Question on alignment.py about dcsr

[Tengfei-Wang]

It is for even-size kernels. Currently, the alignment only supports ever-sized kerkers. For example, when kernel size is 6, this gives a local mesh grid of [-2.5, -1.5, -0.5, 0.5, 1.5, 2.5]^2.
In this case, add (self.kernel_size-1)//2+0.5=2.5 transform the local mesh grid to 0~5. We then add it on p_0 to obtain global mesh coordinates.

[codeslake]

It is for even-size kernels. Currently, the alignment only supports ever-sized kerkers. For example, when kernel size is 6, this gives a local mesh grid of [-2.5, -1.5, -0.5, 0.5, 1.5, 2.5]^2. In this case, add (self.kernel_size-1)//2+0.5=2.5 transform the local mesh grid to 0~5. We then add it on p_0 to obtain global mesh coordinates.

But if you add (self.kernel_size-1)//2+0.5, the range of global mesh coordinate becomes 1~h. It lies between 0~h-1 if we add (self.kernel_size-1)//2-0.5. Isn't it the global mesh coordinate that should be in range 0~h-1?

[Tengfei-Wang]

It is for even-size kernels. Currently, the alignment only supports ever-sized kerkers. For example, when kernel size is 6, this gives a local mesh grid of [-2.5, -1.5, -0.5, 0.5, 1.5, 2.5]^2. In this case, add (self.kernel_size-1)//2+0.5=2.5 transform the local mesh grid to 0~5. We then add it on p_0 to obtain global mesh coordinates.

But if you add (self.kernel_size-1)//2+0.5, the range of global mesh coordinate becomes 1h. It lies between 0h if we add (self.kernel_size-1)//2-0.5. Isn't it the global mesh coordinate that should be in range 0~h-1?

Yes, or you may also modify the range of p_0 to [0, h-1].

[codeslake]

Ok, Thanks for the answer!