## Introduction

There are many implementations of interpolation schema based on the Bezier curve around the web, but they all seem to be either specifically oriented about one curve, or have functions that are not general enough for a wide variety of curves.

The build in Bezier curve is just the quadratic and cubic curve, and should be enough for the simple use, but in many cases this simply won't do. The goal of this article is to give a program with reusable functions that you could easily import into your program.

## Background

Let's say that you wanted to create a smooth curve that went through some points, and you are stuck in the 1950's; What you'd have to do is to place ducts at the points you wanted to stay still, get a spline with the desired stiffness, and climb up on the table like this guy from Boeing:

As Microsoft writes in their MSDN article "*A physical spline is a thin piece of wood or other flexible material*" and voila the term Spline later got coined by Isaac Jacob Schoenberg. Further history behind the development could be read here, and the basic is that it was used many places, but the mathematics behind it didn't take off until the 50's and 60's*, *and I quote*:*

*The use of splines for modeling automobile bodies seems to have several independent beginnings. Credit is claimed on behalf of de Casteljau at Citroen, Bezier at Renault, and Birkhoff, Garabedian, and de Boor at General Motors, all for work occurring in the very early 1960's or late 1950's. At least one of de Casteljau's papers was published, but not widely, in 1959. De Boor's work at GM resulted in a number of papers being published in the early 60's, including some of the fundamental work on B-splines. *

## Bezier curves

The most commonly know way of constructing a Bezier curve is to use De Casteljau's algorithm, which is a linear interpolation to generate the Bezier curve. However, the Bezier curve can also be defined in a different way, through the use of Bernstein polynomials.

### Bernstein polynomial

The Bernstein polynomial is linked with a theorem published by Karl Weierstrass in 1885 (He was 70 years old at the time, so the theory that mathematical innovations are only done by young people (20 or younger) should perhaps be laid to rest?). He basically found out that "*any continuous function on the interval [0,1] may be uniformly approximated, arbitrarily closely, by polynomials".* The Bernstein polynomial is simply a different proof of Weierstrass theorem, and it can be shown trough the standard way; with the use of the binomial function:

\begin{equation} B_{i,n}(t) = {n \choose i} t^i (1-t)^{n-i} \end{equation}

However If you have taken statistics you would (or should) notice that the formula is the exact same as the

Binomial mass distribution. At first, the formula might seem a bit strange and exotic, but in reality, it is really simple. The first thing that one needs to understand is the phrase

\({N \choose k}\), and written out this formula is simply:

\begin{equation} {N \choose k} = \frac{N!}{k! \cdot (N - k) !} \end{equation}

Explain in layman's terms it just explains that if you have N elements and want to pick out k elements from it, there is exactly \({N \choose k} \) ways of doing it. The aforementioned formula really consists of two distinct parts, the first is which explains the number of ways you can pick out k elements from N: \(\frac{N!}{(N-k)!}\), and the second \(\frac{1}{k!}\) simply eliminates permutations, number of different ways you can combine \(k\) elements in.

Why the factorial appears in selecting elements is actually really neat. If you have a card deck of 52 cards and you want to choose 4 cards from it, I can break the statistics of it down of each draw from the deck as follows:

- The first time I choose a card, there is exactly 52 cards to pick from, so I have 52 equally likely choices.
- The next time I want to take a card from the deck, there is now just 51 cards left to choose from. The number of choices I had to get exactly the two first cards is then \(52 \cdot 51\).
- The third draw, 50 cards to choose from, number of combinations is \(52 \cdot 51 \cdot 50\).
- The fourth time the odds are \(52 \cdot 51 \cdot 50 \cdot 49\).

This is quite tedious to write out, but luckily there is a simple general way of creating this series on by simply writing out:

\begin{equation} \frac{N!}{(N-k)!} = \frac{52 \cdot 51 \cdots 2 \cdot 1}{48 \cdot 47 \cdots 2 \cdot 1} \end{equation}

This will give the number of ways to combine 4 cards in an explicit sequence, but normally you don't care about the sequence of events you draw them in, just the resulting 4 cards. It can be shown by this simple analogy; if you have to choose 2 cards, they can either come in the sequence {1,2} or {2,1} and the two ways are called permutations in math. The number of ways to combine the sequence and still get the cards you want is 2! or in our case with 4 cards 4!. We want to eliminate number of permutations in the way of getting k so we modify the formula:

\begin{equation}{N \choose k} = \frac{N!}{k!(N-k)!}\end{equation}

The number of ways to combine them is just half the story when calculating the probability of getting the desired outcome. You also need to determine the probability of getting the desired outcome vs not getting it. The easiest way to do this with the Bernstein polynomial is to choose 2 points and combine them. There is a demand that the probability of the desired outcome plus the probability of undesired outcome must be 1 for all t. It is just one way of achieving this and thus the formula becomes:

$B_{i,n}(t) = {n \choose i} t^i (1-t)^{n-i}$

We generally don't like to take factorial with computers, as the numbers can get frighteningly large very quickly and is thus highly subjectable to roundoff errors. The formula is instead implemented by the use of the recursive relation of the Bernstein polynomial:

$B_{i,n} = (1-t) \cdot B_{i,n-1}+ t \cdot B_{i-1,n-1}$

This can be translated into code as given below:

Private Function AllBernstein(ByVal n As Integer, ByVal u As Double) As Double()
Dim B(n - 1) As Double
B(0) = 1
Dim u1 As Double
u1 = 1 - u
Dim saved, temp As Double
For j As Integer = 1 To n - 1
saved = 0
For k As Integer = 0 To j - 1
temp = B(k)
B(k) = saved + u1 * temp
saved = u * temp
Next
B(j) = saved
Next
Return B
End Function

The code will produce all the Bernstein polynomials that are required given the number of points in the input. The curves are shown below, ranging from 1 point to 6 points given to the function (image from MathWorld):

The implementation, given that you have gotten all your Bernstein polynomials, is pretty straight forward. It is just the sum of the Bernstein polynomial and the point that belongs to it:

Private Function BezierFunction(ByVal p As PointCollection, ByVal StepSize As Double) As PointCollection
Dim result As New PointCollection
Dim B As Double()
For k As Double = 0 To 1 Step StepSize
B = AllBernstein(p.Count, k)
Dim CX, CY As Double
CX = 0
CY = 0
For j As Integer = 0 To p.Count - 1
CX = CX + B(j) * p(j).X
CY = CY + B(j) * p(j).Y
Next
result.Add(New Point(CX, CY))
Next
Return result
End Function

### Derivative of the Beziers Curve

The derivative of the Bernstein polynomial could be used to generate the tangent of any Bezier curve, as is done here in an article on Silverlight. I'm going to do the same thing only with some slight changes.

Remember that it is only the Bernstein polynomial that is dependent on changes in u. This means that the knots as well as the start and endpoint is to be considered a constant. There are several articles about how to get the derivative of the Bezier curve, however few mentions that one can use the Bernstein polynomial to get the derivative. The relation that one is interested in is:

*B'*_{i,n}(u) )= n ( *B*_{i-1,n-1}(u) - *B*_{i,n-1}(u))

One must of course remember that the Bernstein polynomial is defined to be zero if *i* is below zero, or *i* is higher than *n*. In mathematical (Read: In short) terms:

*B*_{-1,n-1}(u) = *B*_{n,n-1}(u) = 0

The code for the derivative of the Bernstein polynomial could then easily be constructed:

Private Function AllDerivateBernstein(ByVal n As Integer, ByVal u As Double) As Double()
Dim B As Double()
B = AllBernstein(n - 1, u)
Dim result(n - 1) As Double
For i As Integer = 0 To n - 1
If i = 0 Then
result(i) = n * (-B(i))
ElseIf i = n - 1 Then
result(i) = n * (B(i - 1))
Else
result(i) = n * (B(i - 1) - B(i))
End If
Next
Return result
End Function

Since it is only the Bernstein polynomial that provides the line in the curve, together with the control points, one can use the series expansion (or power laws) to integrate the polynomial part by part. All we have to do now is to use the derivative to calculate the slope:

Private Function DerivateBezierFunction(ByVal p As PointCollection, ByVal StepSize As Double) As PointCollection
Dim result As New PointCollection
Dim B As Double()
For k As Double = 0 To 1 Step StepSize
B = AllDerivateBernstein(p.Count, k)
Dim test() As Double = AllDerivateBernstein(p.Count, k)
Dim CX, CY As Double
CX = 0
CY = 0
For j As Integer = 0 To p.Count - 1
CX = CX + B(j) * p(j).X
CY = CY + B(j) * p(j).Y
Next
result.Add(New Point(CX, CY))
Next
result.Add(p(p.Count - 1))
Return result
End Function

All you need to do now, is to divide y'/x' and you'll get the slope at a given point. Since this is an analytical method it is valid in every step point from 0 to 1 in the Bezier curve.

So, I decided to nick the arrow picture from the article "Bezier curve angle calculation in Silverlight", and use it in my implementation. There is however, a problem, as the rotation should not revolve around the center, or rather it can, but it is easier to revolve it around its given point at 0,0 and move that point by half the Length/Width with a 90-degree turn. This ensures that the arrow would always be in the right position, and the code makes easy use of trigonometry:

Private Function MovePoint(ByVal StartPoint As Point, ByVal Length As Double, Optional ByVal theta As Double = 90) As Point
Dim EndPoint As New Point
theta *= Math.PI / 180
EndPoint.X = Math.Cos(theta) * (Length) - Math.Sin(theta) * (Length) + StartPoint.X
EndPoint.Y = Math.Sin(theta) * (Length) + Math.Cos(theta) * (Length) + StartPoint.Y
Return EndPoint
End Function

If you see in the comments section there is one that points out that one could use the `Tan2`

method to calculate the angle. It takes in two points, and calculates the angle in relation to a straight line on the X-axis:

Private Function AngleInDegree(ByVal start As Point, ByVal endd As Point) As Double
Return Math.Atan2(start.Y - endd.Y, endd.X - start.X) * 180 / Math.PI
End Function

The code for placing the image and rotating it could also be simplified:

RemoveImage()
Dim image As New Image()
Dim uri As New Uri("arrow.jpg", UriKind.Relative)
Dim img As New System.Windows.Media.Imaging.BitmapImage(uri)
Dim pp As Point = MovePoint(New Point(ll.X1, ll.Y1), 10, angle - 90)
image.Margin = New Thickness(pp.X, pp.Y, 0, 0)
image.Source = img
image.Width = 20
image.Height = 20
image.Stretch = Stretch.Fill
Dim RotTransform As New RotateTransform
RotTransform.Angle = angle
image.RenderTransform = RotTransform
cnvMain.Children.Add(image)

### WPF's Bezier methods

Now, there is also the possibility of using the predefined method in WPF to draw a cubic Bezier segment. MSDN gives the following example in XAML:

<Path Stroke="Black" StrokeThickness="1">
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure StartPoint="10,100">
<PathFigure.Segments>
<PathSegmentCollection>
<BezierSegment Point1="100,0" Point2="200,200" Point3="300,100" />
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>

And there is a different code for the quadratic Bezier segment too (also from MSDN):

<Path Stroke="Black" StrokeThickness="1">
<Path.Data>
<PathGeometry>
<PathGeometry.Figures>
<PathFigureCollection>
<PathFigure StartPoint="10,100">
<PathFigure.Segments>
<PathSegmentCollection>
<QuadraticBezierSegment Point1="200,200" Point2="300,100" />
</PathSegmentCollection>
</PathFigure.Segments>
</PathFigure>
</PathFigureCollection>
</PathGeometry.Figures>
</PathGeometry>
</Path.Data>
</Path>

Both of these versions is included if you decide to use the Bernstein polynomial instead, however one could also set up animations in XAML by using the code above.

## Rational Bezier curves

If one uses the Bernstein polynomial alone, all points are equally important. If one, on the other hand, wants to assign different weight on each of the points (same as different mass in the solar system), one gets more control over the shape of the Bezier curve. With proper usage one can even replicate a circle with just three points.

So, instead of using the Bernstein polynomial directly, one multiply it by a weight, called *w*_{i} . However, to ensure that the curve stays within boundaries it is important that the sum is equal to 1 for all u. This is quite simply called the rational basis function R_{i,n}(u) and is defined:

Private Function RationalBasisFunction(ByVal n As Integer, ByVal u As Double, ByVal weight() As Double) As Double()
If weight.Length <> n Then Return Nothing
Dim B() As Double
B = AllBernstein(n, u)
Dim result(n - 1) As Double
Dim test As Double
test = 0
For j As Integer = 0 To n - 1
test += B(j) * weight(j)
Next
For i As Integer = 0 To n - 1
result(i) = B(i) * weight(i) / test
Next
Return result
End Function

## Catmull-Rom interpolation

I know programmers can be kind of lazy, so they could just see that they need a Catmull-Rom interpolation, do a search, and find one. Like this one and grab the code:

Public Function PointOnCurve(ByVal p0 As System.Windows.Point, ByVal p1 As System.Windows.Point, ByVal p2 As System.Windows.Point, ByVal p3 As System.Windows.Point, ByVal t As Double) As System.Windows.Point
Dim result As New System.Windows.Point()
Dim t2 As Single = t * t
Dim t3 As Single = t2 * t
result.X = 0.5F * ((2.0F * p1.X) + (-p0.X + p2.X) * t + (2.0F * p0.X - 5.0F * p1.X + 4 * p2.X - p3.X) * t2 + (-p0.X + 3.0F * p1.X - 3.0F * p2.X + p3.X) * t3)
result.Y = 0.5F * ((2.0F * p1.Y) + (-p0.Y + p2.Y) * t + (2.0F * p0.Y - 5.0F * p1.Y + 4 * p2.Y - p3.Y) * t2 + (-p0.Y + 3.0F * p1.Y - 3.0F * p2.Y + p3.Y) * t3)
Return result
End Function

There is also another way, and that is to try and understand what the Catmull-Rom spline really is. The polynomial that is used in the code above is actually generated from the derivatives in the figure below. But it only uses the derivative to define two control points between point 1 and point 2.

The derivative is of course defined as usual:

And the points P_{i}^{+} and p_{i+1}^{-} is the control points designed by the derivative:

We can now calculate two new points between the points marked Pi and Pi+1, the clue now is to actually use a 3rd degree Bezier curve by the use of the start point Pi and the end point Pi+1, and two-knot points Pi+ and Pi+1-. We now understand that the Catmull-Rom spline is just a special form of 3rd degree Bezier curve. Useful links that explains this and where the figures are from is here and here.

Public Function PointOnCurve(ByVal p0 As System.Windows.Point, ByVal p1 As System.Windows.Point, ByVal p2 As System.Windows.Point, ByVal p3 As System.Windows.Point, ByVal t As Double) As System.Windows.Point
Dim result As New System.Windows.Point()
Dim lix1, liy1, lix2, liy2 As Double
lix1 = 0.5 * (p2.X - p0.X)
lix2 = 0.5 * (p3.X - p1.X)
liy1 = 0.5 * (p2.Y - p0.Y)
liy2 = 0.5 * (p3.Y - p1.Y)
Dim PointList As New PointCollection
PointList.Add(p1)
PointList.Add(New Point(p1.X + (1 / 3) * lix1, p1.Y + (1 / 3) * liy1))
PointList.Add(New Point(p2.X - (1 / 3) * lix2, p2.Y - (1 / 3) * liy2))
PointList.Add(p2)
Return PointBezierFunction(PointList, t)
End Function

The problem I faced was with a single line, which I often had to deal with, so how should I interpolate those? I choose to use the logic that the weather tomorrow will be the same as the weather today, or rather half a day. I added one point prior and one point past the original line, and I said that there was a point at the same slope as between the two first points, and the two last points, only half as long away. It seem to generate a good agreement as long as you did not go completely mental in the beginning or end.

All that is left to construct is a wrapper that can take the standard PointCollection as input for the Catmull-Rom spline:

Public Function CatmullRomSpline(ByVal Points As PointCollection, Optional ByVal InterpolationStep As Double = 0.1, Optional ByVal IsPolygon As Boolean = False) As PointCollection
Dim result As New PointCollection
If Points.Count <= 2 Then
Return Points
End If
If IsPolygon Then
For i As Integer = 0 To Points.Count - 1
If i = 0 Then
For k As Double = 0 To (1 - InterpolationStep) Step InterpolationStep
result.Add(PointOnCurve(Points(Points.Count - 1), Points(i), Points(i + 1), Points(i + 2), k))
Next
ElseIf i = Points.Count - 1 Then
For k As Double = 0 To (1 - InterpolationStep) Step InterpolationStep
result.Add(PointOnCurve(Points(i - 1), Points(i), Points(0), Points(1), k))
Next
ElseIf i = Points.Count - 2 Then
For k As Double = 0 To (1 - InterpolationStep) Step InterpolationStep
result.Add(PointOnCurve(Points(i - 1), Points(i), Points(i + 1), Points(0), k))
Next
Else
For k As Double = 0 To (1 - InterpolationStep) Step InterpolationStep
result.Add(PointOnCurve(Points(i - 1), Points(i), Points(i + 1), Points(i + 2), k))
Next
End If
Next
Else
Dim yarray, xarray As New List(Of Double)
xarray.Add(Points(0).X - (Points(1).X - Points(0).X) / 2)
yarray.Add(Points(0).Y - (Points(1).Y - Points(0).Y) / 2)
For Each ps As System.Windows.Point In Points
xarray.Add(ps.X)
yarray.Add(ps.Y)
Next
xarray.Add((Points(Points.Count - 1).X - (Points(Points.Count - 2).X) / 2 + Points(Points.Count - 1).X))
yarray.Add((Points(Points.Count - 1).Y - (Points(Points.Count - 2).Y) / 2 + Points(Points.Count - 1).Y))
Dim r As New PointCollection
For i As Integer = 0 To yarray.Count - 1
r.Add(New System.Windows.Point(xarray(i), yarray(i)))
Next
For i As Integer = 3 To r.Count - 1
For k As Double = 0 To (1 - InterpolationStep) Step InterpolationStep
result.Add(PointOnCurve(r(i - 3), r(i - 2), r(i - 1), r(i), k))
Next
Next
result.Add(Points(Points.Count - 1))
End If
Return result
End Function

The Catmull-Rom spline is prone to rather nasty curves if the control points are very unevenly distributed. If you take a look at what happens to that curve if you take 4 points and place the two center points quite close together and have the start and endpoint quite a distance from the two center points. You'll get a sharp bend between the two center points.

It uses the Bessel tangent method or Overhausers method to generate the spline, by the use of a non-uniform weight *t*_{i}. Incidentally, if you have a uniform weight it will produce the Catmull-Rom Spline instead. The values of *t*_{i} is suppose to reduce the "velocity" so that a more smooth curve can appear.

I didn't find any implementation of the curve so it's created from an algorithm in the book "Mathematical tools in computer graphics with C# implementations", although you can find the implementation described on this site or in this pdf master thesis.

Public Function PointOnBesselOverhauserCurve(ByVal p0 As System.Windows.Point, ByVal p1 As System.Windows.Point, ByVal p2 As System.Windows.Point, ByVal p3 As System.Windows.Point, ByVal u As Double, Optional ByVal t() As Double = Nothing) As System.Windows.Point
Dim result As New System.Windows.Point()
If t Is Nothing Then
t = {1, 2, 3, 4}
End If
Dim ViXPlusHalf, VixMinusHalf, ViYPlusHalf, ViYMinusHalf, ViX, ViY As Double
ViXPlusHalf = (p2.X - p1.X) / (t(2) - t(1))
VixMinusHalf = (p1.X - p0.X) / (t(1) - t(0))
ViYPlusHalf = (p2.Y - p1.Y) / (t(2) - t(1))
ViYMinusHalf = (p1.Y - p0.Y) / (t(1) - t(0))
ViX = ((t(2) - t(1)) * VixMinusHalf + (t(1) - t(0)) * ViXPlusHalf) / (t(2) - t(0))
ViY = ((t(2) - t(1)) * ViYMinusHalf + (t(1) - t(0)) * ViYPlusHalf) / (t(2) - t(0))
Dim PointList As New PointCollection
PointList.Add(p1)
PointList.Add(New Point(p1.X + (1 / 3) * (t(2) - t(1)) * ViX, p1.Y + (1 / 3) * (t(2) - t(1)) * ViY))
ViXPlusHalf = (p3.X - p2.X) / (t(3) - t(2))
VixMinusHalf = (p2.X - p1.X) / (t(2) - t(1))
ViYPlusHalf = (p3.Y - p2.Y) / (t(3) - t(2))
ViYMinusHalf = (p2.Y - p1.Y) / (t(2) - t(1))
ViX = ((t(3) - t(2)) * VixMinusHalf + (t(2) - t(1)) * ViXPlusHalf) / (t(3) - t(1))
ViY = ((t(3) - t(2)) * ViYMinusHalf + (t(2) - t(1)) * ViYPlusHalf) / (t(3) - t(1))
PointList.Add(New Point(p2.X - (1 / 3) * (t(3) - t(2)) * ViX, p2.Y - (1 / 3) * (t(3) - t(2)) * ViY))
PointList.Add(p2)
Return PointBezierFunction(PointList, u)
End Function

## Lagrange Interpolation

This is one of the first properly known algorithms for drawing continuous curves , and this implementation uses code from "Mathematical tools in Computer graphics with C# implementations. It uses Neville's algorithm to calculate the basis function for a given t:

Private Function LagrangeBasisFunction(ByVal n As Integer, ByVal k As Integer, ByVal t As Double) As Double
Dim l As Double = 1
Dim tj, tk As Double
tk = k / (n - 1)
For j As Integer = 0 To n - 1
If j <> k Then
tj = j / (n - 1)
l *= (t - tj) / (tk - tj)
End If
Next
Return l
End Function

We must then use the same principle as the Bernstein polynomial to calculate the value at a given point on the curve:

Private Function GetLagrangianAtPointT(ByVal p As PointCollection, ByVal t As Double) As Point
Dim n As Integer = p.Count
Dim rx, ry As Double
rx = 0
ry = 0
For i As Integer = 0 To n - 1
rx += LagrangeBasisFunction(n, i, t) * p(i).X
ry += LagrangeBasisFunction(n, i, t) * p(i).Y
Next
Return New Point(rx, ry)
End Function

Then the wrapper to translate the mathematics into a simple call that returns the curve:

Private Function LargrangianInterpolation(ByVal p As PointCollection, Optional ByVal StepSize As Double = 0.01) As PointCollection
Dim result As New PointCollection
For i As Double = 0 To 1 Step StepSize
result.Add(GetLagrangianAtPointT(p, i))
Next
result.Add(p(p.Count - 1))
Return result
End Function

Now that you have this, I must warn you that by using it you are quite confident that all you points are quite evenly spaced in both X and Y direction. If not, it might start to oscillate quite badly, so consider yourself warned.

## Convex Hull

A convex hull could be explained quite easily by thinking of the two-dimensional points as strikes that firmly planted in the ground. The convex hull would be the rubber band that encloses all the points. The most commonly used computer algorithm for convex hull uses the cross product to evaluate which side a point is relative to a line.

But to use that algorithm most efficiently (not checking the same point twice) one need's to sort the 2D points lexicographically, that means first sort the points in the X direction, and if the X values are equal, the point will be further sorted on the Y value. To do this with an array of point's you first need a class that implements `IComparer`

:

Private Class PointSort
Implements IComparer
Public Enum Mode
X
Y
End Enum
Private currentMode As Mode = Mode.X
Public Sub New(ByVal mode As Mode)
currentMode = mode
End Sub
Private Function IComparer_Compare(ByVal a As Object, ByVal b As Object) As Integer Implements IComparer.Compare
Dim point1 As Point = CType(a, Point)
Dim point2 As Point = CType(b, Point)
If currentMode = Mode.X Then
If point1.X > point2.X Then
Return 1
ElseIf point1.X < point2.X Then
Return -1
Else
If point1.Y > point2.Y Then
Return 1
ElseIf point1.Y < point2.Y Then
Return -1
Else
Return 0
End If
End If
Else
If point1.Y > point2.Y Then
Return 1
ElseIf point1.Y < point2.Y Then
Return -1
Else
If point1.X > point2.X Then
Return 1
ElseIf point1.X < point2.X Then
Return -1
Else
Return 0
End If
End If
End If
End Function
End Class

The `IComparer `

is used with the Array.Sort method, and placed inside a wrapper function:

Public Function SortPoints(ByVal samplepoints As PointCollection, ByVal SortByXdirection As Boolean) As PointCollection
Dim copySamplePoints As Point() = New Point(samplepoints.Count - 1) {}
samplepoints.CopyTo(copySamplePoints, 0)
If SortByXdirection Then
Array.Sort(copySamplePoints, New PointSort(PointSort.Mode.X))
Else
Array.Sort(copySamplePoints, New PointSort(PointSort.Mode.Y))
End If
Dim result As New PointCollection
For Each p As Point In copySamplePoints
result.Add(p)
Next
Return result
End Function

Checking which side a point lies relative to a line is implemented the following way:

Private Function WhichSide(ByVal PointToBeEvaluated As Point, ByVal StartPointOnLine As Point, ByVal EndPointOnLine As Point) As Integer
Dim ReturnvalueEquation As Double
ReturnvalueEquation = ((PointToBeEvaluated.Y - StartPointOnLine.Y) _
* (EndPointOnLine.X - StartPointOnLine.X)) - ((EndPointOnLine.Y - StartPointOnLine.Y) _
* (PointToBeEvaluated.X - StartPointOnLine.X))
If ReturnvalueEquation > 0 Then
Return -1
ElseIf ReturnvalueEquation = 0 Then
Return 0
Else
Return 1
End If
End Function

After creating all the basic functions needed, all that is left is to put it together. The code below should run on n*log n time, but I suspect that several improvements are possible, so that it could be even faster.

The code below does the following thing:
- Removes duplicate points
- Sorts the points
**Creates two arrays one sorts the points in descending order and the collection is called upper, and the other in ascending order and is called lower.** - Removes point in Upper if there is a point above it, and the same is done for lower
- Connect the two arrays again

The code is given below:

Public Function ConvexHull2D(ByVal points As PointCollection) As PointCollection
Dim SortedList As New PointCollection
Dim Upper As New PointCollection
Dim Lower As New PointCollection
Dim tempPoints As New PointCollection
For Each p As Point In points
If Not tempPoints.Contains(p) Then tempPoints.Add(p)
Next
Dim res As New PointCollection
SortedList = SortPoints(tempPoints, True)
Upper = SortedList
For i As Integer = SortedList.Count - 1 To 0 Step -1
Lower.Add(SortedList(i))
Next
Dim j_set As Boolean = False
Dim j As Integer = 0
Dim r As Integer
Do While j < Upper.Count - 2
r = WhichSide(Upper(j + 1), Upper(j), Upper(j + 2))
If r = -1 Or r = 0 Then
Upper.RemoveAt(j + 1)
j = 0
j_set = True
End If
If Not j_set Then
j += 1
End If
j_set = False
Loop
j = 0
j_set = False
Do While j < Lower.Count - 2
r = WhichSide(Lower(j + 1), Lower(j), Lower(j + 2))
If r = -1 Or r = 0 Then
Lower.RemoveAt(j + 1)
j = 0
j_set = True
End If
If Not j_set Then
j += 1
End If
j_set = False
Loop
For Each p As Point In Upper
res.Add(p)
Next
For Each p As Point In Lower
If Not res.Contains(p) Then
res.Add(p)
End If
Next
res.Add(res(0))
Return res
End Function

## Epilogue

There is a lot of information that got lost in the implementation. It is not directly involved but it needs to be addressed.

First off it the term of continuity, as a curve which is piecewise interpolated have, but the question remains how many of the derivative are continuous. A definition of a cure is C0, as it just says that the curve is continuous . But a curve would be c1 continuous if it also satisfied that the first derivative is also conscious. C2 of course, means that the first and second derivative are both continuous in each connecting point.

There are also something called G1 continuity, which demands that the changes are constant, i.a. the derivative is strictly increasing. A typical variation of this requirement would be the clothoid spline which is used to design turns in the road, with an increasing inward acceleration and deceleration.

One function is missing, and that is the general implementation of the B-Spline curve, and it is planned to write that code and include it here as well later.

## References

The bulk of information that is used in this article came from the first chapter in the book:

"The NURBS Book" 2nd Edition"

The code with Catmull-Rom and Bessel-Overhauser was described in the book below, and the Lagrange interpolation is just rewritten from it:

"Mathematical tools in computer graphics with C# implementation"

There are other sources as well, but they are mentioned in the main article.

**"Harmonic interpolation for smooth curves and surfaces"** by Alexandre Hardy (Ph.d thesis). Large parts of the previous book are taken from this thesis.

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