On Tue, 11 Mar 2003, Sarad AV wrote: > > Taking v=3 bit accuracy,the 3 leading bits are > > 000 > 100 > 110 > 111 > > In the example k=3 and v=3 > > So according to definition there are 2^(kv) possible > combinations of bits occur the same > number of times in a period. > i.e 2^(3*3)=512 combinations. > > But where are the 512 combinations. > > We are choosing 3 bits out of 4 bits > hence C(4,3)=4!/(3!*1!)=4 > There are k=4,hence total combinations arising is only > 4*4=16. > > where did i go wrong?
The order of the blocks. You have v=3 bits and k=3 blocks, so 9 bits total. In a block of 9 bits there are 512 possible combinations. The order of each block matters (in this example). > Also why is k-distribution considered as a strong test > for randomness? It's a useful test, also called "chi^2". It's been applied to feedback shift registers for a long time, so tradition is now part of the reason :-) Patience, persistence, truth, Dr. mike
