2

I am trying to write an algorithm to rotate a bitmap image of $n$ by $n$ size by an angle $\alpha$.

I know that I have to find a rotation matrix, then perform matrix multiplication of the rotation matrix by the image matrix data input.

I know that the $2\times 2$ rotation matrix is

$$T_\alpha=\left(\begin{array}{cc}\cos\alpha & -\sin\alpha \\ \sin \alpha & \cos \alpha \end{array}\right)$$

However, I am not sure how to find the appropriate $n \times n$ matrix.

Alexander Gruber
  • 28,037
codigomonstruo
  • 121
  • 2
    @CheungJoonHee Not a duplicate. The O.P. isn’t looking for rotations in an $n$-dimensional vector space. The problem here is to rotate an $n\times n$ pixel array. – amd Apr 30 '19 at 04:57
  • 2
    You’re confusing the size of the image pixel array with the dimension of a vector space. You need an algorithm to rotate a pixel array. Try https://stackoverflow.com/q/484573 for suggestions. – amd Apr 30 '19 at 04:59
  • 1
    This is a 2-d rotation. You need to pick an origin and decide how to map the resulting rotated pixels into another $n \times n$ array. – copper.hat Apr 30 '19 at 05:00
  • The obvious mathematical answer might not be optimal for real work programming with discrete data. It might be better to ask in Computer Science. – badjohn Apr 26 '25 at 16:10

1 Answers1

2

Let $w, h$ be the width & height of the image in pixels, but promote them to floats for further computation.

You take the image center $c = (w, h)/2$

Then treat each pixel coordinate as a coordinate with origin at $c$. In other words subtract $c$ from each pixel coordinate $v' = (x,y) - c$.

Now as you have the rotation matrix, you need only convert $v'$ to a column vector and multiply with $v'$ on the right.

If your rotation matrix was constructed using the right hand rule. And your image space (in whatever coding environment) originally started with $(0,0)$ at the bottom left, then this will rotate the image around its center $\alpha$ radians counter clockwise.

As far as final image quality though, this is probably bad once you discretize into a new image. This is why you should use something already coded.

Daniel Donnelly
  • 22,288
  • The last paragraph is very important. A simple, naïve mathematical solution might give a poor result. The tricky bit is that you will need to round the results to whole pixels. You might get holes in the resulting image. Also, you might than rotating by 30 followed by 60 is not the same as rotating by 90 in one step. As Daniel says: try to find an existing library. – badjohn Apr 26 '25 at 16:13