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

[Pietro Davide Delugas]

Hello

there is a factor of 2 because  the G vectors range from - N_max/ 2  to N_max/2, where N_max in the number of divisions of the FFT grid in a given direction. You may wannto to look at the routines called by fft_type_init to arrange the grid;  but  the simplified expression for the max number of divisions in the FFT grid for a given edge of the lattice box is just:

N_max = int(G_max * A_i / PI)

PI being 3.14.... , A_i the length of the lattice vector for the particular edge and G_max the longest reciprocal vector prescribed by the cutoff, in  QE,

other few thing to keep in mind are that


  *
to be sure that the required cutoff is applied in all the Brillouin zone , G_max is set as:

                      G_max = sqrt(ECUTRHO) + 1/2 * MAX (PI/A_i)


  *
the value reported by pw.x should  be the one for the wave functions   with ECUTRHO set to  4 * ECUTWFC
  *
the final values of the divisions are further mangled to avoid too large prime factors that can spoil  the FFTW performance.

Hope it helps
Pietro




________________________________
From: users <users-bounces at lists.quantum-espresso.org> on behalf of Dyer, Brock <brdyer at ursinus.edu>
Sent: Sunday, March 2, 2025 06:04
To: Quantum ESPRESSO users Forum <users at lists.quantum-espresso.org>
Subject: Re: [QE-users] Details of Automatic FFT Grid Size Calculation

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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