Algorithm for rating the monotonicity of an array

问题:

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EDIT: Wow, many great responses. Yes, I am using this as a fitness function for judging the quality of a sort performed by a genetic algorithm. So cost-of-evaluation is important (i.e., it has to be fast, preferably O(n).)

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As part of an AI application I am toying with, I'd like to be able to rate a candidate array of integers based on its monotonicity, aka its "sortedness". At the moment, I'm using a heuristic that calculates the longest sorted run, and then divides that by the length of the array:

public double monotonicity(int[] array) {
    if (array.length == 0) return 1d;

    int longestRun = longestSortedRun(array);
    return (double) longestRun / (double) array.length;
}

public int longestSortedRun(int[] array) {

    if (array.length == 0) return 0;

    int longestRun = 1;
    int currentRun = 1;

    for (int i = 1; i < array.length; i++) {
        if (array[i] >= array[i - 1]) {
            currentRun++;
        } else {
            currentRun = 1;
        }

        if (currentRun > longestRun) longestRun = currentRun;
    }

    return longestRun;
}

This is a good start, but it fails to take into account the possibility that there may be "clumps" of sorted sub-sequences. E.g.:

{ 4, 5, 6, 0, 1, 2, 3, 7, 8, 9}

This array is partitioned into three sorted sub-sequences. My algorithm will rate it as only 40% sorted, but intuitively, it should get a higher score than that. Is there a standard algorithm for this sort of thing?

回答1:

I expect that the choice of function to use depends very strongly on what you intend to use it for. Based on your question, I would guess that you are using a genetic system to create a sorting program, and this is to be the ranking function. If that is the case, then speed of execution is crucial. Based on that, I bet your longest-sorted-subsequence algorithm would work pretty well. That sounds like it should define fitness pretty well.

回答2:

This seems like a good candidate for Levenshtein Damerau–Levenshtein distance - the number of swaps needed to sort the array. This should be proportional to how far each item is from where it should be in a sorted array.

Here's a simple ruby algorithm that sums the squares of the distances. It seems a good measure of sortedness - the result gets smaller every time two out-of-order elements are swapped.

ap = a.sort
sum = 0
a.each_index{|i| j = ap.index(a[i])-i 
  sum += (j*j)
}
dist = sum/(a.size*a.size)

回答3:

Something like these? http://en.wikipedia.org/wiki/Rank_correlation

回答4:

Here's one I just made up.

For each pair of adjacent values, calculate the numeric difference between them. If the second is greater than or equal to the first, add that to the sorted total, otherwise add to the unsorted total. When done, take the ratio of the two.

回答5:

Compute the lenghts of all sorted sub-sequences, then square them and add them. If you want to calibrate how much enphasis you put on largest, use a power different than 2.

I'm not sure what's the best way to normalize this by length, maybe divide it per length squared?

回答6:

What you're probably looking for is Kendall Tau. It's a one-to-one function of the bubble sort distance between two arrays. To test whether an array is "almost sorted", compute its Kendall Tau against a sorted array.

回答7:

I would suggest looking at the Pancake Problem and the reversal distance of the permutations. These algorithms are often used to find the distance between two permutations (the Identity and the permuted string). This distance measure should take into account more clumps of in order values, as well as reversals (monotonically decreasing instead of increasing subsequences). There are also approximations that are polynomial time[PDF].

It really all depends on what the number means and if this distance function makes sense in your context though.

回答8:

I have the same problem (monotonicity scoring), and I suggest you to try Longest Increasing Subsequence. The most efficient algorithm run in O(n log n), not so bad.

Taking example from the question, the longest increasing sequence of {4, 5, 6, 0, 1, 2, 3, 7, 8, 9} is {0, 1, 2, 3, 7, 8, 9} (length of 7). Maybe it rate better (70%) than your longest-sorted-run algorithm.

回答9:

It highly depends on what you're intending to use the measure for, but one easy way to do this is to feed the array into a standard sorting algorithm and measure how many operations (swaps and/or comparisons) need to be done to sort the array.

回答10:

Some experiments with a modifier Ratcliff & Obershelp

>>> from difflib import SequenceMatcher as sm
>>> a = [ 4, 5, 6, 0, 1, 2, 3, 7, 8, 9 ]
>>> c = [ 0, 1, 9, 2, 8, 3, 6, 4, 7, 5 ]
>>> b = [ 4, 5, 6, 0, 1, 2, 3, 7, 8, 9 ]
>>> b.sort()
>>> s = sm(None, a, b)
>>> s.ratio()
0.69999999999999996
>>> s2 = sm(None, c, b)
>>> s2.ratio()
0.29999999999999999

So kind of does what it needs to. Not too sure how to prove it though.

回答11:

How about counting the number of steps with increasing value vs. the number of total steps. That's O(n).