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
Related
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 want to create a MxN matrix as shown below:
[1 1 1 1 1 0 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 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 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 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 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 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]
I have window size, let's say, 5 and it moves 3 in every row. Is it possible to create such a matrix without using for loops? Or is there any optimum way to do it?
This is a one line solution:
reshape([reshape([ones(5,6);zeros(21,6)], 1,[]), ones(1,5)],[],7).'
note:
The desired matrix can be seen as concatenation of a [6, 5+21] matrix:
1 1 1 1 1 0 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 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 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 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 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
and a [1 ,5] matrix:
1 1 1 1 1
that reshaped to a [7 , 23] matrix.
Other solution using repelem + bsxfun + accumarray:
r = repelem (1:7,5);
c= bsxfun(#plus, ((1:5)-3).',3*(1:7));
out = accumarray([r(:) c(:)] ,1)
Indices of rows and columns of 1 s can be generated and accumarray can be used to create the desired matrix.
I'm using Matlab. I have a 2-D Binary image/array. like this
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 1 1 0 0 1 0 0
0 0 1 1 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 1 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 1 0 0 0 0 0 0
I want to find out center of very first white block/Circle with respect to y-axis
Answer of the above image will be.
0 1 0
1 1 1
0 1 0
Anyone who have have a simplest solution for this.
If you are looking for exact matches of the template, you can use a moving filter, one example is:
H=[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 1 1 0 0 1 0 0;
0 0 1 1 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 1 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 1 0 0 0 0 0 0];
b=[0 1 0;
1 1 1;
0 1 0];
C=filter2(b,H, 'same');
[x,y]=find(C==max(max(C)));
x and y are the locations of your template in the order that it appears from the top left corner of your array.
Edit: if you have the Image Processing Toolbox and are looking for a less strict way of finding objects that have a roughly circular shape you can use regionprops with the 'Centroid' and 'Eccentricity' arguments with the bwconncomp function.
ObjectStats=regionprops(bwconncomp(H,4), 'Centroid', 'Eccentricity');
Objects with an 'Eccentricity' of 0 (or close to 0) will be the circles.
idx=find(cell2mat({ObjectStats.Eccentricity})==0); % Change ==0 to <0.2 or something to make it less strict.
ctrs={ObjectStats.Centroid};
>> ctrs{1,idx(1)}
ans =
7 3
Note that in your case, a lone pixel is an object with an eccentricity of 0, it is the smallest 'circle' that you can find. If you need to define a minimum size, use the 'Area' property of regionprops
You can do this with a simple 2 dimensional convolution. It will "overlay" the filter along a larger matrix and multiply the filter by the values it is overlaying. If the product is equal to the sum of the filter, then you know you found a match.
Here is some simple code.
mat = [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 1 1 0 0 1 0 0
0 0 1 1 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 1 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 1 0 0 0 0 0 0];
filt = [0 1 0
1 1 1
0 1 0];
[row,col] = find(conv2(mat,filt,'same') == sum(filt(:)))
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);
How to replace elements of a matrix by an another matrix in MATLAB?
Ex: let say if we have a matrix A, where
A=[1 0 0; 0 1 0; 1 0 1]
I want to replace all ones by
J=[1 0 0; 0 1 0; 0 0 1]
and zeros by
K=[0 0 0; 0 0 0; 0 0 0]
So that I can get 9x9 matrix. So how we will code it in MATLAB
Thanks
Sounds like you might want to take a look at the kronecker tensor product. This is not a general case but the idea should work for what you want
>> kron(A==1,J)+kron(A==0,K)
ans =
1 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 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 0 1 0 0 0
1 0 0 0 0 0 1 0 0
0 1 0 0 0 0 0 1 0
0 0 1 0 0 0 0 0 1
which, for the example case, would simplify to a simpler command:
>> kron(A,J)
ans =
1 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 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 0 1 0 0 0
1 0 0 0 0 0 1 0 0
0 1 0 0 0 0 0 1 0
0 0 1 0 0 0 0 0 1
You can do:
A2=imresize(A,size(A).*size(J),'nearest');
J2=repmat(J,size(A));
K2=repmat(K,size(A));
A2(A2==1)=J2(A2==1);
A2(A2==0)=K2(A2==0)