One or both of us has got hold of the wrong end of some sticks here;-( Brian Bessley wrote:
> But in any case it's a moot point because the (input) "phase space" values > aren't integers, The checksum refers to input FFT values (not the "phase space" inputs to the IFFT). > and adding (FFT run length) strictly positive values > together is likely to cause a loss of precision even if they were. (Run > length = 2^20 is likely to lose about 20 bits of precision in the sum, > compared with the precision of the inputs, if you're thinking in fixed point > terms. Now have mentioned a simple example of error combination which I wanted to use, namely 2^20 irrational numbers which should sum to an integer. With 20 binary places, the maximum error in the representation of each number is ~2^-20. If all the errors were the same sign then our "integer" result may be in jeapardy. But this is not the case, and random walk/Central Limit Theorem arguments tell us that the total arror is normally distributed about zero with standard deviation 2^-10. (The variance of the sum is the sum of the variances). BTW if we are worried about the integer overflowing, we can always "balance" the representaion by subtracting 2^power as described before. > When working in floating point, you won't actually get a wraparound > type error like integer overflow, but you do lose the least significant bit > of the mantissa every time you add two values with the same sign and the same > exponent as the exponent of the result has to increase.) This is why fixed point is the natural method for addition, while floating point is appropriate for multiplication. > > In fact, as the FFT run length increases, the checksum criterion becomes > increasingly poor as a predictor of accuracy compared with the roundoff error > check criterion, as the imprecision of the checksum is proportional to the > logarithm of the number of elements in the computation whereas the roundoff > error increases much more slowly. To use the checksum, it seems obvious to use as much precision as necessary to obtain an exact integer. > > Also, the "special features" of the transform used for the L-L test give us > the "subtract 2" part of the L-L algorithm for free; this may interfere in > some way with the exact equality mentioned in the basic theory. Both the subtract 2 and modulo operations may be trivially performed in "real space". David Eddy _________________________________________________________________ Be one of the first to try Windows Live Mail. http://ideas.live.com/programpage.aspx?versionId=5d21c51a-b161-4314-9b0e-4911fb2b2e6d _______________________________________________ Prime mailing list [email protected] http://hogranch.com/mailman/listinfo/prime
