3 ms·
The FFT version is interesting G∗K == F⁻¹( F(G)·F(K) ) So if we iterate t times (as in t step in time) this do we get this formula (convolution product bei
by ttoinou 8y ago
The FFT version is interesting
G∗K == F⁻¹( F(G)·F(K) )
So if we iterate t times (as in t step in time) this do we get this formula (convolution product being associative) ?
((G∗K)*K)*K ... t time = G∗K^t == F⁻¹( F(G)·F(K^t) )
- moopling 8y agoIf you assume K is unchanging, yes. Like if it were a fixed point on the board.
- ttoinou 8y agoThe kernel doesn't change in Smooth Life, right ?
- no_identd 8y agoHmm... I now wonder if it'd make sense to make a version of this that uses Constant Q instead of Fourier transforms...
- no_identd 8y ago...and I also wonder why this still uses a square lattice on the bottom layer - shouldn't a continuous version permit the use of a hexagonal lattice, which doesn't work that well within the discrete domain (although people have attempted it not that unsuccessful, but it still had issues.)
- Thorondor 8y agoNo, because the transition function has to be applied at each time step as well as the convolution. In the source code, that's self.rules.s(N_buffer, M_buffer)
- ttoinou 8y agoOkay thanks ! Makes it more interesting. I like it when it's doing something in one space (fourier space here) then something else in another space (original space) and you have to switch between multiple point of view to make something awesome (a fractal for example)