Fortran 中使用 PlanMany 的 cuFFT 双精度误差

Question

在 (answer of JackOLantern) 之后，我正在尝试使用 cufftPlanMany 计算批量 1D FFT。

下面的代码执行 nwfs=23 次 n=256 复杂数组的前向 1D FFT 和后向 1D FFT。这是为了训练我处理常规 cufftPlanMany。作为第二步，nwfs 数组将不同。最后，我检查每个数组的错误。

因为数据分配为：cinput_d(n,nwfs) 我这样使用函数：cufftPlanMany(planmany, 1, fftsize, inembed, nwfs,1, onembed, nwfs,1, CUFFT_C2C, nwfs)

其中：

rank = 1
fftsize = {n} 每个 FFT 的暗淡度相同
inembed = onembed = {0} 忽略
istride = ostride = nwfs 两个连续输入和输出之间的距离
idist = odist = 1 两个信号之间的距离
batch = nwfs 要完成的 fft 数

program fft
use cudafor
use precision_m
use cufft_m
implicit none
integer, allocatable:: kx(:)
complex(fp_kind), allocatable:: matrix(:)
complex(fp_kind), allocatable, pinned :: cinput(:,:),coutput(:,:)
complex(fp_kind), allocatable, device :: cinput_d(:,:),coutput_d(:,:)

integer:: i,j,k,n,nwfs
integer, allocatable :: fftsize(:),inembed(:),onembed(:)
type(c_ptr):: plan,planmany
real(fp_kind):: twopi=8._fp_kind*atan(1._fp_kind),h

integer::clock_start,clock_end,clock_rate,istat
real :: elapsed_time
character*1:: a
real(fp_kind):: w,x,y,z
integer:: nerrors

n=256
nwfs=23
h=twopi/real(n,fp_kind)

! allocate arrays on the host
allocate (cinput(n,nwfs),coutput(n,nwfs))
allocate (kx(n),matrix(n))
allocate (fftsize(nwfs),inembed(nwfs),onembed(nwfs))
! allocate arrays on the device
allocate (cinput_d(n,nwfs),coutput_d(n,nwfs))


fftsize(:) = n
inembed(:) = 0
onembed(:) = 0
!initialize arrays on host
kx =(/ ((i-0.5)*0.1953125, i=1,n/2), ((-n+i-0.5)*0.1953125, i=n/2+1,n) /)

matrix = (/ ... /)

!write(*,*) cinput
!copy arrays to device
do i =1,nwfs
    cinput(:,i)=matrix(:)
end do
cinput_d=cinput

! Initialize the plan for complex to complex transform
if (fp_kind== singlePrecision) call cufftPlan1D(plan,n,CUFFT_C2C,1)
if (fp_kind== doublePrecision) call cufftPlan1D(plan,n,CUFFT_Z2Z,1)
if (fp_kind== doublePrecision)     call cufftPlanMany(planmany, 1, fftsize, inembed, &
          nwfs,1, &
          onembed, &
          nwfs,1,  &
          CUFFT_Z2Z, nwfs)

if (fp_kind== singlePrecision)     call cufftPlanMany(planmany, 1, fftsize, inembed, &
          nwfs,1, &
          onembed, &
          nwfs,1,  &
          CUFFT_C2C, nwfs)
          !c_null_ptr  fftsize,inembed,onembed
         ! cufftPlanMany(plan, rank, n, inembed, istride, idist, &
         ! onembed, ostride, odist,  &
         ! type, batch)
     !subroutine cufftPlan1d(plan, nx, type, batch)
call SYSTEM_CLOCK(COUNT_RATE=clock_rate)

istat=cudaThreadSynchronize()
call SYSTEM_CLOCK(count=clock_start)
    ! Forward transform out of place
     call cufftExec(planmany,cinput_d,coutput_d,CUFFT_FORWARD)
     !$cuf kernel do <<<*,*>>>
     do i=1,n
        do j =1,n
        coutput_d(i,j) = coutput_d(i,j)/real(n,fp_kind)!sqrt(twopi*real(n,fp_kind))*sqrt(2.*pi)/sqrt(real(maxn))
        end do
     end do

     call cufftExec(planmany,coutput_d,coutput_d,CUFFT_INVERSE)


istat=cudaThreadSynchronize()
call SYSTEM_CLOCK(count=clock_end)

! Copy results back to host
coutput=coutput_d

do i=1,n
! write(*,'(i2,1x,2(f8.4),1x,2(f8.4),2x,e13.7)') i,cinput(i),coutput(i),abs(coutput(i)-cinput(i))
end do
nerrors=0
do i=1,n
    !write(*,'(i2,5(1x,2(f8.4),1x,2(f8.4),2x,3(e13.7,2x)))') i,cinput(i,1),coutput(i,1),abs(coutput(i,1)-cinput(i,1)),abs(coutput(i,6)-cinput(i,6)),abs(coutput(i,nwfs)-cinput(i,nwfs))
    do j=1,nwfs
        if (abs(coutput(i,j)-cinput(i,j))>1.d-5) then
            write(*,'(i3,i3,1x,e13.7,2x,4(f8.4))') i,j,abs(coutput(i,j)-cinput(i,j)),cinput(i,j),coutput(i,j)
            nerrors = nerrors + 1
        end if
    end do
end do
elapsed_time = REAL(clock_end-clock_start)/REAL(clock_rate)

write(*,*) 'elapsed_time :',elapsed_time,clock_start,clock_end,clock_rate
if (nerrors .eq. 0) then
   print *, "Test Passed"
else
   print *, "Test Failed"
endif

!release memory on the host and on the device
deallocate (cinput,coutput,kx,cinput_d,coutput_d)

! Destroy the plans
call cufftDestroy(plan)

end program fft

有人能告诉我为什么以下 "many-FFT" 有时在双精度上失败但在单精度上从不失败吗？

单精度："Test Passed"总是！双精度："Test Failed" 有时！

确实，我检查了设备到主机的数据传输。好像不是。

感谢您的帮助。

Answer 1

多亏了利爪。这是 WDDM 超时检测和恢复限制。

见link，到change the TDR

Fortran 中使用 PlanMany 的 cuFFT 双精度误差

Double precision error of cuFFT with PlanMany in Fortran

fortran

cuda

cufft