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The Adaline Neuron. Ranga Rodrigo February 8, 2014. Introduction. In the last lecture, we studied about the perceptron .

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the adaline neuron

The Adaline Neuron

Ranga Rodrigo

February 8, 2014

introduction
Introduction
  • In the last lecture, we studied about the perceptron.
  • By minimizing the sum-of-squares error with respect to the weights, with the assumption of an identity activation function, we derived the Widrow-Hoff Learning Rule.
  • We implemented this in Matlab to carry out linear classification in 2 dimensions.
  • What we have implemented, actually, is the adaline model.
  • Today we consider more such models.
adaline neuron
Adaline Neuron
  • The construction of this neuron is very similar to the perceptron model, and the only difference relates to the learning algorithm.
  • Computation of output signal y is identical to the perceptron.
  • However, the output desired signal d is compared to signal s at the output of the linear part of the neuron (adder).
adaline neuron1
Adaline Neuron

x0 = 1

w0

x1

w1

w2

x2

s

y

x3

w3

wD

xD

-

+

d

slide6

% Adaline Neuron

% Ranga Rodrigo

% February 17, 2014

clc; clear all; close all

c1 = [2,4; 3, 3];

c2 = [4, 8; 8, 4];

scatter(c1(:,1), c1(:,2), \'o\', \'MarkerEdgeColor\', \'b\', \'MarkerFaceColor\', [0, 0.5, 1])

hold on

scatter(c2(:,1), c2(:,2), \'o\', \'MarkerEdgeColor\', \'r\', \'MarkerFaceColor\', [1, 0.5, 0.5])

axis([0,10, 0,10])

hold on

eta = 0.02; % Learning rate (must be carefully selected)

w = [-1, 4, 3]\'; % Randomly initialize [w(1), w(2), theta]

x = [c1; c2];

n = size(x,1);

t = [ones(size(c1,1), 1); -ones(size(c2,1), 1)];

slide7

e = inf;

k = 1; % Maximum number of iterations

kmax = 1000;

while e > 1 && k < kmax

e = 0;

for i = 1:n

s = w\'*[x(i,:), 1]\'

delta = (t(i) - s)

w = w + eta*delta*[x(i, :), 1]\'

e = e + delta^2

end

e

k = k +1;

% Plotting the current decision boundary

p1 = [0, -w(3)/w(2)]\';

p2 = [-w(3)/w(1), 0]\';

line([p1(1), p2(1)], [p1(2), p2(2)])

pause(0.5)

end

slide8

% Plotting the final decision boundary

p1 = [0, -w(3)/w(2)]\';

p2 = [-w(3)/w(1), 0]\';

line([p1(1), p2(1)], [p1(2), p2(2)], \'Color\', \'m\', \'LineWidth\', 4)

% Test point

x = [2,1]\';

y = stepfun(w\'*[x\', 1]\');

if y == 1

scatter(x(1), x(2), \'s\', \'MarkerEdgeColor\', \'b\', \'MarkerFaceColor\', [0, 0.5, 1])

else

scatter(x(1), x(2), \'s\', \'MarkerEdgeColor\', \'r\', \'MarkerFaceColor\', [1, 0.5, 0.5])

end

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