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Posted 29 Dec 2007
Licenced CPOL

Interpolation of BezierSplines and Cubic Splines

, 5 Jan 2008
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Algorithms to compute the Y - Value at an X - Position of a curve, constructed by some support-points


Its easy to draw a BezierSpline with GDI+ ( call Graphics.DrawCurve(Points) ), but how get any arbitrary Point on that drawn curve?

What I'm talking about

Please take a look at what the screenshot shows:
A Bezier-Splines-Curve (brown), constructed by 7 support-points, which subdivide the BezierSplines into 6 Bezier-Segments.
The 19 construction-points, which model the curve (orange).
The "Curve-Pointer" (red). It can be moved along the BezierSplines and its location is displayed, computed by interpolating the BezierSplines.

How BezierSplines is constructed

Each segment between two supportpoints is constructed as BezierCurve with 4 construction-points:
The two support-points themselfes and two additional, which take care, that the grade of the bezier arriving the support-point is the same as the grade of the bezier, which leaves the support-point.

Now interpolate it

To get the Y-value of an X-position I interpolate the BezierSplines in two steps:
First I search the Bezier-Segment, which contains the X-position. This is quickly done by a binary search:

Dim Indx = _Points.BinarySearch(New PointF(X, 0), Function(P1, P2) P1.X.CompareTo(P2.X))

(The Comparison of this search only compares the Point.X-Values.)
Indx will be the bit-complement of the index of the first point with point.X > X.

This data I pass to the second step, to InterpolateSegment()

Protected Overrides Function InterpolateSegment( _
      ByVal X As Single, ByVal Indx As Integer) As PointF
   'Since for Beziers there's no Y = f(X) function available, I aproximate X
   'by binary try and error.
   'note: Indx references the support-point *after* X
   Indx = Indx * 3 'now Indx references the last *construction-point* of the 
   '                   BezierSegment in Me.DrawPath, which is to interpolate
   Dim Pts = Me.DrawPath.PathPoints
   Dim BezierSegment = New PointF() { _
      Pts(Indx - 3), Pts(Indx - 2), Pts(Indx - 1), Pts(Indx)}
   'binary aproximation
   Dim Range = New Single() {0, 0, 1}  'elsewhere known as: Bottom, Mid, Top
   Dim Dlt = -2.0F
   Dim Pt As PointF
      Range(If(Dlt < 0, 0, 2)) = Range(1)          'set Bottom or Top to Mid
      Range(1) = (Range(0) + Range(2)) / 2                    'recompute Mid
      Pt = PointOfFraction(Range(1), BezierSegment)                     'try
      Dlt = Pt.X - X        'for drawing an "error" smaller 0.5 is tolerable
   Loop Until Math.Abs(Dlt) < 0.5
   Return Pt
End Function

Ups! Didn't I mention DrawPath? Its a GraphicsPath, which stores all the construction-points of the BezierSplines, I'd like to get drawn. Very useful! The construction-points are stored in the ".PathPoints" - Property.

Now we come to the Casteljau-algorithm, which computes a point on a Bezier.
I'm lucky: The commentation is sufficient, so I can do without to make still more words.

Private Shared Function PointOfFraction( _
         ByVal Fraction As Single, ByVal ptBeziers() As PointF) As PointF
   'Calculating a point on a BezierCurve bases on a given *Fraction* of the curve (half, 
   'third, else). For each two construction-points the proportionate between-point is 
   'computed, and is taken as new construction-point.
   'So in each loop there will be found one point less, until there's only one left
   'Strongly recommended: search "Casteljau" on Wikipedia. Surve to "Bezier-curve" and to
   '"Casteljau-algorithm" (which is implemented here)
   Dim Pts(ptBeziers.Length - 2) As PointF
   For I = 0 To ptBeziers.Length - 2   'first loop copies the points to a temporary array
      Pts(I) = PointBetween(ptBeziers(I), ptBeziers(I + 1), Fraction)
   For UBord = Pts.Length - 2 To 0 Step -1           'following loops overwrite the array
      For I = 0 To UBord
         Pts(I) = PointBetween(Pts(I), Pts(I + 1), Fraction)
   Return Pts(0)
End Function
Private Shared Function PointBetween( _
         ByVal Pt1 As PointF, ByVal Pt2 As PointF, ByVal Fraction As Single) As PointF
   'compute the to Fraction proportionate between-point between Pt1 and Pt2 
   Return Pt1.Add(Pt2.Subtract(Pt1).Mult(Fraction))      'Pt1 + (Pt2 - Pt1) * Fraction
End Function 


To interpolate Bezier-Spline-Segments as an Y = f(X) - function is mathematically incorrect.
Although I keep the support-points ordered "from left to right" a Segment can take forms, where it shows more than one Y-Value on certain X-positions.
My "interpolation" ignores such, and simply returns the first found Y-Value.
The failure lets itself be seen by some parts of the curve, which cannot be reached by interpolation.

Mathematically correct is to interpolate CubicSplines.
They are mathematically undefined only if two support-points set on the same X-position (as vertical line).

Polygons, Cubic Splines

while interpolating Polygons is trivial, the same on CubicSplines is - ah - untrivial. There's no GDI+ - Function which draws it for you, so there is to deal with linear algebra to solve linear equation-systems - brrr! - I've done that without real understanding. For that:


to Marco Roello.

My exercises on CubicSplines base on his article.


The project deals with several ownerdraw-requirements:
move- and highlight-able points, polygons, bezier-splines, cubic-splines, and a caption

I collected them into a little hierarchy of classes:

DrawObjectBase --DrawPoint
               \         BezierSpline
                \       /

You see: programmically I consider a spline as a polygon with different way to draw and interpolate the line between two support-points. And if there are only two support-points - they actually display the same straight line.


30/12/2007: submit article, sample-solution in VB2008 prof, beta

1/5/08: VB2005-version to zip-file added


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


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Comments and Discussions

QuestionOpen in VS2005? Pin
jronecube4-Jan-08 3:04
memberjronecube4-Jan-08 3:04 
AnswerRe: Open in VS2005? Pin
Mr.PoorEnglish4-Jan-08 12:44
memberMr.PoorEnglish4-Jan-08 12:44 

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