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If I have a simple 2-D matrix with normalized values on x-axis between 0 and 1 and y-axys between 0 and 1, and I have 3 points in this matrix e.g. P1 (x=0.2,y=0.9), P2 (x=0.5,y=0.1) and P3 (x=0.9,y=0.4).

How can I simply calculate a curve thru this points, meaning having a function which is giving me the y for any x.

I now that there are any number of possible curves thru 3 points. But hey, you know what I mean: I want a smooth curve thru it, usable for audio-sample-interpolation, usable for calculation a volume-fade-curve, usable for calculating a monster-walking-path in a game.

Now I have searched the net for this question about 3 days, and I cannot believe that there is no usable solution for this task. All the text dealing about Catmull-rom-Splines, bezier-curves and all that theroretical stuff has all at least one point which doesn't make it for me usable. For example Catmull-Rom-splines need to have a fix distance between the control-points (I would use this code and set the 4. point-y to the 3. point y) :

void CatmullRomSpline(float *x,float *y,float x1,float y1,float x2,float y2,float x3,float y3,float x4,float y4,float u)
{
//x,y are calculated for x1,y1,x2,y2,x3,y3 and x4,y4 if u is the normalized distance (0-1) in relation to the distance between x2 and x3 for my whiched point

float u3,u2,f1,f2,f3,f4;

u3=u*u*u;
u2=u*u;
f1=-0.5f * u3 + u2 -0.5f *u;
f2= 1.5f * u3 -2.5f * u2+1.0f;
f3=-1.5f * u3 +2.0f * u2+0.5f*u;
f4=0.5f*u3-0.5f*u2;

*x=x1*f1+x2*f2+x3*f3+x4*f4;
*y=y1*f1+y2*f2+y3*f3+y4*f4;

}

But I don't see that x1 to x4 have any affect on the calculation of y, so I think x1 to x4 must have the same distance?

...

Or bezier-code doesn't calcuate the curve thru the points. The points (at least the 2. point) seem only to have a force-effect on the line. 

typedef struct Point2D
{
double x;
double y;
} Point2D;

class bezier
{
std::vector<Point2D> points;
bezier();
void PushPoint2D( Point2D point );
Point2D GetPoint( double time );
~bezier();
};

void bezier::PushPoint2D(Point2D point)
{
points.push_back(point);
}

Point2D bezier::GetPoint( double x )
{
int i;
Point2D p;

p.x=0;
p.y=0;

if( points.size() == 1 ) return points[0];
if( points.size() == 0 ) return p;

bezier b;
for (i=0;i<(int)points.size()-1;i++)
{
    p.x = ( points[i+1].x - points[i].x ) * x + points[i].x;
    p.y = ( points[i+1].y - points[i].y ) * x + points[i].y;
    if (points.size()<=2) return p;
    b.PushPoint2D(p);
}

return b.GetPoint(x);
}

double GetLogicalYAtX(double x)
{
bezier bz;
Point2D p;

p.x=0.2;
p.y=0.9;
bz.PushPoint2D(p);

p.x=0.5;
p.y=0.1;
bz.PushPoint2D(p);

p.x=0.9;
p.y=0.4;
bz.PushPoint2D(p);

p=bz.GetPoint(x);

return p.y;
}

This is better than nothing, but it is 1. very slow (recursive) and 2. as I said doesn't really calculate the line thru the 2. point.

Is there a mathematical brain outside which could help me?

4

2 回答 2

0

感谢 TOCS (Scott) 提供您的代码，如果有时间我也会尝试一下。但是我现在测试的是INFACT（答案3）的提示：这个“大范围多项式”非常接近我正在寻找的内容：

我已将我的类贝塞尔曲线重命名为曲线，因为我添加了一些用于拉格朗日插值的代码。我还添加了 3 张图形表示的图片，代码是计算什么。

在图 1 中，您可以看到旧贝塞尔函数的松散中间点。

在图 2 中，您现在可以看到拉格朗日插值的全点结果。

在图 3 中，您可以看到唯一的问题，或者我应该说“我还需要解决的问题”（无论如何这是迄今为止最好的解决方案）：如果我移动中间点，曲线会变快，变快到上限或下限）。我希望它上下更顺畅。这样它看起来更像对数函数。这样它就不会过早地超出 0 和 1 之间的 y 边界。

现在我的代码如下所示：

curve::curve(void)
{
}

void curve::PushPoint2D(Point2D point)
{
    points.push_back(point);
}

Point2D curve::GetPoint( double x )
{
//GetPoint y for x with bezier-mathematics...

//was the only calculating function in old class "bezier"
//now the class is renamed "curve"
int i;
Point2D p;

p.x=0;
p.y=0;

if( points.size() == 1 ) return points[0];
if( points.size() == 0 ) return p;

curve b;
for (i=0;i<(int)points.size()-1;i++)
{
    p.x = ( points[i+1].x - points[i].x ) * x + points[i].x;
    p.y = ( points[i+1].y - points[i].y ) * x + points[i].y;
    if (points.size()<=2) return p;
    b.PushPoint2D(p);
}

return b.GetPoint(x);
}

//THIS IS NEW AND VERY VERY COOL
double curve::LagrangeInterpolation(double x)
{
double y = 0;

for (int i = 0; i <= (int)points.size()-1; i++)
{
    double numerator = 1;
    double denominator = 1;

    for (int c = 0; c <= (int)points.size()-1; c++)
    {
        if (c != i)
        {
            numerator *= (x - points[c].x);
            denominator *= (points[i].x - points[c].x);
        }
    }

    y += (points[i].y * (numerator / denominator));

}

return y;
}

curve::~curve(void)
{
}


double GetLogicalYAtX(double x)
{
curve cv;
Point2D p;

p.x=0; //always left edge
p.y=y1; //now by var
cv.PushPoint2D(p);

p.x=x2; //now by var
p.y=y2; //now by var
cv.PushPoint2D(p);

p.x=1; //always right edge
p.y=y3; //now by var
cv.PushPoint2D(p);

//p=cv.GetPoint(x);

//return p.y;

return cv.LagrangeInterpolation(x);
}

老贝塞尔曲线 拉格朗日，很酷但是... ...剪辑这么快...

你有什么想法我可以让新的解决方案更“软”吗？这样我就可以移动 2. 点在更大的区域而不会使曲线超出边界？谢谢

于 2013-02-22T16:25:22.657 回答
0 
static bezier From3Points(const Point2D &a, const Point2D &b, const Point2D &c)
{
    bezier result;
    result.PushPoint2D(a);

    Point2D middle;
    middle.x = 2*b.x - a.x/2 - c.x/2;
    middle.y = 2*b.y - a.y/2 - c.y/2;
    result.PushPoint2D(middle);

    result.PushPoint2D(c);
    return result;
}

未经测试，但应返回贝塞尔曲线，其中 t=0.5 处曲线通过点“b”。

此外（前面还有更多未经测试的代码），您可以使用像这样的伯恩斯坦基多项式来计算您的分数。

static int binomialcoefficient (int n, int k)
{
    if (k == 0)
        return 1;
    if (n == 0)
        return 0;

    int result = 0;
    for (int i = 1; i <= k; ++i)
    {
        result += (n - (k - i))/i;
    }
    return result;
}

static double bernstein (int v, int n, double t)
{
    return binomialcoefficient(v,n) * pow(t,v) * pow(1 - t,n - v);
}

Point2D GetPoint (double t)
{
    Point2D result;
    result.x = 0;
    result.y = 0;

    for (int i = 0; i < points.size(); ++i)
    {
        double coeff = bernstein (i,points.size(),t);
        result.x += coeff * points[i].x;
        result.y += coeff * points[i].y;
    }

    return result;
}
于 2013-02-21T20:04:26.037 回答