h=0;
i=0.0195314;
m=100
for m=1:512
for j=1:n
for i=1:n-m
h=h+(z(i+m,j)-z(i,j))^2;
end
end
h=1/(n*(n-m))*h;
rr(m)=log(m*i);
hh(m)=log(h);
end
Let's take a look at this section of your code. When m becomes 2, what value should h contain? What value does h contain?
You want to put the "h = 0;" line inside your loop over m.
You also want to handle the case where n is equal to m. In that case, your loop over i executes no times and i is []. This causes problems on your line where you assign to rr(m). You probably want to use (n-m) [the last value i would take when the loop executed, which it doesn't when n is equal to m.]
But anyway, we can probably simplify that code a bit. Since the limits of the loop over i don't depend on j at all, we can swap the order of the loops.
for i=1:n-m
for j=1:n
h=h+(z(i+m,j)-z(i,j))^2;
end
end
If we use array operations and sum rather than operating on each scalar element of z independently and using plus, we can eliminate the innermost loop entirely. We do need to be a little careful, since you used length on a matrix and that may not do what you expect if z is taller than it is wide. Instead you should use size to get the number of rows and columns. for i=1:n-m
v = z(i+m, :)-z(i, :);
h = h + sum(v.^2);
end
You probably can even eliminate that loop if you think of using array operations on submatrices rather than on vectors, but I'll leave that as an exercise for the reader.
