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May 6, 2010 at 22:52 comment added fedja You can lose the top 1 and move one step back getting a new bit on the bottom whose value is very unlikely to be determined by the current state alone even in the sense of Markov chains. On the other hand, most of the time you go forward at linear speed thus making any attempts to play with the bottom bits quite inefficient for controlling the distribution of the top ones. I have put this question on my list of things to think of but, alas, my free time right now is very limited :-(.
May 6, 2010 at 0:25 comment added Did Nice problem. You write that "This vector isn't by itself a Markov chain": why?
May 4, 2010 at 19:08 answer added Alekk timeline score: 4
May 4, 2010 at 17:58 history asked passing by CC BY-SA 2.5