[QE-users] Details of Automatic FFT Grid Size Calculation

Sun Mar 2 11:18:46 CET 2025

Dear Dyer,

    you are right... the factor of two you find is the difference 
between the radius and the diameter of a sphere.

    the bit of code in FFTXlib/src/fft_type.f90 I pointed to earlier 
defines the ranges nr1,nr2,nr3 of a parallelepiped box that contains the 
G-sphere of sqrt(gcutm) radius.

     these are then passed to grid_set( dfft, bg, gcutm, dfft%nr1, 
dfft%nr2, dfft%nr3 ) where the actual size is determined in three nested 
loops that range from - the estimated limit to + the estimated limit

      DO k = -nr3, nr3
           DO j = -nr2, nr2
             DO i = -nr1, nr1

    where the maximum ACTUAL absolute index values are set as 
nb(1),nb(2),nb(3), which should coincide with nr1,nr2,nr3 if they were 
computed properly, and finally set the grid dimensions as

       nr1 = 2 * nb(1) + 1
       nr2 = 2 * nb(2) + 1
       nr3 = 2 * nb(3) + 1

stefano

On 02/03/25 06:04, Dyer, Brock wrote:
> I've been tracing down all the variables that are required to generate 
> a good FFT grid, and I seem to be off by a factor of 2 in the end. I 
> can't quite figure out what the issue is, but my guess may be that 
> while I am storing 'ecutwfc' in Rydbergs there is a conversion to 
> Hartrees somewhere in the code that I haven't seen yet.
>
> My current process looks like this (with some values from a run I had 
> recently so I can compare):
>
> ecutwfc = 100 Ry
> Ecutrho = 400 Ry
>
> # Unit cell dimensions, given in Bohrs
> v1 = [44.09733757, 0.0, 0.0]
> v2 = [0.0, 44.09733757, 0.0]
> v3 = [0.0, 0.0, 44.09733757]
>
> alat = sqrt(v1[0]**2 + v1[1]**2 + v1[2]**2)
> tpiba = (2.0 * pi) / alat
> gcutm = ecutrho / (tpiba**2)
>
> at = [v1/alat, v2/alat, v3/alat]
>
> nr1 = int(sqrt(gcutm) * sqrt(at[0][0]**2 + at[0][1]**2 + at[0][2]**2)) + 1
> nr2 = int(sqrt(gcutm) * sqrt(at[1][0]**2 + at[1][1]**2 + at[1][2]**2)) + 1
> nr3 = int(sqrt(gcutm) * sqrt(at[2][0]**2 + at[2][1]**2 + at[2][2]**2)) + 1
>
> These last lines are where I've noticed the problem. From looking at 
> the output of my run with the given cell sizes, I expect an FFT grid 
> of 288x288x288, however if I were to run this code (and the code that 
> checks if it's a good size) I'd get an FFT grid that is only half 
> that. I'd love some advice on this if it is at all possible. I also 
> can send some more formatted code if it would help (I decided to cut 
> down the python code so it looked a bit more like the original f90 code).
>
> ------------------------------------------------------------------------
> *From:* users on behalf of Stefano de Gironcoli
> *Sent:* Thursday, February 27, 2025 2:50 PM
> *To:* users at lists.quantum-espresso.org
> *Subject:* Re: [QE-users] Details of Automatic FFT Grid Size Calculation
>
> it's in SUBROUTINE realspace_grid_init  in  FFTXlib/src/file fft_types.f90
>
>         !
>          ! ... calculate the size of the real-space dense grid for FFT
>          ! ... first, an estimate of nr1,nr2,nr3, based on the max values
>          ! ... of n_i indices in:   G = i*b_1 + j*b_2 + k*b_3
>          ! ... We use G*a_i = n_i => n_i .le. |Gmax||a_i|
>          !
>          dfft%nr1 = int ( sqrt (gcutm) * sqrt (at(1, 1)**2 + at(2, 
> 1)**2 + at(3, 1)**2) ) + 1
>          dfft%nr2 = int ( sqrt (gcutm) * sqrt (at(1, 2)**2 + at(2, 
> 2)**2 + at(3, 2)**2) ) + 1
>          dfft%nr3 = int ( sqrt (gcutm) * sqrt (at(1, 3)**2 + at(2, 
> 3)**2 + at(3, 3)**2) ) + 1
>
> stefano
>
>
>
> On 27/02/25 20:11, Dyer, Brock wrote:
>
>     Hello all, I have been working quite a bit lately on automating my
>     QE workflow, and as part of that workflow I check the
>     automatically calculated FFT grid sizes for the level of theory
>     that I have been using in order to improve my parallelization.
>
>     I have tried tracing down and reading the code that calculates the
>     FFT grid sizes, however I cannot find/understand the actual code
>     to calculate the grid sizes. My current understanding is that the
>     initial parameters to calculate the grid size are 'ecutwfc' and/or
>     'ecutrho', and the unit cell size, and then there seems to be some
>     more math, and perhaps at the end the final dimensions get
>     calculated in 'fft_ggen.f90'.
>
>     What I am looking for ideally is a mathematical formula that
>     includes all of the input parameters and operations required to
>     calculate the FFT grid sizes so that I can implement it into my
>     workflow and hopefully not have to run double calculations to
>     properly parallelize.
>
>     Thanks in advance for the help!
>
>     Brock Dyer, Ursinus College Class of 2025
>
>
>
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