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Posted 25 May 2007

A C# Implementation of Douglas-Peucker Line Approximation Algorithm

, 6 Jun 2007
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DP Line approximation algorithm is a well-known method to approximate 2D lines. It is quite fast, O(nlog_2(n)) for a n-points line and can drastically compress a data curve. Here, a fully OOP implementation is given.


I have found multiple implementations of the Douglas-Peucker algorithm but not in any .NET language, so I decided to port it over. Jonathan de Halleux has a wonderful explanation here.


I needed to reduce polygons size to display on a map based on zoom levels.

Using the Code

The code included is complete and should run out of the box in Visual Studio 2005. If it does not, please let me know.

/// <summary>
/// Uses the Douglas Peucker algorithm to reduce the number of points.
/// </summary>
/// <param name="Points">The points.</param>
/// <param name="Tolerance">The tolerance.</param>
/// <returns></returns>
public static List<Point> DouglasPeuckerReduction
    (List<Point> Points, Double Tolerance)
    if (Points == null || Points.Count < 3)
    return Points;

    Int32 firstPoint = 0;
    Int32 lastPoint = Points.Count - 1;
    List<Int32> pointIndexsToKeep = new List<Int32>();

    //Add the first and last index to the keepers

    //The first and the last point cannot be the same
    while (Points[firstPoint].Equals(Points[lastPoint]))

    DouglasPeuckerReduction(Points, firstPoint, lastPoint, 
    Tolerance, ref pointIndexsToKeep);

    List<Point> returnPoints = new List<Point>();
    foreach (Int32 index in pointIndexsToKeep)

    return returnPoints;
/// <summary>
/// Douglases the peucker reduction.
/// </summary>
/// <param name="points">The points.</param>
/// <param name="firstPoint">The first point.</param>
/// <param name="lastPoint">The last point.</param>
/// <param name="tolerance">The tolerance.</param>
/// <param name="pointIndexsToKeep">The point index to keep.</param>
private static void DouglasPeuckerReduction(List<Point> 
    points, Int32 firstPoint, Int32 lastPoint, Double tolerance, 
    ref List<Int32> pointIndexsToKeep)
    Double maxDistance = 0;
    Int32 indexFarthest = 0;
    for (Int32 index = firstPoint; index < lastPoint; index++)
        Double distance = PerpendicularDistance
            (points[firstPoint], points[lastPoint], points[index]);
        if (distance > maxDistance)
            maxDistance = distance;
            indexFarthest = index;

    if (maxDistance > tolerance && indexFarthest != 0)
        //Add the largest point that exceeds the tolerance
        DouglasPeuckerReduction(points, firstPoint, 
        indexFarthest, tolerance, ref pointIndexsToKeep);
        DouglasPeuckerReduction(points, indexFarthest, 
        lastPoint, tolerance, ref pointIndexsToKeep);

/// <summary>
/// The distance of a point from a line made from point1 and point2.
/// </summary>
/// <param name="pt1">The PT1.</param>
/// <param name="pt2">The PT2.</param>
/// <param name="p">The p.</param>
/// <returns></returns>
public static Double PerpendicularDistance
    (Point Point1, Point Point2, Point Point)
    //Area = |(1/2)(x1y2 + x2y3 + x3y1 - x2y1 - x3y2 - x1y3)|   *Area of triangle
    //Base = v((x1-x2)²+(x1-x2)²)                               *Base of Triangle*
    //Area = .5*Base*H                                          *Solve for height
    //Height = Area/.5/Base

    Double area = Math.Abs(.5 * (Point1.X * Point2.Y + Point2.X * 
    Point.Y + Point.X * Point1.Y - Point2.X * Point1.Y - Point.X * 
    Point2.Y - Point1.X * Point.Y));
    Double bottom = Math.Sqrt(Math.Pow(Point1.X - Point2.X, 2) + 
    Math.Pow(Point1.Y - Point2.Y, 2));
    Double height = area / bottom * 2;

    return height;
    //Another option
    //Double A = Point.X - Point1.X;
    //Double B = Point.Y - Point1.Y;
    //Double C = Point2.X - Point1.X;
    //Double D = Point2.Y - Point1.Y;
    //Double dot = A * C + B * D;
    //Double len_sq = C * C + D * D;
    //Double param = dot / len_sq;
    //Double xx, yy;
    //if (param < 0)
    //    xx = Point1.X;
    //    yy = Point1.Y;
    //else if (param > 1)
    //    xx = Point2.X;
    //    yy = Point2.Y;
    //    xx = Point1.X + param * C;
    //    yy = Point1.Y + param * D;
    //Double d = DistanceBetweenOn2DPlane(Point, new Point(xx, yy));

Points of Interest

The code is not overly complicated. It was fun to port this algorithm, since all I feel I do nowadays is hold business' hands to help them try and solve business problems that there is no consensus on.


  • Beta 1


This article, along with any associated source code and files, is licensed under The MIT License


About the Author

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Comments and Discussions

Questionrejecting recursion + distance calc acceleration Pin
adevder19-Feb-17 0:12
memberadevder19-Feb-17 0:12 
AnswerRe: rejecting recursion + distance calc acceleration Pin
TerrySpiderman21-Mar-17 3:05
professionalTerrySpiderman21-Mar-17 3:05 
Questionto save stack memory with great amount of points and tolerance near zero i prefer to reject recursion Pin
adevder18-Feb-17 1:22
memberadevder18-Feb-17 1:22 
PraiseNice Pin
veen_rp18-Apr-16 19:41
professionalveen_rp18-Apr-16 19:41 
BugIf (first = last) Bug! Pin
Giovani Luigi14-Dec-15 12:44
memberGiovani Luigi14-Dec-15 12:44 
QuestionDouglasPeuckerN() conversion? Pin
cbordeman11-Dec-15 17:42
membercbordeman11-Dec-15 17:42 
AnswerRe: DouglasPeuckerN() conversion? Pin
lorsco0127-Apr-17 7:44
memberlorsco0127-Apr-17 7:44 
GeneralUseful Pin
Dineshkumar_MCA22-Apr-15 2:18
professionalDineshkumar_MCA22-Apr-15 2:18 
QuestionBUG tolerance Pin
Member 1017584518-Feb-15 2:08
memberMember 1017584518-Feb-15 2:08 
QuestionThanks + pruned code Pin
Stoyan Haralampiev23-Oct-14 3:30
memberStoyan Haralampiev23-Oct-14 3:30 
QuestionUsing this algorithm on data from database Pin
Saket Surya2-Jun-14 3:38
memberSaket Surya2-Jun-14 3:38 
QuestionPoint projected on the line but behind the segment? Bug? Pin
Sergiy Tkachuk25-Mar-14 14:31
memberSergiy Tkachuk25-Mar-14 14:31 
I have not found the check if point is projected to line which contains segment, but actual projection lies behind segment.

Do you perform such check?
In such case tolerance should be calculated as a distance between point and one of the ends of segment, and should _not_ be considered as a length of the projection to the line.

Without such check wrong behaviour could appear if point is close to the line but far behind segment at the same time.
In such case point could be removed by algorithms, but in fact it should be preserved.
QuestionAbout Tolerance Pin
kamlesh Prajapati8-Jan-14 23:55
groupkamlesh Prajapati8-Jan-14 23:55 
QuestionWhy do you create your own Point? Pin
leiyangge3-Jan-14 22:37
memberleiyangge3-Jan-14 22:37 
QuestionLicense Pin
Diego Catalano14-Dec-13 2:42
memberDiego Catalano14-Dec-13 2:42 
Question1 bug, + 1 potential bug Pin
gyuri1023-Apr-12 23:45
membergyuri1023-Apr-12 23:45 
AnswerRe: 1 bug, + 1 potential bug Pin
tplokas26-Mar-13 14:46
membertplokas26-Mar-13 14:46 
QuestionUseful,Thanks! Pin
masy19112-Mar-12 23:43
membermasy19112-Mar-12 23:43 
QuestionGreat Job! Pin
cdrake19-Dec-11 8:32
membercdrake19-Dec-11 8:32 
Generalhow calculate an epsilon value Pin
unravel-the-sky23-May-11 6:52
memberunravel-the-sky23-May-11 6:52 
GeneralSome Issues with your implementation Pin
matthew minke15-Jul-10 11:57
membermatthew minke15-Jul-10 11:57 
GeneralVery useful thanks - Point number reduction (Max Distance relevant tolerance) [modified] Pin
kamyk3-Mar-10 23:05
memberkamyk3-Mar-10 23:05 
GeneralVery useful thanks Pin
dylanhayes17-Nov-09 2:35
memberdylanhayes17-Nov-09 2:35 
GeneralThanks! Pin
Member 337866727-Mar-09 10:28
memberMember 337866727-Mar-09 10:28 
QuestionAnd other functions from Plot Graphic Library? Pin
agorby28-Apr-08 3:28
memberagorby28-Apr-08 3:28 

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