Single-Neuron Perceptron

The single-neuron perceptron is about finding a decision boundary. This method uses supervised learning rule to determine the decision boundary and the form of supervised learning is:

{p1, t1}, {p2, t2}, ..., {pQ, tQ}

where pQ is an input and tQ is the corresponding correct output (target).


The figure below is the model of single-neuron perceptron with two inputs.













After we determine the a, we have to find a new weight and these are the equations to find it:









Let's see this example.

For the logic function AND gate, the input & target pairs are:




The initial weight is all zeros (w^T = [0 0]) and there is no bias (b = 0).

Solution
1st iteration: use {p1, t1}

a = hardlim([0 0][0 0]^T) = 1
e = t - a = t1 - a = -1
w^new = [0 0] - [0 0] = [0 0]
b^new = b^old + e = -1

2nd iteration: use {p2, t2}

a = hardlim([0 0][0 1]^T- 1) = 0
e = t - a = t2 - a = 0
w^new = [0 0]
b^new = b^old + e = -1

3rd iteration: use {p3, t3}

a = hardlim([0 0][1 0]^T- 2) = 0
e = t - a = t3 - a = 0
w^new = [0 0]
b^new = b^old + e = -1

4th iteration: use {p4, t4}

a = hardlim([0 0][1 1]^T- 2) = 0
e = t - a = t4 - a = 1
w^new = [0 0] + [1 1] = [1 1]
b^new = b^old + e = 0

5th iteration: use {p1, t1}

a = hardlim([1 1][0 0]^T) = 1
e = t - a = t1 - a = -1
w^new = [1 1]-[0 0] = [1 1]
b^new = b^old + e = -1

6th iteration: use {p2, t2}

a = hardlim([1 1][0 1]^T - 1) = 1
e = t - a = t2 - a = -1
w^new = [1 1] - [0 1] = [1 0]
b^new = b^old + e = -2

7th iteration: use {p3, t3}

a = hardlim([1 0][1 0]^T - 2) = 0
e = t - a = t3 - a = 0
w^new = [1 0]
b^new = b^old + e = -2

8th iteration: use {p4, t4}

a = hardlim([1 0][1 1]^T- 2) = 0
e = t - a = t4 - a = 1 - 0 = 1
w^new = [1 0] + [1 1] = [2 1]
b^new = b^old + e = -2 + 1 = -1

9th iteration: use {p1, t1}

a = hardlim([2 1][0 0]^T- 1)= 0
e = t - a = t1 - a = 0
w^new = [2 1]
b^new = b^old + e = -1

10th iteration: use {p2, t2}

a = hardlim([2 1][0 1]^T- 1) = 0
e = t - a = t2 - a = 0
w^new = [2 1]
b^new = b^old + e = -1

11th iteration: use {p3, t3}

a = hardlim([2 1][1 0]^T- 1) = 1
e = t - a = t3 - a = -1
w^new = [2 1]-[1 0] = [1 1]
b^new = b^old + e = -2

12th iteration: use {p4, t4}

a = hardlim([1 1][1 1]^T- 2) = 1
e = t - a = t4 - a = 1 - 1 = 0
w^new = [1 1]
b^new = b^old + e = -2

13th iteration: use {p1, t1}

a = hardlim([1 1][0 0]^T- 1) = 0
e = t - a = t1 - a = 0
w^new = [1 1]
b^new = b^old + e = -2

14th iteration: use {p2, t2}

a = hardlim([1 1][0 1]^T- 1) = 1
e = t - a = t2 - a = -1
w^new = [1 1] - [0 1] = [1 0]
b^new = b^old + e = -3

15th iteration: use {p3, t3}

a = hardlim([1 0][1 0]^T- 3) = 0
e = t - a = t3 - a = 0
w^new = [1 0]
b^new = b^old + e = -3

16th iteration: use {p4, t4}

a = hardlim([1 0][1 1]^T- 3) = 0
e = t - a = t4 - a = 1
w^new = [1 0] + [1 1] = [2 1]
b^new = b^old + e = -2

17th iteration: use {p1, t1}

a = hardlim([2 1][0 0]^T- 2) = 0
e = t - a = t1 - a = 0
w^new = [2 1]
b^new = b^old + e = -2

18th iteration: use {p2, t2}

a = hardlim([2 1][0 1]^T- 2) = 0
e = t - a = t2 - a = 0
w^new = [2 1]
b^new = b^old + e = -2

19th iteration: use {p3, t3}

a = hardlim([2 1][1 0]^T- 2) = 1
e = t - a = t3 - a = 1
w^new = [2 1] + [1 0] = [3 1]
b^new = b^old + e = -1

20th iteration: use {p4, t4}

a = hardlim([3 1][1 1]^T- 1) = 1
e = t - a = t4 - a = 0
w^new = [3 1]
b^new = b^old + e = -1

21st iteration: use {p1, t1}

a = hardlim([3 1][0 0]^T- 1) = 0
e = t - a = t1 - a = 0
w^new = [3 1]
b^new = b^old + e = -1

22nd iteration: use {p2, t2}

a = hardlim([3 1][0 1]^T- 1) = 1
e = t - a = t2 - a = -1
w^new = [3 1] - [0 1] = [3 0]
b^new = b^old + e = -2

23rd iteration: use {p3, t3}

a = hardlim([3 0][1 0]^T- 2) = 1
e = t - a = t3 - a = -1
w^new = [3 0] - [1 0] = [2 0]
b^new = b^old + e = -3


24th iteration: use {p4, t4}

a = hardlim([2 0][1 1]^T- 3) = 0
e = t - a = t4 - a = 1
w^new = [2 0] + [1 1] = [3 1]
b^new = b^old + e = -2

25th iteration: use {p1, t1}

a = hardlim([3 1][0 0]^T- 2) = 0
e = t - a = t1 - a = 0
w^new = [3 1]
b^new = b^old + e = -2

26th iteration: use {p2, t2}

a = hardlim([3 1][0 1]^T- 2) = 0
e = t - a = t2 - a = 0
w^new = [3 1]
b^new = b^old + e = -2

27th iteration: use {p3, t3}

a = hardlim([3 1][1 0]^T- 2) = 1
e = t - a = t3 - a = -1
w^new = [3 1] - [1 0] = [2 1]
b^new = b^old + e = -3

28th iteration: use {p4, t4}

a = hardlim([2 1][1 1]^T- 3) = 1
e = t - a = t4 - a = 0
w^new = [2 1]
b^new = b^old + e = -3

29th iteration: use {p1, t1}

a = hardlim([2 1][0 0]^T- 3) = 0
e = t - a = t1 - a = 0
w^new = [2 1]
b^new = b^old + e = -3

30th iteration: use {p2, t2}

a = hardlim([2 1][0 1]^T- 3) = 0
e = t - a = t2 - a = 0
w^new = [2 1]
b^new = b^old + e = -3

31st iteration: use {p3, t3}

a = hardlim([2 1][1 0]^T- 3) = 0
e = t - a = t3 - a = 0
w^new = [2 1]
b^new = b^old + e = -3

32nd iteration: use {p4, t3}

a = hardlim([2 1][1 1]^T- 3) = 1
e = t - a = t4 - a = 0
w^new = [2 1]
b^new = b^old + e = -3

Thus, w = [2 1] and b = -3.

Let's solve the question using MATLAB
>> p = [0 0 1 1; 0 1 0 1];
>> t = [0 0 0 1];
>> plotpv(p, t)





















>> sample = newp(minmax(p), 1)
>> sample.trainParam.epochs = 20;
>> sample = train(sample, p, t)
>> y = sim(sample, p)
>> plotpv(p, y)
>> sample.iw{1,1}

Answer [2 1]

>> sample.b{1}

Answer -3

>> plotpc(sample.iw{1,1}, sample.b{1})