Catmull-rom curve with no cusps and no self-intersections
Asked Answered
D

5

45

I have the following code to calculate points between four control points to generate a catmull-rom curve:

CGPoint interpolatedPosition(CGPoint p0, CGPoint p1, CGPoint p2, CGPoint p3, float t)
{
    float t3 = t * t * t;
    float t2 = t * t;

    float f1 = -0.5 * t3 + t2 - 0.5 * t;
    float f2 = 1.5 * t3 - 2.5 * t2 + 1.0;
    float f3 = -1.5 * t3 + 2.0 * t2 + 0.5 * t;
    float f4 = 0.5 * t3 - 0.5 * t2;

    float x = p0.x * f1 + p1.x * f2 + p2.x * f3 + p3.x * f4;
    float y = p0.y * f1 + p1.y * f2 + p2.y * f3 + p3.y * f4;

    return CGPointMake(x, y);
}

This works fine, but I want to create something I think is called centripetal parameterization. This means that the curve will have no cusps and no self-intersections. If I move one control point really close to another one, the curve should become "smaller". I have Googled my eyes off trying to find a way to do this. Anyone know how to do this?

Downes answered 28/2, 2012 at 21:9 Comment(1)
I've found several very mathematical papers on the subject, for instance: scholar.lib.vt.edu/theses/available/etd-04192001-172731/… cs.bgu.ac.il/~leonid/na105/Splines/Lee.pdf and faculty.cs.tamu.edu/schaefer/research/catmull_rom.pdfAkins
D
84

I needed to implement this for work as well. The fundamental concept you need to start with is that the main difference between the regular Catmull-Rom implementation and the modified versions is how they treat time.

Catmull-Rom Time

In the unparameterized version from your original Catmull-Rom implementation, t starts at 0 and ends with 1 and calculates the curve from P1 to P2. In the parameterized time implementation, t starts with 0 at P0, and keeps increasing across all four points. So in the uniform case, it would be 1 at P1 and 2 at P2, and you would pass in values ranging from 1 to 2 for your interpolation.

The chordal case shows |Pi+1 - P| as the time span change. This just means that you can use the straight line distance between the points of each segment to calculate the actual length to use. The centripetal case just uses a slightly different method for calculating the optimal length of time to use for each segment.

Catmull-Rom Parameterization

So now we just need to know how to come up with equations that will let us plug in our new time values. The typical Catmull-Rom equation only has one t in it, the time you are trying to calculate a value for. I found the best article for describing how those parameters are calculated here: http://www.cemyuksel.com/research/catmullrom_param/catmullrom.pdf. They were focusing on a mathematical evaluation of the curves, but in it lies the crucial formula from Barry and Goldman.(1)

Cubic Catmull-Rom curve formulation

In the diagram above, the arrows mean "multiplied by" the ratio given in the arrow.

This then gives us what we need to actually perform a calculation to get the desired result. X and Y are calculated independently, although I used the "Distance" factor for modifying time based on the 2D distance, and not the 1D distance.

Test results:

Catmull-Rom Test Results

(1) P. J. Barry and R. N. Goldman. A recursive evaluation algorithm for a class of catmull-rom splines. SIGGRAPH Computer Graphics, 22(4):199{204, 1988.

The source code for my final implementation in Java looks as follows:

/**
 * This method will calculate the Catmull-Rom interpolation curve, returning
 * it as a list of Coord coordinate objects.  This method in particular
 * adds the first and last control points which are not visible, but required
 * for calculating the spline.
 *
 * @param coordinates The list of original straight line points to calculate
 * an interpolation from.
 * @param pointsPerSegment The integer number of equally spaced points to
 * return along each curve.  The actual distance between each
 * point will depend on the spacing between the control points.
 * @return The list of interpolated coordinates.
 * @param curveType Chordal (stiff), Uniform(floppy), or Centripetal(medium)
 * @throws gov.ca.water.shapelite.analysis.CatmullRomException if
 * pointsPerSegment is less than 2.
 */
public static List<Coord> interpolate(List<Coord> coordinates, int pointsPerSegment, CatmullRomType curveType)
        throws CatmullRomException {
    List<Coord> vertices = new ArrayList<>();
    for (Coord c : coordinates) {
        vertices.add(c.copy());
    }
    if (pointsPerSegment < 2) {
        throw new CatmullRomException("The pointsPerSegment parameter must be greater than 2, since 2 points is just the linear segment.");
    }

    // Cannot interpolate curves given only two points.  Two points
    // is best represented as a simple line segment.
    if (vertices.size() < 3) {
        return vertices;
    }

    // Test whether the shape is open or closed by checking to see if
    // the first point intersects with the last point.  M and Z are ignored.
    boolean isClosed = vertices.get(0).intersects2D(vertices.get(vertices.size() - 1));
    if (isClosed) {
        // Use the second and second from last points as control points.
        // get the second point.
        Coord p2 = vertices.get(1).copy();
        // get the point before the last point
        Coord pn1 = vertices.get(vertices.size() - 2).copy();

        // insert the second from the last point as the first point in the list
        // because when the shape is closed it keeps wrapping around to
        // the second point.
        vertices.add(0, pn1);
        // add the second point to the end.
        vertices.add(p2);
    } else {
        // The shape is open, so use control points that simply extend
        // the first and last segments

        // Get the change in x and y between the first and second coordinates.
        double dx = vertices.get(1).X - vertices.get(0).X;
        double dy = vertices.get(1).Y - vertices.get(0).Y;

        // Then using the change, extrapolate backwards to find a control point.
        double x1 = vertices.get(0).X - dx;
        double y1 = vertices.get(0).Y - dy;

        // Actaully create the start point from the extrapolated values.
        Coord start = new Coord(x1, y1, vertices.get(0).Z);

        // Repeat for the end control point.
        int n = vertices.size() - 1;
        dx = vertices.get(n).X - vertices.get(n - 1).X;
        dy = vertices.get(n).Y - vertices.get(n - 1).Y;
        double xn = vertices.get(n).X + dx;
        double yn = vertices.get(n).Y + dy;
        Coord end = new Coord(xn, yn, vertices.get(n).Z);

        // insert the start control point at the start of the vertices list.
        vertices.add(0, start);

        // append the end control ponit to the end of the vertices list.
        vertices.add(end);
    }

    // Dimension a result list of coordinates. 
    List<Coord> result = new ArrayList<>();
    // When looping, remember that each cycle requires 4 points, starting
    // with i and ending with i+3.  So we don't loop through all the points.
    for (int i = 0; i < vertices.size() - 3; i++) {

        // Actually calculate the Catmull-Rom curve for one segment.
        List<Coord> points = interpolate(vertices, i, pointsPerSegment, curveType);
        // Since the middle points are added twice, once for each bordering
        // segment, we only add the 0 index result point for the first
        // segment.  Otherwise we will have duplicate points.
        if (result.size() > 0) {
            points.remove(0);
        }

        // Add the coordinates for the segment to the result list.
        result.addAll(points);
    }
    return result;

}

/**
 * Given a list of control points, this will create a list of pointsPerSegment
 * points spaced uniformly along the resulting Catmull-Rom curve.
 *
 * @param points The list of control points, leading and ending with a 
 * coordinate that is only used for controling the spline and is not visualized.
 * @param index The index of control point p0, where p0, p1, p2, and p3 are
 * used in order to create a curve between p1 and p2.
 * @param pointsPerSegment The total number of uniformly spaced interpolated
 * points to calculate for each segment. The larger this number, the
 * smoother the resulting curve.
 * @param curveType Clarifies whether the curve should use uniform, chordal
 * or centripetal curve types. Uniform can produce loops, chordal can
 * produce large distortions from the original lines, and centripetal is an
 * optimal balance without spaces.
 * @return the list of coordinates that define the CatmullRom curve
 * between the points defined by index+1 and index+2.
 */
public static List<Coord> interpolate(List<Coord> points, int index, int pointsPerSegment, CatmullRomType curveType) {
    List<Coord> result = new ArrayList<>();
    double[] x = new double[4];
    double[] y = new double[4];
    double[] time = new double[4];
    for (int i = 0; i < 4; i++) {
        x[i] = points.get(index + i).X;
        y[i] = points.get(index + i).Y;
        time[i] = i;
    }

    double tstart = 1;
    double tend = 2;
    if (!curveType.equals(CatmullRomType.Uniform)) {
        double total = 0;
        for (int i = 1; i < 4; i++) {
            double dx = x[i] - x[i - 1];
            double dy = y[i] - y[i - 1];
            if (curveType.equals(CatmullRomType.Centripetal)) {
                total += Math.pow(dx * dx + dy * dy, .25);
            } else {
                total += Math.pow(dx * dx + dy * dy, .5);
            }
            time[i] = total;
        }
        tstart = time[1];
        tend = time[2];
    }
    double z1 = 0.0;
    double z2 = 0.0;
    if (!Double.isNaN(points.get(index + 1).Z)) {
        z1 = points.get(index + 1).Z;
    }
    if (!Double.isNaN(points.get(index + 2).Z)) {
        z2 = points.get(index + 2).Z;
    }
    double dz = z2 - z1;
    int segments = pointsPerSegment - 1;
    result.add(points.get(index + 1));
    for (int i = 1; i < segments; i++) {
        double xi = interpolate(x, time, tstart + (i * (tend - tstart)) / segments);
        double yi = interpolate(y, time, tstart + (i * (tend - tstart)) / segments);
        double zi = z1 + (dz * i) / segments;
        result.add(new Coord(xi, yi, zi));
    }
    result.add(points.get(index + 2));
    return result;
}

/**
 * Unlike the other implementation here, which uses the default "uniform"
 * treatment of t, this computation is used to calculate the same values but
 * introduces the ability to "parameterize" the t values used in the
 * calculation. This is based on Figure 3 from
 * http://www.cemyuksel.com/research/catmullrom_param/catmullrom.pdf
 *
 * @param p An array of double values of length 4, where interpolation
 * occurs from p1 to p2.
 * @param time An array of time measures of length 4, corresponding to each
 * p value.
 * @param t the actual interpolation ratio from 0 to 1 representing the
 * position between p1 and p2 to interpolate the value.
 * @return
 */
public static double interpolate(double[] p, double[] time, double t) {
    double L01 = p[0] * (time[1] - t) / (time[1] - time[0]) + p[1] * (t - time[0]) / (time[1] - time[0]);
    double L12 = p[1] * (time[2] - t) / (time[2] - time[1]) + p[2] * (t - time[1]) / (time[2] - time[1]);
    double L23 = p[2] * (time[3] - t) / (time[3] - time[2]) + p[3] * (t - time[2]) / (time[3] - time[2]);
    double L012 = L01 * (time[2] - t) / (time[2] - time[0]) + L12 * (t - time[0]) / (time[2] - time[0]);
    double L123 = L12 * (time[3] - t) / (time[3] - time[1]) + L23 * (t - time[1]) / (time[3] - time[1]);
    double C12 = L012 * (time[2] - t) / (time[2] - time[1]) + L123 * (t - time[1]) / (time[2] - time[1]);
    return C12;
}   
Domiciliate answered 9/10, 2013 at 21:52 Comment(11)
I'm gonna mark this a the correct answer, your code is in Java which will easily convert to other languages. I'm unable to test your code, but I take your word for it that it works, this will probably be useful to other, thanks!Coagulant
@ted what if I have a y = 5 line and want to calculate the intersection point with the spline? Is this possible with interpolation? ThanksCm
Another question: why is the alpha coefficient 0.25 for the centripetal equation?Cm
When calculating the euclidean distance, you already take the square root, like Sqrt(dxdx + dydy). This is the definition of the chordal case, where the euclidean distance is used to calculate the relative time content. This is the same as (dxdx+dydy)^.5. But this actually tends to make the spline too stiff, and gives quite a bit of distortion to the spline. So the centripital case then takes the square root of THAT, making it (dxdx+dydy)^.25, and is a compromise between the stiff chordal and floppy uniform case.Domiciliate
As for intersection, I would simply go ahead and create the spline and calculate the intersections with the given y of the resulting LineString. You might be able to restructure your equations using algebra to solve for t given a value of y, but based on the complexity, I'm not sure it would be worth the effort to do that.Domiciliate
In java you can use jts (vividsolutions.com/jts/JTSHome.htm) to do linestring intersection. In C# you can use DotSpatial (dotspatial.codeplex.com) or NTS (code.google.com/p/nettopologysuite). To do it yourself, just cycle through the line segments in the resulting linestring and test each segment for an intersection.Domiciliate
As for the alpha coefficient, you made me realise how long is since I use Math skills :) ^.25 = 1/2 * ^0.5 = 1/2 * ^1/2 = 1/2 * sqrtCm
A question: what is z?Cypher
In my case, z was tracking the elevation of the lines. This is not required for the spline calculation itself.Domiciliate
if I will set different segment points, for example, between p1 and p2 set 10 segment points, and between p3 and p4 set 50 - then does the curve smoothness will be fine?Epiclesis
The curve shape is controlled by the mathematical spline curve and adding more points means that your LineString will have a better and better representation of the mathematical curve. So as far as smoothness goes, the curve with 50 points will more perfectly represent the curve than the part with 10 points. But it will depend on the context of the curve how many points you need.Domiciliate
S
47

There is a much easier and more efficient way to implement this which only requires you to compute your tangents using a different formula, without the need to implement the recursive evaluation algorithm of Barry and Goldman.

If you take the Barry-Goldman parametrization (referenced in Ted's answer) C(t) for the knots (t0,t1,t2,t3) and the control points (P0,P1,P2,P3), its closed form is pretty complicated, but in the end it's still a cubic polynomial in t when you constrain it to the interval (t1,t2). So all we need to describe it fully are the values and tangents at the two end points t1 and t2. If we work out these values (I did this in Mathematica), we find

C(t1)  = P1
C(t2)  = P2
C'(t1) = (P1 - P0) / (t1 - t0) - (P2 - P0) / (t2 - t0) + (P2 - P1) / (t2 - t1)
C'(t2) = (P2 - P1) / (t2 - t1) - (P3 - P1) / (t3 - t1) + (P3 - P2) / (t3 - t2)

We can simply plug this into the standard formula for computing a cubic spline with given values and tangents at the end points and we have our nonuniform Catmull-Rom spline. One caveat is that the above tangents are computed for the interval (t1,t2), so if you want to evaluate the curve in the standard interval (0,1), simply rescale the tangents by multiplying them with the factor (t2-t1).

I put a working C++ example on Ideone: http://ideone.com/NoEbVM

I'll also paste the code below.

#include <iostream>
#include <cmath>

using namespace std;

struct CubicPoly
{
    float c0, c1, c2, c3;

    float eval(float t)
    {
        float t2 = t*t;
        float t3 = t2 * t;
        return c0 + c1*t + c2*t2 + c3*t3;
    }
};

/*
 * Compute coefficients for a cubic polynomial
 *   p(s) = c0 + c1*s + c2*s^2 + c3*s^3
 * such that
 *   p(0) = x0, p(1) = x1
 *  and
 *   p'(0) = t0, p'(1) = t1.
 */
void InitCubicPoly(float x0, float x1, float t0, float t1, CubicPoly &p)
{
    p.c0 = x0;
    p.c1 = t0;
    p.c2 = -3*x0 + 3*x1 - 2*t0 - t1;
    p.c3 = 2*x0 - 2*x1 + t0 + t1;
}

// standard Catmull-Rom spline: interpolate between x1 and x2 with previous/following points x0/x3
// (we don't need this here, but it's for illustration)
void InitCatmullRom(float x0, float x1, float x2, float x3, CubicPoly &p)
{
    // Catmull-Rom with tension 0.5
    InitCubicPoly(x1, x2, 0.5f*(x2-x0), 0.5f*(x3-x1), p);
}

// compute coefficients for a nonuniform Catmull-Rom spline
void InitNonuniformCatmullRom(float x0, float x1, float x2, float x3, float dt0, float dt1, float dt2, CubicPoly &p)
{
    // compute tangents when parameterized in [t1,t2]
    float t1 = (x1 - x0) / dt0 - (x2 - x0) / (dt0 + dt1) + (x2 - x1) / dt1;
    float t2 = (x2 - x1) / dt1 - (x3 - x1) / (dt1 + dt2) + (x3 - x2) / dt2;

    // rescale tangents for parametrization in [0,1]
    t1 *= dt1;
    t2 *= dt1;

    InitCubicPoly(x1, x2, t1, t2, p);
}

struct Vec2D
{
    Vec2D(float _x, float _y) : x(_x), y(_y) {}
    float x, y;
};

float VecDistSquared(const Vec2D& p, const Vec2D& q)
{
    float dx = q.x - p.x;
    float dy = q.y - p.y;
    return dx*dx + dy*dy;
}

void InitCentripetalCR(const Vec2D& p0, const Vec2D& p1, const Vec2D& p2, const Vec2D& p3,
    CubicPoly &px, CubicPoly &py)
{
    float dt0 = powf(VecDistSquared(p0, p1), 0.25f);
    float dt1 = powf(VecDistSquared(p1, p2), 0.25f);
    float dt2 = powf(VecDistSquared(p2, p3), 0.25f);

    // safety check for repeated points
    if (dt1 < 1e-4f)    dt1 = 1.0f;
    if (dt0 < 1e-4f)    dt0 = dt1;
    if (dt2 < 1e-4f)    dt2 = dt1;

    InitNonuniformCatmullRom(p0.x, p1.x, p2.x, p3.x, dt0, dt1, dt2, px);
    InitNonuniformCatmullRom(p0.y, p1.y, p2.y, p3.y, dt0, dt1, dt2, py);
}


int main()
{
    Vec2D p0(0,0), p1(1,1), p2(1.1,1), p3(2,0);
    CubicPoly px, py;
    InitCentripetalCR(p0, p1, p2, p3, px, py);
    for (int i = 0; i <= 10; ++i)
        cout << px.eval(0.1f*i) << " " << py.eval(0.1f*i) << endl;
}
Shedevil answered 1/6, 2014 at 13:40 Comment(17)
How exactly did you create that simplified formula? Where did t go?Greensickness
@ssb: The values C(t1), C(t2), C'(t1), C'(t2) that I computed are the result of evaluating C(t) and its derivative, C'(t), at the two end points of the interval (t1,t2). I computed them in Mathematica. Since the curve is a cubic polynomial in this interval and therefore has four degrees of freedom, these four values are enough to determine the curve completely. So we just need to plug these four values into the standard formula for a cubic polynomial with given values and derivatives.Shedevil
So you put the giant expanded pyramid formula into mathematica with t1 and t2 substituted and it gave you those two formulas? @eriatarka84Greensickness
@ssb: Basically, yes. Look here: gist.github.com/anonymous/0eedb67914f554ee9cb5 Sorry it came out a bit ugly, but you can paste it into Mathematica if you have it.Shedevil
I know it will sound silly, but how did you derive the equation for tangent using 3 points? I am referring to: t = (x1 - x0) / dt0 - (x2 - x0) / (dt0 + dt1) + (x2 - x1) / dt1? It seems as if you are doing simple vector sum (a + b = c; a + b - c = 0) but what is left (0) you consider as a tangent.Prickly
It's just the formula I explained how to derive above, C'(t1) = (P1 - P0) / (t1 - t0) - (P2 - P0) / (t2 - t0) + (P2 - P1) / (t2 - t1), rewritten in different variables.Shedevil
Cool. I've used <uniform> Catmull-Rom splines before, and they're great for creating curves where all you have is points on the curve. The centripetal variant seems better-behaved, since it doesn't introduce kinks like the uniform version does. However, it seems like it's more computationally expensive than uniform Catmull-Rom splines since it requires several divisions, plus the VERY expensive pow function. Has anybody benchmarked centripetal C-R splines against the original uniform splines?Thaumaturgy
@DuncanC You only need the pow() function when you initialize the coefficients, not when you actually evaluate the spline, so the overhead should be quite minor.Shedevil
Interestingly, there is an another commonly used parametrization detailed here: 15462.courses.cs.cmu.edu/fall2015content/misc/… However I cannot match it mathematically with the Barry-Goldman version.Daiseydaisi
Can we generalize like float dt0 = powf(VecDistSquared(p0, p1), 0.5f*alpha) with alpha from (0.0,1.0) ? For all kinds of Catmull-Rom splines...Biparietal
@Shedevil Indeed. Just checked it for alpha=0.0, 0.5, 1.0 .Biparietal
This was precisely what I needed thanks. If anyone wants it in C# for Unity, you can find it hereRosemarierosemary
@Daiseydaisi The approach from your link is simply wrong for the non-uniform case, it just happens to reduce to the correct solution if all time intervals are 1. And that's why you cannot match it with Barry-Goldman. The left and right time interval must appear separately in the equation, otherwise it can never work correctly!Roughandready
@Roughandready Check the before the last equation in the paper, not the last one. According to the paper, that is for the general case. Later, it also mentions the special case when the knots are uniform, which corresponds to the last equation.Daiseydaisi
@Daiseydaisi The problem is that the very first equation in the linked paper is already wrong (except in the uniform case), so are all following equations that are derived from the first, including the penultimate. The correct equations for the tangents are shown in the answer above. An alternative (but equivalent) form of the equation would be C'(t1) = ((t2 - t1)^2 * (P1 - P0) + (t1 - t0)^2 * (P2 - P1)) / ((t2 - t1) * (t1 - t0) * (t2 - t0)). The two chords (P1 - P0 and P2 - P1) must have different factors, the (non-uniform) equation cannot be correct if it has only P2 - P0 in it.Roughandready
@Roughandready Okay, I also use the Barry-Goldman algorithm for Catmull-Rom splines, so I didn't question the correctness of it. I just said there is an also common method for the parametrization. I didn't test those equations analytically though. It is good that it turned out that they are wrong and thus useless, thanks for sharing it.Daiseydaisi
@Shedevil You say above that "C(t1), C(t2), C'(t1), C'(t2) ... are the result of evaluating C(t) and its derivative, C'(t), at the two end points of the interval (t1,t2)". What is C exactly here? If it is the equation of the Catmull-Rom in the case of uniform parametrization, then I don't get, for example, why you get C(t1) = P1. Furthermore, why would you even try to evaluate the equation of a curve that was constructed assuming uniform parametrization at non-uniform parameters (t1 or t2)?Agglomeration
L
4

Here is an iOS version of Ted's code. I excluded the 'z' parts.

.h

typedef enum {
    CatmullRomTypeUniform,
    CatmullRomTypeChordal,
    CatmullRomTypeCentripetal
} CatmullRomType ;

.m

-(NSMutableArray *)interpolate:(NSArray *)coordinates withPointsPerSegment:(NSInteger)pointsPerSegment andType:(CatmullRomType)curveType {

    NSMutableArray *vertices = [[NSMutableArray alloc] initWithArray:coordinates copyItems:YES];

    if (pointsPerSegment < 3)
        return vertices;

    //start point
    CGPoint pt1 = [vertices[0] CGPointValue];
    CGPoint pt2 = [vertices[1] CGPointValue];

    double dx = pt2.x - pt1.x;
    double dy = pt2.y - pt1.y;

    double x1 = pt1.x - dx;
    double y1 = pt1.y - dy;

    CGPoint start = CGPointMake(x1*.5, y1);

    //end point
    pt2 = [vertices[vertices.count-1] CGPointValue];
    pt1 = [vertices[vertices.count-2] CGPointValue];

    dx = pt2.x - pt1.x;
    dy = pt2.y - pt1.y;

    x1 = pt2.x + dx;
    y1 = pt2.y + dy;

    CGPoint end = CGPointMake(x1, y1);

    [vertices insertObject:[NSValue valueWithCGPoint:start] atIndex:0];
    [vertices addObject:[NSValue valueWithCGPoint:end]];

    NSMutableArray *result = [[NSMutableArray alloc] init];

    for (int i = 0; i < vertices.count - 3; i++) {
        NSMutableArray *points = [self interpolate:vertices forIndex:i withPointsPerSegment:pointsPerSegment andType:curveType];

        if ([points count] > 0)
            [points removeObjectAtIndex:0];

        [result addObjectsFromArray:points];
    }

    return result;
}

-(double)interpolate:(double*)p  time:(double*)time t:(double) t {
    double L01 = p[0] * (time[1] - t) / (time[1] - time[0]) + p[1] * (t - time[0]) / (time[1] - time[0]);
    double L12 = p[1] * (time[2] - t) / (time[2] - time[1]) + p[2] * (t - time[1]) / (time[2] - time[1]);
    double L23 = p[2] * (time[3] - t) / (time[3] - time[2]) + p[3] * (t - time[2]) / (time[3] - time[2]);
    double L012 = L01 * (time[2] - t) / (time[2] - time[0]) + L12 * (t - time[0]) / (time[2] - time[0]);
    double L123 = L12 * (time[3] - t) / (time[3] - time[1]) + L23 * (t - time[1]) / (time[3] - time[1]);
    double C12 = L012 * (time[2] - t) / (time[2] - time[1]) + L123 * (t - time[1]) / (time[2] - time[1]);
    return C12;
    }

-(NSMutableArray*)interpolate:(NSArray *)points forIndex:(NSInteger)index withPointsPerSegment:(NSInteger)pointsPerSegment andType:(CatmullRomType)curveType {
    NSMutableArray *result = [[NSMutableArray alloc] init];

    double x[4];
    double y[4];
    double time[4];

    for (int i=0; i < 4; i++) {
        x[i] = [points[index+i] CGPointValue].x;
        y[i] = [points[index+i] CGPointValue].y;
        time[i] = i;
    }

    double tstart = 1;
    double tend = 2;

    if (curveType != CatmullRomTypeUniform) {
        double total = 0;

        for (int i=1; i < 4; i++) {
            double dx = x[i] - x[i-1];
            double dy = y[i] - y[i-1];

            if (curveType == CatmullRomTypeCentripetal) {
                total += pow(dx * dx + dy * dy, 0.25);
            }
            else {
                total += pow(dx * dx + dy * dy, 0.5); //sqrt
            }
            time[i] = total;
        }
        tstart = time[1];
        tend = time[2];
    }

    int segments = pointsPerSegment - 1;

    [result addObject:points[index+1]];

    for (int i =1; i < segments; i++) {
        double xi = [self interpolate:x time:time t:tstart + (i * (tend - tstart)) / segments];
        double yi = [self interpolate:y time:time t:tstart + (i * (tend - tstart)) / segments];
        NSLog(@"(%f,%f)",xi,yi);
        [result addObject:[NSValue valueWithCGPoint:CGPointMake(xi, yi)]];
    }
    [result addObject:points[index+2]];

    return result;
}

Also, here is a method for turning an array of points into a Bezier path for drawing, using the above

-(UIBezierPath*)bezierPathFromPoints:(NSArray *)points withGranulaity:(NSInteger)granularity
{
    UIBezierPath __block *path = [[UIBezierPath alloc] init];

    NSMutableArray *curve = [self interpolate:points withPointsPerSegment:granularity andType:CatmullRomTypeCentripetal];

    CGPoint __block p0 = [curve[0] CGPointValue];

    [path moveToPoint:p0];

    //use this loop to draw lines between all points
    for (int idx=1; idx < [curve count]; idx+=1) {
        CGPoint c1 = [curve[idx] CGPointValue];

        [path addLineToPoint:c1];
    };

    //or use this loop to use actual control points (less smooth but probably faster)
//    for (int idx=0; idx < [curve count]-3; idx+=3) {
//        CGPoint c1 = [curve[idx+1] CGPointValue];
//        CGPoint c2 = [curve[idx+2] CGPointValue];
//        CGPoint p1 = [curve[idx+3] CGPointValue];
//
//        [path addCurveToPoint:p1 controlPoint1:c1 controlPoint2:c2];
//    };

    return path;
}
Li answered 13/5, 2014 at 20:57 Comment(2)
Don't know why this is downvoted, it seems to work as it should. I'm going to do a patch to CorePlot (github.com/core-plot/core-plot), I'll post it here eventually.Hermie
Ya I wondered about that too:) Thanks for the upvote. Looking forward to the CorePlot version!Li
C
1

Thanks for the reply of Ted and cfh.

Sorry to my poor English and I am not very sure if my understanding is right.

It confused me before that what is the relation between τ in http://graphics.cs.cmu.edu/nsp/course/15-462/Fall04/assts/catmullRom.pdf with α in "Parameterization of Catmull-Rom Curves"[Yuksel et al. 2009].

It finally seems that τ has little relation to α.

In Ted's reply we can find that as long as t(i+1)-t(i)=1, we call it "uniform" Catmull-Rom curve. So, all curves in http://graphics.cs.cmu.edu/nsp/course/15-462/Fall04/assts/catmullRom.pdf are "uniform" curves, while parameterized Catmull-Rom can only produce one "uniform" curve when α=0.τ just affects how sharply the curve bends at the (interpolated) control points and can produce a series of "uniform" curves.

Here, if we take the equation from cfh into consideration, i.e.:

C(t1)  = P1
C(t2)  = P2
C'(t1) = (P1 - P0) / (t1 - t0) - (P2 - P0) / (t2 - t0) + (P2 - P1) / (t2 - t1)
C'(t2) = (P2 - P1) / (t2 - t1) - (P3 - P1) / (t3 - t1) + (P3 - P2) / (t3 - t2)

When α=0, we know that t(i+1)-t(i)=1. Substituting t(i+1)-t(i)=1 to C'(t1) and C'(t2), we can get:

C'(t1)=1/2(P2-P0)
C'(t2)=1/2(P3-P1).

That is to say, the only one "uniform" curve generated by parameterized Catmull-Rom curve meets the special case τ=1/2 http://graphics.cs.cmu.edu/nsp/course/15-462/Fall04/assts/catmullRom.pdf. More different "uniform" curves can be generated by the changes of τ, while α pays more attention to something else.

Chromoprotein answered 26/2, 2021 at 12:58 Comment(0)
H
0

I coded something in Python (adapted form Catmull-Rom Wikipedia page) that compares uniform, centripedal, and chordial CR Splines (though you can set alpha to whatever you'd like) using random data (you can use your own data and the fucntions work fine). Note that for the endpoints I just stuck in a quick 'hack' that maintains the slope from the first and last 2 points, although the distance between this point and the first/lost known point is arbitrary (I set it to 1% of the domain... for no reason at all. So keep that in mind before applying to something important):

# coding: utf-8

# In[1]:

import numpy
import matplotlib.pyplot as plt
get_ipython().magic(u'pylab inline')


# In[2]:

def CatmullRomSpline(P0, P1, P2, P3, a, nPoints=100):
  """
  P0, P1, P2, and P3 should be (x,y) point pairs that define the Catmull-Rom spline.
  nPoints is the number of points to include in this curve segment.
  """
  # Convert the points to numpy so that we can do array multiplication
  P0, P1, P2, P3 = map(numpy.array, [P0, P1, P2, P3])

  # Calculate t0 to t4
  alpha = a
  def tj(ti, Pi, Pj):
    xi, yi = Pi
    xj, yj = Pj
    return ( ( (xj-xi)**2 + (yj-yi)**2 )**0.5 )**alpha + ti

  t0 = 0
  t1 = tj(t0, P0, P1)
  t2 = tj(t1, P1, P2)
  t3 = tj(t2, P2, P3)

  # Only calculate points between P1 and P2
  t = numpy.linspace(t1,t2,nPoints)

  # Reshape so that we can multiply by the points P0 to P3
  # and get a point for each value of t.
  t = t.reshape(len(t),1)

  A1 = (t1-t)/(t1-t0)*P0 + (t-t0)/(t1-t0)*P1
  A2 = (t2-t)/(t2-t1)*P1 + (t-t1)/(t2-t1)*P2
  A3 = (t3-t)/(t3-t2)*P2 + (t-t2)/(t3-t2)*P3

  B1 = (t2-t)/(t2-t0)*A1 + (t-t0)/(t2-t0)*A2
  B2 = (t3-t)/(t3-t1)*A2 + (t-t1)/(t3-t1)*A3

  C  = (t2-t)/(t2-t1)*B1 + (t-t1)/(t2-t1)*B2
  return C

def CatmullRomChain(P,alpha):
  """
  Calculate Catmull Rom for a chain of points and return the combined curve.
  """
  sz = len(P)

  # The curve C will contain an array of (x,y) points.
  C = []
  for i in range(sz-3):
    c = CatmullRomSpline(P[i], P[i+1], P[i+2], P[i+3],alpha)
    C.extend(c)

  return C


# In[8]:

# Define a set of points for curve to go through
Points = numpy.random.rand(12,2)
x1=Points[0][0]
x2=Points[1][0]
y1=Points[0][1]
y2=Points[1][1]
x3=Points[-2][0]
x4=Points[-1][0]
y3=Points[-2][1]
y4=Points[-1][1]
dom=max(Points[:,0])-min(Points[:,0])
rng=max(Points[:,1])-min(Points[:,1])
prex=x1+sign(x1-x2)*dom*0.01
prey=(y1-y2)/(x1-x2)*dom*0.01+y1
endx=x4+sign(x4-x3)*dom*0.01
endy=(y4-y3)/(x4-x3)*dom*0.01+y4
print len(Points)
Points=list(Points)
Points.insert(0,array([prex,prey]))
Points.append(array([endx,endy]))
print len(Points)


# In[9]:

#Define alpha
a=0.

# Calculate the Catmull-Rom splines through the points
c = CatmullRomChain(Points,a)

# Convert the Catmull-Rom curve points into x and y arrays and plot
x,y = zip(*c)
plt.plot(x,y,c='green',zorder=10)

# Plot the control points
px, py = zip(*Points)
plt.plot(px,py,'or')

a=0.5
c = CatmullRomChain(Points,a)
x,y = zip(*c)
plt.plot(x,y,c='blue')

a=1.
c = CatmullRomChain(Points,a)
x,y = zip(*c)
plt.plot(x,y,c='red')


plt.grid(b=True)
plt.show()


# In[10]:

Points


# In[ ]:

original code: https://en.wikipedia.org/wiki/Centripetal_Catmull%E2%80%93Rom_spline

Heterophony answered 25/8, 2016 at 16:20 Comment(0)

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