Exploiting symmetry in multidimensional arrays
1 view (last 30 days)
Show older comments
Patrick Mboma
on 13 Jul 2014
Commented: Roger Stafford
on 14 Jul 2014
Dear all, In a symmetric matrix A of size [n,n], for two indices i and j such that 1<=i<=n and 1<=j<=n we have that A(i,j)=A(j,i). Suppose I know A(i,j) for i>=j how can I efficiently set A(j,i)?
More generally, if I have a hypercube A of size n^m, how can I, on the basis of A(i1,i2,...,im) with i1>=i2>=i3....>=im, set all A for all permutations of i1,i2,...,im instead of recalculating the entry for each permutation.
Thanks,
0 Comments
Accepted Answer
Roger Stafford
on 13 Jul 2014
For a general m-dimensional n^m array, A, do this:
[I1,I2,...,Im] = ndgrid(1:n);
P = [I1(:),I2(:),...,Im(:)];
Q = sort(P,2,'descend');
A(sub2ind(size(P),P(:,1),P(:,2),...,P(:,m))) = ...
A(sub2ind(size(Q),Q(:,1),Q(:,2),...,Q(:,m)))
Note: The four three-dot ellipses (...) shown above (but not the one following the '=' sign) are to be filled in by the user.
4 Comments
Roger Stafford
on 14 Jul 2014
@Patrick. Yes, you are quite right. I should have used size(A). I must have had a mental black out at that point! :-)
More Answers (1)
Roger Stafford
on 13 Jul 2014
For a 2D array it's easy:
B = tril(A,-1)
A = B+B.'+diag(A);
0 Comments
See Also
Categories
Find more on Matrix Indexing in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!