Interpolation of a set of $(x, y)$ coordinates using a parametrization $(x(t), y(t))$ where $0\leq t \leq1$? The question pretty much states it all, but it must also satisfy these conditions:


*

*$x(0) = a$ and $x(1) = b$ where a is the $x$ coordinate of a specified "start point" that is in the set of $(x, y)$ points and b is the x coordinate of a specified "end point" that is in the set of $(x, y)$ points, and similarly, $y(0) = c$ and $y(1) = d$ where c is the y coordinate of a specified "start point" that is in the set of $(x, y)$ points and d is the y coordinate of a specified "end point" that is in the set of $(x, y)$ points. (so basically, plugging in 0 for t gives the start point and plugging in 1 gives the end point)

*The parametrization must be continuous (I'm not sure how else to say this. Basically, the parametrization must be a continuous "path" from the start point to the end point that passes through all points in between)
I'm not fluent in the language of mathematics, so I probably could have stated those conditions much better. In fact, some of them probably don't make sense, so I would be happy to clarify them.
 A: You want to interpolate a finite number of points
$$(x_0,y_0),(x_1,y_1),(x_2,y_2),\ldots,(x_n,y_n)$$
with a continuous curve $(x(t),y(t))$. [I am thinking of $(x_0,y_0)=(a,b)$ as the starting point and of $(x_n,y_n)=(c,d)$ as the ending point. It doesn't matter how you call them.]
A first easy solution would be to glue segments from one point to the next: The straight segment
$$
\ell_j(t)=(x_{j-1}+t(x_{j}-x_{j-1}),y_{j-1}+t(y_{j}-y_{j-1})),\,\,0\leq t\leq1
$$
joins the points $(x_{j-1},y_{j-1})$ and $(x_{j},y_{j})$. Choosing times $t_j$ in $(0,1]$ when we want to arrive at $(x_j,y_j)$ we have the solution
$$
\ell(t):=\left\{\begin{array}{lc}\ell_1(t)&,0\leq t\leq t_1\\
\ell_2(t)&,t_1\leq t\leq t_2\\
\,\,\,\,\vdots&\,\,\,\vdots\\
\ell_n(t)&,t_{n-1}\leq t\leq t_n=1.\end{array}\right.
$$
We actually have a lot of margin here to come up with other solutions. As a start I would have a look at this Wikipedia article. Using Lagrange polynomials you might find polynomials $p(t)$ and $q(t)$ such that $p(t_j)=x_j$ and $q(t_j)=y_j$, and then another solution would be
$$
(p(t),q(t)),\,\,0\leq t\leq1.
$$
Finally notice that if it is the case that $a=x_0<x_j<c=x_n$ for every $j$ and if you don't mind in which order you visit the points $(x_j,y_j)$ then you might reorder your points so that the first coordinate does not decrease from one point to the next, take a function (e.g. a Lagrange polynomial) such that $f(x_j)=y_j$ and then consider
$$
\gamma(t)=(t,f(t)),\,\,x_0\leq t\leq x_n.
$$
This strictly is not a solution as the time domain is not $[0,1]$, but that can be quickly fixed considering $\tilde{\gamma}(t)=\gamma(x_0+t(x_n-x_0)),\,0\leq t\leq0$.
[Obviously this last solution can also be used if $b=y_0<y_j<d=y_n$ for every $j$ and if you don't mind in which order you visit the points $(x_j,y_j)$. The same argument applies, only this time we will have something like $\gamma(t)=(f(t),t)$. Or if "the case that  $a=x_0<x_j<c=x_n$ for every $j$ and if you don't mind ..." holds with switched inequality signs, for if $\gamma(t),\,0\leq t\leq1$, is a path interpolating $(x_n,y_n),\ldots,(x_1,y_1),(x_0,y_0)$ then $\tilde{\gamma}(t):=\gamma(1-t),\,0\leq t\leq 1$ does the same work backwards.]
