rotating a matrix without interpolation

Question

Oded Scharf 2020-12-1

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/672048-rotating-a-matrix-without-interpolation

评论： Oded Scharf 2020-12-6

Hey,

I want to rotate a mtrix by a certain degree without interpolating the intermediate pixels and just place NaN in them.

I tried to use imrotate but there is no option to do it without interpolating.

Any suggestions?

Thank in advence.

Oded

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2020-12-2

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/672048-rotating-a-matrix-without-interpolation#answer_562213

When you rotate a matrix, then the only pixels that can be calculated without any interpolation are the ones whose coordinates are such that the ratio of coordinates is exactly an integer multiple of tan() of the rotation angle. For example if the rotation is 42.7093899573615 degrees then there are some pixels whose coordinates happen to fall on the famous (5, 12, 13) pythagorean triple, and those single pixels could be copied without any interpolation. Every other pixel would require interpolation.

Are you sure you want to construct a matrix of nan that is the appropriate size to hold the rotated matrix, with the entire matrix nan except for the few scattered pixels that just happen to be at exactly the right coordinates to be carried from integer position to integer position by the rotation?

7 个评论
显示 5更早的评论隐藏 5更早的评论

Walter Roberson 2020-12-2

在 MATLAB Online 中打开

M=[1:3;4:6;7:9]
M = 3×3
     1     2     3
     4     5     6
     7     8     9
●
M_after_45_rot=[0 0 1 0 0; 0 4 NaN 2 0; 7 NaN 5 NaN 3; 0 8 NaN 6 0; 0 0 9 0 0]
M_after_45_rot = 5×5
     0     0     1     0     0
     0     4   NaN     2     0
     7   NaN     5   NaN     3
     0     8   NaN     6     0
     0     0     9     0     0
●

You appear to be rotating around the center, 5.

Rotations preserve distances.

The 1 starts at at distance sqrt(2) away from the 5. After the rotation, it has to end up at distance sqrt(2). The points that are distance sqrt(2) away from the central 5 in your result matrix are

   0     1     0     0
   *   NaN     *     0
 NaN     5   NaN     3
   *   NaN     *     0
   0     9     0     0

Those points (marked with *) are 0 degrees, 90 degrees, 180 degrees, and 270 degrees rotation relative to the original 1. Therefore, by symmetry, the original 1, 3, 9, and 7 must map to NaN -- if they go anywhere discrete they do not preserve distances as rotations must.

OKay, how about the original 2, where does it map to? Rotations preserve distances, and it starts 1 unit away, and the output locations that are 1 unit away from the 5 are

   0     1     0     0
   4     *     2     0
   *     5     *     3
   8     *     6     0
   0     9     0     0

but those are 0, 90, 180, 270 degrees from the original 2 position. Therefore, by symmetry, the original 2, 6, 8, 4 must go to NaN as the locations you would have them move to do not preserve distances.

In the expanded 5 x 5 square, the outer values correspond to distances 2, sqrt(5) and sqrt(8) from the center 5. No points are that distance from the central 5 in the original image, so all of them would require interpolation, so you require that all of them would be NaN as well.

Thus, the 3 x 3 case would have to expand to 5 x 5 all nan except for the central 5 that does not move.

The rotation matrix is

[cos(theta), -sin(theta); sin(theta), cos(theta)]

with theta = pi/2 that would be

[sqrt(2), -sqrt(2); sqrt(2), sqrt(2)]

For cordinates (x, y) relative to center (xc, yc), and translating back after rotation, the coordinates would map to

[xc + sqrt(2)/2*(x - xc) + sqrt(2)/2*(y - yc),
 yc - sqrt(2)/2*(x - xc) + sqrt(2)/2*(y - yc)]

By examination, we can see that the first coordinate is an integer either when x == xc and y == yc, or when (x - xc = -(y - yc). The first of those conditions is the condition of being located right at the center. For the second possibilty, let us consider 3 x 3 with center (2,2), then for example consider (1,3), then (x-xc) = -1 and (y-yc) = 1, and we see that [1,3] rotates around to a position with x coordinate exactly 2.

But, the second coordinate we can see by examination is an integer either when x == xc and y == yc, or when (x - xc) = (y - yc) -- that (1,3) original coordinate in the 3 x 3 is going to rotate around to y coordinate 2 + sqrt(2) which is not an integer.

Is it possible for (x-xc) = -(y - yc) to be an integer in the second coordinate?

yc - sqrt(2)/2*(x - xc) + sqrt(2)/2*(-(x-xc)) ==>
yc - 2*sqrt(2)/2*(x-xc) ==>
yc - sqrt(2)*(x-xc)

and that can only be an integer with integer coordinates if x - xc = 0. But if x - xc = 0 then with x-xc = -(y-yc) then 0 = -(y-yc) implies y - yc = 0... which is back to the coordinates of the center only.

Therefore if you rotate by 45 degrees or -45 degrees, the only integer coordinates that can map to integer coordinates are the coordinates of the center of rotation. And thus the output has to be nan except for the center piece as no interpolation is permitted.

Rotating by 90 or -90 degrees is, of course, a very different matter, rot90(A,1), rot90(A,-1) give those to you, and every integer coordinate maps to an integer coordinate.

Walter Roberson 2020-12-5

I notice that in your desired result that each location ends up sqrt(2) times the original distance. Items that were 1 unit end up on the diagonal sqrt(2) away. Items that were sqrt(2) away end up 2 away. Is that the pattern? We might be able to work with that.

Oded Scharf 2020-12-6

Yeah, I think it is the right pattern

请先登录，再进行评论。

rotating a matrix without interpolation

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

回答（1 个）

7 个评论
显示 5更早的评论隐藏 5更早的评论

另请参阅

类别

标签

Community Treasure Hunt

rotating a matrix without interpolation

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

回答（1 个）

7 个评论 显示 5更早的评论隐藏 5更早的评论

另请参阅

类别

标签

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

7 个评论
显示 5更早的评论隐藏 5更早的评论