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<!--This HTML is auto-generated from an m-file.Your changes will be overwritten.--><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:x-large">Linearly Dependent Problem</p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">A linear neuron is trained to find the minimum error solution for a problemwith linearly dependent input vectors. If a linear dependence in inputvectors is not matched in the target vectors, the problem is nonlinear anddoes not have a zero error linear solution.</p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">Copyright 1992-2002 The MathWorks, Inc.$Revision: 1.12 $ $Date: 2002/03/29 19:36:15 $</p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:medium; page-break-before: auto;"><a name=""></a></p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">P defines three 2-element input patterns (column vectors). Note that 0.5times the sum of (column) vectors 1 and 3 results in vector 2. This is calledlinear dependence.</p><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="position: relative; left:30px">P = [ 1.0 2.0 3.0; <span style="color:blue">...</span> 4.0 5.0 6.0];</pre><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:medium; page-break-before: auto;"><a name=""></a></p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">T defines an associated 1-element target (column vectors). Note that 0.5times the sum of -1.0 and 0.5 does not equal 1.0. Because the lineardependence in P is not matched in T this problem is nonlinear and does nothave a zero error linear solution.</p><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="position: relative; left:30px">T = [0.5 1.0 -1.0];</pre><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:medium; page-break-before: auto;"><a name=""></a></p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">MAXLINLR finds the fastest stable learning rate for TRAINWH. NEWLIN creates alinear neuron. NEWLIN takes these arguments: 1) Rx2 matrix of min and maxvalues for R input elements, 2) Number of elements in the output vector, 3)Input delay vector, and 4) Learning rate.</p><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="position: relative; left:30px">maxlr = maxlinlr(P,<span style="color:#B20000">'bias'</span>);net = newlin([0 10;0 10],1,[0],maxlr);</pre><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:medium; page-break-before: auto;"><a name=""></a></p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">TRAIN uses the Widrow-Hoff rule to train linear networks by default. We willdisplay each 50 epochs and train for a maximum of 500 epochs.</p><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="position: relative; left:30px">net.trainParam.show = 50; <span style="color:green">% Frequency of progress displays (in epochs).</span>net.trainParam.epochs = 500; <span style="color:green">% Maximum number of epochs to train.</span>net.trainParam.goal = 0.001; <span style="color:green">% Sum-squared error goal.</span></pre><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:medium; page-break-before: auto;"><a name=""></a></p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">Now the network is trained on the inputs P and targets T. Note that, due tothe linear dependence between input vectors, the problem did not reach theerror goal represented by the black line.</p><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="position: relative; left:30px">[net,tr] = train(net,P,T);</pre><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:gray; font-style:italic;">TRAINB, Epoch 0/500, MSE 0.75/0.001.TRAINB, Epoch 50/500, MSE 0.53429/0.001.TRAINB, Epoch 100/500, MSE 0.434609/0.001.TRAINB, Epoch 150/500, MSE 0.388044/0.001.TRAINB, Epoch 200/500, MSE 0.366292/0.001.TRAINB, Epoch 250/500, MSE 0.35613/0.001.TRAINB, Epoch 300/500, MSE 0.351384/0.001.TRAINB, Epoch 350/500, MSE 0.349166/0.001.TRAINB, Epoch 400/500, MSE 0.34813/0.001.TRAINB, Epoch 450/500, MSE 0.347646/0.001.TRAINB, Epoch 500/500, MSE 0.34742/0.001.TRAINB, Maximum epoch reached.</pre><img xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" src="demolin6_img06.gif"><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:medium; page-break-before: auto;"><a name=""></a></p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">We can now test the associator with one of the original inputs, [1; 4] , andsee if it returns the target, 0.5. The result is not 0.5 as the linearnetwork could not fit the nonlinear problem caused by the linear dependencebetween input vectors.</p><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="position: relative; left:30px">p = [1.0; 4];a = sim(net,p)</pre><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:gray; font-style:italic;">a = 0.8971</pre><originalCode xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" code="%% Linearly Dependent Problem
% A linear neuron is trained to find the minimum error solution for a problem
% with linearly dependent input vectors. If a linear dependence in input
% vectors is not matched in the target vectors, the problem is nonlinear and
% does not have a zero error linear solution.
%
% Copyright 1992-2002 The MathWorks, Inc.
% $Revision: 1.12 $ $Date: 2002/03/29 19:36:15 $

%%
% P defines three 2-element input patterns (column vectors). Note that 0.5
% times the sum of (column) vectors 1 and 3 results in vector 2. This is called
% linear dependence.

P = [ 1.0 2.0 3.0; ...
 4.0 5.0 6.0];

%%
% T defines an associated 1-element target (column vectors). Note that 0.5
% times the sum of -1.0 and 0.5 does not equal 1.0. Because the linear
% dependence in P is not matched in T this problem is nonlinear and does not
% have a zero error linear solution.

T = [0.5 1.0 -1.0];

%%
% MAXLINLR finds the fastest stable learning rate for TRAINWH. NEWLIN creates a
% linear neuron. NEWLIN takes these arguments: 1) Rx2 matrix of min and max
% values for R input elements, 2) Number of elements in the output vector, 3)
% Input delay vector, and 4) Learning rate.

maxlr = maxlinlr(P,'bias');
net = newlin([0 10;0 10],1,[0],maxlr);

%%
% TRAIN uses the Widrow-Hoff rule to train linear networks by default. We will
% display each 50 epochs and train for a maximum of 500 epochs.

net.trainParam.show = 50; % Frequency of progress displays (in epochs).
net.trainParam.epochs = 500; % Maximum number of epochs to train.
net.trainParam.goal = 0.001; % Sum-squared error goal.

%%
% Now the network is trained on the inputs P and targets T. Note that, due to
% the linear dependence between input vectors, the problem did not reach the
% error goal represented by the black line.

[net,tr] = train(net,P,T);

%%
% We can now test the associator with one of the original inputs, [1; 4] , and
% see if it returns the target, 0.5. The result is not 0.5 as the linear
% network could not fit the nonlinear problem caused by the linear dependence
% between input vectors.

p = [1.0; 4];
a = sim(net,p)
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