How can I iterate across the opposite diagonal?
4 views (last 30 days)
Show older comments
Hi everyone, I am new to matlab and trying to build a matrix in it. I am wanting to know how to iterate across the other diagonal, opposite to where the 6's are. If anyone can do this with a similar for loop, all help would be appreciated. Thank you!
N1=2;
N2=2;
nband=1;
n= (N1* N2)*2*nband;
eps_s=1;
eps_sp=2;
eps_spp=3;
tp=4;
tps=5;
HM = zeros(n);
for ii = 1:n
HM(ii,ii) = 6; % (ii,ii) = eps_s
HM(ii,ii+1) = 4;
HM(ii+1,ii) = 4
end
3 Comments
Torsten
on 25 Dec 2021
If you set
for ii = 1:n
HM(ii,n+1-ii) = 6;
end
you set the opposite diagonal to 6.
For ii=1, you set HM(1,n) = 6.
For ii=2, you set HM(2,n-1) = 6.
...
For ii=n, you set HM(n,1) = 6.
Answers (3)
Image Analyst
on 25 Dec 2021
At the end of your loop, just add
HM = fliplr(HM);
or else here is one way:
N1=2;
N2=2;
nband=1;
n= (N1* N2)*2*nband;
eps_s=1;
eps_sp=2;
eps_spp=3;
tp=4;
tps=5;
HM = zeros(n);
for ii = 1:n
col = n-ii+1;
if col <= n && col >= 1
HM(ii, col) = 6; % (ii,ii) = eps_s
end
col = n-ii+2;
if col <= n && col >= 1
HM(ii, col) = 4;
end
col = n-ii;
if col <= n && col >= 1
HM(ii, col) = 4;
end
end
HM
0 Comments
DGM
on 25 Dec 2021
Edited: DGM
on 25 Dec 2021
You can also do:
HM = fliplr(toeplitz([6 4 zeros(1,7)]))
That said, it's unclear whether you're actually intending for the output to be 8x8 or 9x9. You assert n=8, and you preallocate an 8x8 array, but your indexing forces the array to be expanded to 9x9. Consequently, you're missing a 6 in the corner element. If the desired output is only 8x8, then that's as simple as changing the number of zeros.
HM = fliplr(toeplitz([6 4 zeros(1,6)]))
0 Comments
See Also
Categories
Find more on Operating on Diagonal Matrices 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!