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#1 2013-10-21 10:13:13

White_Owl
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Number of changes

We 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 000-111).
In another number, each digit inverts on random.
Each inversion requires some work to be done. So for the first number, in 10 steps we require 3*10=30 energy.
If none of the digits in the second number was inverted (this can happen), then in the same 10 steps we require 0*10=0 energy. If all of the digits in the second number was inverted (this also can happen), then we will require 3*10=30 energy.
But since the second number is under randomizer we will usually have less energy spent on it in comparison with the first number.
The question is: what is the average reduction of energy which would be required by the second number in comparison with the first one?

My current train of thought is:

There are four possible cases:
0 inverted - 2^3 combinations
1 inverted - (2^3)^3 combinations
2 inverted - (2^3)^3 combinations
3 inverted - 2^3 combinations

At each step one of those four cases can happen.
So the probability to spend get any single from combination A to combination B is

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 2013-10-21 10:18:37

White_Owl
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Re: Number of changes

After 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.
Is this correct or am I oversimplifying it?

#3 2013-10-21 11:54:35

anonimnystefy
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Re: Number of changes

That is correct. The random one will on average require half as much energy as the first one.

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