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Licence CPOL
First Posted 20 Oct 2005
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Interpolation with Polynomials and Splines

By | 19 Oct 2005 | Article
A port of the Java sample project written by Tobias von Petersdorff
Sample Image - SplineInterpolation.jpg

Introduction

First of all, take a look here, where Tobias von Petersdorff has published his interpolation form written in Java.

Have you tried to find some sample code written in C# that helps you to draw a Spline? I've tried with no success.

This is a port of his sample project.

switch (mode) { 
case(InterpolationMode.POLY):
// use divided difference algorithm to compute Newton form coefficients
for (int k=1; k<=np-1; k++) 
{
for (int i=0; i<=np-1-k; i++) 
{
yCoords[i]=(yCoords[i+1]-yCoords[i])/(xCoords[i+k]-xCoords[i]);
}
}

// for equidistant points along x-axis evaluate polynomial and draw line
float dt = ((float) this.graphPanel.Width-1)/npp;
for (int k=0; k<=npp; k++) 
{
x=k*dt;
// evaluate polynomial at t
y = yCoords[0];
for (int i=1; i<=np-1; i++) 
{ 
y=y*(x-xCoords[i])+yCoords[i];
}
// draw line

try
{
g.DrawLine(linePen, (int)oldx,(int)oldy,(int)x,(int)y);
}
catch(Exception ex)
{
Console.WriteLine(ex.Message + " x1={0} y1={1} x2={2} y2={3}", oldx, oldy, x, y);
}
oldx=x;
oldy=y;
}
break;
case(InterpolationMode.NAT_SPL): 
if (np>1)
{ 
float[] a = new float[np];
float x1;
float x2;
float[] h = new float[np];
for (int i=1; i<=np-1; i++)
{
h[i] = xCoords[i] - xCoords[i-1];
}
if (np>2)
{
float[] sub = new float[np-1];
float[] diag = new float[np-1];
float[] sup = new float[np-1];
for (int i=1; i<=np-2; i++)
{
diag[i] = (h[i] + h[i+1])/3;
sup[i] = h[i+1]/6;
sub[i] = h[i]/6;
a[i] = (yCoords[i+1]-yCoords[i])/h[i+1]-(yCoords[i]-yCoords[i-1])/h[i];
}
solveTridiag(sub,diag,sup,ref a,np-2);
}
// note that a[0]=a[np-1]=0
// draw
oldx=xCoords[0];
oldy=yCoords[0];
try
{
g.DrawLine(linePen, (int)oldx,(int)oldy,(int)oldx,(int)oldy);
}
catch(Exception ex)
{
Console.WriteLine(ex.Message + " x1={0} y1={1} x2={0} y2={1}", oldx, oldy);
}

for (int i=1; i<=np-1; i++) 
{ // loop over intervals between nodes
for (int j=1; j<=precision; j++)
{
x1 = (h[i]*j)/precision;
x2 = h[i] - x1;
y = ((-a[i-1]/6*(x2+h[i])*x1+yCoords[i-1])*x2 +
(-a[i]/6*(x1+h[i])*x2+yCoords[i])*x1)/h[i];
x=xCoords[i-1]+x1;
if ((lastPen == null) || (lastPen == linePen2))
linePen = linePen1;
else
linePen = linePen2;
lastPen = linePen;

try
{
g.DrawLine(linePen, (int)oldx,(int)oldy,(int)x,(int)y);
}
catch(Exception ex)
{
Console.WriteLine(ex.Message + " x1={0} y1={1} x2={2} y2={3}", oldx, oldy, x, y);
}

oldx=x;
oldy=y;
}
}
}
break;
}

History

  • 20th October, 2005: Initial post

License

This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

About the Author

Marco Roello

Web Developer

Italy Italy

Member



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GeneralNot works when points are close one to another PinmemberRetro_011:27 28 May '10  
GeneralIt works only in one dimension Pinmemberkuratowski12:52 10 Nov '09  
Generalquestion for Marco Roello [modified] Pinmemberflashenoise0:42 30 May '06  
QuestionNice. Did you try Graphics.DrawCurve() ? Pinmemberaxelriet0:31 21 Oct '05  
AnswerRe: Nice. Did you try Graphics.DrawCurve() ? PinmemberMarco Roello1:39 21 Oct '05  
GeneralRe: Nice. Did you try Graphics.DrawCurve() ? Pinmemberaxelriet1:55 21 Oct '05  
GeneralBad source Link PinmemberMarco Roello21:15 20 Oct '05  
QuestionThanks. How about the sources? Pinsuss_MMM_16:15 20 Oct '05  
AnswerRe: Thanks. How about the sources? PinsussAnonymous20:32 20 Oct '05  

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