I have a matrix A which is
A=[1 0 0 1 0;
0 1 1 0 0;
0 0 1 1 0;
1 1 1 0 0]
And a given vector v=[ 0 0 1 1 0] which has two elements one. I have to change the position of element one such that the new vector v is orthogonal to all the rows in the matrix A.
How can I do it in Matlab?
To verify the correct answer, just check gfrank([A;v_new]) is 5 (i.e v_new=[0 1 0 0 1]).
Note that: Two vectors uand v whose dot product is u.v=0 (i.e., the vectors are perpendicular) are said to be orthogonal.
As AVK also mentioned in the comments, v_new = [0 1 0 0 1] is not orthogonal to all rows of A.
Explanation:-
A=[1 0 0 1 0;
0 1 1 0 0;
0 0 1 1 0;
1 1 1 0 0]
For A(1,:).*v = 0 to A(4,:).*v = 0,
0 x x 0 x % elements of v so that it's orthagonal to the 1st row of A
x 0 0 x x % -------------------------------------------- 2nd row of A
x x 0 0 x % -------------------------------------------- 3rd row of A
0 0 0 x x % -------------------------------------------- 4th row of A
where 0 represents the terms which have to be 0 and x represents the terms which can be either 0 or 1.
If you look as a whole, first 4 columns of v have to be zero so that the output is orthagonal to all rows of A. The 5th column can either be zero or 1.
So,
v_new can either be: v_new = [0 0 0 0 1] or v_new = [0 0 0 0 0]
From above explanation, you can also see that [0 1 0 0 1] is not orthagonal to 2nd and 4th row of A
Solution:-
To find v_new, you can use null function as: v_new = null(A).'
which gives: v_new = [0 0 0 0 1] for which gfrank([A;v_new]) also gives 5.
Maybe this will help you see the orthogonality between two vectors in N dimension.
N=100;
B1 = ones(1,N);
B2 = -1*ones(1,N/2);
B2 = [ones(1,N/2) B2];
B2 = transpose(B2);
B3 = dot(B1,B2);
The above code generates two vectors in N dimension. To check for orthogonality just transpose one of the vectors and multiply with the other one. You should get zero if they are Orthogonal.
The example I used makes sure that I get zero indeed.
Related
I need to create all possible permutation matrices for a matrix where every permutation matrix contains only one 1 in each column and each row, and 0 in all other places.
For example, below example in (1) is all possible permutation matrices for 2x2 matrix and in (2) is a all possible permutation matrices for 3x3 matrix and so on
So how can I get these matrices of a matrix NxN in MATLAB and store them into one three-dimensional matrix?
Here's my solution, using implicit expansion (tested with Octave 5.2.0 and MATLAB Online):
n = 3;
% Get all permutations of length n
p = perms(1:n);
% Number of permutations
n_p = size(p, 1);
% Set up indices, where to set elements to 1
p = p + (0:n:n^2-1) + (0:n^2:n^2*n_p-1).';
% Set up indices, where to set elements to 1 (for MATLAB R2016a and before)
%p = bsxfun(#plus, bsxfun(#plus, p, (0:n:n^2-1)), (0:n^2:n^2*n_p-1).');
% Initialize 3-dimensional matrix
a = zeros(n, n, n_p);
% Set proper elements to 1
a(p) = 1
The output for n = 3:
a =
ans(:,:,1) =
0 0 1
0 1 0
1 0 0
ans(:,:,2) =
0 1 0
0 0 1
1 0 0
ans(:,:,3) =
0 0 1
1 0 0
0 1 0
ans(:,:,4) =
0 1 0
1 0 0
0 0 1
ans(:,:,5) =
1 0 0
0 0 1
0 1 0
ans(:,:,6) =
1 0 0
0 1 0
0 0 1
Using repelem, perms and reshape:
n = 3; % matrix size
f = factorial(n); % number of permutation
rep = repelem(eye(n),1,1,f) % repeat n! time the diagonal matrix
res = reshape(rep(:,perms(1:n).'),n,n,f) % indexing and reshaping
Where res is:
res =
ans(:,:,1) =
0 0 1
0 1 0
1 0 0
ans(:,:,2) =
0 1 0
0 0 1
1 0 0
ans(:,:,3) =
0 0 1
1 0 0
0 1 0
ans(:,:,4) =
0 1 0
1 0 0
0 0 1
ans(:,:,5) =
1 0 0
0 0 1
0 1 0
ans(:,:,6) =
1 0 0
0 1 0
0 0 1
And according to your comment:
What I need to do is to multiply a matrix i.e Z with all possible
permutation matrices and choose that permutation matrix which
resulting a tr(Y) minimum; where Y is the results of multiplication of
Z with the permutation matrix. I Think I don't need to generate all
permutation matrices and store them in such variable, I can generate
them one by one and get the result of multiplication. Is that possible
?
You're trying to solve the assignment problem, you can use the well known hungarian algorithm to solve this task in polynomial time. No needs to generate a googleplex of permutation matrix.
I have a logical 1-by-n vector with sum m. Now, I need to convert it into a matrix m-by-n in a way that the row sum is equal 1.
vector (1-by-8) with sum 4
[0 1 0 0 1 0 1 1]
matrix (4-by-8) with row sum 1
[0 1 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 1]
Is there a mathematically efficient way without calculating the sum, creating a empty matrix, loop through the vector and adding the 1s row by row?
I think that in that case, given your input, you don't even need to calculate the sum.
You can define an identity matrix of size n, then use your input vector to sample the required rows out of it:
I = eye(n);
y = I(x, :) ; % Output Matrix. x is the input vector
Here's another method, using sparse:
matrix = full(sparse(1:m, find(vector), 1, m, n));
I want to obtain all the possible permutations of one vector elements by another vector elements. For example one vector is A=[0 0 0 0] and another is B=[1 1]. I want to replace the elements of A by B to obtain all the permutations in a matrix like this [1 1 0 0; 1 0 1 0; 1 0 0 1; 0 1 1 0; 0 1 0 1; 0 0 1 1]. The length of real A is big and I should be able to choose the length of B_max and to obtain all the permutations of A with B=[1], [1 1], [1 1 1],..., B_max.
Thanks a lot
Actually, since A and B are always defined, respectively, as a vector of zeros and a vector of ones, this computation is much easier than you may think. The only constraints you should respect concerns B, which shoud not be empty and it's elements cannot be greater than or equal to the number of elements in A... because after that threshold A will become a vector of ones and calculating its permutations will be just a waste of CPU cycles.
Here is the core function of the script, which undertakes the creation of the unique permutations of 0 and 1 given the target vector X:
function p = uperms(X)
n = numel(X);
k = sum(X);
c = nchoosek(1:n,k);
m = size(c,1);
p = zeros(m,n);
p(repmat((1-m:0)',1,k) + m*c) = 1;
end
And here is the full code:
clear();
clc();
% Define the main parameter: the number of elements in A...
A_len = 4;
% Compute the elements of B accordingly...
B_len = A_len - 1;
B_seq = 1:B_len;
% Compute the possible mixtures of A and B...
X = tril(ones(A_len));
X = X(B_seq,:);
% Compute the unique permutations...
p = [];
for i = B_seq
p = [p; uperms(X(i,:).')];
end
Output for A_len = 4:
p =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
1 1 0 0
1 0 1 0
1 0 0 1
0 1 1 0
0 1 0 1
0 0 1 1
1 1 1 0
1 1 0 1
1 0 1 1
0 1 1 1
I have a 3D matrix A (size m*n*k) where m=latitude, n*longitude and k=time.
I want only specific values from first and second dimension, specified by a logical matrix B (size m*n), and I want only the timesteps specified by vector C (size k).
In the end this should become a 2D matrix D, since the first two dimesions will colapse to one.
What is the most easy approach to do this?
And also is it possible to combine logical with linear indizes here? For example B is logical and C is linear?
Sample code with rand:
A=rand(10,10,10);
B=randi([0 1], 10,10);
C=randi([0 1], 10,1);
D=A(B,C) %This would be my approach which doesnt work. The size of D should be sum(B)*sum(c)
Another example without rand:
A=reshape([1:27],3,3,3);
B=logical([1,0,0;1,0,0;0,0,0]);
C=(1,3); %get data from timestep 1 and 5
D=A(B,C);%What I want to do, but doesnÄt work that way
D=[1,19;2,20];%Result should look like this! First dimension is now all data from dimesion 1 and 2. New dimesion 2 is now the time.
A = rand(4,4,4);
B = randi([0 1], 4,4)
B =
1 1 0 1
1 0 1 1
0 0 1 0
1 0 1 1
>> C = randi([0 1],1,1,4);
>> C(:)
ans =
0
1
1
0
Then use bsxfun or implicit expansion expansion whith .* if newer Matlab version to generate a matrix of logical for you given coordinates.
>> idx = logical(bsxfun(#times,B,C))
idx(:,:,1) =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
idx(:,:,2) =
1 1 0 1
1 0 1 1
0 0 1 0
1 0 1 1
idx(:,:,3) =
1 1 0 1
1 0 1 1
0 0 1 0
1 0 1 1
idx(:,:,4) =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
Then your output is D = A(idx). However, note that this D is now an Nx1 array. Where N is number of true elements is B times number of true elements in C. 10x True in B and 2x True in C:
>> size(D)
ans =
20 1
An easy way to do it is to first reshape A into an m*n-by-k matrix, then do your indexing:
result = reshape(A, [], size(A, 3));
result = result(B, C);
In this case C can be either a logical vector or vector of indices.
I have a A which is 640x1 cell. where the value of each cell A(i,1) varies from row to row, for example A(1,1) =[], while A(2,1)=[1] and A(3,1)=[1,2,3].
There is another matrix B of size 480x640, where the row_index (i) of vector A corresponds to the col_index of matrix B. While the cell value of each row in vector A corresponds to the row_index in matrix B. For example, A(2,1)=[1] this means col_2 row_1 in matrix B, while A(3,1)=[1,2,3] means col_3 rows 1,2&3 in matrix B.
What I'm trying to do is to for each non-zero value in matrix B that are referenced from vector A, I want to check whether there are at least 4 other neighbors that are also referenced from vector A. The number neighbors of each value are determined by a value N.
For example, this is a part of matrix B where all the zeros"just to clarify, as in fact they may be non-zeros" are the neighbors of pixel X when N=3:
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 X 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
As shown, because N=3, all these zeros are pixel X's neighbors. So if more than 4 neighbor pixels are found in vector A then do something e.g G=1 if not then G=0;
So if anyone could please advise. And please let me know if any more clarification is needed.
The first thing I would do is to convert your cell of indices A to a logic matrix Amat. This makes it easier to check how many neighbours are included in A.
Here is a solution that uses this conversion. I hope the comments are enough to make it understandable.
clear all
clc
nCols = 7;
nRows = 6;
N = 3; %// Number of neighbours
M = 4; %// Minimum number of wanted connections
%// Create cell of indices A
A = cell(nCols,1);
A{1} = [];
A{2} = 1;
A{3} = [1 2 3];
A{4} = [2 5];
A{5} = 3;
A{6} = [3 5];
A{7} = [1 4 6];
%// Generate radom data B
%// (There is a 50% probability for each element of B to be zero)
Bmax = 17;
B = (randi(2,nRows,nCols)-1).*(randi(Bmax,nRows,nCols));
%// Convert the cell A to a logic matrix Amat
Amat = zeros(size(B));
for ii = 1:nCols
Amat(A{ii},ii) = 1;
end
A
B
Amat
for ii = 1:nCols
for jj = A{ii}
if B(jj,ii)>0
%// Calculate neighbour indices with a lower bound of 1
%// and an upper bound of nCols or nRows
col_lim_low = max(1,ii-N);
col_lim_high = min(nCols,ii+N);
row_lim_low = max(1,jj-N);
row_lim_high = min(nRows,jj+N);
%// Get the corresponding neighbouring-matrix from Amat
A_neighbours = ...
Amat(row_lim_low:row_lim_high,col_lim_low:col_lim_high);
%// Check the number of neighbours against the wanted number M
if sum(A_neighbours(:)) > 1 + M
%# do something
fprintf('We should do something here at (%d,%d)\n',jj,ii)
end
end
end
end
The following is a printout from one run of the code.
A =
[]
[ 1]
[1x3 double]
[1x2 double]
[ 3]
[1x2 double]
[1x3 double]
B =
1 5 0 0 11 0 16
0 13 13 0 0 0 9
0 0 0 5 0 0 0
3 8 16 16 0 2 12
0 0 5 0 9 9 0
12 13 0 6 0 15 0
Amat =
0 1 1 0 0 0 1
0 0 1 1 0 0 0
0 0 1 0 1 1 0
0 0 0 0 0 0 1
0 0 0 1 0 1 0
0 0 0 0 0 0 1
We should do something here at (1,2)
We should do something here at (2,3)
We should do something here at (5,6)
We should do something here at (4,7)
Since you have a one-to-one correspondence between A and B, there is no need to work on A. B is a logical matrix (0 if not referenced in A, 1 if referenced). You can therefore apply a simple filter2 function counting the number of active neighbors within the 8 closest elements.
Here is the code
B = rand(10,10); %generate binary matrix
h = [1 1 1;1 0 1;1 1 1]; %filter to be applied
filter2(h,B,'same')>=4 & B>0 %apply filter on B, count minimum of 4 neighbors, if only B>1
EDIT
To transform a cell array B into binary presence (0=empty, 1=not empty), use of cellfunis straightforward
B = ~cellfun(#isempty,B);
And see Armo's response to your previous question for how to create B based on A.