How can I iterate across the opposite diagonal?

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

[Torsten]

HM(ii,n+1-ii) = ...

[Karen Smith]

pls elaborate if possible!!

[Torsten]

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.

Answer 1

[Image Analyst]

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

Answer 2

[Sean de Wolski]

full(flip(gallery('tridiag', 8, 4, 6, 4), 2))

ans = 8×8

0 0 0 0 0 0 4 6 0 0 0 0 0 4 6 4 0 0 0 0 4 6 4 0 0 0 0 4 6 4 0 0 0 0 4 6 4 0 0 0 0 4 6 4 0 0 0 0 4 6 4 0 0 0 0 0 6 4 0 0 0 0 0 0

Answer 3

[DGM]

You can also do:

HM = fliplr(toeplitz([6 4 zeros(1,7)]))

HM = 9×9

0 0 0 0 0 0 0 4 6 0 0 0 0 0 0 4 6 4 0 0 0 0 0 4 6 4 0 0 0 0 0 4 6 4 0 0 0 0 0 4 6 4 0 0 0 0 0 4 6 4 0 0 0 0 0 4 6 4 0 0 0 0 0 4 6 4 0 0 0 0 0 0 6 4 0 0 0 0 0 0 0

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)]))