%% SOLUTIONS TO THE HOMEWORK, DEC 9
% Just as an overview, the mapping is taking points in R^2 and producing
% points in R^3 using 5 centers.  
%
% What are the sizes of our weights and bias vectors?  The weight matrix W
% takes a vector from R^5 and produces a vector in R^3, so W is 5 x 3, and
% the bias vector (if there is one) is in R^3.

% First the data:
X=[1 0;0 1;1 1;-1 0];
T=[2 1 -1;1 1 1;1 1 0;-1 0 1];
C=2*randn(5,2);

% Build the RBF
E1=edm(X,C);
Phi=rbf1(E1,1,1);  % Use the Gaussian and width 1
W=pinv(Phi)*T     % This weight matrix is 5 x 3
Y=Phi*W           % Y is the model output.


% Modification of the previous code to put in a bias:
Phi01=[Phi,ones(4,1)];  % This is transposed from the usual to match sizes.
W01=pinv(Phi01)*T       % The bias is the last row of W01 (which is 6 x 3)

% We can break it up:
W=W01(1:5,:); b=W01(6,:)';
Y01=Phi*W+repmat(b',4,1)

%% Problem 2:

P=[1 2 3];
T=[2.0 4.1 5.9];
net=newrb(P,T);
tt=linspace(1,3);
yy=sim(net,tt);
plot(P,T,'k*',tt,yy);

% Find the weights and biases and centers.  Verify the previous answer.
Centers=net.IW{1,1};
W=net.LW{2,1}
b=net.b{2};
scale1=net.b{1};

%Re-compute the output of the RBF and compare to Matlab
A=edm(tt',Centers);
A1=A.*repmat(scale1',100,1);
Phi=rbf1(A1,1,1);
Yout=W*Phi'+b;

max(abs(yy-Yout))  %Biggest difference is?