At 23:26 14.07.99 -0400, Lucas Wiman wrote:
>> 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?

/* Steinar */


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