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Create a Voronoi diagram 1 of 2

, 26 Jul 2012 CPOL
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A detailed description for creating Voronoi diagrams, review of basic functions


Voronoi diagrams are quite useful tools in computational geometry and have a wide range of uses such as, calculating the area per tree in the forest, or figuring out where the poisoned wells were in a city (based on victims' addresses), and so on. In general it is useful for finding "who is closest to whom." A collection of problems where Voronoi diagrams are used is shown below:

  • Collision detection
  • Pattern recognition
  • Geographical optimization
  • Geometric clustering
  • Closest pairs algorithms
  • k-nearest-neighbor queries

If you are interested in even more examples you can go to the website here.

The history behind the Voronoi diagram is quite complicated as it has been rediscovered several times. The first written description of a Voronoi diagram was done by Descartes in the middle if the 17th century. The general idea was first conceived by Dirichlet in 1850, but was not given a rigid mathematical treatment before Voronoi's article in 1908. For a complete history behind it you should read the article by Aurenhammer. As an important side note, Delaunay (1934) proved that the Voronoi diagram had a sister (or brother if you will), called Delaunay triangulation, which (usually) can be constructed from a Voronoi diagram and vice versa. (Fun fact: Delaunay actually did his Ph.d. thesis under the supervision of Voronoi.) 

In the decades following the initial discovery there were several approaches for constructing a Voronoi diagram, but most of them were usually quite slow O(n^2). There was an algorithm that could do it in O(n*log n) time, but it was quite difficult to implement so it was barely used. It was not until Fortune discovered the sweep line algorithm in 1985 (published in 1987) - which also runs on O(n*log n) time - that a generally accepted method for constructing a Voronoi diagram was implemented. In this article I am going to try explain the basic functions for this algorithm. However, a brief overview of different algorithms for constructing the Voronoi diagram are shown below:

  • Sweep line
  • Divide-and-conquer
  • Spiral-search
  • Three dimensional convex hull
  • Incremental

The reason for writing this article about Fortune's sweep line algorithm is to clarify the implementation of a Voronoi diagram. This is mostly to clarify things for me as I try to implement the algorithm, with the hope that it is interesting for others as well and will hopefully increase your understanding of the basics. And yes, I am familiar with Ben's article on CodeProject. Some of the code used in this article came from his implementation.


What is a Voronoi diagram anyway? Well, there are several different types of Voroni diagrams, but they all share some common attributes that are quite general in style and can be explained by using simple metaphors. It is normally described as the area around a point, where the center point is the closest one in the surrounding polygon, as illustrated below:

There is a general underlying pattern, that can be illustrated by finding the halfwayline between just two points. We find the line that is equally far from the two points by constructing two circles that has a radius that is greater than half the distance between the two points. An illustration of the halfway line between the two points A and B: 

If we were just interested in finding the halfway point on the line between points A and B, there is one more possible way to calculate the point. If you take a look at the illustration below where a rope is tied to the points A and B, you immediately see that the halfway point is where the rope is at its lowest point. Therefore, this a complementary way of finding the halway distance between the two points as in the illustration of the two circles above.

If we replace the rope by a blanket and set up wooden sticks that all have the same height, you could imagine that all the valleys would give you all the Voronoi lines in the diagram, assuming that the sticks are high enough (or the tension in the blanket is high enough) so the blanket doesn’t touch the ground.

Hopefully the aforementioned image is simple enough for everyone to grasp, but it is not convenient to construct the Voronoi diagram from this analogy, or in Steven Fortune's words: "It is notoriously difficult to obtain a practical implementation of an abstractly described geometric algorithm."

A more mathematically sound way of looking at the Voronoi diagram is to place cones (or a pointy hat if you’d like - see picture below) with the slopes at the side tilted in a 45 degree angle, with the center point as the highest point. This would give the same overlapping sides that generates the straight lines between the points in the Voronoi diagram as seen in the first picture.

This would, in fact, mean that you will always find the overlapping line exactly at the half way point between any of the closest points. The cone is a collection of many circles with different radiuses so the analogy to the halfway line is still valid. The result from the example picture above gives the intersection lines below:

Fortune's sweep line algorithm is based on these principles, but instead of placing the cones with the point at the top, the cones are turned upside down. The sides of the cone are tilted at 45 degrees and an intersection sweep line (the plane is named pi in the picture below) is tilted at 45 degrees opposite to the tilted cones. You will also, I hope, notice that there is no height limitation on the cones by turning them upside down. All angles are given relative to the plain at z = 0. This approach is illustrated below where the sweep line constantly moves from 0 to x = infinite (or rather the boundary rectangle that is certain to complete all the lines in the Voronoi diagram for the area that we are interested in).

For anyone familiar with different geometrical curves (see the article about Conic slides from Wikipedia) it is clear that the intersection line between the sweep plain pi and the cone would form an parabolic equation. This intersection line is dependent on the relative position at the center of the cone P(x,y) and the position of the sweep line L. We will call the curved intersection line generated form the different arcs the beach line from now on.

The algorithm must at all times maintain the current location (y-coordinate) of the sweep line. It stores, in left-to-right (or top to bottom, as you can perform sweeps in both X and Y direction) the order of the set of sites (center point on the cones) that define the beach line. It is, however, important to know that the algorithm never needs to store all the parabolic arcs of the beach line. We will only maintain a sweep-line status and we are interested in simulating the discrete event points where there is a "significant event" occurs. That is, any event that changes the topological structure of the Voronoi diagram and the beach line.

Two kinds of events appear that could change the beach line. The first is a site event and the second is a circle event.

Site Events  

When the sweep line passes over a new site a new arc will be inserted into the beach line.

The equation for any parabola given Fortune's sweep line algorithm is given by the equation below were Pjx and Pjy is x and y position to the center point of the cone above the sweep line, Ly is the position of the sweep line, and X is the location of the new intersection point on the sweep line:

The formula must be used on all the points passed by the sweep line in order to find the one that gives you the shortest line from the point to the sweep line. You can now easily generate all the arcs from all the points, by calculation beta j for all the points as the sweep line descend. However, the first intersection point is not that interesting when you are trying to compute the Voronoi diagram. What is interesting to find are the two points in the above picture to the left, where the beach line forms two intersection points that would form the beginning of a new line in the Voronoi diagram.

So we assume that we have to points that form two parabolas, with two intersection points and each individual parabola would be governed by the parabolic equation above. We can now set the two parabolic equations equal each other and calculate the values where the two intersections occur by using the quadratic formula. The result is two different x values, that you could plug into any of the two parabolic equations to get the resulting y value, and thereby the two cross points. The equation is given below where Pi and Pj is the two center site points of the cones:

I do not want to clean up the equation as it is quite a long, but the idea is to make it into an equation that could be written on the form:

With the two solutions for X being found from the equation below:

But there is a pretty big problem in the equation because of the site events, when are you supposed to make the guess about the line in a Voronoi diagram? If you increment it a little before the calculation the line you will get a large error and if you wait too long you get a mess as you have to shorten the line between two points, as new intersections would break the calculated lines.

You might be tempted to ask how long you would have to keep checking the parabolic line for any given point and the answer is that you have to keep checking the parabola of a point as long as the center point is not enclosed in lines, meaning no event further down the sweep line would change the boundaries for any of the lines surrounding the point. This is normally not a problem as the checks are event driven by two conditions and a parabolic arc is terminated when a circle event (also called vertex event) occurs.

Circle Event  

This event could only occur if two parabolas swallow a third parabola, meaning that we only need to check if three (or more) points will form a circle that does not contain any other point.  

"Which arc would dissapear," you might ask. The answer is that if you have the three points that form a circle like the picture above, it is the middle arc that would dissapear.

There is a formula for getting the center and the radius of a circle here and if you want to know if another point is inside the circle you could simply calculate the distance between the center point of the circle and the potential point. If the distance is less than the radius, the point is inside the circle. If not the center of the circle would give you another vertex point and one parabola would disappear. The calculation is done by using the general equation of a circle:

(X - Xc)*2 + (Y - Yc)*2 = R*2

And substitute the three points for X and Y, you'll have three equations with three unknown variables which then can be solved for Xc, Yc and R. There are other ways of calculating (which are faster) the center point, but I will leave that investigation up to you. 

Elementary Functions Used in the Code

Please note, this code uses the native library in .NET which may have some limited uses for an arbitrary collection of points. This is done so the celerity of the code is not compromised.  

The first thing we need to do with an arbitrary collection of points is to sort them in the sweep line direction. A general algorithem sorting points in either x or y direction, with lexographically ordered points, meaning that if you sort by y value, and get equal y's, you would sort the two points by the x direction after the y check. The code is given below: 

''' <summary>
''' Returns a sorted pointcollection based on Lexografically sorting
''' </summary>
''' <param name="samplepoints"></param>
''' <param name="SortByXdirection"></param>
''' <returns></returns>
''' <remarks></remarks>
Private Function SortPoints(ByVal samplepoints As PointCollection, _
        ByVal SortByXdirection As Boolean) As PointCollection
    'Create another array so we can keep the original array out of order
    Dim copySamplePoints As Point() = New Point(samplepoints.Count - 1) {}
    samplepoints.CopyTo(copySamplePoints, 0)

    If SortByXdirection Then
        Array.Sort(copySamplePoints, New PointSort(PointSort.Mode.X))
        Array.Sort(copySamplePoints, New PointSort(PointSort.Mode.Y))
    End If

    Dim result As New PointCollection

    For Each p As Point In copySamplePoints

    Return result
End Function

Private Class PointSort
    Implements IComparer
    Public Enum Mode
    End Enum

    Private currentMode As Mode = Mode.X

    Public Sub New(ByVal mode As Mode)
        currentMode = mode
    End Sub

    'Comparing function
    'Returns one of three values - 0 (equal), 1 (greater than), -1 (less than)
    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
            'Compare X values
            If point1.X > point2.X Then
                Return 1
            ElseIf point1.X < point2.X Then
                Return -1
                If point1.Y > point2.Y Then
                    Return 1
                ElseIf point1.Y < point2.Y Then
                    Return -1
                Else 'Identical points
                    Return 0
                End If
            End If
            If point1.Y > point2.Y Then
                'Compare Y Values
                Return 1
            ElseIf point1.Y < point2.Y Then
                Return -1
            Else 'Y1 = Y2
                If point1.X > point2.X Then
                    'Compare Y Values
                    Return 1
                ElseIf point1.X < point2.X Then
                    Return -1
                Else 'Identical points
                    Return 0
                End If
            End If
        End If
    End Function
End Class

A function for finding the intersection parabola of a new site event:  

''' <summary>
''' Returns a polyline between the newpoint and the contructed arc of the oldpoint
''' </summary>
''' <param name="NewPoint">New occuring point on the sweep line</param>
''' <param name="OldPoint">Point (above the sweep line) that generates the prarabolic arc</param>
''' <param name="SweepLine">Position of the sweep line (Y) </param>
''' <returns>A straight line between the NewPoint to the Intersection point on the parabolic arc</returns>
''' <remarks></remarks>
Private Function Intersection(ByVal NewPoint As Point, ByVal OldPoint As Point, _
        ByVal SweepLine As Double) As Polyline
    Dim result As New Polyline

    Dim endpoint As New Point
    endpoint = NewPoint
    Dim res As Double
    res = (1 / (2 * ((OldPoint.Y) - SweepLine)))
    res *= (NewPoint.X ^ 2 - 2 * OldPoint.X * NewPoint.X + OldPoint.X ^ 2 + OldPoint.Y ^ 2 - SweepLine ^ 2)
    endpoint.Y = res
    result.Stroke = Brushes.Red
    result.StrokeThickness = 2
    Return result
End Function

This next function is used just to calculate the same as the function above, but this just returns the y value in the arc created by the oldpoint. 

''' <summary>
''' Returns the y value construcked by the event from the newpoint and the contructed arc of the oldpoint
''' </summary>
''' <param name="NewPoint">This is the point on the sweep line</param>
''' <param name="OldPoint">This is one point above the sweep line</param>
''' <param name="SweepLine">Y position of the sweep line</param>
''' <returns>Calculates the distance fromn a point (NewPoint) to the Parabolic
''' arc generated by the OldPoint and the Sweep line</returns>
''' <remarks>The Function only gives you the Y value of the position
''' on the parabolic intersection given the X location</remarks>
Private Function Intersection2(ByVal NewPoint As Point, _
        ByVal OldPoint As Point, ByVal SweepLine As Double) As Double
    Dim res As Double
    res = (1 / (2 * ((OldPoint.Y) - SweepLine)))
    res *= (NewPoint.X ^ 2 - 2 * OldPoint.X * NewPoint.X + OldPoint.X ^ 2 + OldPoint.Y ^ 2 - SweepLine ^ 2)
    Return (res)
End Function

The next function we need to handle is the so-called site event, or as I called it a Parabolic cut (the sweep line is set at the location ys + 500 in the code below):

''' <summary>
''' Calculates the line between two Parabolic intersections
''' </summary>
''' <param name="Point1">A point in the Voronoi diagram</param>
''' <param name="point2">A Point in the Voronoi diagram (different from point 1)</param>
''' <param name="ys">The position of the sweep line</param>
''' <returns>A staight line between the two intersection poitns</returns>
''' <remarks>It would give wrong values if the two points have the same or
''' close to the same Y coordinate, as the double has limited significant storage</remarks>
Private Function ParabolicCut(ByVal Point1 As Point, ByVal point2 As Point, ys As Double) As Polyline

    'Stores the values in double format, as I didnt bother to rewrite Benjamin Dittes quadratic equation.
    Dim x1, y1, x2, y2 As Double

    'Inizialize Point 1
    x1 = Point1.X
    y1 = Point1.Y

    'Inizialize Point 2 
    x2 = point2.X
    y2 = point2.Y

    'Sweep line, added 500 to make sure the two calculated points are well of the boundaries
    ys = ys + 500

    'Setting ut calculation of the quadratic equation
    Dim a1 As Double = 1 / (2 * (y1 - ys))
    Dim a2 As Double = 1 / (2 * (y2 - ys))

    'The two resulting values of x from the quadratic equation
    Dim xs1 As Double = 0.5 / (2 * a1 - 2 * a2) * (4 * a1 * x1 - 4 * a2 * x2 + 2 * _
        Math.Sqrt(-8 * a1 * x1 * a2 * x2 - 2 * a1 * y1 + 2 * a1 * y2 + 4 * a1 * a2 _
        * x2 * x2 + 2 * a2 * y1 + 4 * a2 * a1 * x1 * x1 - 2 * a2 * y2))
    Dim xs2 As Double = 0.5 / (2 * a1 - 2 * a2) * (4 * a1 * x1 - 4 * a2 * x2 - 2 * _
        Math.Sqrt(-8 * a1 * x1 * a2 * x2 - 2 * a1 * y1 + 2 * a1 * y2 + 4 * a1 * _
        a2 * x2 * x2 + 2 * a2 * y1 + 4 * a2 * a1 * x1 * x1 - 2 * a2 * y2))

    'Generate two points to store the two intersection points
    Dim p1, p2 As New Point
    p1.X = xs1
    p2.X = xs2
    p1.Y = 0
    p2.Y = 0

    'Find the resulting Y values calculated from the quadratic equation
    ' (It dosent matter that the Y is 0 as the function Intersection2
    ' only uses the X value for computation)
    p1.Y = Intersection2(p1, Point1, ys)
    p2.Y = Intersection2(p2, Point1, ys)

    'Make a polyline to store the result
    Dim result As New Polyline

    'Add the two calculated poitns

    'Making the line visible
    result.Stroke = Brushes.Black
    result.StrokeThickness = 2

    'Return the polyline
    Return result
End Function

I have also created an own circle class with all the nessecary functions inside:

''' <summary>
''' A class for preforming circle tests
''' </summary>
''' <remarks></remarks>
Public Class Circle
    Private _CenterPoint As Point
    Public Property CenterPoint() As Point
            Return _CenterPoint
        End Get
        Set(ByVal value As Point)
            _CenterPoint = value
        End Set
    End Property

    Private _radius As Double
    Public Property Radius() As Double
            Return _radius
        End Get
        Set(ByVal value As Double)
            _radius = value
        End Set
    End Property

    'Calculates the Eucladian distence between two poitns
    Private Function Distance(ByVal p1 As Point, ByVal p2 As Point) As Double
        Return Math.Sqrt((p1.X - p2.X) ^ 2 + (p1.Y - p2.Y) ^ 2)
    End Function

    ''' <summary>
    ''' Checks if a Point1 is inside the circle
    ''' </summary>
    ''' <param name="p1">Test point</param>
    ''' <returns>If point is on the circle it is considered to be outside,
    ''' as three or more points is checked for</returns>
    ''' <remarks></remarks>
    Public Function DoesCircleContainPoint(ByVal p1 As Point) As Boolean
        Dim d As Double
        d = Distance(p1, CenterPoint)
        'Replace this with <= if you just want three points to form a circle
        If d < Radius Then
            Return True
            Return False
        End If
    End Function

    ''' <summary>
    ''' Returns the centerpoint of the created circle
    ''' </summary>
    ''' <param name="can"></param>
    ''' <remarks></remarks>
    Public Sub ReturnCenterPoint(ByVal can As Canvas)
        Dim p As New Ellipse
        p.Fill = Brushes.Blue
        p.StrokeThickness = 2
        p.Stroke = Brushes.Blue

        ' Set the width and height of the Ellipse.
        p.Width = 5
        p.Height = 5

        Canvas.SetLeft(p, CenterPoint.X - 2.5)
        Canvas.SetTop(p, CenterPoint.Y - 2.5)
    End Sub

    ''' <summary>
    ''' Returns the complete circle created from the three points
    ''' </summary>
    ''' <param name="can"></param>
    ''' <remarks></remarks>
    Public Sub ReturnEllipse(ByVal can As Canvas)
        Dim p As New Ellipse
        p.Fill = Brushes.Yellow
        p.StrokeThickness = 2
        p.Stroke = Brushes.Black

        ' Set the width and height of the Ellipse.
        p.Width = Radius * 2
        p.Height = Radius * 2

        Canvas.SetLeft(p, CenterPoint.X - Radius)
        Canvas.SetTop(p, CenterPoint.Y - Radius)
    End Sub

    Public Sub CreatCircleFromThreePoints(ByVal Point1 As Point, _
           ByVal Point2 As Point, ByVal Point3 As Point)
        Dim Pt(3) As Point
        Pt(0) = Point1
        Pt(1) = Point2
        Pt(2) = Point3

    End Sub

    Private Sub GetCircleRectFromPoints(ByVal Pts() As Point) 'As RectangleF
        '(X - Cx)^2 + (Y - Cy)^2 = R^2
        '(R^2 - Cx^2 - Cy^2) + 2X*Cx + 2Y*Cy = X^2 +  Y^2
        'Cr + A*X + B*Y = X^2 + Y^2
        'Solve matrix using gaussian elimination.
        Dim N As Integer = Pts.Length - 1
        Dim X(N, N) As Double, Y(N) As Double
        For I As Integer = 0 To N
            X(I, 0) = 1 : X(I, 1) = Pts(I).X : X(I, 2) = Pts(I).Y
            Y(I) = X(I, 1) * X(I, 1) + X(I, 2) * X(I, 2)
        Dim MatInv As New Elimination
        MatInv.ComputeCoefficents(X, Y)
        Dim A As Single = CSng(Y(1) / 2)
        Dim B As Single = CSng(Y(2) / 2)
        Dim R As Single = CSng(Math.Sqrt(Y(0) + A * A + B * B))
        CenterPoint = New Point(A, B)
        Radius = R
    End Sub

    Public Class Elimination
        Sub ComputeCoefficents(ByVal X(,) As Double, ByVal Y() As Double)
            Dim I, J, K, K1, N As Integer
            N = Y.Length
            For K = 0 To N - 1
                K1 = K + 1
                For I = K To N - 1
                    If X(I, K) <> 0 Then
                        For J = K1 To N - 1
                            X(I, J) /= X(I, K)
                        Next J
                        Y(I) /= X(I, K)
                    End If
                Next I
                For I = K1 To N - 1
                    If X(I, K) <> 0 Then
                        For J = K1 To N - 1
                            X(I, J) -= X(K, J)
                        Next J
                        Y(I) -= Y(K)
                    End If
                Next I
            Next K
            For I = N - 2 To 0 Step -1
                For J = N - 1 To I + 1 Step -1
                    Y(I) -= X(I, J) * Y(J)
                Next J
            Next I
        End Sub
    End Class
End Class

The Main Code

As I was writing the article I soon realized that it would become a monster if I included all the necessary steps for creating a complete Voronoi diagram. So I stopped with a version that does not, by any means, contain the creation of a complete Voronoi diagram, but rather give you a chance to understand the most necessary functions for creating it. The final complete implementation and creation will have to happen in the next article.


First release: 29.06.2012.

Second release: 01.07.2012, Minor changes in languange, added some references inside the article.



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


About the Author

Kenneth Haugland
Norway Norway
I hope that you like the stuff I have created and if you do wish to say thank you then a donation is always appreciated.
You can donate here[^].

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Comments and Discussions

GeneralMy vote of 5 Pin
ISpliter5-Mar-15 15:24
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GeneralRe: My vote of 5 Pin
Kenneth Haugland12-Mar-15 22:22
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GeneralRe: My vote of 5 Pin
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GeneralMy vote of 5 Pin
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QuestionVoronoi.vb Pin
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AnswerRe: Voronoi.vb Pin
Kenneth Haugland22-Mar-13 2:51
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GeneralRe: Voronoi.vb Pin
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GeneralRe: Voronoi.vb Pin
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GeneralMy vote of 5 Pin
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GeneralRe: My vote of 5 Pin
Kenneth Haugland18-Mar-13 6:16
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GeneralMy vote of 5 Pin
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GeneralRe: My vote of 5 Pin
Kenneth Haugland10-Mar-13 1:26
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GeneralWell done, Kenneth Pin
Espen Harlinn15-Aug-12 9:03
mvpEspen Harlinn15-Aug-12 9:03 
Well done, Kenneth - "Best VB.NET article of June 2012" - congratulations Big Grin | :-D
Espen Harlinn
Principal Architect, Software - Goodtech Projects & Services AS

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GeneralRe: Well done, Kenneth Pin
Kenneth Haugland15-Aug-12 9:06
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Questionvery nice Pin
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AnswerRe: very nice Pin
Kenneth Haugland19-Jul-12 7:32
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GeneralMy vote of 5 Pin
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