For the following letter, I wish to add noise to it by changing 5 percent of the 1's into 0's. So far, I have the following code which turns them all into 0's. Can someone please point me in the right direction? Thank you!
letterA = [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...
0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 ...
0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 ...
0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 ...
0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 ...
0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 ...
0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 ...
0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 ...
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 ...
0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 ...
0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 ...
0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 ...
0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 ...
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 ...
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0];
for i=1:numel(letterA)
if letterA(i)==1
letterA(i)=0;
end
end
disp(letterA)
try this:
letterA( letterA == 1 & rand(size(letterA)) <= 0.05 ) = 0;
In fact you could also do
letterA( rand(size(letterA)) <= 0.05 ) = 0;
which sets each element with probability of 5% to zero. The already zero elements are not affected. I think what causes confusion here is that you have to recognize that each element is independently handled from each other. It makes no difference if you do the first or the second version.
You can check it:
letterA = (rand(1e5,1) < 0.2); N1 = nnz(letterA);
letterA( rand(size(letterA)) <= 0.05 ) = 0;
(N1 - nnz(letterA))/N1
which gives values around 0.05, i.e. 5%. And it is not true what EitanT says, that it will flip at maximum 5%. It can be more than 5% or less, but on average it is 5%.
EitanTs version flippes exactly 5%, so which version to select depends on the application. For EitanT version the noise is correlated to the signal (because it is exact), which may or may not be what you want.
The basic approach is to find the indices of the 1's and count them, randomly pick a desired amount of indices out of them, and then operate on them:
one_flip_ratio = 0.05;
idx_ones = find(letterA == 1); %// Indices of 1's
flips = round(one_flip_ratio * numel(idx_ones)); %// Number of flips
idx_flips = idx_ones(randperm(numel(idx_ones), flips)); %// Indices of elements
letterA(idx_flips) = 0; %// Flip elements
This will flip 5% of the 1's to 0's.
Thanks for throwing out all these ideas, but eventually I came up with this. It will allow me to easily control both letter and background noise, which is what I intend to do. I'm just a novice, so this may not be the most efficient code, but it gets the job done! (I'm not looking for exactly 5%, the naked eye display value is what I'm more worried about.) PLEASE let me know how this can be improved! Thank you.
background_noise_intensity=0.05;
letter_noise_intensity=0.05;
for i=1:numel(letterA)
if letterA(i)==0
if rand < background_noise_intensity
letterA(i)=1;
end
elseif letterA(i)==1
if rand < letter_noise_intensity
letterA(i)=0;
end
end
end
noisy_letters=letterA;
reshaped_noisy_letters=reshape(noisy_letters,37,19)';
imshow(reshaped_noisy_letters);
Related
i have a matrix 20 rows and 20 columns,
If the value 1 in the row 5 the column take 0
matric=[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ;
0 1 1 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 1 1;
0 1 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 1 0 1;
0 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 1 0 1;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;];
if (matric(5,:)==1)
matric(1:5,1:end)=0;end
I try to compare the second row and the 5 row
If we have "1" in row 2 and row 5
The row 2 take 0
if (matric(5,:)==matric(2,:)==1)
matric(2,1:end)=0;end
do you have an idea
Thank you
The desired output is:
matric=[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
1 0 1 1 0 0 1 1 1 1 0 1 0 1 1 1 1 0 1 0 ;2row will change
0 1 1 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 1 1;
0 1 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 1 0 1;
0 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 1 0 1; % 5 row
You can use logical indexing.
I strongly encourage you to read the following references:
Mathworks documentation page: Find Array Elements That Meet a Condition
Loren on the Art of MATLAB blog entry: Logical Indexing – Multiple
Conditions
Use the following lines of code:
Put 0 in all those columns that have a value of 1 in row 5:
matric(:, matric(5, :) == 1) = 0;
Put 0 in all those columns of row 2 that have a value of 1 both in rows 2 and 5:
matric(2, matric(2, :) == matric(5, :)) = 0;
You can use logical indexing to achieve this. Now I must say I am a bit confused by what exactly you want to achieve based on your description, but based on the your code the first statement can be done as follow:
matric(1:5,matric(5,:)==1) = 0;
and the second would look like:
matric(2,matric(5,:)==1 & matric(2,:)==1)=0;
I am attempting to solve the system Ax = b in MATLAB, where A is a 30x30 triangular matrix with (nonzero) values ranging from 1e-14 to 0.7, and b is a 30-element column vector with values ranging from 1e-3 to 1e3. When I enter x = A\b, I get an answer and no warning messages, but the answer is not reasonable (looks like just random numbers at the bottom of the vector). I presume this is due to numerical errors.
Message 5 on this page suggests decomposing/scaling the matrix in order to avoid numerical errors, but I haven't been able to figure out how to calculate the scaling matrices.
So the question is: Is this indeed an example of numerical instability, and if so, how can I rescale my matrix A, or change how MATLAB is performing the calculation, to avoid it?
Here is the matrix and vector that are producing the issue:
A =
Columns 1 through 15
0.69 0.4278 0.19893 0.082223 0.031861 0.011852 0.0042866 0.0015187 0.00052965 0.00018243 6.221e-05 2.1038e-05 7.0653e-06 2.3587e-06 7.8344e-07
0 0.4761 0.44277 0.27452 0.14183 0.065953 0.028624 0.011831 0.0047156 0.0018273 0.00069233 0.00025755 9.4356e-05 3.4126e-05 1.2206e-05
0 0 0.32851 0.40735 0.3157 0.19573 0.10618 0.052668 0.02449 0.010846 0.004623 0.0019108 0.00077007 0.00030383 0.00011773
0 0 0 0.22667 0.35134 0.32675 0.23635 0.14653 0.081766 0.042246 0.02058 0.0095696 0.0042851 0.0018597 0.00078615
0 0 0 0 0.1564 0.29091 0.31564 0.26093 0.182 0.11284 0.064129 0.03408 0.017168 0.0082788 0.0038496
0 0 0 0 0 0.10792 0.23418 0.29039 0.27006 0.2093 0.14274 0.088499 0.05095 0.02764 0.014281
0 0 0 0 0 0 0.074464 0.18467 0.25761 0.2662 0.22694 0.16884 0.1134 0.070311 0.040868
0 0 0 0 0 0 0 0.05138 0.14335 0.22219 0.25256 0.23488 0.18931 0.13694 0.090965
0 0 0 0 0 0 0 0 0.035452 0.1099 0.18738 0.23235 0.2341 0.2032 0.15748
0 0 0 0 0 0 0 0 0 0.024462 0.083415 0.15515 0.20842 0.22614 0.21031
0 0 0 0 0 0 0 0 0 0 0.016879 0.062789 0.12652 0.18303 0.21277
0 0 0 0 0 0 0 0 0 0 0 0.011646 0.046935 0.10185 0.15786
0 0 0 0 0 0 0 0 0 0 0 0 0.008036 0.034876 0.081087
0 0 0 0 0 0 0 0 0 0 0 0 0 0.0055448 0.025783
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0038259
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 16 through 30
2.5906e-07 8.5327e-08 2.8007e-08 9.1646e-09 2.9906e-09 9.7342e-10 3.1613e-10 1.0246e-10 3.3142e-11 1.0702e-11 3.4504e-12 1.1108e-12 3.5709e-13 1.1465e-13 3.6767e-14
4.3246e-06 1.5194e-06 5.2988e-07 1.8359e-07 6.3236e-08 2.1667e-08 7.3883e-09 2.5085e-09 8.4833e-10 2.8585e-10 9.5998e-11 3.214e-11 1.073e-11 3.5726e-12 1.1866e-12
4.492e-05 1.6909e-05 6.2902e-06 2.3156e-06 8.445e-07 3.0543e-07 1.0963e-07 3.9084e-08 1.3847e-08 4.8779e-09 1.7094e-09 5.9615e-10 2.0698e-10 7.1568e-11 2.4651e-11
0.00032494 0.00013173 5.2503e-05 2.0616e-05 7.9887e-06 3.0592e-06 1.1591e-06 4.3497e-07 1.6181e-07 5.9715e-08 2.1877e-08 7.9615e-09 2.8794e-09 1.0354e-09 3.7037e-10
0.0017358 0.00076232 0.00032721 0.00013766 5.69e-05 2.3151e-05 9.2878e-06 3.679e-06 1.4406e-06 5.5824e-07 2.1426e-07 8.1515e-08 3.0763e-08 1.1523e-08 4.2867e-09
0.0070833 0.0033935 0.001578 0.00071495 0.00031662 0.00013741 5.8573e-05 2.4566e-05 1.0154e-05 4.1418e-06 1.6691e-06 6.6527e-07 2.6248e-07 1.0259e-07 3.9755e-08
0.022523 0.01187 0.0060211 0.0029554 0.0014095 0.00065541 0.00029799 0.00013279 5.8116e-05 2.5022e-05 1.0615e-05 4.4423e-06 1.8361e-06 7.5031e-07 3.0339e-07
0.056398 0.033024 0.018428 0.0098671 0.005098 0.0025529 0.0012436 0.00059114 0.00027488 0.00012531 5.6113e-05 2.4719e-05 1.0728e-05 4.5926e-06 1.9414e-06
0.11158 0.073506 0.045573 0.026843 0.01513 0.0082078 0.0043059 0.0021929 0.0010877 0.00052686 0.00024979 0.00011615 5.3064e-05 2.3852e-05 1.0563e-05
0.17385 0.13089 0.091294 0.059747 0.037043 0.021923 0.012459 0.0068335 0.0036315 0.0018763 0.00094518 0.00046536 0.00022441 0.00010618 4.9374e-05
0.21107 0.18539 0.14778 0.10881 0.074955 0.048796 0.030253 0.017976 0.010288 0.0056949 0.0030601 0.0016008 0.00081734 0.00040822 0.00019981
0.19575 0.20632 0.19188 0.16145 0.12513 0.090508 0.061727 0.04001 0.024806 0.014788 0.0085139 0.0047507 0.0025773 0.0013629 0.00070418
0.13406 0.17663 0.19712 0.1935 0.17139 0.13947 0.10569 0.075354 0.050967 0.032916 0.020408 0.012201 0.0070603 0.003967 0.0021702
0.063943 0.11233 0.1567 0.18459 0.19074 0.17739 0.15122 0.1198 0.089133 0.062798 0.042179 0.027157 0.016837 0.010091 0.0058655
0.018977 0.050003 0.093006 0.13695 0.16982 0.18425 0.17952 0.15999 0.13226 0.1025 0.075107 0.052387 0.034978 0.022461 0.013926
0.0026399 0.013912 0.038815 0.076207 0.11812 0.15379 0.17481 0.17806 0.16559 0.14259 0.11493 0.087452 0.063257 0.043745 0.029059
0 0.0018215 0.010164 0.029933 0.061862 0.10068 0.13733 0.16319 0.17345 0.16803 0.15048 0.12595 0.099387 0.074457 0.053266
0 0 0.0012569 0.0074028 0.022949 0.049799 0.084907 0.12108 0.15014 0.16622 0.16747 0.15575 0.13519 0.11049 0.085626
0 0 0 0.00086723 0.0053768 0.017502 0.039787 0.07092 0.10553 0.13631 0.15695 0.16421 0.15837 0.14237 0.12037
0 0 0 0 0.00059839 0.0038955 0.013284 0.031571 0.058722 0.091019 0.12227 0.1462 0.15862 0.15845 0.14736
0 0 0 0 0 0.00041289 0.0028159 0.010039 0.024896 0.048236 0.077756 0.10847 0.1345 0.15115 0.15618
0 0 0 0 0 0 0.00028489 0.0020313 0.0075564 0.019521 0.039334 0.065845 0.095256 0.12234 0.14222
0 0 0 0 0 0 0 0.00019658 0.0014625 0.0056673 0.015226 0.031861 0.05531 0.082873 0.1101
0 0 0 0 0 0 0 0 0.00013564 0.0010512 0.0042363 0.011819 0.025648 0.046115 0.071478
0 0 0 0 0 0 0 0 0 9.359e-05 0.00075433 0.0031569 0.0091339 0.020528 0.038183
0 0 0 0 0 0 0 0 0 0 6.4577e-05 0.00054051 0.0023458 0.0070296 0.016344
0 0 0 0 0 0 0 0 0 0 0 4.4558e-05 0.00038676 0.0017385 0.0053894
0 0 0 0 0 0 0 0 0 0 0 0 3.0745e-05 0.0002764 0.0012852
0 0 0 0 0 0 0 0 0 0 0 0 0 2.1214e-05 0.00019729
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.4638e-05
b =
3712
246.89
43.304
22.55
14.897
10.066
6.8138
4.6131
3.1232
2.1146
1.4316
0.96927
0.65623
0.44429
0.3008
0.20365
0.13788
0.093351
0.063202
0.04279
0.02897
0.019614
0.013279
0.0089906
0.006087
0.0041211
0.0027902
0.001889
0.0012789
0.00086589
A .mat file with the full-precision variables may be found here.
Here are the results I'm getting on my machine (Matlab R2013a on OS X 10.10.5):
>> x=A\b
x =
5087.6
433.99
64.166
27.995
19.494
14.546
10.934
8.2265
6.1834
4.6933
3.2779
3.8272
-3.5375
23.953
-79.278
254.22
-702.1
1713.2
-3658.2
6822.7
-11046
15412
-18349
18393
-15244
10181
-5273.4
1992.3
-489.85
59.155
Although norm(A*x-b) returns a value on the order of 1e-13, the results are not physically reasonable given the problem I am trying to solve (values in x should be monotonically decreasing, and none should be negative). As an example, here is a similar dataset that returns a correct (looking) solution with the same matrix A:
>> c
c =
5142.1
339.52
22.417
1.4802
0.097731
0.0064529
0.00042607
2.8132e-05
1.8575e-06
1.2265e-07
8.0979e-09
5.3469e-10
3.5304e-11
2.331e-12
1.5391e-13
1.0162e-14
6.7099e-16
4.4304e-17
2.9253e-18
1.9315e-19
1.2753e-20
8.4205e-22
5.5598e-23
3.671e-24
2.4239e-25
1.6004e-26
1.0567e-27
6.9771e-29
4.6068e-30
3.0418e-31
>> x = A\c
x =
7029.1
653.25
60.709
5.642
0.52434
0.04873
0.0045287
0.00042087
3.9114e-05
3.635e-06
3.3782e-07
3.1395e-08
2.9177e-09
2.7116e-10
2.52e-11
2.342e-12
2.1765e-13
2.0227e-14
1.8798e-15
1.747e-16
1.6236e-17
1.5089e-18
1.4023e-19
1.3033e-20
1.21e-21
1.1339e-22
9.9766e-24
1.1858e-24
2.3902e-26
2.078e-26
I need to construct the tech cycle constraint matrix Aa and the right side ba. The aim is building the technology cycle matrices in order to solve the scheduling linear problem constrained by Ax<=b. In this case -1 and +1 in A refers to the coefficients of the constraints of the problem such as starting times and precedences
TC = [1,2,3,4,6,7;1,2,5,4,6,7;2,5,6,7,0,0]; % Technology cycle
CT = [100,60,200,160,80,120;100,60,150,120,60,150;50,120,40,30,0,0]; % Cycle time
n_jb = size(TC,1); % number of jobs
n_op = sum(TC~=0,2); % number of operations for each job
N_op = sum(n_op); % total number of operations
c=1; % indice for constraints in Aa
Op=1; % counter for overall operation
n_tf = N_op - n_jb- sum(n_op==1); % number of job transfer between machines (also number of tech cycle constraint numbers)
Aa = zeros(n_tf,N_op); % Constraint matrx for tech cycle
ba = zeros(n_tf,1); % The right vector of the constraint function: Aa*x<=ba
for j=1:n_jb
if n_op(j)>1
for op=1:n_op(j)-1
Aa(c,Op)=-1;
Aa(c,Op+1)=1;
ba(c,1)=CT(j,op);
c=c+1;
Op=Op+1;
end
else
Op=Op+1;
end
Op=Op+1;
end
The output, like Aa is 3 """diagonal""" -1/+1 matrices:
-1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1
In order to be more precise in the following there is an image: showing the 3 different part of the matrix Aa. My question is: Is there a way to build the same this avoiding loops since A is not a 3x1 but will definitely become 30-50x1?
You can use diag to create the positive and negative ones. The second input to diag is to shift the diagonal to the side. In this case, 1 to the right.
Use cumsum to find the rows you want to remove. For n = [6, 6, 4], you want to remove the 6th, 12th and 16th row.
n = [6, 6, 4];
cols = sum(n);
A = -eye(cols) + diag(ones(cols-1,1), 1);
A(cumsum(n),:) = []
A =
-1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1
So I tried the matrix exponential function using MATLAB's Coder toolkit, and I got it to build. I went on to test to see if the results were reliable and more efficient. While the code was faster, the answer it produced was very slight.
I ran the original function and got an answer of:
p =
1 0 0 0 0 0 0 0 0 0 0 0
-0.05 1 -1.25e-07 5e-06 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0 0 0
1.25e-07 -5e-06 -0.05 1 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0
0 0 0 0 -0.05 1 0 0 0 0 0 0
-2.0833e-05 0.00125 -5.2083e-11 4.1667e-09 0 0 1 0.05 0 1.25e-07 0 0
-0.00125 0.05 -4.1667e-09 2.5e-07 0 0 0 1 0 5e-06 0 0
5.2083e-11 -4.1667e-09 -2.0833e-05 0.00125 0 0 0 -1.25e-07 1 0.05 0 0
4.1667e-09 -2.5e-07 -0.00125 0.05 0 0 0 -5e-06 0 1 0 0
0 0 0 0 -2.0833e-05 0.00125 0 0 0 0 1 0.05
0 0 0 0 -0.00125 0.05 0 0 0 0 0 1
And I then ran the mexed version of the function with the same input:
p2 =
1 0 0 0 0 0 0 0 0 0 0 0
-0.05 1 -1.25e-07 5e-06 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0 0 0
1.25e-07 -5e-06 -0.05 1 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0
0 0 0 0 -0.05 1 0 0 0 0 0 0
-2.0833e-05 0.00125 -5.2083e-11 4.1667e-09 0 0 1 0.05 0 1.25e-07 0 0
-0.00125 0.05 -4.1667e-09 2.5e-07 0 0 0 1 0 5e-06 0 0
5.2083e-11 -4.1667e-09 -2.0833e-05 0.00125 0 0 0 -1.25e-07 1 0.05 0 0
4.1667e-09 -2.5e-07 -0.00125 0.05 0 0 0 -5e-06 0 1 0 0
0 0 0 0 -2.0833e-05 0.00125 0 0 0 0 1 0.05
0 0 0 0 -0.00125 0.05 0 0 0 0 0 1
At first glance, these two matrices are equal, but they are actually VERY SLIGHTLY off:
p-p2
ans =
0 0 0 0 0 0 0 0 0 0 0 0
-6.9389e-18 0 0 8.4703e-22 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
-3.3881e-21 -2.1684e-19 -3.2312e-26 8.2718e-25 0 0 0 0 0 5.294e-23 0 0
-2.1684e-19 0 -8.2718e-25 5.294e-23 0 0 0 0 0 8.4703e-22 0 0
6.4623e-27 0 -3.3881e-21 2.1684e-19 0 0 0 0 0 0 0 0
0 0 0 6.9389e-18 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
Most of the result is equivalent to the original function, but some of it is not. Also, the difference between the two is so small that I cant possibly believe it would be a mathematical error, rather than perhaps a precision error. And the reason I am so concerned with this, is because this does cause issues with the overlaying reason I am using the function.
Is there a reason why the mex function is off by so little, and is there a way to fix this?
The differences you observe are so little that you can consider the results actually are the same.
The way they are computed are different and that is why you do not get exactly the same result. Yet the difference is roughly machine epsilon and is just due to the fact that computers do not work with an infinite precision but with some discrete representation of the numbers.
With the code below:
t=10;
period=9;
A=zeros(36,36,t);
for j=1:t
A(7:period:end,7:period:end,j)=rand(1,1);
A(8:period:end,8:period:end,j)=rand(1,1);
A(9:period:end,9:period:end,j)=rand(1,1);
end
Edit: This is A. I've split them up into 9x9 visually so it's easier to see the pattern.
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
I'd like to make every single number which is labelled 1, a random number between 0 to 1. Is it necessary to create a vector(48,1) and then plug in the values where the 1s are?
Of course the numbers are not entirely random, the same three random numbers are being reproduced every ''period''.
Is there a simple way that can create random numbers so every single randomnly generated value is random?
Many thanks.
It was slightly trickier than I thought at first. Here's my solution though:
A = zeros(36,36,t);
idx = sub2ind([36 36 t],repmat(34:36,1,t),repmat(34:36,1,t),repmat(1:t,1,3));
A(idx) = rand(length(idx),1);
The solution with logical indexing is:
A(logical(A)) = rand(nnz(A),1);
But you have to prepare A as the binary matrix given in the question.