Having a bit of trouble generating an NxN matrix in Mathematica. Given the value of N, I need to construct the NxN matrix that looks like the following:
N = Input["Enter value for N:"];
matrix = ConsantArray[0,{N,N}];
Do[matrix[[i,j]] = **"???"** ,{i,N}, {j,N}]
matrix // Matrix Form
Not sure in what should go as my statement in Do-Loop. Any help would appreciate it.
You could create a 1D array [1 ... n2] and then reshape or partition it to a matrix.
matrix = ArrayReshape[Range[n^2], {n, n}]
(* also works: *)
matrix = Partition[Range[n^2], n]
a couple more ways.
matrix=Table[j+(i-1) n,{i,n},{j,n}]
matrix=Array[#2+(#1-1) n &,{n,n}]
the Table form should give a clue how to fix your Do as well, but that's usually a poor approach performance-wise.
do not use capital N by the way its a reserved symbol.
Related
I have a nX2 matrix A and a 3D matrix K. I would like to take element-wise multiplication specifying 2 indices in 3rd dimension of K designated by each row vector in A and take summation of them.
For instance of a simplified example when n=2,
A=[1 2;3 4];%2X2 matrix
K=unifrnd(0.1,0.1,2,2,4);%just random 3D matrix
L=zeros(2,2);%save result to here
for t=1:2
L=L+prod(K(:,:,A(t,:)),3);
end
Can I get rid of the for loop in this case?
How's this?
B = A.'; %'
L = squeeze(sum(prod(...
reshape(permute(K(:,:,B(:)),[3 1 2]),2,[],size(K,1),size(K,2)),...
1),...
2));
Although your test case is too simple, so I can't be entirely sure that it's correct.
The idea is that we first take all the indices in A, in column-major order, then reshape the elements of K such that the first two dimensions are of size [2, n], and the second two dimensions are the original 2 of K. We then take the product, then the sum along the necessary dimensions, ending up with a matrix that has to be squeezed to get a 2d matrix.
Using a bit more informative test case:
K = rand(2,3,4);
A = randi(4,4,2);
L = zeros(2,3);%save result to here
for t=1:size(A,1)
L = L+prod(K(:,:,A(t,:)),3);
end
B = A.'; %'
L2 = squeeze(sum(prod(reshape(permute(K(:,:,B(:)),[3 1 2]),2,[],size(K,1),size(K,2)),1),2));
Then
>> isequal(L,L2)
ans =
1
With some reshaping magic -
%// Get sizes
[m1,n1,r1] = size(K);
[m2,n2] = size(A);
%// Index into 3rd dim of K; perform reductions and reshape back
Lout = reshape(sum(prod(reshape(K(:,:,A'),[],n2,m2),2),3),m1,n1);
Explanation :
Index into the third dimension of K with a transposed version of A (transposed because we are using rows of A for indexing).
Perform the prod() and sum() operations.
Finally reshape back to a shape same as K but without the third dimension as that was removed in the earlier reduction steps.
I have a sparse matrix and I need to fill certain entries with a specific value, I am using a for loop right now but I know its not the correct way to do it so I was wondering if its possible to vectorise this for loop?
K = sparse(N);
for i=vectorofrandomintegers
K(i,i) = 1;
end
If I vectorise it normally as so:
K(A,A) = 1;
then it fills all the entries in each row denoted by A whereas I want individual entries (i.e. K(1,1) = 1 or K(6,6)=1).
Also, the entries are not diagonally adjacent so I can't plop the identity matrix into it.
If you are going to use a vectorized method, you would need to get the linear indices to be set. The issue is that if you define your sparse matrix as K = sparse(N) and then linearly index into K, it would extend the size of it in one direction only and not along both row and column. Thus, you need to specify to MATLAB that you are
looking to use this sparse to store a 2D array. Thus, it would be -
K = sparse(N,N);
Get the linear indices to index into K using sub2ind and set them -
ind1 = sub2ind([N N],vectorofrandomintegers,vectorofrandomintegers);
K(ind1) = 1;
It's fairly simple
i'd use
K((A-1)*N+A))=1;
i believe that should fix your problem by treating the matrix as a vector
Instead of declaring and then filling a sparse matrix, you can fill it at the same time you define it:
i = vectorofrandomintegers; j = i;
K = sparse(i,j,1,N,N)
I have a nxn singular matrix. I want to add k rows (which must be from the standard basis e1, e2, ..., en) to this matrix such that the new (n+k)xn matrix is full column rank. The number of added rows k must be minimum and they can be added in any order (not just e1, e2 ,..., it can be e4, e10, e1, ...) as long as k is minimum.
Does anybody know a simple way to do this? Any help is appreciated.
You can achieve this by doing a QR decomposition with column pivoting, then taking the transpose of the last n-rank(A) columns of the permutation matrix.
In matlab, this is achieved by the qr function(See the matlab documentation here):
r=rank(A);
[Q,R,E]=qr(A);
newA=[A;transpose(E(:,end-r+1:end))];
Each row of transpose(E(:,end-r+1:end)) will be a member of standard basis, rank of newA will be n, and this is also the minimal number of standard basis you will need to do so.
Here is how this works:
QR decomposition with column pivoting is a standard procedure to decompose a matrix A into products:
A*E==Q*R
where Q is an orthogonal matrix if A is real, or an unitary matrix if A is complex; R is upper triangular matrix, and E is a permutation matrix.
In short, the permutations are chosen so that the diagonal elements are larger than the off-diagonals in the same row, and that size of the diagonal elements are non-increasing. More detailed description can be found on the netlib QR factorization page.
Since Q and E are both orthogonal (or unitary) matrices, the rank of R is the same as the rank of A. To bring up the rank of A, we just need to find ways to increase the rank of R; and this is much more straight forward thanks to the structure of R as the result of pivoting and the fact that it is upper-triangular.
Now, with the requirement placed on pivoting procedure, if any diagonal element of R is 0, the entire row has to be 0. The n-rank(A) rows of 0s in the bottom if R is responsible for the nullity. If we replace the lower right corner with an identity matrix, the that new matrix would be full rank. Well, we cannot really do the replacement, but we can append the rows matrix to the bottom of R and form a new matrix that has the same rank:
B==[ 0 I ] => newR=[ R ; B ]
Here the dimensionality of I is the nullity of A and that of R.
It is readily seen that rank(newR)=n. Then we can also define a new unitary Q matrix by expanding its dimensionality in a trivial manner:
newQ=[Q 0 ; 0 I]
With that, our new rank n matrix can be obtained as
newA=newQ*newR.transpose(E)=[Q*R ; B ]*transpose(E) =[A ; B*transpose(E)]
Note that B is [0 I] and E is a permutation matrix, so B*transpose(E) is simply the transpose
of the last n-rank(A) columns of E, and thus a set of rows made of standard basis, and that's just what you wanted!
Is n very large? The simplest solution without using any math would be to try adding e_i and seeing if the rank increases. If it does, keep e_i. proceed until finished.
I like #Xiaolei Zhu's solution because it's elegant, but another way to go (that's even more computationally efficient is):
Determine if any rows, indexed by i, of your matrix A are all zero. If so, then the corresponding e_i must be concatenated.
After that process, you can simply concatenate any subset of the n - rank(A) columns of the identity matrix that you didn't add in step 1.
rows/cols from Identity matrix can be added in any order. it does not need to be added in usual order as e1,e2,... in general situation for making matrix full rank.
I am a new student learning to use Matlab.
Could anyone please tell me is there a faster way possibly without loops:
to assign for each row only two values 1, -1 into different positions of a big sparse matrix.
My code to build a bimatrix or bibimatrix for the MILP problem of condition :
f^k_{ij} <= y_{ij} for every arc (i,j) and all k ~=r; in a multi-commodity flow model.
Naive approach:
bimatrix=[];
% create each row and then add to bimatrix
newrow4= zeros(1,n*(n+1)^2);
for k=1:n
for i=0:n
for j=1: n
if j~=i
%change value of some positions to -1 and 1
newrow4(i*n^2+(j-1)*n+k)=1;
newrow4((n+1)*n^2+i*n+j)=-1;
% add to bimatrix
bimatrix=[bimatrix; newrow4];
% change newrow4 back to zeros row.
newrow4(i*n^2+(j-1)*n+k)=0;
newrow4((n+1)*n^2+i*n+j)=0;
end
end
end
end
OR:
% Generate the big sparse matrix first.
bibimatrix=zeros(n^3 ,n*(n+1)^2);
t=1;
for k=1:n
for i=0:n
for j=1: n
if j~=i
%Change 2 positions in each row to -1 and 1 in each row.
bibimatrix(t,i*n^2+(j-1)*n+k)=1;
bibimatrix(t,(n+1)*n^2+i*n+j)=-1;
t=t+1
end
end
end
end
With these above code in Matlab, the time to generate this matrix, with n~12, is more than 3s. I need to generate a larger matrix in less time.
Thank you.
Suggestion: Use sparse matrices.
You should be able to create two vectors containing the column number where you want your +1 and -1 in each row. Let's call these two vectors vec_1 and vec_2. You should be able to do this without loops (if not, I still think the procedure below will be faster).
Let the size of your matrix be (max_row X max_col). Then you can create your matrix like this:
bibimatrix = sparse(1:max_row,vec_1,1,max_row,max_col);
bibimatrix = bibimatrix + sparse(1:max_row, vec_2,-1,max_row,max_col)
If you want to see the entire matrix (which you don't, since it's huge) you can write: full(bibimatrix).
EDIT:
You may also do it this way:
col_vec = [vec_1, vec_2];
row_vec = [1:max_row, 1:max_row];
s = [ones(1,max_row), -1*ones(1,max_row)];
bibimatrix = sparse(row_vec, col_vec, s, max_row, max_col)
Disclaimer: I don't have MATLAB available, so it might not be error-free.
I don't have enough memory to simply create a diagonal D-by-D matrix, since D is large. I keep getting an 'out of memory' error.
Instead of performing M x D x D operations in the first multiplication, I do M x D operations, but still my code takes ages to run.
Can anybody find a more effective way to perform the multiplication A'*B*A? Here's what I've attempted so far:
D=20000
M=25
A = floor(rand(D,M)*10);
B = floor(rand(1,D)*10);
for i=1:D
for j=1:M
result(i,j) = A(i,j) * B(1,j);
end
end
manual = result * A';
auto = A*diag(B)*A';
isequal(manual,auto)
One option that should solve your problem is using sparse matrices. Here's an example:
D = 20000;
M = 25;
A = floor(rand(D,M).*10); %# A D-by-M matrix
diagB = rand(1,D).*10; %# Main diagonal of B
B = sparse(1:D,1:D,diagB); %# A sparse D-by-D diagonal matrix
result = (A.'*B)*A; %'# An M-by-M result
Another option would be to replicate the D elements along the main diagonal of B to create an M-by-D matrix using the function REPMAT, then use element-wise multiplication with A.':
B = repmat(diagB,M,1); %# Replicate diagB to create an M-by-D matrix
result = (A.'.*B)*A; %'# An M-by-M result
And yet another option would be to use the function BSXFUN:
result = bsxfun(#times,A.',diagB)*A; %'# An M-by-M result
Maybe I'm having a bit of a brainfart here, but can't you turn your DxD matrix into a DxM matrix (with M copies of the vector you're given) and then .* the last two matrices rather than multiply them (and then, of course, normally multiply the first with the found product quantity)?
You are getting "out of memory" because MATLAB can not find a chunk of memory large enough to accommodate the entire matrix. There are different techniques to avoid this error described in MATLAB documentation.
In MATLAB you obviously do not need programming explicit loops in most cases because you can use operator *. There exists a technique how to speed up matrix multiplication if it is done with explicit loops, here is an example in C#. It has a good idea how (potentially large) matrix can be split into smaller matrices. To contain these smaller matrices in MATLAB you can use cell matrix. It is much more probably that system finds enough RAM to accommodate two smaller sub-matrices then the resulting large matrix.