Lambda$ and a suitable value of $c$ is therefore $\lfloor d \rfloor$. We can refine this a little by asking for the diagonal matrix $D$ such that $\mathcal{V} \subseteq DP$. Then we have $n$ different $c$'s, specifically $c_i = \lfloor d_i \rfloor$ where $d_i$ are the diagonal elements of $D$.

The problem is now to find the value $d$ (or matrix $D$). Perhaps if $R$ is suitably reduced, say LLL reduced or Korkine–Zolotareff reduced, then bounds on $d$ can be found?