Using perceptron model for classification : an illustrative approach

In this post, we are going to devise a measurement tool (perceptron model) in order to classify : whether a person is infected by a diseases or not.

In binary terms, the output will be { 1 if infected 0 not infected }

To build inputs for our neural network, we take readings from the patients and we will treat readings as follows :

  body temperature = {
                          1   if body temperator > 99'F
                         -1   if body temperator = 99'F
                     }
  
  heart rate = {
                      1   if heart rate > 60 to 100
                     -1   if heart rate = 60 to 100
                 }
  
   blood pressure = {
                          1   if heart rate > 120/80
                         -1   if heart rate = 120/80
                     }
 

So, input from each patient will be represented as a three dimensional vector:

  input = (body temperatur, heart rate, blood pressure)

So, a person can now be represented as : (1, -1, 1) i.e (body temperator > 99'F, heart rate = 60 to 100, heart rate > 120/80)

Let us create two inputs with desired output value

      x1 = (1, 1, 1), d1 = 1 (infected)
       x2 = (-1, -1, -1), d2 = 0 (not infected)

Let us take initial values for weights and biases: weights, w0 = (-1, 0.5, 0) bias, b0 = 0.5

And, activation function:

         A(S)   = {
                    1 if S >=0
                    0 otherwise
                  }
STEP 1

Feed x1 = (1, 1, 1) into the network.

weighted_sum:

S = (-1, 0.5, 0) * (1, 1, 1)^T + 0
  = -1 + 0.5 + 0 + 0
  = -0.5

When passed through activation function A(-0.5) = 0 = y1 We passed an infected input vector, but our perceptron classified it as not infected. Let's calculate the error term: e = d1 - y1 = 1 - 0 = 1 Update weight as:

             w1 = w0 + e * x1 = (-1, 0.5, 0) + 1 * (1, 1, 1) = (0, 1.5, 1)

And, update bias as: b1 = b0 + e = 1

STEP 2

Now, we feed second input (-1, -1, -1) into our network.

weighted_sum : S = w1 * x2^T + b1 = (0, 1.5, 1) * (-1, -1, -1)^T + 1 = -1.5 - 1 + 1 = -1.5 When passed through activation function A(-1.5) = 0 = y2 We passed an not infected input vector, and our perceptron successfully classified it as not infected.

STEP 3

Since, our first input is mis-classified, so we will go for it.

weighted_sum :

S = w1 * x1^T + b1 
  = (0, 1.5, 1) * (1, 1, 1)^T + 1
  = 1.5 + 1 + 1
  = 3.5

When passed through activation function A(3.5) = 1 = y3 We passed an infected input vector, and our perceptron successfully classified it as infected.

Here, both input vectors are correctly classified. i.e algorithm is converged to a solution point.

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