Equivalence testing of n objects

问题

Assume that we are given 'n' objects and a subroutine that takes two inputs and says if they are equivalent or not (e.g. it can give output as 1 if they are equal).

I need to come up with an algorithm that calls the above function O(n log n) times and decides if the input has more than 'n/2' items that are equivalent to each other.

回答1:

You can use the Boyer-Moore majority voting algorithm, which does at most n-1 comparisons.

function find_majority(A)
    majority = None
    count = 0
    for a in A:
        if count == 0
            majority = a
        else if a == majority
            count += 1
        else
            count -= 1
    return majority

It returns the most common element if appears more than n/2 times in the array.

If you need to know if there is a majority item, then you can make a second pass through the array counting how many times the value returned from the find_majority function appears. This adds another n comparisons.

回答2:

Here is a classical divide and conquer solution which gives O(n log n)

split into two subsets, A1 and A2, ..., and show T(n) is O(n log n). If A has a majority element v, v must also be a majority element of A1 or A2 or both. The equivalent contra-positive restatement is immediate: (If v is <= half in each, it is <= half in the total.) If both parts have the same majority element, it is automatically the majority element for A. If one of the parts has a majority element, count the number of repetitions of that element in both parts (in O(n) time) to see if it is a majority element. If both parts have a majority, you may need to do this count for each of the two candidates, still O(n). This splitting can be done recursively. The base case is when n = 1. A recurrence relation is T(n) = 2T(n/2) + O(n), so T(n) is O(n log n) by the Master Theorem.

http://anh.cs.luc.edu/363/handouts/MajorityProblem.pdf

回答3:

Paul Hankin's answer is a $\mathcal{O}(n)$ solution and very fancy.

If do not use Boyer-Moore, we can also solve this problem in $\mathcal{O}(nlogn)$. Every time after comparision, we "drop" the different pair of cards and "merge" the equal pair to a group, and so on, every time we drop the same number of cards, the final left card or card group is the "Majority", then we can to run the comparision again to check is card is True or not.

As every time we drop, we at least drop a majority card and another, so it will not influence the result, the process of drop is a "offset" operation. Not matter drop or merge, help us reduce at least half of the card. So the recursive times is $logn$, the total time is $nlogn$.

回答4:

Considering that the sequence is ordered you can use binary search, which takes O(log n), and since you'll have to do it for every element, and you have n elements it will take O(n*log n).

来源：https://stackoverflow.com/questions/26157855/equivalence-testing-of-n-objects