>Are algorithms already known, that generate (arbitrarily good approximations of) *random* curves, w.l.o.g. with unit length, and joining endpoints $(0,0)$ and $(\alpha,0)$ with $\alpha \lt1$ given?   

The fixed distance between the endpoints is essential for the question, because otherwise a simple rescaling of an arbitrary curve would work.  
<br>
**edit**  

In view of the comments and the answer of [Bjørn Kjos-Hanssen](https://mathoverflow.net/users/4600/bj%c3%b8rn-kjos-hanssen), I see the need for some clarification:  

- By *random curve of unit length connecting $(0,0)$ and $(\alpha,0)$*, I mean a random sample from the space of all such curves; that means, that the algorithm should be capable to approximate *every* such curve to arbitrary precision with a finite number of steps.  
So "random" is not restricted to the appearance of the curve.

- Being able to generate Brownian Bridges is not sufficient, because I would like the algorithm to be able to generate *curves* (ideally in any $\mathbb{R}^n$) and not only *functions*.  

So my apologies for not being precise enough.  
<br>
I have used the formulation "are algorithms *already known*", because I have found one, that seems to be able to produce all those curves.  
  
I will provide details in a later edit.  

<br/>
**Here are the promised details:**


 

the algorithm, that motivated this question is essentially based on realizing, that *no* point of the curve can lie outside the [ellipse](https://en.wikipedia.org/wiki/Ellipse) centered at $\left(\frac{\alpha}{2},0\right)$, foci $p$ at $\left(0,0\right)$ and $q$ at $\left(\alpha,0\right)$, for which the length of the semi-major axis equals $\frac{1}{2}$ and, $\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{\alpha}{2}\right)^2}$ for the semi-minor axis.  
If the intermediate curve-point $r$ is chosen from the boundary of that ellipse, then the "length-stock" is used up and the algorithm terminates with a curve consisting of two line-segments and exact length $1$, joining $p$ and $q$ as demanded.  

Otherwise the length-stock is split up and assigned to two newly generated line-segments and the original problem of finding a curve of fixed length with endpoints at fixed has to be solved recursively for both segments separately.  

>**Pseudo code:**

>$\text{expand}$(Point $p$, Point $q$, Length $\ell$, Curve curve)  
<br>$\quad$Point $r\in \lbrace x\in\mathbb{R}^n\ |\ \| r-p \| + \|q-r\|\ \le \ell\rbrace$;
<br>$\quad$Length $\ell_{pr}$ := $\|r-p\|$;
<br>$\quad$Length $\ell_{rq}$ := $\|q-r\|$;
<br>$\quad$Length $\Delta\ell$ := $\ell-\left(\ell_{pr}+\ell_{rq}\right)$;
<br>$\quad$Scalar $a\in\left[0,1\right]$
<br>
<br>$\quad$if (a < threshold)
<br>$\quad\quad$curve.append($r-p$);
<br>$\quad$else
<br>$\quad\quad\text{expand}$($p$,$\ r$,$\ \ell_{pr}$+$a$ * $\Delta\ell$);
<br>
<br>$\quad$if ($1-a$ < threshold)
<br>$\quad\quad$curve.append($q-r$);
<br>$\quad$else
<br>$\quad\quad\text{expand}$($r$,$\ q$,$\ \ell_{rq}$+(1-$a$) * $\Delta\ell$);  

<br/ >
Some remarks:  

the pseudo code is aimed at full generality and also covers "degenerate" cases; those need to be ruled out by further checks. One such case is the collinearity of $p$, $\ q$ and $r$ with positive $\Delta\ell$.  

Selecting $r$ from the mentioned elliptical regions with foci $p$ and $q$ can also be interpreted as chosing one of the intersection points of a circle around $p$ and $q$. That covers the algorithm of Matt F. as a special case.  

The followup challenge is now to  control further properties of the curve via taylored rules for selecting $r$ and distributing $\Delta\ell$ on each recursion level.  
Or, play with the options to discover interesting curves and fractals.