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RE: Fun with Arnold's cat map transform
Let B be the matrix associated to the Arnold map, so B =[1 1;1 2] . Then with some effort you can show that B^48 mod 576 is equal to the identity matrix. This explains the result.
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Oh ok. That is a nice find. But is there a general rule to find periodicity of any number sized image?
So you can write B as B=invXDX where X is the eigen matrix with corresponding eigenvalues on the diagonal of the diagonal matrix D. It is then easy compute B^m you then need to find an m such that B^m mod size of your matrix is equal to the identity matrix. So this gives you equations to find the periodicity
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Again to find m I will have to iterate till I see identity matrix right? So that means it is sometimes computationally intensive right?
When you iterate you only have to compute D^m D is a diagonal matrix so that is pretty easy to compute. X and invX stay the same. More specifically B^m=invX D^m X
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