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 <[email protected]> on behalf of Dyer, Brock
<[email protected]>
Sent: Sunday, March 2, 2025 06:04
To: Quantum ESPRESSO users Forum <[email protected]>
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: [email protected]
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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