Related
I have written a code that stores data in a matrix, but I want to shorten it so it iterates over itself.
The number of matrices created is the known variable. If it was 3, the code would be:
for i = 1:31
if idx(i) == 1
C1 = [C1; Output2(i,:)];
end
if idx(i) == 2
C2 = [C2; Output2(i,:)];
end
if idx(i) == 3
C3 = [C3; Output2(i,:)];
end
end
If I understand correctly, you want to extract rows from Output2 into new variables based on idx values? If so, you can do as follows:
Output2 = rand(5, 10); % example
idx = [1,1,2,2,3];
% get rows from Output which numbers correspond to those in idx with given value
C1 = Output2(find(idx==1),:);
C2 = Output2(find(idx==2),:);
C3 = Output2(find(idx==3),:);
Similar to Marcin i have another solution. Here i predefine my_C as a cell array. Output2 and idx are random generated and instead of find i just use logical adressing. You have to convert the data to type cell {}
Output2 = round(rand(31,15)*10);
idx = uint8(round(1+rand(1,31)*2));
my_C = cell(1,3);
my_C(1,1) = {Output2(idx==1,:)};
my_C(1,2) = {Output2(idx==2,:)};
my_C(1,3) = {Output2(idx==3,:)};
If you want to get your data back just use e.g. my_C{1,1} for the first group.
If you have not 3 but n resulting matrices you can use:
Output2 = round(rand(31,15)*10);
idx = uint8(round(1+rand(1,31)*(n-1)));
my_C = cell(1,n);
for k=1:n
my_C(1,k) = {Output2(idx==k,:)};
end
Where n is a positive integer number
I would recommend a slighty different approach. Except for making the rest of the code more maintainable it may also slightly speed up the execution. This due to that matlab uses a JIT compiler and eval must be recompiled every time. Try this:
nMatrices = 3
for k = 1:nMatrices
C{k} = Output2(idx==k,:);
end
As patrik said in the comments, naming variables like this is poor practice. You would be better off using cell arrays M{1}=C1, or if all the Ci are the same size, even just a 3D array M, for example, where M(:,:,1)=C1.
If you really want to use C1, C2, ... as you variable names, I think you will have to use eval, as arielnmz mentioned. One way to do this in matlab is
for i=1:3
eval(['C' num2str(idx(i)) '=[C' num2str(idx(i)) ';Output2(' num2str(i) ',:)];'])
end
Edited to add test code:
idx=[2 1 3 2 2 3];
Output2=rand(6,4);
C1a=[];
C2a=[];
C3a=[];
for i = 1:length(idx)
if idx(i) == 1
C1a = [C1a; Output2(i,:)];
end
if idx(i) == 2
C2a = [C2a; Output2(i,:)];
end
if idx(i) == 3
C3a = [C3a; Output2(i,:)];
end
end
C1=[];
C2=[];
C3=[];
for i=1:length(idx)
eval(['C' num2str(idx(i)) '=[C' num2str(idx(i)) ';Output2(' num2str(i) ',:)];'])
end
all(C1a(:)==C1(:))
all(C2a(:)==C2(:))
all(C3a(:)==C3(:))
The following is an M-file for generating an MD5 digest on a file of ASCII values. I have a string of HEX values and am converting it to ASCII, then writing these values to a file which I pass to the md5 function. I am using two different online MD5 calculators to validate the script's solution. Here is one of them.
The method by which I am converting from HEX string to ACSII is by pairing two hex values and then converting the pairs to ASCII chars which are then written to the file. I know this is being done correctly because I can pass either the HEX string to the above mentioned online calculators or upload their equivalent ASCII representation and they produce the same (valid) result even though the MATLAB script gives the wrong digest.
For some reason, whether the digest is computed correctly or not depends on the number of ASCII values being read from the file. I've tried to understand if there is any pattern to this behavior but I cannot find any. It produces a valid digest for messages with 200,300-to-322, 500,996,998,1008,1010,1050,1070,1076 HEX characters which are first converted to the ASCII file. But not for 1000,1002,1004,1006,1078,1100 HEX characters. In short, I see no method to this madness... any help would be much appreciated.
% md5 Compute MD5 hash function for files
%
% d = md5(FileName)
%
% md5() computes the MD5 hash function of
% the file specified in the string FileName
% and returns it as a 64-character array d.
% The MD5 message-digest algorithm is specified
% in RFC 1321.
% The code below is for instructional and illustrational
% purposes only. It is very clear, but very slow.
% (C) Stefan Stoll, ETH Zurich, 2006
function Digest = md5(FileName)
% Guard against old Matlab versions
MatlabVersion = version;
if MatlabVersion(1)<'7'
error('md5() requires Matlab 7.0 or later!');
end
% Run autotest if no parameters are given
if (nargin==0)
md5autotest;
return;
end
% Read in entire file into uint32 vector
[Message,nBits] = readmessagefromfile(FileName);
%--------------------------------------------------
% Append a bit-1 to the last bit read from file
BytesInLastInt = mod(nBits,32)/8;
if BytesInLastInt
Message(end) = bitset(Message(end),BytesInLastInt*8+8);
else
Message = [Message; uint32(128)];
end
% Append zeros
nZeros = 16 - mod(numel(Message)+2,16);
Message = [Message; zeros(nZeros,1,'uint32')];
% Append bit length of original message as uint64, lower significant uint32 first
Lower32 = uint32(nBits);
Upper32 = uint32(bitshift(uint64(nBits),-32));
Message = [Message; Lower32; Upper32];
%--------------------------------------------------
% 64-element transformation array
T = uint32(fix(4294967296*abs(sin(1:64))));
% 64-element array of number of bits for circular left shift
S = repmat([7 12 17 22; 5 9 14 20; 4 11 16 23; 6 10 15 21].',4,1);
S = S(:).';
% 64-element array of indices into X
idxX = [0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
1 6 11 0 5 10 15 4 9 14 3 8 13 2 7 12 ...
5 8 11 14 1 4 7 10 13 0 3 6 9 12 15 2 ...
0 7 14 5 12 3 10 1 8 15 6 13 4 11 2 9] + 1;
% Initial state of buffer (consisting of A, B, C and D)
A = uint32(hex2dec('67452301'));
B = uint32(hex2dec('efcdab89'));
C = uint32(hex2dec('98badcfe'));
D = uint32(hex2dec('10325476'));
%--------------------------------------------------
Message = reshape(Message,16,[]);
% Loop over message blocks each 16 uint32 long
for iBlock = 1:size(Message,2)
% Extract next block
X = Message(:,iBlock);
% Store current buffer state
AA = A;
BB = B;
CC = C;
DD = D;
% Transform buffer using message block X and the
% parameters from S, T and idxX
k = 0;
for iRound = 1:4
for q = 1:4
A = Fun(iRound,A,B,C,D,X(idxX(k+1)),S(k+1),T(k+1));
D = Fun(iRound,D,A,B,C,X(idxX(k+2)),S(k+2),T(k+2));
C = Fun(iRound,C,D,A,B,X(idxX(k+3)),S(k+3),T(k+3));
B = Fun(iRound,B,C,D,A,X(idxX(k+4)),S(k+4),T(k+4));
k = k + 4;
end
end
% Add old buffer state
A = bitadd32(A,AA);
B = bitadd32(B,BB);
C = bitadd32(C,CC);
D = bitadd32(D,DD);
end
%--------------------------------------------------
% Combine uint32 from buffer to form message digest
Str = lower(dec2hex([A;B;C;D]));
Str = Str(:,[7 8 5 6 3 4 1 2]).';
Digest = Str(:).';
%==================================================
function y = Fun(iRound,a,b,c,d,x,s,t)
switch iRound
case 1
q = bitor(bitand(b,c),bitand(bitcmp(b),d));
case 2
q = bitor(bitand(b,d),bitand(c,bitcmp(d)));
case 3
q = bitxor(bitxor(b,c),d);
case 4
q = bitxor(c,bitor(b,bitcmp(d)));
end
y = bitadd32(b,rotateleft32(bitadd32(a,q,x,t),s));
%--------------------------------------------
function y = rotateleft32(x,s)
y = bitor(bitshift(x,s),bitshift(x,s-32));
%--------------------------------------------
function sum = bitadd32(varargin)
sum = varargin{1};
for k = 2:nargin
add = varargin{k};
carry = bitand(sum,add);
sum = bitxor(sum,add);
for q = 1:32
shift = bitshift(carry,1);
carry = bitand(shift,sum);
sum = bitxor(shift,sum);
end
end
function [Message,nBits] = readmessagefromfile(FileName)
[hFile,ErrMsg] = fopen(FileName,'r');
error(ErrMsg);
%Message = fread(hFile,inf,'bit32=>uint32');
Message = fread(hFile,inf,'ubit32=>uint32');
%Message = fread(hFile);
fclose(hFile);
d = dir(FileName);
nBits = d.bytes*8;
%============================================
function md5autotest
disp('Running md5 autotest...');
Messages{1} = '';
Messages{2} = 'a';
Messages{3} = 'abc';
Messages{4} = 'message digest';
Messages{5} = 'abcdefghijklmnopqrstuvwxyz';
Messages{6} = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
Messages{7} = char(128:255);
CorrectDigests{1} = 'd41d8cd98f00b204e9800998ecf8427e';
CorrectDigests{2} = '0cc175b9c0f1b6a831c399e269772661';
CorrectDigests{3} = '900150983cd24fb0d6963f7d28e17f72';
CorrectDigests{4} = 'f96b697d7cb7938d525a2f31aaf161d0';
CorrectDigests{5} = 'c3fcd3d76192e4007dfb496cca67e13b';
CorrectDigests{6} = 'd174ab98d277d9f5a5611c2c9f419d9f';
CorrectDigests{7} = '16f404156c0500ac48efa2d3abc5fbcf';
TmpFile = tempname;
for k=1:numel(Messages)
[h,ErrMsg] = fopen(TmpFile,'w');
error(ErrMsg);
fwrite(h,Messages{k},'char');
fclose(h);
Digest = md5(TmpFile);
fprintf('%d: %s\n',k,Digest);
if ~strcmp(Digest,CorrectDigests{k})
error('md5 autotest failed on the following string: %s',Messages{k});
end
end
delete(TmpFile);
disp('md5 autotest passed!');
This is quite old, but I looked through the code and I think that you are not handling the message length properly. I also found some issues with way that certain maths were done in MatLAB differently - for instance bitshifting was not used as I had expected, and also integer maths of uint32 variables not always handled the way you'd expect. You need to handle modular math and truncating math manually/explicitly or read the help documents a little more closely regarding those topics. Also, check the character base in your options, char128:255 might not be translating correctly, try using the command to create the message and then paste it into a different interface. Even in matlab if you process that variable even once it will get messed up bad, it comes out as ???????????yadda yadda. If you were using decimal maths you might have introduced an epsilon error as well, but check out the actual length...
i want to clarify how to control returning of variables from function in matlab,for exmaple let us consider this code
function [x y z]=percentage(a)
n=length(a);
maximum=0;
minimum=0;
subst=0;
minus=0;
plus=0;
minus_perc=0;
plus_perc=0;
for i=1:1:n
if a(i)>0
plus=plus+1;
else
minus=minus+1;
end
end
minuc_perc=minus/n;
plus_perc=plus/n;
maximum=max(minus_perc,plus_perc);
minimum=min(minus_perc,plus_perc);
subst=maximum-minimum;
x=plus_perc;
y=minus_perc;
z=subst*100;
if plus_perc>minus_perc
disp('among the successful people,relevant propession was choosen by');
disp(z)
disp('% people');
else
disp('among the successful people,irrelevant propession was choosen by');
disp(z);
disp('% people');
end
end
what i want to return is plus_proc,min_proc and subst,but when i run following command,get result like this
[c d e]=percentage(a)
among the successful people,relevant propession was choosen by
58.3333
% people
c =
0.5833
d =
0
e =
58.3333
so i think something is wrong,array is like this
a =
1 -1 1 1 -1 1 -1 -1 1 1 1 -1
so ones again,i want to return plus_proc,minus_proc,and subst
To return a variable in matlab you just assign into one of the specified return parameters. For example: to return the number five I would use:
function [foo] = gimmeFive()
foo = 5;
end
Your code is not giving you the right answer because you have a typo:
minuc_perc=minus/n;
should be
minus_perc=minus/n;
You could greatly simplify the function by taking advantage of the find function, like so:
Find the indeces of any element of a > 0, count them.
plus = length(find(a > 0));
plus_perc = plus ./ length(a);
Or if you want to cut even more out:
a > 0 gives us a vector of 0 and 1, so sum up the 1's
plus = sum(a > 0);
plus_perc = plus ./ length(a);
I am trying to implement decision tree with recursion: So far I have written the following:
From a give data set, find the best split and return the branches, to give more details lets say I have data with features as columns of matrix and last column indicate the class of the data 1, -1.
Based on 1. I have a best feature to split along with the branches under that split, lets say based on Information gain I get feature 9 is the best split and unique values in feature 9 {1,3,5} are the branches of 9
I have figured how to get the data related to ach branch, then I need to iterate over each branch's data to get the next set of split. I am having trouble figuring this recursion.
Here is the code that I have so far, the recursion that I am doing right now doesn't look right: How can I fix this?
function [indeces_of_node, best_split] = split_node(X_train, Y_train)
%cell to save split information
feature_to_split_cell = cell(size(X_train,2)-1,4);
%iterate over features
for feature_idx=1:(size(X_train,2) - 1)
%get current feature
curr_X_feature = X_train(:,feature_idx);
%identify the unique values
unique_values_in_feature = unique(curr_X_feature);
H = get_entropy(Y_train); %This is actually H(X) in slides
%temp entropy holder
%Storage for feature element's class
element_class = zeros(size(unique_values_in_feature,1),2);
%conditional probability H(X|y)
H_cond = zeros(size(unique_values_in_feature,1),1);
for aUnique=1:size(unique_values_in_feature,1)
match = curr_X_feature(:,1)==unique_values_in_feature(aUnique);
mat = Y_train(match);
majority_class = mode(mat);
element_class(aUnique,1) = unique_values_in_feature(aUnique);
element_class(aUnique,2) = majority_class;
H_cond(aUnique,1) = (length(mat)/size((curr_X_feature),1)) * get_entropy(mat);
end
%Getting the information gain
IG = H - sum(H_cond);
%Storing the IG of features
feature_to_split_cell{feature_idx, 1} = feature_idx;
feature_to_split_cell{feature_idx, 2} = max(IG);
feature_to_split_cell{feature_idx, 3} = unique_values_in_feature;
feature_to_split_cell{feature_idx, 4} = element_class;
end
%set feature to split zero for every fold
feature_to_split = 0;
%getting the max IG of the fold
max_IG_of_fold = max([feature_to_split_cell{:,2:2}]);
%vector to store values in the best feature
values_of_best_feature = zeros(size(15,1));
%Iterating over cell to get get the index and the values under best
%splited feature.
for i=1:length(feature_to_split_cell)
if (max_IG_of_fold == feature_to_split_cell{i,2});
feature_to_split = i;
values_of_best_feature = feature_to_split_cell{i,4};
end
end
display(feature_to_split)
display(values_of_best_feature(:,1)')
curr_X_feature = X_train(:,feature_to_split);
best_split = feature_to_split
indeces_of_node = unique(curr_X_feature)
%testing
for k = 1 : length(values_of_best_feature)
% Condition to stop the recursion, if clases are pure then we are
% done splitting, if both classes have save number of attributes
% then we are done splitting.
if (sum(values_of_best_feature(:,2) == -1) ~= sum(values_of_best_feature(:,2) == 1))
if((sum(values_of_best_feature(:,2) == -1) ~= 0) || (sum(values_of_best_feature(:,2) == 1) ~= 0))
mat1 = X_train(X_train(:,5)== values_of_best_feature(k),:);
[indeces_of_node, best_split] = split_node(mat1, Y_train);
end
end
end
end
Here is the out of my code: and looks like some in my recursion I am only going depth of one branch and after that I never go back to rest of the branches
feature_to_split =
5
ans =
1 2 3 4 5 6 7 8 9
feature_to_split =
9
ans =
3 5 7 8 11
feature_to_split =
21
feature_to_split =
21
feature_to_split =
21
feature_to_split =
21
if you are interest in running this code: git
After multiple rounds of debug, I figured the answers, I hope someone will benefit from this:
for k = 1 : length(values_of_best_feature)
% Condition to stop the recursion, if clases are pure then we are
% done splitting, if both classes have save number of attributes
% then we are done splitting.
if((sum(values_of_best_feature(:,2) == -1) ~= 0) || (sum(values_of_best_feature(:,2) == 1) ~= 0))
X_train(:,feature_to_split) = [];
mat1 = X_train(X_train(:,5)== values_of_best_feature(k),:);
%if(level >= curr_level)
split_node(mat1, Y_train, 1, 2, level-1);
%end
end
end
return;
Please help me to improve the following Matlab code to improve execution time.
Actually I want to make a random matrix (size [8,12,10]), and on every row, only have integer values between 1 and 12. I want the random matrix to have the sum of elements which has value (1,2,3,4) per column to equal 2.
The following code will make things more clear, but it is very slow.
Can anyone give me a suggestion??
clc
clear all
jum_kel=8
jum_bag=12
uk_pop=10
for ii=1:uk_pop;
for a=1:jum_kel
krom(a,:,ii)=randperm(jum_bag); %batasan tidak boleh satu kelompok melakukan lebih dari satu aktivitas dalam satu waktu
end
end
for ii=1:uk_pop;
gab1(:,:,ii) = sum(krom(:,:,ii)==1)
gab2(:,:,ii) = sum(krom(:,:,ii)==2)
gab3(:,:,ii) = sum(krom(:,:,ii)==3)
gab4(:,:,ii) = sum(krom(:,:,ii)==4)
end
for jj=1:uk_pop;
gabh1(:,:,jj)=numel(find(gab1(:,:,jj)~=2& gab1(:,:,jj)~=0))
gabh2(:,:,jj)=numel(find(gab2(:,:,jj)~=2& gab2(:,:,jj)~=0))
gabh3(:,:,jj)=numel(find(gab3(:,:,jj)~=2& gab3(:,:,jj)~=0))
gabh4(:,:,jj)=numel(find(gab4(:,:,jj)~=2& gab4(:,:,jj)~=0))
end
for ii=1:uk_pop;
tot(:,:,ii)=gabh1(:,:,ii)+gabh2(:,:,ii)+gabh3(:,:,ii)+gabh4(:,:,ii)
end
for ii=1:uk_pop;
while tot(:,:,ii)~=0;
for a=1:jum_kel
krom(a,:,ii)=randperm(jum_bag); %batasan tidak boleh satu kelompok melakukan lebih dari satu aktivitas dalam satu waktu
end
gabb1 = sum(krom(:,:,ii)==1)
gabb2 = sum(krom(:,:,ii)==2)
gabb3 = sum(krom(:,:,ii)==3)
gabb4 = sum(krom(:,:,ii)==4)
gabbh1=numel(find(gabb1~=2& gabb1~=0));
gabbh2=numel(find(gabb2~=2& gabb2~=0));
gabbh3=numel(find(gabb3~=2& gabb3~=0));
gabbh4=numel(find(gabb4~=2& gabb4~=0));
tot(:,:,ii)=gabbh1+gabbh2+gabbh3+gabbh4;
end
end
Some general suggestions:
Name variables in English. Give a short explanation if it is not immediately clear,
what they are indented for. What is jum_bag for example? For me uk_pop is music style.
Write comments in English, even if you develop source code only for yourself.
If you ever have to share your code with a foreigner, you will spend a lot of time
explaining or re-translating. I would like to know for example, what
%batasan tidak boleh means. Probably, you describe here that this is only a quick
hack but that someone should really check this again, before going into production.
Specific to your code:
Its really easy to confuse gab1 with gabh1 or gabb1.
For me, krom is too similar to the built-in function kron. In fact, I first
thought that you are computing lots of tensor products.
gab1 .. gab4 are probably best combined into an array or into a cell, e.g. you
could use
gab = cell(1, 4);
for ii = ...
gab{1}(:,:,ii) = sum(krom(:,:,ii)==1);
gab{2}(:,:,ii) = sum(krom(:,:,ii)==2);
gab{3}(:,:,ii) = sum(krom(:,:,ii)==3);
gab{4}(:,:,ii) = sum(krom(:,:,ii)==4);
end
The advantage is that you can re-write the comparsisons with another loop.
It also helps when computing gabh1, gabb1 and tot later on.
If you further introduce a variable like highestNumberToCompare, you only have to
make one change, when you certainly find out that its important to check, if the
elements are equal to 5 and 6, too.
Add a semicolon at the end of every command. Having too much output is annoying and
also slow.
The numel(find(gabb1 ~= 2 & gabb1 ~= 0)) is better expressed as
sum(gabb1(:) ~= 2 & gabb1(:) ~= 0). A find is not needed because you do not care
about the indices but only about the number of indices, which is equal to the number
of true's.
And of course: This code
for ii=1:uk_pop
gab1(:,:,ii) = sum(krom(:,:,ii)==1)
end
is really, really slow. In every iteration, you increase the size of the gab1
array, which means that you have to i) allocate more memory, ii) copy the old matrix
and iii) write the new row. This is much faster, if you set the size of the
gab1 array in front of the loop:
gab1 = zeros(... final size ...);
for ii=1:uk_pop
gab1(:,:,ii) = sum(krom(:,:,ii)==1)
end
Probably, you should also re-think the size and shape of gab1. I don't think, you
need a 3D array here, because sum() already reduces one dimension (if krom is
3D the output of sum() is at most 2D).
Probably, you can skip the loop at all and use a simple sum(krom==1, 3) instead.
However, in every case you should be really aware of the size and shape of your
results.
Edit inspired by Rody Oldenhuis:
As Rody pointed out, the 'problem' with your code is that its highly unlikely (though
not impossible) that you create a matrix which fulfills your constraints by assigning
the numbers randomly. The code below creates a matrix temp with the following characteristics:
The numbers 1 .. maxNumber appear either twice per column or not at all.
All rows are a random permutation of the numbers 1 .. B, where B is equal to
the length of a row (i.e. the number of columns).
Finally, the temp matrix is used to fill a 3D array called result. I hope, you can adapt it to your needs.
clear all;
A = 8; B = 12; C = 10;
% The numbers [1 .. maxNumber] have to appear exactly twice in a
% column or not at all.
maxNumber = 4;
result = zeros(A, B, C);
for ii = 1 : C
temp = zeros(A, B);
for number = 1 : maxNumber
forbiddenRows = zeros(1, A);
forbiddenColumns = zeros(1, A/2);
for count = 1 : A/2
illegalIndices = true;
while illegalIndices
illegalIndices = false;
% Draw a column which has not been used for this number.
randomColumn = randi(B);
while any(ismember(forbiddenColumns, randomColumn))
randomColumn = randi(B);
end
% Draw two rows which have not been used for this number.
randomRows = randi(A, 1, 2);
while randomRows(1) == randomRows(2) ...
|| any(ismember(forbiddenRows, randomRows))
randomRows = randi(A, 1, 2);
end
% Make sure not to overwrite previous non-zeros.
if any(temp(randomRows, randomColumn))
illegalIndices = true;
continue;
end
end
% Mark the rows and column as forbidden for this number.
forbiddenColumns(count) = randomColumn;
forbiddenRows((count - 1) * 2 + (1:2)) = randomRows;
temp(randomRows, randomColumn) = number;
end
end
% Now every row contains the numbers [1 .. maxNumber] by
% construction. Fill the zeros with a permutation of the
% interval [maxNumber + 1 .. B].
for count = 1 : A
mask = temp(count, :) == 0;
temp(count, mask) = maxNumber + randperm(B - maxNumber);
end
% Store this page.
result(:,:,ii) = temp;
end
OK, the code below will improve the timing significantly. It's not perfect yet, it can all be optimized a lot further.
But, before I do so: I think what you want is fundamentally impossible.
So you want
all rows contain the numbers 1 through 12, in a random permutation
any value between 1 and 4 must be present either twice or not at all in any column
I have a hunch this is impossible (that's why your code never completes), but let me think about this a bit more.
Anyway, my 5-minute-and-obvious-improvements-only-version:
clc
clear all
jum_kel = 8;
jum_bag = 12;
uk_pop = 10;
A = jum_kel; % renamed to make language independent
B = jum_bag; % and a lot shorter for readability
C = uk_pop;
krom = zeros(A, B, C);
for ii = 1:C;
for a = 1:A
krom(a,:,ii) = randperm(B);
end
end
gab1 = sum(krom == 1);
gab2 = sum(krom == 2);
gab3 = sum(krom == 3);
gab4 = sum(krom == 4);
gabh1 = sum( gab1 ~= 2 & gab1 ~= 0 );
gabh2 = sum( gab2 ~= 2 & gab2 ~= 0 );
gabh3 = sum( gab3 ~= 2 & gab3 ~= 0 );
gabh4 = sum( gab4 ~= 2 & gab4 ~= 0 );
tot = gabh1+gabh2+gabh3+gabh4;
for ii = 1:C
ii
while tot(:,:,ii) ~= 0
for a = 1:A
krom(a,:,ii) = randperm(B);
end
gabb1 = sum(krom(:,:,ii) == 1);
gabb2 = sum(krom(:,:,ii) == 2);
gabb3 = sum(krom(:,:,ii) == 3);
gabb4 = sum(krom(:,:,ii) == 4);
gabbh1 = sum(gabb1 ~= 2 & gabb1 ~= 0)
gabbh2 = sum(gabb2 ~= 2 & gabb2 ~= 0);
gabbh3 = sum(gabb3 ~= 2 & gabb3 ~= 0);
gabbh4 = sum(gabb4 ~= 2 & gabb4 ~= 0);
tot(:,:,ii) = gabbh1+gabbh2+gabbh3+gabbh4;
end
end