I'm testing the y = SinC(x) function with single hidden layer feedforward neural networks (SLFNs) with 20 neurons.
With a SLFN, in the output layer, the output weight(OW) can be described by
OW = pinv(H)*T
after adding regularized parameter gamma, which
OW = pinv(I/gamma+H'*H)*H'*T
with
gamma -> Inf, pinv(H'*H)*H'*T == pinv(H)*T, also pinv(H'*H)*H' == pinv(H).
But when I try to calculate pinv(H'*H)*H' and pinv(H), I find a huge difference between these two when neurons number is over 5 (under 5, they are equal or almost the same).
For example, when H is 10*10 matrix, cond(H) = 21137561386980.3, rank(H) = 10,
H = [0.736251410036783 0.499731137079796 0.450233920602169 0.296610970576716 0.369359425954153 0.505556211442208 0.502934880027889 0.364904559142718 0.253349959726753 0.298697900877265;
0.724064281864009 0.521667364351399 0.435944895257239 0.337878535128756 0.364906002569385 0.496504064726699 0.492798607017131 0.390656915261343 0.289981152837390 0.307212326718916;
0.711534656474153 0.543520341487420 0.421761457948049 0.381771374416867 0.360475582262355 0.487454209236671 0.482668250979627 0.417033287703137 0.329570921359082 0.315860145366824;
0.698672860220896 0.565207057974387 0.407705930918082 0.427683127210120 0.356068794706095 0.478412571446765 0.472552121296395 0.443893207685379 0.371735862991355 0.324637323886021;
0.685491077062637 0.586647027111176 0.393799811411985 0.474875155650945 0.351686254239637 0.469385056318048 0.462458480695760 0.471085139463084 0.415948455902421 0.333539494486324;
0.672003357663056 0.607763454504209 0.380063647372632 0.522520267708374 0.347328559602877 0.460377531907542 0.452395518357816 0.498449772544129 0.461556360076788 0.342561958147251;
0.658225608290477 0.628484290731116 0.366516925684188 0.569759064961507 0.342996293691614 0.451395814182317 0.442371323528726 0.525823695636816 0.507817005881821 0.351699689941632;
0.644175558300583 0.648743139215935 0.353177974096445 0.615761051907079 0.338690023332811 0.442445652121229 0.432393859824045 0.553043275759248 0.553944175102542 0.360947346089454;
0.629872705346690 0.668479997764613 0.340063877672496 0.659781468051379 0.334410299080102 0.433532713184646 0.422470940392161 0.579948548513999 0.599160649563718 0.370299272759337;
0.615338237874436 0.687641820315375 0.327190410302607 0.701205860709835 0.330157655029498 0.424662569229062 0.412610204098877 0.606386924575225 0.642749594844498 0.379749516620049];
T=[-0.806458764562879 -0.251682808380338 -0.834815868451399 -0.750626822371170 0.877733363571576 1 -0.626938984683970 -0.767558933097629 -0.921811074815239 -1]';
There is a huge difference between pinv(H'*H)*H*T and pinv(H)*T, where
pinv(H'*H)*H*T = [-4803.39093243484 3567.08623820149 668.037919243849 5975.10699147077
1709.31211566970 -1328.53407325092 -1844.57938928594 -22511.9388736373
-2377.63048959478 31688.5125271114]';
pinv(H)*T = [-19780274164.6438 -3619388884.32672 -76363206688.3469 16455234.9229156
-135982025652.153 -93890161354.8417 283696409214.039 193801203.735488
-18829106.6110445 19064848675.0189]'.
I also find that if I round H , round(H,2), pinv(H'*H)*H*T and pinv(H)*T return the same answer. So I guess one of the reason might be the float calculation issue inside the matlab.
But since cond(H) is large, any small change of H may result in large difference in the inverse of H. I think the round function may not be a good option to test. As Cris Luengo mentioned, with large cond,the numerical imprecision will affect the accuracy of inverse.
In my test, I use 1000 training samples Input:[-10,10], with noise between [-0.2,0.2], and test samples are noise free. 20 neurons are selected. The OW = pinv(H)*Tcan give reasonable results for SinC training, while the performance for OW = pinv(H'*H)*T is worse. Then I try to increase the precision of H'*H by pinv(vpa(H'*H)), there's no significant improvement.
Does anyone know how to solve this?
After some research, the answer is that ELM is very sentive to scaling and activation function.
Please refer to this paper for details: https://dl.acm.org/citation.cfm?id=2797143.2797161
And paper: https://ieeexplore.ieee.org/document/8533625 demonstrated a noval algorithm to improve the perforamance of ELM for scaling.
I'm having some serious issues fitting an exponential function (Beer-Lambert law) to my data. The optimization toolset function that I'm using produces terrible fits:
function [ Coefficients ] = fitting_new( Modified_Spectrum_Data,trajectory )
x_axis = trajectory;
fun = #(x,x_axis) (x(1)*exp((-x(2))*x_axis));
start = [Modified_Spectrum_Data(1) 0.05];
nlm = nlinfit(x_axis,Modified_Spectrum_Data,fun,start,opts);
Coefficients = nlm;
end
Data:
Modified_Spectrum_Data = [1.11111111111111, 1.08784976353957, 1.06352170731165, 1.04099672033640, 1.02649723285838, 1.00423806910703, 0.994116452961827, 0.975928861361604, 0.963081773802984, 0.953191520906905, 0.940636278551651, 0.930360007604054, 0.922259178548511, 0.916659345499171, 0.909149956799775, 0.901241601559703, 0.895375741449218, 0.893308346234150, 0.887985459843162, 0.884657500398024, 0.883852990694089, 0.877158499678129, 0.874817832833850, 0.875428444059047, 0.873170360623947, 0.871461252768665, 0.867913776631497, 0.866459074988087, 0.863819528471106, 0.863228815347816 ,0.864369045426273 ,0.860602502500599, 0.862653463581049, 0.861169231463016, 0.858658616425390, 0.864588421841755, 0.858668693409622, 0.857993365648639]
trajectory = [0.0043, 0.9996, 2.0007, 2.9994, 3.9996, 4.9994, 5.9981, 6.9978, 7.9997, 8.9992, 10.0007, 10.9993, 11.9994, 12.9992, 14.0001, 14.9968, 15.9972, 16.9996, 17.9996, 18.999, 19.9992, 20.9996, 21.9994, 23.0003, 23.9992, 24.999, 25.9987, 26.9986, 27.999, 28.9991, 29.999, 30.9987, 31.9976, 32.9979, 33.9983, 34.9988, 35.999, 36.9991]
I've tried using multiple different fitting functions and messing around with the options, but they don't seem to make too big of a difference. Additionally, I've tried changing the initial guess, but again that doesn't really make a difference.
Excel seems to be able to fit the data perfectly fine, but I have 900 rows of data I want to fit so doing it in Excel is not possible.
Any help would be greatly appreciated, thank you.
You'll want to use the cftool. Your data looks to follow a power law. Then choose 'Modified Spectrum Data' as your x axis and 'Trajectory' as your y. Select 'Power' from the drop down menu towards the top of the GUI.
Modified_Spectrum_Data = [1.11111111111111, 1.08784976353957, 1.06352170731165, 1.04099672033640, 1.02649723285838, 1.00423806910703, 0.994116452961827, 0.975928861361604, 0.963081773802984, 0.953191520906905, 0.940636278551651, 0.930360007604054, 0.922259178548511, 0.916659345499171, 0.909149956799775, 0.901241601559703, 0.895375741449218, 0.893308346234150, 0.887985459843162, 0.884657500398024, 0.883852990694089, 0.877158499678129, 0.874817832833850, 0.875428444059047, 0.873170360623947, 0.871461252768665, 0.867913776631497, 0.866459074988087, 0.863819528471106, 0.863228815347816 ,0.864369045426273 ,0.860602502500599, 0.862653463581049, 0.861169231463016, 0.858658616425390, 0.864588421841755, 0.858668693409622, 0.857993365648639]
trajectory = [0.0043, 0.9996, 2.0007, 2.9994, 3.9996, 4.9994, 5.9981, 6.9978, 7.9997, 8.9992, 10.0007, 10.9993, 11.9994, 12.9992, 14.0001, 14.9968, 15.9972, 16.9996, 17.9996, 18.999, 19.9992, 20.9996, 21.9994, 23.0003, 23.9992, 24.999, 25.9987, 26.9986, 27.999, 28.9991, 29.999, 30.9987, 31.9976, 32.9979, 33.9983, 34.9988, 35.999, 36.9991]
cftool
Screenshot:
For more information on the curve fitting (cftool), see: https://www.mathworks.com/help/curvefit/curvefitting-app.html
I have the following code that generates a matrix of 15 blocks that will then be used in a Montecarlo approach as multiple starting points. How can I get the same exact result in a smarter way?
assume that J=15*100 are the total simulation and paramNum the number of parameters
[10^-10*ones(paramNum,round(J/15)) 10^-9*ones(paramNum,round(J/15)) 10^-8*ones(paramNum,round(J/15)) 10^-7*ones(paramNum,round(J/15)) 10^-6*ones(paramNum,round(J/15)) 10^-5*ones(paramNum,round(J/15)) rand*10^-5*ones(paramNum,round(J/15)) 10^-4*ones(paramNum,round(J/15)) rand*10^-4*ones(paramNum,round(J/15)) 10^-3*ones(paramNum,round(J/15)) 10^-2*ones(paramNum,round(J/15)) 10^-1*ones(paramNum,round(J/15)) 10^-abs(randn/2)*ones(paramNum,round(J/15))];
you could do
v = 10.^[-10:-5 rand*10^-5 -4:-1 10^-abs(randn/2)];
repmat(repelem(v, 1, round(J/15)), paramNum) .* ...
repmat(ones(paramNum,round(J/15)), numel(v))
Or mimic the repmat/repelem functionality with a for loop. The first is shorter, the later is more understandable.
By the way... it's less than 15 blocks...
I am porting a Matlab/Octave code to C. One of the implementations I need to port is the DST transform.
In Matlab/Octave, this is done using
A =[0.00000 4.24264 5.65685 7.07107 8.48528 0.00000]
res = dst(A)
ans =
22.3599 -4.7499 -4.4287 2.9465 -6.8864 1.9383
In C, I have done what the FFTW3 documentation says:
n = 6;
fftw_plan plan = fftw_plan_r2r_1d( n, in, out, FFTW_RODFT10, FFTW_ESTIMATE);
fftw_execute ( plan );
where in is the input array. The result of this is, out:
42.588457
-9.899495
0.000000
2.449490
-11.411543
5.656854
Why are those results so different? Any ideas of what I am doing wrong here, if any?
If you are looking for exactly the same tranform, you may try the RODFT00 flag instead.
According to docs of fftw, matlab, octave, it seems to fit, except for a 2 factor. Output of fftw may be twice the one of matlab or octave.
http://www.mathworks.fr/fr/help/pde/ug/dst.html
http://octave.sourceforge.net/signal/function/dst.html
http://www.fftw.org/doc/1d-Real_002dodd-DFTs-_0028DSTs_0029.html
Bye,
tried this x=[ 0
-0.041389004581424
-0.049979175343607
-0.057007496876302
-0.062473969179509
-0.066378592253228
-0.068721366097459
-0.069502290712203
-0.068721366097459
-0.066378592253228
-0.062473969179509
-0.057007496876302
-0.049979175343607
-0.041389004581424
0.500000000000000
-0.500000000000000]
matlab's dst gives :
dst(x)=[-0.524649937006448
-0.208549082037091
0.175438242940290
-0.201365706563574
0.122632326416342
0.023410695189363
-0.170916792532103
0.404635063593086
-0.557336293807673
0.766827077578399
-0.834391996423464
0.916008576298954
-0.836787527646708
0.743037852780101
-0.522660424688284
0.285317021943821]
using FFTW with "FFTW_RODFT00" flag
gives
[-1.0493
-0.417098
0.350876
-0.402731
0.245265
0.0468214
-0.341834
0.80927
-1.11467
1.53365
-1.66878
1.83202
-1.67358
1.48608
-1.04532
0.570634]
Basically a factor of two is the difference, even if one uses same flag