Average Case Time Complexity Analysis of Binary Counter Increment

问题

I am attempting to find the average case time complexity, not amortized analysis, of a binary counter. As I am not entirely confident in my time complexity analyzing skills, I would like to confirm that my average case analysis of the pseudocode provided below is correct.

Let k be the length of the array.

Increment(Array)
    i = 0
    while i < k and Array[i] == 1
        Array[i] = o
        i = i + 1
    if i < k
        Array[i] = 1

In order to find the average time taken, I find the average amount of bits flipped per run. As a result, I found this to be O(2+k/(2^k)), which equals O(1) for a large k.

Is this the correct average case running time? If not, how would I begin to approach this problem?

回答1:

I am assuming each input has the same probability to occur

This means that each bit is independently on or off with probability 1/2.

The geometric distribution is the relevant distribution for the complexity: you flip coins, and end the experiment on the first tail outcome (there is nothing further to carry).

The mean of the geometric distribution here is exactly 2 (see above link, or derive it from basic principles), so the average complexity is indeed O(1).

来源：https://stackoverflow.com/questions/40570866/average-case-time-complexity-analysis-of-binary-counter-increment