Intel® oneAPI Math Kernel Library
Ask questions and share information with other developers who use Intel® Math Kernel Library.
Announcements
This community is designed for sharing of public information. Please do not share Intel or third-party confidential information here.
6590 Discussions

## Incosistent in-place and out-of-place results in mkl_dfti

Beginner
144 Views

Hi,

I am trying to transfer from out-of-place calculation to in-place using MKL 10.3.12. I don't see any problem when doing forward FFT. However in backward FFT, for dimensions larger than 4, I get inconsistent results for in-place and out-of-place calculations. This happens when I use backward scaling of 1.0 (which I need in my problem), and the issue is resolved when scaling of 1/(K1*K2*K3) is used instead!. I have attached a minimal code for reproducing the results. I compiled it with:

`gfortran -fcray-pointer -I\$myMKLINC main.f90 -L\$MKLROOT/lib/intel64/ -L/opt/intel/composer_xe_2011_sp1.12.361/compiler/lib/intel64 -lmkl_intel_lp64 -lmkl_core -lmkl_sequential -lpthread -liomp5 -o amain`

Thanks for any help

Amir

1 Solution
Employee
144 Views

Amir,

Backward C->R transform expects the input is conjugate-even symmetric. In your case it is not. Please make sure the input is CCE.

For instance, add the following fragment right after definition of C_matip

```! SYMMETRIZE INPUT
do k = 0, K3-1
do j = 0, K2-1
do i = 0, K1/2
if (i == mod(K1-i,K1)) then
c1 = C_matip(i+(K1/2+1)*j+(K1/2+1)*K2*k)
c2 = C_matip(i+(K1/2+1)*mod(K2-j,K2)+(K1/2+1)*K2*mod(K3-k,K3))
if (c1 /= conjg(c2)) then
c3 = 0.5*(c1 + conjg(c2))
print '("Symmetrizing",3(I3),"("2G12.4")","("2G12.4")"," -> ","("2G12.4")")',i,j,k,c1,c2,c3
C_matip((i+(K1/2+1)*j+(K1/2+1)*K2*k)) = c3
C_matip(i+(K1/2+1)*mod(K2-j,K2)+(K1/2+1)*K2*mod(K3-k,K3)) = conjg(c3)
endif
endif
enddo
enddo
enddo
```

Thanks
Dima

3 Replies
Employee
144 Views

Thanks for providing a test code. We could reproduce this issue. It needs to be fixed. You'll be notified when a fix is available.

Employee
145 Views

Amir,

Backward C->R transform expects the input is conjugate-even symmetric. In your case it is not. Please make sure the input is CCE.

For instance, add the following fragment right after definition of C_matip

```! SYMMETRIZE INPUT
do k = 0, K3-1
do j = 0, K2-1
do i = 0, K1/2
if (i == mod(K1-i,K1)) then
c1 = C_matip(i+(K1/2+1)*j+(K1/2+1)*K2*k)
c2 = C_matip(i+(K1/2+1)*mod(K2-j,K2)+(K1/2+1)*K2*mod(K3-k,K3))
if (c1 /= conjg(c2)) then
c3 = 0.5*(c1 + conjg(c2))
print '("Symmetrizing",3(I3),"("2G12.4")","("2G12.4")"," -> ","("2G12.4")")',i,j,k,c1,c2,c3
C_matip((i+(K1/2+1)*j+(K1/2+1)*K2*k)) = c3
C_matip(i+(K1/2+1)*mod(K2-j,K2)+(K1/2+1)*K2*mod(K3-k,K3)) = conjg(c3)
endif
endif
enddo
enddo
enddo
```

Thanks
Dima

Beginner
144 Views

Thank you so much for quick replies.

First I did not know that besides the CCE storage for the results of R->C, the conjugate symmetry of the stored elements will also be checked which seems to be particularly important for i=0 and i=K1/2 when K1 is even.

The array that I used as input was in fact the resulting array after some manipulation of the result of the forward R->C transform. The reason that the symmetry in some of the elements , i.e. the elements with i=K1/2, was breaking was directly related to that manipulation in case of K1 even (which I was expecting it from the mathematical point of view). In practice, the transform in this case becomes a C->C transform, with the exception that the CCE storage (the fact that the elements that are not stored can be obtained by conjugate symmetry of the stored elements) is still valid! Interestingly, after performing the C->C backward, I noticed that the C->C and in-place C->R give identical real parts. In general, for my problem, the real part of the results were important. Nevertheless, the very nice suggestion of @Dmitry guaranteed both the C->R transform and the correct real parts.

Many thanks

Amir