> > Jason writes: > > > > > > Without balanced representation, if you have a transform of size N > > > and each element has B bits, then each element of the product must > > > have room for an integer with 2*B*log2(N) bits. Ordinary double-precision > > > floating point has a 52-bit mantissa, so ignoring roundoff error > > > B and N have limits to avoid getting a corrupted answer. > > > > B=20 , N=1024K, log2(N)=20 and you are asking for 800 bits??? > > Grr, that was a mistake. It should be 2*B+log2(N) and 2*(B-2)+log2(N) > for ordinary and balanced representation, respectively.
56 bits is nearer the mark! I assume statistics along the lines I suggested (normal distribution) explains the 4 bit discrpancy. Or is it the varying integer length of a weighted transform? David _________________________________________________________________ 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
