Index 2D matrix along line rotated about the matrix midpoint

Hi,

I have an N x N matrix that I would like to sequentially index along a line that slices the middle of the matrix and is then rotated by 180/(2(N-1)) degrees until it completes half a full rotation. In the end, we want to end up with 2(N-1) distinct lines. Picture attached for a 5x5 case for clarity, since it's easier to describe by a picture than by words, I think. If N were even, then the "equivalent" center point of the matrix could be arbitrarily decided by ceil(N/2).

So for example, if the matrix is:

a = [1 2 3 4 5; 6 7 8 9 10; 11 12 13 14 15; 16 17 18 19 20]

Then the elements returned for each iteration would be the middle row, middle column, diagonals as well as:

[2 13 24]

[4 13 22]

[6 13 20]

[10 13 16]

The only way I can think of implementing this is to rotate the 2D matrix by the angle 180/(2(N-1)) and then always slice the middle column of the matrix but I'm not sure this is the most optimal way to do so for large matrices. Is there any other, more efficient way to achieve this?

[Daniel M]

Regarding the updated post with the figure for N = 5:

If this is all you want, that isn't difficult. But what about the generalized case? For example, I have no idea what elements you would want if N = 7. And I still don't know how you want to handle even-numbered N.

[Roberto Brenes]

Hopefully the new updated post makes sense. In the end, we would want 2(N-1) lines in total but making sure that every element in the matrix appears at least once. In that case, it might be best to think of the lines in between the diagonals and the mid row or column as a staircase or the linear interpolation between the start pixel and end pixel due to the discreteness of the matrix.

[Daniel M]

Sorry I can't help because what you're asking for is still unclear. You have not been able to specify what you want in the general case. My post above still applies regarding if N = 7 (let alone if N = 99 or higher), or if N is even. What you want is unknown, so how can I help?

Answer 1

[Daniel M]

360/(N-1) doesn't make sense. What if N = 8? I think you just mean 180. But it's irrelevant. It would not be efficient to recreate the matrix by shifting the elements around.

It seems all you want to do is get the middle column, middle row, and the diagonals. Well that is simple enough for odd N (you can provide more clarity for even-numbered N if you wish).

A = magic(5);
sz = size(A,1);
mid = ceil(sz/2);
midcol = A(:,mid);
midrow = A(mid,:);
maindiagonal = diag(A);
oppositediagonal = A(sz:sz-1:end-1);

Other than that I'm not sure what order or arrangement you want the values it, but it should be pretty straightforward from here. Use commands like flip and transpose.

[Roberto Brenes]

You're right, it should be 180, I fixed that on the original answer.

Your solution gets me close but not quite there. It's true that the middle column, row and the diagonals are part of what I need, but I do need to index every single entry of the array. Maybe the 3x3 was the wrong example because it lacked the lines in between the column and the diagonal.

For clarification, in the 5x5 (ie. magic(5)) case apart from the middle column, row and diagonals there would also be the entries:

[2 13 24]

[4 13 22]

[6 13 20]

[10 13 16]

which represent the two lines between the middle column and the two diagonals. As you can tell, these would normally contain less entries than the diagonals and the rows or columns.

I don't really care about the order and, as you said, I can take care of that after the fact if need be. I'll update the above question with the 5x5 case.

[Daniel M]

Those are an odd pairing of indices for sure. I hope when you rewrite your question, you pose it in a very clear way that generalizes it for a matrix of size N.

[Roberto Brenes]

Just edited the question above with a picture, hopefully that explains it. Also, disregard the first set of elements that I replied with the first time, I have no idea what I wrote and they didn't make any sense. I have edited them again to hopefully clear up the confusion.

Answer 2

[Daniel M]

Here is another idea, perhaps you can play with it. You'll have to implement this in a loop.

[cx,cy,c] = improfile(I,xi,yi,n);

xi are the line endpoints x values. yi are the endpoints y values. n is the number of samples along the line that you want to extract. c are those image values, and cx,cy are the exact points where the measurements were made.

For example, the green line in your image:

x = magic(5);
[cx,cy,c] = improfile(x,[2,4],[1 5],3)