Vector Walk

$\mathbf{P} = n\begin{pmatrix}2\\1\end{pmatrix} + m\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}2n\\n+m\end{pmatrix}$

The end destination tells you how many of each vector were used, but does not include cancelled steps: from the $x$ coordinate we can deduce how many net $b_1$ steps there were; this then allows us to determine how many net $b_2$ steps there must have been.

The vectors can together reach any point with integer $y$ coordinate and even $x$ coordinate: Thus, the set of possible destinations are the points, and only the points, with coordiantes $\begin{pmatrix}2p\\q\end{pmatrix}$ for any integers $p, q$.

Two other base vectors which give rise to exactly the same set of points are
$$
\begin{pmatrix}2\\0\end{pmatrix} \,,\begin{pmatrix}0\\1\end{pmatrix}
$$

Imagine now that we have two other more general base vectors
$$
\mathbf{a} = \begin{pmatrix} x_a \\ y_a\end{pmatrix} \quad \mathbf{b} = \begin{pmatrix} x_b \\y_b\end{pmatrix}
$$
If these are to reach exactly the same set of points we must have, for any integers $N$ and $M$ that
$$
N\begin{pmatrix} x_a \\ y_a\end{pmatrix} +M\begin{pmatrix} x_b\\ y_b\end{pmatrix}\mathbf{b} = \begin{pmatrix} 2p \\ q\end{pmatrix}
$$
Clearly to reach these, and only these points, I would need both $a$ and $b$ to have even $x$ component. Furthermore, if I can reach the points $(2, 0)$ and $(0, 1)$ with combinations of my base vectors then I would be able to reach all of the points $(2n, m)$ by adding together $n$ and $m$ lots of these combinations.

Algebra quickly becomes very complicated, as there are too many variable to give me anything to 'solve'.

So, look at some more particular cases. Look at a base vector with an $x$-component of $4$.
$$
\begin{pmatrix} 4 \\ 0\end{pmatrix}
$$
Clearly, to be able to reach the point $(2, 0)$ the other base vector would need to be $\begin{pmatrix} \pm 2 \\ 0\end{pmatrix}$, but this would not allow us to reach the point $(0, 1)$. Thus, any other possible pairs of base vectors would have to have non-zero $x$ and $y$ components.
A suitable pair would be
$$
\mathbf{a}= \begin{pmatrix} 4 \\ 1\end{pmatrix}\,, \mathbf{b} =\begin{pmatrix}2 \\ 1\end{pmatrix}
$$
To see why, note that the combinations $\mathbf{a}-\mathbf{b}$ and $2\mathbf{b}-\mathbf{a}$ take us to the points $(2, 0)$ and $(0, 1)$.
Another suitable pair would be
$$ \mathbf{a}= \begin{pmatrix}6 \\ 2\end{pmatrix}\,, \mathbf{b} =\begin{pmatrix}4 \\ 1\end{pmatrix} $$ To see why, note that the combinations $2\mathbf{b}-\mathbf{a}$ and $2\mathbf{a}-3\mathbf{a}$ take us to the points $(2, 0)$ and $(0, 1)$.


For base vectors which yield none of the points in the original lattice, we would need irrational components, because if I had a base vector with rational components $\frac{n}{m}$ and $\frac{p}{q}$ then taking $2mq$ lots of this vector would take us to the point $(2qn, 2pm)$, which is in our original lattice of points.