Performing what toeplitz does without using pre-installed function
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Losoupbowla
el 29 de En. de 2021
Comentada: Losoupbowla
el 1 de Feb. de 2021
I currently have 8x8 matrix of NaNs. The assignment is to replace all elements of the NaN with numbers 1-8 and then flip them around the way toeplitz function would, without actually using it. We are supposed to use nested for loops and end up with this result:
1 2 3 4 5 6 7 8
2 1 2 3 4 5 6 7
3 2 1 2 3 4 5 6
4 3 2 1 2 3 4 5
5 4 3 2 1 2 3 4
6 5 4 3 2 1 2 3
7 6 5 4 3 2 1 2
8 7 6 5 4 3 2 1
Any idea how to do this? I am completely stuck. So far I only have this
toeplitzSize=8
toeplitz=nan(toeplitzSize)
for a=1:toeplitzSize
for b=1:toeplitzSize
if isnan(toeplitz(a,b))
toeplitz(a,b)= b;
end
end
end
1 comentario
Stephen23
el 1 de Feb. de 2021
Note that you should NOT name any variable toeplitz, as this shadows the inbuilt function.
Respuesta aceptada
Sindhu Karri
el 1 de Feb. de 2021
Hiii Losoupbowla,
Hope the below attached code helps
toeplitSize=8;
toeplitz=nan(toeplitSize);
for i=1:toeplitSize
k=i;
for j=1:toeplitSize
if isnan(toeplitz(i,j))
if(j<=i)
toeplitz(i,j)=k;
k=k-1;
else
toeplitz(i,j)=j-i+1;
end
end
end
end
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