I asked this question in Math Stackexchange, but it seems it didn't get enough attention there so I am asking it here. https://math.stackexchange.com/questions/1729946/why-do-we-say-svd-can-handle-singular-matrx-when-doing-least-square-comparison?noredirect=1#comment3530971_1729946
I learned from some tutorials that SVD should be more stable than QR decomposition when solving Least Square problem, and it is able to handle singular matrix. But the following example I wrote in matlab seems to support the opposite conclusion. I don't have a deep understanding of SVD, so if you could look at my questions in the old post in Math StackExchange and explain it to me, I would appreciate a lot.
I use a matrix that have a large condition number(e+13). The result shows SVD get a much larger error(0.8) than QR(e-27)
% we do a linear regression between Y and X
data= [
47.667483331 -122.1070832;
47.667483331001 -122.1070832
];
X = data(:,1);
Y = data(:,2);
X_1 = [ones(length(X),1),X];
%%
%SVD method
[U,D,V] = svd(X_1,'econ');
beta_svd = V*diag(1./diag(D))*U'*Y;
%% QR method(here one can also use "\" operator, which will get the same result as I tested. I just wrote down backward substitution to educate myself)
[Q,R] = qr(X_1)
%now do backward substitution
[nr nc] = size(R)
beta_qr=[]
Y_1 = Q'*Y
for i = nc:-1:1
s = Y_1(i)
for j = m:-1:i+1
s = s - R(i,j)*beta_qr(j)
end
beta_qr(i) = s/R(i,i)
end
svd_error = 0;
qr_error = 0;
for i=1:length(X)
svd_error = svd_error + (Y(i) - beta_svd(1) - beta_svd(2) * X(i))^2;
qr_error = qr_error + (Y(i) - beta_qr(1) - beta_qr(2) * X(i))^2;
end
You SVD-based approach is basically the same as the pinv function in MATLAB (see Pseudo-inverse and SVD). What you are missing though (for numerical reasons) is using a tolerance value such that any singular values less than this tolerance are treated as zero.
If you refer to edit pinv.m, you can see something like the following (I won't post the exact code here because the file is copyrighted to MathWorks):
[U,S,V] = svd(A,'econ');
s = diag(S);
tol = max(size(A)) * eps(norm(s,inf));
% .. use above tolerance to truncate singular values
invS = diag(1./s);
out = V*invS*U';
In fact pinv has a second syntax where you can explicitly specify the tolerance value pinv(A,tol) if the default one is not suitable...
So when solving a least-squares problem of the form minimize norm(A*x-b), you should understand that the pinv and mldivide solutions have different properties:
x = pinv(A)*b is characterized by the fact that norm(x) is smaller than the norm of any other solution.
x = A\b has the fewest possible nonzero components (i.e sparse).
Using your example (note that rcond(A) is very small near machine epsilon):
data = [
47.667483331 -122.1070832;
47.667483331001 -122.1070832
];
A = [ones(size(data,1),1), data(:,1)];
b = data(:,2);
Let's compare the two solutions:
x1 = A\b;
x2 = pinv(A)*b;
First you can see how mldivide returns a solution x1 with one zero component (this is obviously a valid solution because you can solve both equations by multiplying by zero as in b + a*0 = b):
>> sol = [x1 x2]
sol =
-122.1071 -0.0537
0 -2.5605
Next you see how pinv returns a solution x2 with a smaller norm:
>> nrm = [norm(x1) norm(x2)]
nrm =
122.1071 2.5611
Here is the error of both solutions which is acceptably very small:
>> err = [norm(A*x1-b) norm(A*x2-b)]
err =
1.0e-11 *
0 0.1819
Note that use mldivide, linsolve, or qr will give pretty much same results:
>> x3 = linsolve(A,b)
Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 2.159326e-16.
x3 =
-122.1071
0
>> [Q,R] = qr(A); x4 = R\(Q'*b)
x4 =
-122.1071
0
SVD can handle rank-deficiency. The diagonal matrix D has a near-zero element in your code and you need use pseudoinverse for SVD, i.e. set the 2nd element of 1./diag(D) to 0 other than the huge value (10^14). You should find SVD and QR have equally good accuracy in your example. For more information, see this document http://www.cs.princeton.edu/courses/archive/fall11/cos323/notes/cos323_f11_lecture09_svd.pdf
Try this SVD version called block SVD - you just set the iterations equal to the accuracy you want - usually 1 is enough. If you want all the factors (this has a default # selected for factor reduction) then edit the line k= to the size(matrix) if I recall my MATLAB correctly
A= randn(100,5000);
A=corr(A);
% A is your correlation matrix
tic
k = 1000; % number of factors to extract
bsize = k +50;
block = randn(size(A,2),bsize);
iter = 2; % could set via tolerance
[block,R] = qr(A*block,0);
for i=1:iter
[block,R] = qr(A*(A'*block),0);
end
M = block'*A;
% Economy size dense SVD.
[U,S] = svd(M,0);
U = block*U(:,1:k);
S = S(1:k,1:k);
% Note SVD of a symmetric matrix is:
% A = U*S*U' since V=U in this case, S=eigenvalues, U=eigenvectors
V=real(U*sqrt(S)); %scaling matrix for simulation
toc
% reduced randomized matrix for simulation
sims = 2000;
randnums = randn(k,sims);
corrrandnums = V*randnums;
est_corr_matrix = corr(corrrandnums');
total_corrmatrix_difference =sum(sum(est_corr_matrix-A))
num = zeros(1,freq);
den = zeros(1,freq);
for R = 1:freq
[num(R), den(R)]=butter(4, [0.1 0.9]);
end
I thought it was quite trivial but once I run it, I get:
In an assignment A(I) = B, the number of elements in B and I must be the same.
What am I doing wrong?
What you are doing wrong is that both num and den will contain multiple coefficients:
[b,a] = butter(n,Wn) returns the transfer function coefficients of an nth-order lowpass digital Butterworth filter with normalized cutoff frequency Wn.
b,a — Transfer function coefficients
row vectors
As copied from the documentation
The way to get your code working would be to either set num and den to a matrix, or to a cell array:
num = zeros(freq,4);
den = zeros(freq,4);
for R = 1:freq
[num(R,:), den(R,:)]=butter(4, [A(R) B(R)]); % matrix
end
for R = 1:freq
[num{R}, den{R}]=butter(4, [A(R) B(R)]); % cell
end
Probably the matrix is better suited for your purposes.
I'm looking for a way to speed up some simple two port matrix calculations. See the below code example for what I'm doing currently. In essence, I create a [Nx1] frequency vector first. I then loop through the frequency vector and create the [2x2] matrices H1 and H2 (all functions of f). A bit of simple matrix math including a matrix left division '\' later, and I got my result pb as a [Nx1] vector. The problem is the loop - it takes a long time to calculate and I'm looking for way to improve efficiency of the calculations. I tried assembling the problem using [2x2xN] transfer matrices, but the mtimes operation cannot handle 3-D multiplications.
Can anybody please give me an idea how I can approach such a calculation without the need for looping through f?
Many thanks: svenr
% calculate frequency and wave number vector
f = linspace(20,200,400);
w = 2.*pi.*f;
% calculation for each frequency w
for i=1:length(w)
H1(i,1) = {[1, rho*c*k(i)^2 / (crad*pi); 0,1]};
H2(i,1) = {[1, 1i.*w(i).*mp; 0, 1]};
HZin(i,1) = {H1{i,1}*H2{i,1}};
temp_mat = HZin{i,1}*[1; 0];
Zin(i,1) = temp_mat(1,1)/temp_mat(2,1);
temp_mat= H1{i,1}\[1; 1/Zin(i,1)];
pb(i,1) = temp_mat(1,1); Ub(i,:) = temp_mat(2,1);
end
Assuming that length(w) == length(k) returns true , rho , c, crad, mp are all scalars and in the last line is Ub(i,1) = temp_mat(2,1) instead of Ub(i,:) = temp_mat(2,1);
temp = repmat(eyes(2),[1 1 length(w)]);
temp1(1,2,:) = rho*c*(k.^2)/crad/pi;
temp2(1,2,:) = (1i.*w)*mp;
H1 = permute(num2cell(temp1,[1 2]),[3 2 1]);
H2 = permute(num2cell(temp2,[1 2]),[3 2 1]);
HZin = cellfun(#(a,b)(a*b),H1,H2,'UniformOutput',0);
temp_cell = cellfun(#(a,b)(a*b),H1,repmat({[1;0]},length(w),1),'UniformOutput',0);
Zin_cell = cellfun(#(a)(a(1,1)/a(2,1)),temp_cell,'UniformOutput',0);
Zin = cell2mat(Zin);
temp2_cell = cellfun(#(a)({[1;1/a]}),Zin_cell,'UniformOutput',0);
temp3_cell = cellfun(#(a,b)(pinv(a)*b),H1,temp2_cell);
temp4 = cell2mat(temp3_cell);
p(:,1) = temp4(1:2:end-1);
Ub(:,1) = temp4(2:2:end);
I trained a neural network using the MATLAB Neural Network Toolbox, and in particular using the command nprtool, which provides a simple GUI to use the toolbox features, and to export a net object containing the informations about the NN generated.
In this way, I created a working neural network, that I can use as classifier, and a diagram representing it is the following:
There are 200 inputs, 20 neurons in the first hidden layer, and 2 neurons in the last layer that provide a bidimensional output.
What I want to do is to use the network in some other programming language (C#, Java, ...).
In order to solve this problem, I try to use the following code in MATLAB:
y1 = tansig(net.IW{1} * input + net.b{1});
Results = tansig(net.LW{2} * y1 + net.b{2});
Assuming that input is a monodimensional array of 200 elements, the previous code would work if net.IW{1} is a 20x200 matrix (20 neurons, 200 weights).
The problem is that I noticed that size(net.IW{1}) returns unexpected values:
>> size(net.IW{1})
ans =
20 199
I got the same problem with a network with 10000 input. In this case, the result wasn't 20x10000, but something like 20x9384 (I don't remember the exact value).
So, the question is: how can I obtain the weights of each neuron? And after that, can someone explain me how can I use them to produce the same output of MATLAB?
I solved the problems described above, and I think it is useful to share what I've learned.
Premises
First of all, we need some definitions. Let's consider the following image, taken from [1]:
In the above figure, IW stands for initial weights: they represent the weights of neurons on the Layer 1, each of which is connected with each input, as the following image shows [1]:
All the other weights, are called layer weights (LW in the first figure), that are also connected with each output of the previous layer. In our case of study, we use a network with only two layers, so we will use only one LW array to solve our problems.
Solution of the problem
After the above introduction, we can proceed by dividing the issue in two steps:
Force the number of initial weights to match with the input array length
Use the weights to implement and use the neural network just trained in other programming languages
A - Force the number of initial weights to match with the input array length
Using the nprtool, we can train our network, and at the end of the process, we can also export in the workspace some information about the entire training process. In particular, we need to export:
a MATLAB network object that represents the neural network created
the input array used to train the network
the target array used to train the network
Also, we need to generate a M-file that contains the code used by MATLAB to create the neural network, because we need to modify it and change some training options.
The following image shows how to perform these operations:
The M-code generated will be similar to the following one:
function net = create_pr_net(inputs,targets)
%CREATE_PR_NET Creates and trains a pattern recognition neural network.
%
% NET = CREATE_PR_NET(INPUTS,TARGETS) takes these arguments:
% INPUTS - RxQ matrix of Q R-element input samples
% TARGETS - SxQ matrix of Q S-element associated target samples, where
% each column contains a single 1, with all other elements set to 0.
% and returns these results:
% NET - The trained neural network
%
% For example, to solve the Iris dataset problem with this function:
%
% load iris_dataset
% net = create_pr_net(irisInputs,irisTargets);
% irisOutputs = sim(net,irisInputs);
%
% To reproduce the results you obtained in NPRTOOL:
%
% net = create_pr_net(trainingSetInput,trainingSetOutput);
% Create Network
numHiddenNeurons = 20; % Adjust as desired
net = newpr(inputs,targets,numHiddenNeurons);
net.divideParam.trainRatio = 75/100; % Adjust as desired
net.divideParam.valRatio = 15/100; % Adjust as desired
net.divideParam.testRatio = 10/100; % Adjust as desired
% Train and Apply Network
[net,tr] = train(net,inputs,targets);
outputs = sim(net,inputs);
% Plot
plotperf(tr)
plotconfusion(targets,outputs)
Before start the training process, we need to remove all preprocessing and postprocessing functions that MATLAB executes on inputs and outputs. This can be done adding the following lines just before the % Train and Apply Network lines:
net.inputs{1}.processFcns = {};
net.outputs{2}.processFcns = {};
After these changes to the create_pr_net() function, simply we can use it to create our final neural network:
net = create_pr_net(input, target);
where input and target are the values we exported through nprtool.
In this way, we are sure that the number of weights is equal to the length of input array. Also, this process is useful in order to simplify the porting to other programming languages.
B - Implement and use the neural network just trained in other programming languages
With these changes, we can define a function like this:
function [ Results ] = classify( net, input )
y1 = tansig(net.IW{1} * input + net.b{1});
Results = tansig(net.LW{2} * y1 + net.b{2});
end
In this code, we use the IW and LW arrays mentioned above, but also the biases b, used in the network schema by the nprtool. In this context, we don't care about the role of biases; simply, we need to use them because nprtool does it.
Now, we can use the classify() function defined above, or the sim() function equally, obtaining the same results, as shown in the following example:
>> sim(net, input(:, 1))
ans =
0.9759
-0.1867
-0.1891
>> classify(net, input(:, 1))
ans =
0.9759
-0.1867
-0.1891
Obviously, the classify() function can be interpreted as a pseudocode, and then implemented in every programming languages in which is possibile to define the MATLAB tansig() function [2] and the basic operations between arrays.
References
[1] Howard Demuth, Mark Beale, Martin Hagan: Neural Network Toolbox 6 - User Guide, MATLAB
[2] Mathworks, tansig - Hyperbolic tangent sigmoid transfer function, MATLAB Documentation center
Additional notes
Take a look to the robott's answer and the Sangeun Chi's answer for more details.
Thanks to VitoShadow and robott answers, I can export Matlab neural network values to other applications.
I really appreciate them, but I found some trivial errors in their codes and want to correct them.
1) In the VitoShadow codes,
Results = tansig(net.LW{2} * y1 + net.b{2});
-> Results = net.LW{2} * y1 + net.b{2};
2) In the robott preprocessing codes,
It would be easier extracting xmax and xmin from the net variable than calculating them.
xmax = net.inputs{1}.processSettings{1}.xmax
xmin = net.inputs{1}.processSettings{1}.xmin
3) In the robott postprocessing codes,
xmax = net.outputs{2}.processSettings{1}.xmax
xmin = net.outputs{2}.processSettings{1}.xmin
Results = (ymax-ymin)*(Results-xmin)/(xmax-xmin) + ymin;
-> Results = (Results-ymin)*(xmax-xmin)/(ymax-ymin) + xmin;
You can manually check and confirm the values as follows:
p2 = mapminmax('apply', net(:, 1), net.inputs{1}.processSettings{1})
-> preprocessed data
y1 = purelin ( net.LW{2} * tansig(net.iw{1}* p2 + net.b{1}) + net.b{2})
-> Neural Network processed data
y2 = mapminmax( 'reverse' , y1, net.outputs{2}.processSettings{1})
-> postprocessed data
Reference:
http://www.mathworks.com/matlabcentral/answers/14517-processing-of-i-p-data
This is a small improvement to the great Vito Gentile's answer.
If you want to use the preprocessing and postprocessing 'mapminmax' functions, you have to pay attention because 'mapminmax' in Matlab normalizes by ROW and not by column!
This is what you need to add to the upper "classify" function, to keep a coherent pre/post-processing:
[m n] = size(input);
ymax = 1;
ymin = -1;
for i=1:m
xmax = max(input(i,:));
xmin = min(input(i,:));
for j=1:n
input(i,j) = (ymax-ymin)*(input(i,j)-xmin)/(xmax-xmin) + ymin;
end
end
And this at the end of the function:
ymax = 1;
ymin = 0;
xmax = 1;
xmin = -1;
Results = (ymax-ymin)*(Results-xmin)/(xmax-xmin) + ymin;
This is Matlab code, but it can be easily read as pseudocode.
Hope this will be helpful!
I tried to implement a simply 2-layer NN in C++ using OpenCV and then exported the weights to Android which worked quiet well. I wrote a small script which generates a header file with the learned weights and this is used in the following code snipped.
// Map Minimum and Maximum Input Processing Function
Mat mapminmax_apply(Mat x, Mat settings_gain, Mat settings_xoffset, double settings_ymin){
Mat y;
subtract(x, settings_xoffset, y);
multiply(y, settings_gain, y);
add(y, settings_ymin, y);
return y;
/* MATLAB CODE
y = x - settings_xoffset;
y = y .* settings_gain;
y = y + settings_ymin;
*/
}
// Sigmoid Symmetric Transfer Function
Mat transig_apply(Mat n){
Mat tempexp;
exp(-2*n, tempexp);
Mat transig_apply_result = 2 /(1 + tempexp) - 1;
return transig_apply_result;
}
// Map Minimum and Maximum Output Reverse-Processing Function
Mat mapminmax_reverse(Mat y, Mat settings_gain, Mat settings_xoffset, double settings_ymin){
Mat x;
subtract(y, settings_ymin, x);
divide(x, settings_gain, x);
add(x, settings_xoffset, x);
return x;
/* MATLAB CODE
function x = mapminmax_reverse(y,settings_gain,settings_xoffset,settings_ymin)
x = y - settings_ymin;
x = x ./ settings_gain;
x = x + settings_xoffset;
end
*/
}
Mat getNNParameter (Mat x1)
{
// convert double array to MAT
// input 1
Mat x1_step1_xoffsetM = Mat(1, 48, CV_64FC1, x1_step1_xoffset).t();
Mat x1_step1_gainM = Mat(1, 48, CV_64FC1, x1_step1_gain).t();
double x1_step1_ymin = -1;
// Layer 1
Mat b1M = Mat(1, 25, CV_64FC1, b1).t();
Mat IW1_1M = Mat(48, 25, CV_64FC1, IW1_1).t();
// Layer 2
Mat b2M = Mat(1, 48, CV_64FC1, b2).t();
Mat LW2_1M = Mat(25, 48, CV_64FC1, LW2_1).t();
// input 1
Mat y1_step1_gainM = Mat(1, 48, CV_64FC1, y1_step1_gain).t();
Mat y1_step1_xoffsetM = Mat(1, 48, CV_64FC1, y1_step1_xoffset).t();
double y1_step1_ymin = -1;
// ===== SIMULATION ========
// Input 1
Mat xp1 = mapminmax_apply(x1, x1_step1_gainM, x1_step1_xoffsetM, x1_step1_ymin);
Mat temp = b1M + IW1_1M*xp1;
// Layer 1
Mat a1M = transig_apply(temp);
// Layer 2
Mat a2M = b2M + LW2_1M*a1M;
// Output 1
Mat y1M = mapminmax_reverse(a2M, y1_step1_gainM, y1_step1_xoffsetM, y1_step1_ymin);
return y1M;
}
example for a bias in the header could be this:
static double b2[1][48] = {
{-0.19879, 0.78254, -0.87674, -0.5827, -0.017464, 0.13143, -0.74361, 0.4645, 0.25262, 0.54249, -0.22292, -0.35605, -0.42747, 0.044744, -0.14827, -0.27354, 0.77793, -0.4511, 0.059346, 0.29589, -0.65137, -0.51788, 0.38366, -0.030243, -0.57632, 0.76785, -0.36374, 0.19446, 0.10383, -0.57989, -0.82931, 0.15301, -0.89212, -0.17296, -0.16356, 0.18946, -1.0032, 0.48846, -0.78148, 0.66608, 0.14946, 0.1972, -0.93501, 0.42523, -0.37773, -0.068266, -0.27003, 0.1196}};
Now, that Google published Tensorflow, this became obsolete.
Hence the solution becomes (after correcting all parts)
Here I am giving a solution in Matlab, but if you have tanh() function, you may easily convert it to any programming language. It is for just showing the fields from network object and the operations you need.
Assume you have a trained ann (network object) that you want to export
Assume that the name of the trained ann is trained_ann
Here is the script for exporting and testing.
Testing script compares original network result with my_ann_evaluation() result
% Export IT
exported_ann_structure = my_ann_exporter(trained_ann);
% Run and Compare
% Works only for single INPUT vector
% Please extend it to MATRIX version by yourself
input = [12 3 5 100];
res1 = trained_ann(input')';
res2 = my_ann_evaluation(exported_ann_structure, input')';
where you need the following two functions
First my_ann_exporter:
function [ my_ann_structure ] = my_ann_exporter(trained_netw)
% Just for extracting as Structure object
my_ann_structure.input_ymax = trained_netw.inputs{1}.processSettings{1}.ymax;
my_ann_structure.input_ymin = trained_netw.inputs{1}.processSettings{1}.ymin;
my_ann_structure.input_xmax = trained_netw.inputs{1}.processSettings{1}.xmax;
my_ann_structure.input_xmin = trained_netw.inputs{1}.processSettings{1}.xmin;
my_ann_structure.IW = trained_netw.IW{1};
my_ann_structure.b1 = trained_netw.b{1};
my_ann_structure.LW = trained_netw.LW{2};
my_ann_structure.b2 = trained_netw.b{2};
my_ann_structure.output_ymax = trained_netw.outputs{2}.processSettings{1}.ymax;
my_ann_structure.output_ymin = trained_netw.outputs{2}.processSettings{1}.ymin;
my_ann_structure.output_xmax = trained_netw.outputs{2}.processSettings{1}.xmax;
my_ann_structure.output_xmin = trained_netw.outputs{2}.processSettings{1}.xmin;
end
Second my_ann_evaluation:
function [ res ] = my_ann_evaluation(my_ann_structure, input)
% Works with only single INPUT vector
% Matrix version can be implemented
ymax = my_ann_structure.input_ymax;
ymin = my_ann_structure.input_ymin;
xmax = my_ann_structure.input_xmax;
xmin = my_ann_structure.input_xmin;
input_preprocessed = (ymax-ymin) * (input-xmin) ./ (xmax-xmin) + ymin;
% Pass it through the ANN matrix multiplication
y1 = tanh(my_ann_structure.IW * input_preprocessed + my_ann_structure.b1);
y2 = my_ann_structure.LW * y1 + my_ann_structure.b2;
ymax = my_ann_structure.output_ymax;
ymin = my_ann_structure.output_ymin;
xmax = my_ann_structure.output_xmax;
xmin = my_ann_structure.output_xmin;
res = (y2-ymin) .* (xmax-xmin) /(ymax-ymin) + xmin;
end
I am using Gonzalez frdescp function to get Fourier descriptors of a boundary. I use this code, and I get two totally different sets of numbers describing two identical but different in scale shapes.
So what is wrong?
im = imread('c:\classes\a1.png');
im = im2bw(im);
b = bwboundaries(im);
f = frdescp(b{1}); // fourier descriptors for the boundary of the first object ( my pic only contains one object anyway )
// Normalization
f = f(2:20); // getting the first 20 & deleting the dc component
f = abs(f) ;
f = f/f(1);
Why do I get different descriptors for identical - but different in scale - two circles?
The problem is that the frdescp code (I used this code, that should be the same as referred by you) is written also in order to center the Fourier descriptors.
If you want to describe your shape in a correct way, it is mandatory to mantain some descriptors that are symmetric with respect to the one representing the DC component.
The following image summarize the concept:
In order to solve your problem (and others like yours), I wrote the following two functions:
function descriptors = fourierdescriptor( boundary )
%I assume that the boundary is a N x 2 matrix
%Also, N must be an even number
np = size(boundary, 1);
s = boundary(:, 1) + i*boundary(:, 2);
descriptors = fft(s);
descriptors = [descriptors((1+(np/2)):end); descriptors(1:np/2)];
end
function significativedescriptors = getsignificativedescriptors( alldescriptors, num )
%num is the number of significative descriptors (in your example, is was 20)
%In the following, I assume that num and size(alldescriptors,1) are even numbers
dim = size(alldescriptors, 1);
if num >= dim
significativedescriptors = alldescriptors;
else
a = (dim/2 - num/2) + 1;
b = dim/2 + num/2;
significativedescriptors = alldescriptors(a : b);
end
end
Know, you can use the above functions as follows:
im = imread('test.jpg');
im = im2bw(im);
b = bwboundaries(im);
b = b{1};
%force the number of boundary points to be even
if mod(size(b,1), 2) ~= 0
b = [b; b(end, :)];
end
%define the number of significative descriptors I want to extract (it must be even)
numdescr = 20;
%Now, you can extract all fourier descriptors...
f = fourierdescriptor(b);
%...and get only the most significative:
f_sign = getsignificativedescriptors(f, numdescr);
I just went through the same problem with you.
According to this link, if you want invariant to scaling, make the comparison ratio-like, for example by dividing every Fourier coefficient by the DC-coefficient. f*1 = f1/f[0], f*[2]/f[0], and so on. Thus, you need to use the DC-coefficient where the f(1) in your code is not the actual DC-coefficient after your step "f = f(2:20); % getting the first 20 & deleting the dc component". I think the problem can be solved by keeping the value of the DC-coefficient first, the code after adjusted should be like follows:
% Normalization
DC = f(1);
f = f(2:20); % getting the first 20 & deleting the dc component
f = abs(f) ; % use magnitudes to be invariant to translation & rotation
f = f/DC; % divide the Fourier coefficients by the DC-coefficient to be invariant to scale