Generating random curves with fixed length and endpoint distance

Question

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.

Edit

In view of the comments and the answer of Bjørn 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.

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.

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 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)

$\quad$Point $r\in \lbrace x\in\mathbb{R}^n\ |\ \| r-p \| + \|q-r\|\ \le \ell\rbrace$;
$\quad$Length $\ell_{pr}$ := $\|r-p\|$;
$\quad$Length $\ell_{rq}$ := $\|q-r\|$;
$\quad$Length $\Delta\ell$ := $\ell-\left(\ell_{pr}+\ell_{rq}\right)$;
$\quad$Scalar $a\in\left[0,1\right]$

$\quad$if (a < threshold)
$\quad\quad$curve.append($r-p$);
$\quad$else
$\quad\quad\text{expand}$($p$,$\ r$,$\ \ell_{pr}$+$a$ * $\Delta\ell$);

$\quad$if ($1-a$ < threshold)
$\quad\quad$curve.append($q-r$);
$\quad$else
$\quad\quad\text{expand}$($r$,$\ q$,$\ \ell_{rq}$+(1-$a$) * $\Delta\ell$);

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.

Answer 1

Here are some random "scribbles," based on Bjørn Kjos-Hanssen's idea (but not following his specifications exactly), mixing with Izaak Meckler's comment:

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What you see is points of a random walk fit with cubic Bézier curves with $C^1$ continuity.

Answer 2

One approach is to let the midpoint of the curve be a random point within (curve length)/2 of both of the starting point and the end point, and then iterate. For curves of length 1 between (0,0) and (1/2,0) that gives results like this:

In more detail, to connect a starting point and ending point by a curve of length $2c$, draw circles of radius $c$ about both points, and randomly pick a point in the intersection to be the midpoint of the curve. So from the points at distance 0 and 1 on the curve, calculate the point at distance 1/2, and then the points at distance 1/4 and 3/4, etc. For the pictures above, I went down to distance 1/1024, picking points in the inscribed rhombus rather than the curved shape to simplify the algebra.

I have no reference for this, but I believe that with probability 1 it generates curves of length 1. I've attached the Mathematica code for a curve in a comment if you want to play with it.

Answer 3

How about starting with a Brownian bridge $B_t$; then make it smooth and having finite length by convolution $g_t=B_t*f$; and then multiplying by a constant to get a unit length curve $c g_t$?

Answer 4

Defining $z_t(s)=\int_0^s e^{itu(\sigma)}\ d\sigma$ transforms any function $u$ : $(0,1) \to \mathbb R$ into a one-parameter family of length $1$ curves $C_t=\{z_t(s):\ 0\leq s\leq1\}$ whose distance between end points should initially decrease from $1$, and shrink to $0$ as $t\to\infty$ (provided $u\neq0$ a.e.).

This may not qualify as an "algorithm" though, since the choice of $t$ to get the proper length remains implicit.