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Connected Component Labeling Algorithm

, 2 Feb 2014
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Implementation of Connected Component Labeling.

Contents

Overview

In five seconds 

Detection of connected objects in an image, mainly used in image analysis and OCR.

In five minutes

Connected-component labeling (alternatively connected-component analysis, blob extraction, region labeling, blob discovery, or region extraction) is an algorithmic application of graph theory, where subsets of connected components are uniquely labeled based on a given heuristic. Connected-component labeling is not to be confused with segmentation.

Connected-component labeling is used in computer vision to detect connected regions in binary digital images, although color images and data with higher-dimensionality can also be processed.[1][2] When integrated into an image recognition system or human-computer interaction interface, connected component labeling can operate on a variety of information.[3][4] Blob extraction is generally performed on the resulting binary image from a thresholding step. Blobs may be counted, filtered, and tracked.

Blob extraction is related to but distinct from blob detection.

What to expect

Input

An image containing two shapes:

336915/input.png

Output

Now each is separated into single images:

336915/1.png

336915/2.png

Code

The Interface IConnectedComponentLabeling holds one function That takes a Bitmap as an input an returns a collection of discovered objects. 

public interface IConnectedComponentLabeling
{
    IDictionary<int, Bitmap> Process(Bitmap input);
}

Usage:

IConnectedComponentLabeling target = new CCL();
Bitmap input = new Bitmap(AppDomain.CurrentDomain.BaseDirectory + @"\Test.bmp");

var images= target.Process(input);
foreach (var image in images)
{
    image.Value.Save(savePath + image.Key + ".bmp");
}

The implementation class contains the virtual function CheckIsBackGround(), so you can extend the class, and override this method it to suit the background condition of your image:

#region Protected Functions

protected virtual bool CheckIsBackGround(Pixel currentPixel)
{
    return currentPixel.color.A == 255 && currentPixel.color.R == 255 && 
             currentPixel.color.G == 255 && currentPixel.color.B == 255;
}

#endregion

How it works

First pass (assigning labels)

Second pass (aggregation)

Step by step walkthrough

In the beginning, we have this image, we start with currentLabelCount = 1.

336915/Start.png

We found our non-background pixel:

336915/step_1.png

get its non-background neighbors:

336915/step_2.png

None of the neighbors is labeled yet, we set the current pixel to the currentLabelCount and increment it, we also set the label's parent to itself (we'll get into that in a second):

336915/step_3.png

on to the next pixel, this one has a neighbour which is already labeled:

336915/step_4.png

assigns the pixel's parent label to that of the neighbour:

336915/step_5.png

We continue on, none of the neighbours of this pixel is labeled:

336915/step_6.png

We increment currentLabelCount and assign it to the pixel, again its parent is set to itself: 

336915/step_7.png

It gets interesting here, when neighbors have different labels:

336915/step_8.png

  1. We choose main label, i.e.: that would be the smallest label in the discovered list--> (1)
  2. We set it to be the parent of the other labels

336915/step_9.png

A few more rounds and we should end up with this. Notice the blue number in the upper right corner, that's the parent label, the de facto one upon which we aggregate later.

336915/step_10.png

That's it, now all we have to do is pass the image again pixel by pixel, getting the root of each (if labeled) and store it in our patterns' list.

private Dictionary<int, List<Pixel>> AggregatePatterns(Dictionary<int, 
          Label> allLabels, int width, int height)
{
    var patterns = new Dictionary<int, List<Pixel>>();

    for (int i = 0; i < height; i++)
    {
        for (int j = 0; j < width; j++)
        {
            int patternNumber = _board[j, i];

            if (patternNumber != 0)
            {
                patternNumber = allLabels[patternNumber].GetRoot().Name;

                if (!patterns.ContainsKey(patternNumber))
                {
                    patterns.Add(patternNumber, new List<Pixel>());
                }

                patterns[patternNumber].Add(new Pixel(new Point(j, i), Color.Black));
            }
        }
    }

    return patterns;
}

Tricky part: Merging labels  

To join labels in a same set, we have the following class (which implements the union find algorithm):

using System;
using System.Collections.Generic;
using System.Text;

namespace ConnectedComponentLabeling
{
    internal class Label
    {
        #region Public Properties

        public int Name { get; set; }

        public Label Root { get; set; }

        public int Rank { get; set; }

        #endregion

        #region Constructor

        public Label(int Name)
        {
            this.Name = Name;
            this.Root = this;
            this.Rank = 0;
        }

        #endregion

        #region Public Methods

        internal Label GetRoot()
        {
            if (this.Root != this)
            {
                this.Root = this.Root.GetRoot();//Compact tree
            }

            return this.Root;
        }

        internal void Join(Label root2)
        {
            if (root2.Rank < this.Rank)//is the rank of Root2 less than that of Root1 ?
            {
                root2.Root = this;//yes! then Root1 is the parent of Root2 (since it has the higher rank)
            }
            else //rank of Root2 is greater than or equal to that of Root1
            {
                this.Root = root2;//make Root2 the parent

                if (this.Rank == root2.Rank)//both ranks are equal ?
                {
                    root2.Rank++;//increment Root2, we need to reach a single root for the whole tree
                }
            }
        }

        #endregion
    }
}

Pay special attention to the recursive function GetRoot(), that's how we reach the parent of any label.

Remember this part? This is what the function Join(Label root) does. Now let's say we have three labels, 1, 2, and 3, and we picked 1 to be our current label; all we have to do is loop the other labels, if their roots don't match, set their root to that of the label we just picked.  

Conclusion

Hope I delivered a clear explanation, feel free to comment or ask Wink | <img src=. Drawings by zwibbler

License

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

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About the Author

Omar Gameel Salem
Software Developer
Australia Australia
Enthusiastic programmer/researcher, passionate to learn new technologies, interested in problem solving,data structures, algorithms and automation.
 
If you have a question\suggestion about one of my articles, or you want an algorithm implemented in C#, feel free to contact me.
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Comments and Discussions

 
QuestionInquiry PinmemberMember 1056383724-Apr-14 13:20 
AnswerRe: Inquiry PinpremiumOmar Gameel Salem24-Apr-14 20:01 
Questiongood work Pinmemberame1312-Mar-14 7:22 
GeneralMy vote of 5 Pinmemberscrchhrpn17-May-13 21:30 
GeneralMy vote of 5 PinmemberUilleam12-Feb-13 11:36 
QuestionSome problem PinmemberMember 917258413-Dec-12 5:13 
AnswerRe: Some problem PinmemberOmar Gamil13-Dec-12 5:49 
Questionsome issues [modified] Pinmemberfeversky30-Oct-12 1:59 
QuestionText detection Pinmemberbutterflyzx412-Aug-12 6:42 
AnswerRe: Text detection PinmemberOmar Gamil12-Aug-12 11:13 
QuestionIt's not working properly Pinmemberriyad parvez24-Jun-12 1:54 
AnswerRe: It's not working properly PinmemberOmar Gamil24-Jun-12 4:25 
QuestionHow to use your dll file? Pinmembernprabhu8-Apr-12 19:30 
AnswerRe: How to use your dll file? PinmemberOmar Gamil8-Apr-12 23:03 
GeneralMy vote of 5 PinmemberAbinash Bishoyi6-Apr-12 3:00 
QuestionYou could do this much faster and in 1 step Pinmemberleon de boer1-Apr-12 21:21 
AnswerRe: You could do this much faster and in 1 step PinmemberDewey4-Apr-12 13:23 
GeneralRe: You could do this much faster and in 1 step PinmemberJP van Mackelenbergh5-Apr-12 1:58 
GeneralRe: You could do this much faster and in 1 step Pinmemberleon de boer20-Jun-12 16:29 
AnswerRe: You could do this much faster and in 1 step Pinmemberyvdh16-Dec-12 23:58 
GeneralMy vote of 5 Pinmembermanoj kumar choubey28-Mar-12 0:21 
QuestionHow should I use in my project? PinmemberRayar081223-Mar-12 0:21 
AnswerRe: How should I use in my project? PinmemberOmar Gamil23-Mar-12 5:25 
using this code (as mentioned in the article)
Bitmap input = new Bitmap(path to your image);
            PreProcess target = new PreProcess(input);
            List<Bitmap> actual = target.Extract();
            for (int i = 0; i < actual.Count; i++)
            {
                actual[i].Save(savePath + (i + 1).ToString() + ".bmp");
            }

GeneralNice :) PinmemberHoshiKata8-Mar-12 0:48 
GeneralRe: Nice :) PinmemberOmar Gamil8-Mar-12 3:45 

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