Find a parametrization $\textbf{r}(u)$ of the unit circle $C$ in the plane such that the parameter values $u=0,1,2$ correspond to the points $(1,0), (0,1), (0,-1)$ on $C$. Indicate corresponsing "re-parametrization function" $u(\theta)$ where $r(\theta) = <\cos > \theta, \sin \theta>$ is the standard parametrization of $C$.
I started by finding:
$\theta = 0$ for $(1,0)$
$\theta = \frac{\pi}{2}$ for $(0,1)$
$\theta = \frac{3\pi}{2}$ for $(0,-1)$
I thought of a simple factor $\textbf{r}(u)=<\cos (u\theta),...>$ but failed as $u$ does not increase linearly with $\theta$.
To map $0\to 0$, $1\to \pi/2$, $2\to 3\pi/2$ one could set $\theta= \frac{\pi}{4}u(1+u)$, the interpolating polynomial of lowest order. This is for example obtained by Lagrange interpolation, see https://en.wikipedia.org/wiki/Polynomial_interpolation#Constructing_the_interpolation_polynomial . $x=\cos(\theta)$, $y=\sin(\theta)$.
A different adaptation is that $\theta$ should be a periodic function of $u$. So we may specify in addition that $u=3$ wraps around to $\theta=0$, assuming that the values at $u=0,1,2,3,4,5,\ldots$ are $\theta=0,\pi/2,3\pi/3,0,\pi/2,3\pi/2,\ldots$ . A discrete Fourier analysis of that yields $$ \theta=\frac{\pi}{2}[\frac{4}{3} - \frac{4}{3}\cos(2\pi u/3)-\frac{2}{3}\sqrt{3}\sin(2\pi u/3)]. $$ See https://en.wikipedia.org/wiki/Discrete_Fourier_transform for a tutorial.