C++ Arcs made from Control Points

Distance from a Point to the Arc

Use halfspace tests about the arc arms to handle cases where the minimum distance is to an end point.

The case where the point is in the arcs angle range is then handled.

For computation purposes the square root is avoided and the distance measure is the standard distance squared.

The points q_i are the control points. To associate a control point uniquely with an arc only the minimum distance is considered between a control point and all the arcs.

Handling a large number of Arcs

Lets say that the program can easily solve for a small number of arcs say 10. How can this be applied to solve for 1000.

Maybe use line segment to find a path and order the control points.

Pre order the control points along the path so they divide the path into K control points with I intervals. Let their be N2 arcs for each interval. and that N2 be an integer.

Use the solver to minimized for K control points. Sequentually solve the path.

The client could choose K and N2.

I have suggested this rather than solving for the full N arcs to keep the complexity linear as the number of arcs is increased.

Model Connected Arcs in Cartesian Coordinates

Here the points connecting the arcs are allowed to vary in x and y. The arcs end points are continuous.

Minimization Algorithm

Algorithm1 uses a derivative free minimizer.

Mike suggested the Levenberg-Marquadt method (also called marquardt method) see [1] has become the standard of nonlinear least squares problems.

Arc Formats

There are two formats, one for drawing the arcs and the other for solving for k arcs given n control points.

The idea is to feed the problem to the solver through the arc solver format. It then produces a file in the arc drawing format. To get the solution the application reads the output of the solver through the arc drawing format.

The arcs are convex. The sign of the radius corresponds to anticlockwise winding for a positive radius and clockwise winding for a negative radius.

Arc Drawing Format

Number of points
List of 2D points
Number of arcs
List of arcs

Here is an example of the arc format, with three arcs.

4
0.0 0.0
1.0 1.1
1.5 2.8
2.8 4.2
3
0 1 2.5
1 2 -2.5 
2 3 -2.5

An arc is defined by indexing into the points vector and then the radius. A positive radius has anti clockwise winding and a negative radius has clockwise winding.

The order of the points is important. 0 1 2.5 and 1 0 2.5 have mirrored arcs. Note that changing the point order is equivalent to multiplying the radius by -1.

Arc Solver Format

Initial angle
Number of arcs
Number of control points
List of control points

Doxygen

main.cpp
Makefile
arc
arcdraw
arcprob
arcsconnected
farcmin01
openwindowresource
pathlineseg
pathlinesegdraw
pathlinesegtest
pathlinesegvec

TODO

Develope a format for the problem.
Initial arc angle, Number of arcs, Number of control points, control points.
Have a test function to read in and display the control points in purple. The arc points are displayed in green.
draw arcs correctly
arc format defined and documented.
Initial condition
- implement
Write an algorithm to Handeling a large number of Arcs.

Constructing an Arc through Two Points given an Angle

The problem is that given an initia angle and two points construct a new arc.

Given $ϕ_{0}$ , $p_{0}$ , $p_{1}$
find
radius and center c and $ϕ_{1}$

Let r be the positive valued radius and is a scalar variable.
$a = (\cos ϕ_{0}, \sin ϕ_{0})$
$- c + p_{0} = a r$

Since the arc is on a circle the distance between the center and the points is equal.

$d (p_{0}, c) = d (p_{1}, c)$
$(p_{0} - c)^{2} = (p_{1} - c)^{2}$
$p_{0}^{2} - p_{1}^{2} = 2 (p_{0} - p_{1}) c$

Rearrange c and substitute it into the above expression.

$p_{0}^{2} - p_{1}^{2} = 2 (p_{0} - p_{1}) (p_{0} - a r)$

r is the only scalar, solve for it.

$p_{0}^{2} - p_{1}^{2} = 2 (p_{0}^{2} - p_{1} p_{0}) - 2 (p_{0} - p_{1}) a r$
$(\pm (p_{0} - p_{1}))^{2} = 2 (p_{0} - p_{1}) a r$
$r = \frac{(p_{0} - p_{1})^{2}}{2 (p_{0} - p_{1}) a}$

Use this to calculate the center c.

$c = p_{0} - a r$

Calculate the second angle by shifting the point about the center and using the atan2 routine to get the second angle.

$ϕ_{1} = \tan^{- 1} (\frac{1}{r} (p_{1} - c))$

The arc is interpreted as being convex. However the sign is used to indicate the anticlockwise (positive) winding or clockwise(negative) winding. A test based on the values of the angles was used.

bool const arc::isAntiClockwise2() const
{
  double dphi = phi1-phi0;
  if (dphi<0.0)
    dphi += 2.0*PI;

  if (dphi<=PI)
    return true;

  return false;
}

if arc.isAntiClockwise2()
$radius = r$
else
$radius = - r$

Algorithm Complexity

This is to model and compare different algorithms. I do not know the complexity of the minimizers on a particular problem.

If the problem is divided into N then let this be the measure.

For example I do not know if 1..N^2 or 1..N^3 iterations of the minimizer are needed, but can speculate.

Let PM be the problem minimizer where P is the order of the problem and M the order of the minimizer.

$O (PM) = N^{P - M}$

Adding the inner loop with the distance function used by the minimizer.

$O (PM) = N^{P - M} (inner loop)$

For example using exploratory minimizer M=1, assume arcs minimizing arcs is P=2. O(N) control points.

$O (N^{2 - 1}) O (N^{2}) = N^{3}$

Local arcs would take the distance function from O(N^2) to O(N). A quadratic minimizer would reduce by another order too.

This line of reasoning is speculative. The inner loop is true but the outer look is a guess of what might be happening.

Line Segments

Find N connected line segments best fitted to W control points where W is larger than N.

./main prog=30 file=path002.txt

Minimizing the distance of all the line segments to all the control points produced a solution the I did not expect but is understandable. min=0.273

$./main prog=33 file2=temp3.txt distfn=1 $./main prog=30 file=temp3.txt

Adding the segment length to the distance function. min=2.540

A convexity condition for successive line segments could have stopped chopping and may help other minimization algorithm implementations.

Changing the step length from 0.3 to 0.1 showed the target solution was a local minimum and not a global minimum. min=3.239

$./main prog=33 file2=temppath2.txt h0=0.1 imax=15 distfn=2 $./main prog=30 file=temppath2.txt

This is still not the full story. An initial curve was given which was already a path with convex angles greater than 90 degrees.

Algorithm1

Let w_i be N+1 points connecting the N arcs.

Initial approximation found by fitting N+1 control points which include the fixed ends of the curve and fit the arcs through satisfying the initial angle.

Solve arc_k
Given w_k and w_k+1 and θ
arc_k.p0 = w_k
arc_k.p1 = w_k+1
arc_k.ϕ₀ = θ or θ + π
Solve arc_k by finding arc_k.ϕ₁.

Updating connected arcs f.
f(w₁.x, w₁.y, ...
w_N-1.x, w_N-1.y )
k = 1..N
solve arc_k

$dist01 = \sum_{i} \sum_{k} d (q_{i}, arc_{k})$
$+ \sum_{k} arc_{k} . length ()$

Minimize f with dist01.

References

Press Teukolsky Vetterling Flannery (1992) second edition. Numerical Recipes in C. ISBN 0-521-431088. Page 683

arcs

Intro

Solution Strategies

Problem

Incremental Algorithm

Continuity Condition