Are there always finitely many sinks in a lattice with a piece metric?

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I should preface that I don’t know if this fits within some established theory or not, so the terminology I use may not be canonical.

Terms and definitions:

A Piece $P$ is an $n\times m$ matrix.

A Move $M$ is a $m\times 1$ with integral coefficients.

A Lattice $L$ is the span of the columns of $P$, or $\{PM:M\in \mathbb{Z}^m\}$

The Distance $D$ of a vector $V$ in $L$ is $\min(\{|M|:PM=v\})$ Where $|M|$ is the taxicab metric on $M$

A Sink $S$ in $L$ is a vector $v$ such that $D(V+PM)<D(V)+D(PM)$ $\forall M \neq 0$

Motivation:

If all the vectors in $P$ are integral linearly independent, then L has no sinks because it is isomorphic to $\mathbb{Z}^{m}$ with the taxicab metric.

If $P$ is the set of Knight’s moves, as in chess, then the only sinks are at (2,2), (2,-2), (-2,2), and (-2,-2)

If there is some $K\in \mathbb{Z}$ such that $\forall M$ such that $|M|=K$, $PM$ is not a sink, then $PM:|M|>K$ is not a sink.

The only issue is that there may not be such a $K$. If there is such a case with infinitely many sinks, what is the smallest value of M where this is the case?