"क्रमबद्धता" कैसे मापें

Question

मैं सोच रहा हूँ कि क्या किसी सरणी के "क्रमबद्धता" को मापने का एक मानक तरीका है? एक सरणी जो संभव व्युत्क्रमों की औसत संख्या है, को अधिकतम रूप से अनसोल्ड माना जाएगा? मेरा मतलब है कि यह मूल रूप से जहाँ तक संभव हो या तो छँटा हुआ या उल्टा छँटा हुआ है।

Answer 1

नहीं, यह आपके आवेदन पर निर्भर करता है। छँटाई के उपायों को अक्सर विकार के उपायों के रूप में संदर्भित किया जाता है , जो से आर , जहां एन < $N^{<N}$$\mathbb{R}$$N^{<N}$ अलग-अलग nonnegative पूर्णांकों के सभी परिमित अनुक्रमों का संग्रह है। एस्टिविल-कास्त्रो और वुड द्वारा सर्वेक्षण [1] अनुकूली सॉर्ट एल्गोरिदम के संदर्भ में विकार के 11 विभिन्न उपायों को सूचीबद्ध करता है और उनकी चर्चा करता है।

आक्रमण की संख्या कुछ मामलों के लिए काम कर सकती है, लेकिन कभी-कभी अपर्याप्त होती है। [1] में दिया गया एक उदाहरण अनुक्रम है

$$\langle \lfloor n/2 \rfloor + 1, \lfloor n/2 \rfloor + 2, \ldots, n, 1, \ldots, \lfloor n/2 \rfloor \rangle$$

that has a quadratic number of inversions, but only consists of two ascending runs. It is nearly sorted, but this is not captured by inversions.

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[1] Estivill-Castro, Vladmir, and Derick Wood. "A survey of adaptive sorting algorithms." ACM Computing Surveys (CSUR) 24.4 (1992): 441-476.

Comments

The context is trying to understand why quicksort performs relatively poorly on random permutations of n elements where the number of inversions is close to the median.

— Robert S. Barnes

Great example, that's exactly the info I was looking for.

— Robert S. Barnes

Estivill-Castro and Wood is THE reference for this for sure.

— Pedro Dusso

Answer 2

Mannila [1] axiomatizes presortedness (with a focus on comparison-based algorithms) as follows (paraphrasing).

Let $\Sigma$ a totally ordered set. Then a mapping $m$ from $\Sigma^{\star}$ (the sequences of distinct elements from $\Sigma$) to the naturals is a measure of presortedness if it satisfies below conditions.

1. If $X \in \Sigma^{\star}$ is sorted then $m(X) = 0$.

2. If $X,Y \in \Sigma^{\star}$ with $X = x_1 \dots x_n$, $Y = y_1 \dots y_n$ and $x_i < x_i \iff y_i < y_j$ for all $i,j \in [1..n]$, then $m(X) = m(Y)$.

3. If $X$ is a subsequence of $Y \in \Sigma^{\star}$, then $m(X) \leq m(Y)$.

4. If $x_i < y_j$ for all $i \in [1..|X|]$ and $j \in [1..|Y|]$ for some $X,Y \in \Sigma^{\star}$, then $m(X \cdot Y) \leq m(X) + m(Y)$.

5. $m(a \cdot X) \leq |X| + m(X)$ for all $X \in \Sigma^{\star}$ and $a \in E \setminus X$.

Examples of such measures are the

• number of inversions,
• number of swaps,
• the number of elements that are not left-to-right maxima, and
• the length of a longest increasing subsequence (subtracted from the input length).

Note that random distributions using these measures have been defined, i.e. such that make sequences that are more/less sorted more or less likely. These are called Ewens-like distributions [2, Ch. 4-5; 3, Example 12; 4], a special case of which is the so-called Mallows distribution. The weights are parametric in a constant $\theta > 0$ and fulfill

$\qquad\displaystyle \operatorname{Pr}(X) = \frac{\theta^{\,m(X)}}{\sum_{Y \in \Sigma^{\star} \cap \Sigma^{|X|}} \theta^{\,m(Y)}}$.

Note how $\theta = 1$ defines the uniform distribution (for all $m$).

Since it is possible to sample permutations w.r.t. these measures efficiently, this body of work can be useful in practice when benchmarking sorting algorithms.

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1. Measures of Presortedness and Optimal Sorting Algorithms by H. Mannila (1985)
2. Logarithmic combinatorial structures: a probabilistic approach by R. Arratia, A.D. Barbour and S. Tavaré (2003)
3. On adding a list of numbers (and other one-dependent determinantal processes) by A. Borodin, P. Diaconis and J. Fulman (2010)
4. Ewens-like distributions and Analysis of Algorithms by N. Auger et al. (2016)

Answer 3

I have my own definition of "sortedness" of a sequence.

Given any sequence [a,b,c,…] we compare it with the sorted sequence containing the same elements, count number of matches and divide it by the number of elements in the sequence.

For example, given sequence [5,1,2,3,4] we proceed as follows:

1) sort the sequence: [1,2,3,4,5]

2) compare the sorted sequence with the original by moving it one position at a time and counting the maximal number of matches:

        [5,1,2,3,4]
[1,2,3,4,5]                            one match

        [5,1,2,3,4]
  [1,2,3,4,5]                          no matches

        [5,1,2,3,4]
    [1,2,3,4,5]                        no matches

        [5,1,2,3,4]
      [1,2,3,4,5]                      no matches

        [5,1,2,3,4]
        [1,2,3,4,5]                    no matches

        [5,1,2,3,4]
          [1,2,3,4,5]                  4 matches

        [5,1,2,3,4]
            [1,2,3,4,5]                no matches

                ...

         [5,1,2,3,4]
                 [1,2,3,4,5]            no matches

3) The maximal number of matches is 4, we can calculate the "sortedness" as 4/5 = 0.8.

Sortedness of a sorted sequence would be 1, and sortedness of a sequence with elements placed in reversed order would be 1/n.

The idea behind this definition is to estimate the minimal amount of work we would need to do to convert any sequence to the sorted sequence. In the example above we need to move just one element, the 5 (there are many ways, but moving 5 is the most efficient). When the elements would be placed in reversed order, we would need to move 4 elements. And when the sequence were sorted, no work is needed.

I hope my definition makes sense.

Comments

Nice idea. A similar definition is Exc, the third definition of disorder in the paper mentioned in Juho's answer. Exc is the number of operations required to rearrange a sequence into sorted order.

— Apass.Jack

Well, may be, I just applied my understanding of entropy and disorder to the sequence of elements :-)

— Andrushenko Alexander

Answer 4

If you need something quick and dirty (summation signs scare me) I wrote a super easy disorder function in C++ for a Class named Array which generates int arrays filled with randomly generated numbers:

void Array::disorder() {
    double disorderValue = 0;
    int counter = this->arraySize;
    for (int n = 0; n < this->arraySize; n++) {
        disorderValue += abs(((n + 1) - array[n]));
//      cout << "disorderValue variable test value = " << disorderValue << endl;
        counter++;
    }
    cout << "Disorder Value = " << (disorderValue / this->arraySize) / (this->arraySize / 2) << "\n" << endl;
}

Function simply compares the value in each element to the index of the element + 1 so that an array in reverse order has a disorder value of 1, and a sorted array has a disorder value of 0. Not sophisticated, but working.

Michael

Comments

This is not a programming site. It would have sufficed to define the disorder notion, and to mention that it can be computed in linear time.

— Yuval Filmus