Regarding binary counter amortized analysis

问题

This is regarding amortized analysis of binary counter. All the entries in the Array start at 0 and at each step we will be simply incrementing the counter.

Here author is mentioning as below.

We are using a potential function that is equal to the number of 1-bits in the current count.

Question 1: What does above statement mean?

And other question i have is in the analysis it is mentioned as

n+(n/2)+(n/4)+--- is atmost 2n. how we got result as atmost 2n?

Thanks!

回答1:

The costs for incrementing in this example are the number of bit-flips to execute.

I.e. incrementing 3 (0b11) to 4 (0b100) has cost of 3 (all three positions flipped).

Now with that you couldn't say, that the algorithm is amortized constant, because the amount of time depends on the number of bit-flips and thus varies on the number.

To work around that one use the potential method on a sequence of increment operations starting with 0. The potential is now the number of bits that are 1.

φ(0) = 0
φ(1) = 1
φ(2) = 1
φ(3) = 2
φ(4) = 1 etc

This makes sense, since for each bit that is 1, an increment operation in the future will have to change it to 0 at some point. So whenever in increment occurs that needs to flip more than the last bit, it makes use of the potential of the value.

Now continuing the amortized analysis, you will find, that the potential always increases by 1 and in each increment operation you decrease the potential for every 1 bit you flipped. Combining this in each operation you have costs of 2: a) flipping a 0 to a 1, b) saving 1 for the potential. Flipping all the ones before the 0 is paid using the potential.

See also: http://en.wikipedia.org/wiki/Potential_method

回答2:

This is quite a common question and you can see an example of such here in PDF explained quite good in terms of cost.

It is easier to understand from the PDF with figure but all it means is that we can use a separate function which corresponds to the smallest unit that is a bit in this case and make it equivalent to some sensible value coming out of the function for example 1-bit flip in pdf is equivalent to $1. Potential function just makes it easy to perform analysis. It starts at 0 and should be non-negative.

回答3:

Just use geometric progression where a = 1 and r = 1/2. The sum of geometric progression is

     a(1-r^{n})/(1-r).

    Here it is (1-(1/2)^{n})/(1-(1/2)) = 2*(1- (1/2)^n).
As n goes to infinity, it becomes 2.

来源：https://stackoverflow.com/questions/7319377/regarding-binary-counter-amortized-analysis

标签

algorithm

analysis