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You are not logged in. #1 20131021 10:13:13
Number of changesWe have two three digit binary numbers. In one number all digits do invert every single step (does not matter what the initial value was, lets say we have a 000111). We have probability to spend 0 energy at one step We have probability to spend 1 energy at one step same for 2 and 3. So on average we supposed to get In other words, the second number will spend 1.5 energy to invert random number of bits and on average we will spend 1.5*10=15 energy on the second number. Is this correct? #2 20131021 10:18:37
Re: Number of changesAfter a second look at all these... I have a feeling that on average, we would have exactly half of the bits inverted. So we can say that if we increase number of bits, on average, the randomized inversion will require half the power than full inversion. #3 20131021 11:54:35
Re: Number of changesThat is correct. The random one will on average require half as much energy as the first one. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment 