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Single Layer Perceptron as Linear Classifier

By , 7 Nov 2010
 
PerceptronForm.jpg

Introduction

Perceptron is the simplest type of feed forward neural network. It was designed by Frank Rosenblatt as dichotomic classifier of two classes which are linearly separable. This means that the type of problems the network can solve must be linearly separable. Basic perceptron consists of 3 layers:

  • Sensor layer
  • Associative layer
  • Output neuron

There are a number of inputs (xn) in sensor layer, weights (wn) and an output. Sometimes w0 is called bias and x0 = +1/-1 (In this case is x0=-1).

percept.jpg

For every input on the perceptron (including bias), there is a corresponding weight. To calculate the output of the perceptron, every input is multiplied by its corresponding weight. Then weighted sum is computed of all inputs and fed through a limiter function that evaluates the final output of the perceptron.

The output of neuron is formed by activation of the output neuron, which is function of input:

(1) eq1.JPG

The activation function F can be linear so that we have a linear network, or nonlinear. In this example, I decided to use threshold (signum) function:

(2) eq2.JPG

Output of network in this case is either +1 or -1 depending on the input. If the total input (weighted sum of all inputs) is positive, then the pattern belongs to class +1, otherwise to class -1. Because of this behavior, we can use perceptron for classification tasks.

Let's consider we have a perceptron with 2 inputs and we want to separate input patterns into 2 classes. In this case, the separation between the classes is straight line, given by equation:

(3) eq3.gif

When we set x0=-1 and mark w0=?, then we can rewrite equation (3) into form:

(4) eq4.gif

Here I will describe the learning method for perceptron. Learning method of perceptron is an iterative procedure that adjust the weights. A learning sample is presented to the network. For each weight, the new value is computed by adding a correction to the old value. The threshold is updated in the same way:

(5) eq5.gif

eq5b.gif

where y is output of perceptron, d is desired output and ? is the learning parameter.

Using the Program

When you run the program, you see area where you can input samples. Clicking by left button on this area, you will add first class sample (blue cross). Clicking by right button on this area, you will add first class sample (red cross). Samples are added to the samples list. You can also set learning rate and number of iterations. When you have set all these values, you can click on Learn button to start learning.

Using the Code

All samples are stored in generic list samples which holds only Sample class objects.

public class Sample
{
    double x1;
    double x2;
    double cls;

    public Sample(double x1, double x2, int cls)
    {
        this.x1 = x1;
        this.x2 = x2;
        this.cls = cls;
    }

    public double X1
    {
        get { return x1; }
        set { this.x1 = value; }
    }

    public double X2
    {
        get { return x2; }
        set { this.x2 = value; }
    }

    public double Class
    {
        get { return cls; }
        set { this.cls = value; }
    }
}

Before running a learning of perceptron is important to set learning rate and number of iterations. Perceptron has one great property. If solution exists, perceptron always find it but problem occurs, when solution does not exist. In this case, perceptron will try to find the solution in infinity loop and to avoid this, it is better to set maximum number of iterations.

The next step is to assign random values for weights (w0, w1 and w2).

Random rnd = new Random();

w0 = rnd.NextDouble();
w1 = rnd.NextDouble();
w2 = rnd.NextDouble();

When random values are assigned to weights, we can loop through samples and compute output for every sample and compare it with desired output.

double x1 = samples[i].X1;
double x2 = samples[i].X2;
int y;

if (((w1 * x1) + (w2 * x2) - w0) < 0)
{
    y = -1;
}
else
{
    y = 1;
}

I decided to set x0=-1 and for this reason, the output of perceptron is given by equation: y=w1*w1+w2*w2-w0. When perceptron output and desired output doesn’t match, we must compute new weights:

if (y != samples[i].Class)
{
    error = true;

    w0 = w0 + alpha * (samples[i].Class - y) * x0 / 2;
    w1 = w1 + alpha * (samples[i].Class - y) * x1 / 2;
    w2 = w2 + alpha * (samples[i].Class - y) * x2 / 2;
}

Y is output of perceptron and samples[i].Class is desired output. The last 2 steps (looping through samples and computing new weights), we must repeat while the error variable is <> 0 and current number of iterations (iterations) is less than maxIterations.

int i;
int iterations = 0;
bool error = true;

maxIterations = int.Parse(txtIterations.Text);

Random rnd = new Random();

w0 = rnd.NextDouble();
w1 = rnd.NextDouble();
w2 = rnd.NextDouble();

alpha = (double)trackLearningRate.Value / 1000;

while (error && iterations < maxIterations)
{
    error = false;

    for (i = 0; i <= samples.Count - 1; i++)
    {
        double x1 = samples[i].X1;
        double x2 = samples[i].X2;
        int y;

        if (((w1 * x1) + (w2 * x2) - w0) < 0)
        {
            y = -1;
        }
        else
        {
            y = 1;
        }

        if (y != samples[i].Class)
        {
            error = true;

            w0 = w0 + alpha * (samples[i].Class - y) * x0 / 2; 
            w1 = w1 + alpha * (samples[i].Class - y) * x1 / 2;
            w2 = w2 + alpha * (samples[i].Class - y) * x2 / 2;
        }
    }
    objGraphics.Clear(Color.White);
    DrawSeparationLine();
    iterations++;
}

Function DrawSeparationLine draws separation line of 2 classes.

History

  • 07 Nov 2010 - Original version posted

License

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

About the Author

Kanasz Robert
Architect The Staffing Edge & Marwin Cassovia Soft
Slovakia Slovakia
Member
My name is Robert Kanasz and I have been working with ASP.NET, WinForms and C# for several years.
 
MCTS - .NET Framework 3.5, ASP.NET Applications
- SQL Server 2008, Database Development
- SQL Server 2008, Implementation and Maintenance
- .NET Framework 4, Data Access
- .NET Framework 4, Service Communication Applications
- .NET Framework 4, Web Applications
MCPD - ASP.NET Developer 3.5
- Web Developer 4
MCITP - Database Administrator 2008
- Database Developer 2008
 
Open source projects: DBScripter - Library for scripting SQL Server database objects
 

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Comments and Discussions

 
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QuestionIncorrectly dividing by 2 [modified]membertwocs25 Nov '12 - 16:39 
According to equation 5, you should update the weight by adding the learning rate * error. But in the implementation, you then divide this number by 2. Although halving the learning rate will surely work, I don't understand why the code is different from the equation.
 
Edit: saw this response in another answer
"Hello,

I'm so sorry, that I reply to late, but I had no time.
In this equation you must divide by number 2 because I decided to use -1 an 1 values to distinct between 2 classes.
If you set 0 for first class and 1 for second, than you don't need to divide by number 2.

Regards
Robert"
 
Edit: Now I have a different question on the same topic
The explanation suggests that we need to normalize the learning rate by the size of the space. This could be fine for two classes, but what if we introduce a third class, with a value such as -10. How would we know the normalization value?
 
Second question, why not use 0 and 1? When I train the network with 0 and 1, the line ends up sticking to some of the points, but when -1 and 1 are used, the line ends up between the points. Obviously -1 and 1 are better, but a book on pattern recognition I'm reading had us use 0 and 1, and my results were not good. That's why I searched the Internet and found this tutorial. Can you explain why -1 and 1 are better than 0 and 1?

modified 25 Nov '12 - 22:49.

QuestionNice one!!!memberstrucker_luc18 Nov '12 - 3:17 
Very Well!!!
AnswerRe: Nice one!!!mvpKanasz Robert18 Nov '12 - 3:21 
thank you very much for your comment
QuestionInteresting article and very helpfulmemberkr1234564 Nov '12 - 3:58 
Good job!
AnswerRe: Interesting article and very helpfulmvpKanasz Robert4 Nov '12 - 4:03 
Thank you very much. I'm very glad that you liked this article.
Questionvery wellmembersuperdevX151 Nov '12 - 6:51 
thank's for sharing
AnswerRe: very wellmvpKanasz Robert1 Nov '12 - 6:55 
i'm glad that you liked it
Questionvery well written articlememberhakon12331 Oct '12 - 5:32 
5. well done
AnswerRe: very well written articlemvpKanasz Robert31 Oct '12 - 5:37 
thank you so much
Questionnicemembermemlon mulas29 Oct '12 - 5:15 
5
AnswerRe: nicemvpKanasz Robert29 Oct '12 - 5:19 
thank you
Questiongood and well written articlememberjackhoal27 Oct '12 - 3:56 
nice one
AnswerRe: good and well written articlemvpKanasz Robert27 Oct '12 - 4:00 
thank you
QuestionGoodmemberrobkaan27 Oct '12 - 3:29 
excellent article
AnswerRe: GoodmvpKanasz Robert27 Oct '12 - 3:31 
Thanks
Questionnot badmemberwindevvv21 Oct '12 - 6:52 
Good article
AnswerRe: not badmvpKanasz Robert21 Oct '12 - 6:59 
thank you
Questiongoodmemberkaslaninovic2 Oct '12 - 22:51 
5
AnswerRe: goodmvpKanasz Robert3 Oct '12 - 6:49 
thank's
Questioninterestingmemberdeveloper88123 Sep '12 - 3:05 
good one
AnswerRe: interestingmvpKanasz Robert23 Sep '12 - 23:12 
Thank's
QuestionGoodmemberbikerius19 Sep '12 - 2:13 
great article, but it would be a great if you provide code for multiple classes perceptron classifier.
AnswerRe: GoodmvpKanasz Robert19 Sep '12 - 4:21 
Thank you
QuestionWell donememberpukson11 Sep '12 - 3:43 
It helped me to finish my project Smile | :)
AnswerRe: Well donemvpKanasz Robert11 Sep '12 - 3:51 
Thank you for your vote.

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