Calculate only first half of IFFT

When calculating the inverse fourier transform using IFFT on a data set where all phases are 0 you obtain a waveform that is symmetrical - the first half of the waveform is the same as the second half mirrored and inverted.

Thus it would be trivial to calculate the second half, given the first half.

Given these conditions, is it possible to calculate IFFT in a way that generates only the first half of the output data in order to reduce computation time?
JasonMewesAsked:
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ozoCommented:
Are you now calculating it in a way that prevents you from generating only the first half?
Can you do only a sin transform and ignore cosine?
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JasonMewesAuthor Commented:
I am using an implementation very similar to https://github.com/xcore/sc_dsp_transforms/blob/master/module_fft_simple/src/fftcomplex.xc

I cannot see a straight forward way to modify the algorithm for the purpose described, but given your response I assume it should be possible.
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ozoCommented:
It is possible.  Whether a given implementation offers such a possibility depends on the implementation.
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JasonMewesAuthor Commented:
I understand.

Question as stated is considered answered.

Any chance though you would consider looking at the linked algorithm and give me a pointer in the right direction (knowing macs is multiply accumulate)?
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