On Tuesday 28 May 2002 02:43, you wrote: > 6 is -2 mod 8 > 6*6 = 36 > 36 = -4 mod 8 > 2^2 = 4 > > if the mod of the represented as a negative is much less than the positive, > could we square the negative and save some time ?
Sure we could. However on average we would save 1 bit 25% of the time, 2 bits 12.5% of the time, 3 bits 6.25% of the time .... on average the saving is 0.5 bits. Out of several million. In fact the _practical_ saving is zero, as we aren't ever going to save enough bits to justify using a shorter FFT run length, even just for one iteration. Regards Brian Beesley _________________________________________________________________________ Unsubscribe & list info -- http://www.ndatech.com/mersenne/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
