>>> Speaking of Q2.6, I've heard that with Crandall's DWT, the subtraction >>> 2 step costs nothing at all. It's done automatically within the >>> transformation. Try checking this with George Woltman. >>Is this true? >Not knowing for certain; I thought the DWT did the modulo for you, not the >subtraction? Yes, it does the mod. I read in the archives that x^2-2 could be interpreted as a "delta polynomial" which would give it a valid Fourier transform. I was wondering if this was what was actually performed. I doubt it is as optimizing the -2 step would hardly produce amazing results. Optimizing the -2 step down to nothing (with the amazingly high estimate of 1/50,000th of a sec) would gain only .3 CPU years for all exponents in the 7million range. I was just making sure. -Lucas Wiman ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
