So, for more clarity, my algorithm would segment the following bit pattern....
00000000000000000000000000010010110011010000000000000000000000000 ...into this: 000000000000000000000000000 ---> log2(27) = ~4.754 1 ---> 1 00 ---> 1 1 ---> 1 0 ---> 1 11 ---> 1 00 ---> 1 11 ---> 1 0 ---> 1 1 ---> 1 0000000000000000000000000 ---> log2(25) = ~4.64 ...yielding a total entropy esimate of 18.4 bits. There's no 'human pattern recognition skills' involved, it's fully algorithmic. Sorry if I was not clear. The fact that the central part has high entropy as a whole only follows from the fact that there are many transitioning bits there, and those bits have an entropy of precisely one (since you cannot express a transitioning bit as a smaller number of bits). -- dupswapdrop -- the music-dsp mailing list and website: subscription info, FAQ, source code archive, list archive, book reviews, dsp links http://music.columbia.edu/cmc/music-dsp http://music.columbia.edu/mailman/listinfo/music-dsp
