On Wed, 14 Jul 1999, Lucas Wiman wrote:
All,
I recieved a message pointing out a possible error in my FAQ:
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
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
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?
All,
I recieved a message pointing out a possible error in my FAQ:
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?
-Lucas
