Positioning random points on a 2D plane - geometry

Question

So here's a little bit of geometry for you. I've been stuck on this for a while now:
I need to write a script (in C#, but feel free to answer in whatever script you'd like) that generates random points. A points has to values, x and y.
I must generate N points total (where N > 1 and is also randomly up to 100).
point 1 must be x = 0, y = 0. point 2 must be of distance 1 from point 1. So that Root(x2 + y2) = 1.
point 3 must be of distance 1 from point 2 and so on and so forth.
Now here's the tricky part - point N must be of distance 1 from point 1. So if you were to connect all points into a single shape, you'd get a closed shape with each vertices being the same length.
(vertices may cross and you may even have two points at exactly the same location. As long as it's random).
Any idea how you'd do that?

Answer 1

I would do it with simulation of chain there are 2 basic ways one is start from regular polygon and then randomize one point a bit (rotate a bit) then iterate the rest to maintain the segment size=1.
The second one is start with full random open chain (like in MBo answer) and then iteratively change the angles until the last point is on desired distance from first point. I think the second approach is a bit simpler to code...
If you want something more complicated then you can generate M random points and handle them as closed Bezier curve cubic patches loop control points. Then just find N equidistant points on it (this is hard task) and rescale the whole thing to match segment line size = 1
If you want to try first approach then
Regular polygon start (closed loop)
Start with regular polygon (equidistant points on circle). So divide circle to N angular segments. Select radius r so line length match l=1
so r=0.5/cos(pi/N) ... from half angle triangle
Make function to rotate i-th point by some single small step
So just rotate the i-th point around (i-1)th point with radius 1 and then iteratively change the {i+1,...N} points to match segments sizes
you can exploit symmetry to avoid bullet #2
but this will lead not to very random result for small N. Just inverse rotation of 2 touching segments for random point p(i) and loop this many times.
to make it more random you can apply symmetry on whole parts (between 2 random points) instead of on 2 lines only
The second approach is like this:
create randomized open chain (like in MBo's answer)
so all segments are already with size=1.0. Remember also the angle not just position
i-th point iteration
for simplicity let the points be called p1,p2,...pn
compute d0=||pn-p1|-1.0|
rotate point pi left by some small da angle step
compute dl=||pn-p1|-1.0|
rotate point pi right by 2.0*da
compute dr=||pn-p1|-1.0|
rotate point pi to original position ... left by da
now chose direction closer to the solution (min dl,dr,d0) so:
if d0 is minimal do not change this point at all and stop
if dl is minimal then rotate left by da while dl is lowering
if dr is minimal then rotate right by da while dr is lowering
solution
loop bullet #2 while the d=||pn-p0|-1.0| is lowering then change da to da*=0.1 and loop again. Stop if da step is too small or no change in d after loop iteration.
[notes]
Booth solutions are not precise your distances will be very close to 1.0 but can be +/- some error dependent on the last da step size. If you rotate point pi then just add/sub angle to all pi,pi+1,pi+2,..pn points

Answer 2

Edit: This is not an answer, closeness has not been taken into account.
It is known that Cos(Fi)^2 + Sin(Fi)^2 = 1 for any angle Fi
So you may use the next approach:
P[0].X = 0
P[0].Y = 0
for i = 1 .. N - 1:
RandomAngle = 2 * Pi * Random(0..1)
P[i].X = P[i-1].X + Cos(RandomAngle)
P[i].Y = P[i-1].Y + Sin(RandomAngle)