I want to make the density matrix on qiskit.
The state is ρ=[7/9,0 ; 0, 2/9].
At first, I'm gonna make that matrix by unitrary transform from |0>.
But I realize that it cannot happend.
Because there is no unitary transform which satisfy U|0><0|U(Dagger) = [7/9,0 ; 0, 2/9].
How to do it on qiskit very simply?
making the arbitrary density matrix on qiskit.
Related
I was looking to find a way to perform a symmetric sparse matrix - matrix multiplication:
X = A B
where the sparse matrix A was previously stored in CSR3 format (upper triangular), while the matrix B is a dense non-symmetric matrix. Is there a routine inside the MKL libraries to do it? or do they all need the full sparse matrix in CSR format (to get the handle from) instead of the triangular one (I built the triangular matrix because I need to use it in MKL Pardiso)? I know about the mkl_sparse_d_mv(...) routine, but I couldn't find a way to get the sparse matrix handle from a symmetric sparse matrix previously stored like an upper triangular matrix in CSR format.
thank you in advance,
Daniele
Could you try the mkl_sparse_?_mm, where [?] == s,d,c and z data types.
This routine performs a matrix-matrix operation:
Y := alpha*op(A)X + betaY
where alpha and beta are scalars, A is a sparse matrix, op is a matrix modifier for matrix A, and X and Y are dense matrices.
In most cases you can easily feed a CSR3-stored matrix into sparse_d_create_csr by passing appropriately offset pointers to your row index for pointerB and pointerE.
You can then tell mkl_sparse_d_mm the sparse matrix is triangular and you'd like it to be filled (I have never done this and can't promise that it works).
ok, I can now say the routine needs the full matrix in CSR format, the matrix struct description it will only tells the routine to take one triangle(upper/lower) from the input full CSR matrix, but still it needs it all.
I am assembling the jacobian of a coupled multi physics system. The jacobian consists of a blockmatrix on the diagonal for each system and off diagonal blocks for the coupling.
I find it the best to assemble to block separatly and then sum over them with projection matrices to get the complete jacobian.
pseudo-code (where J[i] are the diagonal elements and C[ij] the couplings, P are the projections to the complete matrix).
// diagonal blocks
J.setZero();
for(int i=0;i<N;++i){
J+=P[i]J[i]P[i].transpose()
}
// off diagonal elements
for(int i=0;i<N;++i){
for(int j=i+1;j<N;++j){
J+=P[i]C[ij]P[j].transpose()
J+=P[j]C[ji]P[i].transpose()
}
}
This takes a lot performance, around 20% of the whole programm, which is too much for some assembling. I have to recalculate the jacobian every time step since the system is highly nonlinear.
Valgrind indicates that the ressource consuming method is Eigen::internal::assign_sparse_to_sparse and in this method the call to Eigen::SparseMatrix<>::InsertBackByOuterInner.
Is there a more efficient way to assemble such a matrix?
(I also had to use P*(JP.transpose()) instead of PJ*J.transpose() to make the programm compile, may be there is already something wrong)
P.S: NDEBUG and optimizations are turned on
Edit: by storing P.transpose in a extra matrix ,I get a bit better performance, but the summation accounts still for 15% of the programm
Your code will be much faster by working inplace. First, estimate the number of non-zeros per column in the final matrix and reserve space (if not already done):
int nnz_per_col = ...;
J.reserve(VectorXi::Constant(n_cols, nnz_per_col));
If the number of nnz per column is highly non-uniform, then you can also compute it per column:
VectorXi nnz_per_col(n_cols);
for each j
nnz_per_col(j) = ...;
J.reserve(nnz_per_col);
Then manually insert elements:
for each block B[k]
for each elements i,j
J.coeffRef(foo(i),foo(j)) += B[k](i,j)
where foo implement the appropriate mapping of indices.
And for the next iteration, no need to reserve, but you need to set coefficient values to zero while preserving the structure:
J.coeffs().setZero();
I have a three-dimensional matrix of these sizes, approximately
A = rand(20, 1000, 20);
where the first and third dimensions are always the same length. I want to zero the elements in a main diagonal slice. This does what I mean
for ii = 1:size(A, 1)
A(ii, :, ii) = 0;
end
Is there a vectorized or otherwise faster way to do this? This code runs about 100,000 times, with these approximate sizes, but not the exact same sizes each time.
You can use logical indexing for multible tailing dimensions while using subscript indexing for all previous dimensions individually. This way you can easily do the operation on an 1000 20 20 matrix. To apply this to your matrix, permute is required which might be slow:
n=size(A,3)
A=permute(A,[2,1,3]);
A(:,diag(true(n,1)))=0;
A=permute(A,[2,1,3]);
If it would be possible to permanently swap the dimensions of A in your code and avoid the permute, this would lead to the fastest solution.
Alternatively you can use repmat to expand the index to the dimensions of A
ix=repmat(reshape(diag(true(n,1)),n,1,n),[1,size(A,2),1])
A(ix)=0
For matrices of the same size you could keep ix. Not having access to MATLAB right now, I don't know which solution is faster.
You can use bsxfun to build a linear index of the elements to be zeroed:
ind = bsxfun(#plus, (0:size(A,2)-1).'*size(A,1), 1:size(A,1)*size(A,2)+1:numel(A) );
A(ind) = 0;
I want to orient my matrix so that the Up vector it facing the same direction as another vector. The orientation of the Forward and Right vectors do not matter.
For example:
matrix4 m; // m.Up = 0, 1, 0
vector3 v = V3(1,0,0);
// Then I think I have to get the rotation from m.Up and v
// And then rotate the matrix accordingly
But I don't know how todo this and I may be wrong.
This is one of the rotation problems that matricies are particularly useful for
First, split your matrix into the three component vectors (up, forward, and right).
Set your up vector to what you want it to be. Then, adjust your forward and right vectors so that they are at right angles, an easy way to do this would be through the use of cross products.
For example:
//Gets a perpendicular vector
V3 V3::Perp() {
V3 perp = v.Cross(NewV3(-1, 0, 0));
if(perp.Length() == 0) {
// If v is too close to -x try -y
perp = v.Cross(NewV3(0, -1, 0));
}
return perp.Unit();
}
//up = Whatever you need
forward = up.Perp()
right = cross(up, forward);
After that, plug your vectors back into the matrix and voila :D.
If I understand correctly, simply set your up axis in the matrix to your chosen vector. As you say that the forward and right vectors do not matter, set them to anything as long as they are orthonormal to your new up axis.
I have this matrix A, representing similarities of pixel intensities of an image. For example: Consider a 10 x 10 image. Matrix A in this case would be of dimension 100 x 100, and element A(i,j) would have a value in the range 0 to 1, representing the similarity of pixel i to j in terms of intensity.
I am using OpenCV for image processing and the development environment is C on Linux.
Objective is to compute the Eigenvectors of matrix A and I have used the following approach:
static CvMat mat, *eigenVec, *eigenVal;
static double A[100][100]={}, Ain1D[10000]={};
int cnt=0;
//Converting matrix A into a one dimensional array
//Reason: That is how cvMat requires it
for(i = 0;i < affnDim;i++){
for(j = 0;j < affnDim;j++){
Ain1D[cnt++] = A[i][j];
}
}
mat = cvMat(100, 100, CV_32FC1, Ain1D);
cvEigenVV(&mat, eigenVec, eigenVal, 1e-300);
for(i=0;i < 100;i++){
val1 = cvmGet(eigenVal,i,0); //Fetching Eigen Value
for(j=0;j < 100;j++){
matX[i][j] = cvmGet(eigenVec,i,j); //Fetching each component of Eigenvector i
}
}
Problem: After execution I get nearly all components of all the Eigenvectors to be zero. I tried different images and also tried populating A with random values between 0 and 1, but the same result.
Few of the top eigenvalues returned look like the following:
9805401476911479666115491135488.000000
-9805401476911479666115491135488.000000
-89222871725331592641813413888.000000
89222862280598626902522986496.000000
5255391142666987110400.000000
I am now thinking on the lines of using cvSVD() which performs singular value decomposition of real floating-point matrix and might yield me the eigenvectors. But before that I thought of asking it here. Is there anything absurd in my current approach? Am I using the right API i.e. cvEigenVV() for the right input matrix (my matrix A is a floating point matrix)?
cheers
Note to readers: This post at first may seem unrelated to the topic, but please refer to the discussion in the comments above.
The following is my attempt at implementing the Spectral Clustering algorithm applied to image pixels in MATLAB. I followed exactly the paper mentioned by #Andriyev:
Andrew Ng, Michael Jordan, and Yair Weiss (2002).
On spectral clustering: analysis and an algorithm.
In T. Dietterich, S. Becker, and Z. Ghahramani (Eds.),
Advances in Neural Information Processing Systems 14. MIT Press
The code:
%# parameters to tune
SIGMA = 2e-3; %# controls Gaussian kernel width
NUM_CLUSTERS = 4; %# specify number of clusters
%% Loading and preparing a sample image
%# read RGB image, and make it smaller for fast processing
I0 = im2double(imread('house.png'));
I0 = imresize(I0, 0.1);
[r,c,~] = size(I0);
%# reshape into one row per-pixel: r*c-by-3
%# (with pixels traversed in columwise-order)
I = reshape(I0, [r*c 3]);
%% 1) Compute affinity matrix
%# for each pair of pixels, apply a Gaussian kernel
%# to obtain a measure of similarity
A = exp(-SIGMA * squareform(pdist(I,'euclidean')).^2);
%# and we plot the matrix obtained
imagesc(A)
axis xy; colorbar; colormap(hot)
%% 2) Compute the Laplacian matrix L
D = diag( 1 ./ sqrt(sum(A,2)) );
L = D*A*D;
%% 3) perform an eigen decomposition of the laplacian marix L
[V,d] = eig(L);
%# Sort the eigenvalues and the eigenvectors in descending order.
[d,order] = sort(real(diag(d)), 'descend');
V = V(:,order);
%# kepp only the largest k eigenvectors
%# In this case 4 vectors are enough to explain 99.999% of the variance
NUM_VECTORS = sum(cumsum(d)./sum(d) < 0.99999) + 1;
V = V(:, 1:NUM_VECTORS);
%% 4) renormalize rows of V to unit length
VV = bsxfun(#rdivide, V, sqrt(sum(V.^2,2)));
%% 5) cluster rows of VV using K-Means
opts = statset('MaxIter',100, 'Display','iter');
[clustIDX,clusters] = kmeans(VV, NUM_CLUSTERS, 'options',opts, ...
'distance','sqEuclidean', 'EmptyAction','singleton');
%% 6) assign pixels to cluster and show the results
%# assign for each pixel the color of the cluster it belongs to
clr = lines(NUM_CLUSTERS);
J = reshape(clr(clustIDX,:), [r c 3]);
%# show results
figure('Name',sprintf('Clustering into K=%d clusters',NUM_CLUSTERS))
subplot(121), imshow(I0), title('original image')
subplot(122), imshow(J), title({'clustered pixels' '(color-coded classes)'})
... and using a simple house image I drew in Paint, the results were:
and by the way, the first 4 eigenvalues used were:
1.0000
0.0014
0.0004
0.0002
and the corresponding eigenvectors [columns of length r*c=400]:
-0.0500 0.0572 -0.0112 -0.0200
-0.0500 0.0553 0.0275 0.0135
-0.0500 0.0560 0.0130 0.0009
-0.0500 0.0572 -0.0122 -0.0209
-0.0500 0.0570 -0.0101 -0.0191
-0.0500 0.0562 -0.0094 -0.0184
......
Note that there are step performed above which you didn't mention in your question (Laplacian matrix, and normalizing its rows)
I would recommend this article. The author implements Eigenfaces for face recognition. On page 4 you can see that he uses cvCalcEigenObjects to generate the eigenvectors from an image. In the article the whole pre processing step necessary for this computations are shown.
Here's a not very helpful answer:
What does theory (or maths scribbled on a piece of paper) tell you the eigenvectors ought to be ? Approximately.
What does another library tell you the eigenvectors ought to be ? Ideally what does a system such as Mathematica or Maple (which can be persuaded to compute to arbitrary precision) tell you the eigenvectors ought to be ? If not for a production-sixed problem at least for a test-sized problem.
I'm not an expert with image processing so I can't be much more helpful, but I spend a lot of time with scientists and experience has taught me that a lot of tears and anger can be avoided by doing some maths first and forming an expectation of what results you ought to get before wondering why you got 0s all over the place. Sure it might be an error in the implementation of an algorithm, it might be loss of precision or some other numerical problem. But you don't know and shouldn't follow up those lines of inquiry yet.
Regards
Mark