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There is given sequence $a_1,...a_n$ such that there are $O(n^{\frac{3}{2}}) $ inversions in this sequence. I am thinking about sorting algorithm for that.

I know lower bound for number of comparisons - it is $O(n)$ - on the contrary, there would be a minimum finding algorithm faster than $O(n)$.

Nevertheless, I don't have idea how sort it in linear time ? What doy you think ?

Inversion is a pair $(i, j)$ such that $i < j$ and $a_i > a_j$

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  • $\begingroup$ What is an inversion? $\endgroup$
    – usul
    Commented Oct 6, 2015 at 14:16
  • $\begingroup$ inversion is a pair $(i, j)$ such that $i < j$ and $a_i > a_j$ $\endgroup$
    – user40545
    Commented Oct 6, 2015 at 14:17
  • $\begingroup$ Hmm, can you expand on the consequences of this $n^{\frac{3}{2}}$ upper bound? It seems like a large part of the sequence must already be sorted? $\endgroup$
    – usul
    Commented Oct 6, 2015 at 15:38
  • $\begingroup$ Yes, intuiton is that large part is arleady sorted. However I don't know how to use this fact. $\endgroup$
    – user40545
    Commented Oct 6, 2015 at 16:47

2 Answers 2

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You can't sort it in linear time.

Suppose you have $n$ items, and you divide them into $\sqrt{n}$ consecutive blocks of $\sqrt{n}$ items each.

You need to take $\sqrt{n} \log \sqrt{n}$ comparisons to sort each one. And there are $\sqrt{n}$ of them, giving $\theta(n \log n)$ time total. And it's easy to see that there can't be more than $n^{3/2}$ inversions in the sequence, since there can't be more than $n$ inversions in each subsequence.

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  • $\begingroup$ Thanks for reply. However, I don't understand your algorithm. Tell more please. And what's is lower bound ? $\endgroup$
    – user40545
    Commented Oct 6, 2015 at 21:38
  • $\begingroup$ @user40545 : $\:$ He didn't give an algorithm. $\;\;\;\;$ $\endgroup$
    – user6973
    Commented Oct 6, 2015 at 22:13
  • $\begingroup$ I am not sure if I understand it correctly. You shown that it impossible to sort faster than $\Omega(n \log n)$. yes ? $\endgroup$
    – user40545
    Commented Oct 7, 2015 at 14:05
  • $\begingroup$ That's right. Even if there are only $n^{3/2}$ inversions, you can't sort using fewer than $\frac{1}{2} n \log n$ comparisons. $\endgroup$ Commented Oct 7, 2015 at 17:10
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    $\begingroup$ @user40545 No, it's not. If you had $\Theta$ on both side, or $O$ on both sides, yes -- but what you wrote, with the $O(\cdot)$ and one side and the $\Theta(\cdot)$ on the other is wrong. $\endgroup$
    – Clement C.
    Commented Oct 8, 2015 at 13:57
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This is a topic of "adaptive sorting." As a starter, see the wikipedia page https://en.wikipedia.org/wiki/Adaptive_sort .

It is known that we can sort a sequence of length $n$ with $k$ inversions with $O(n \log (2+k/n))$ comparisons. When $k=O(n^{3/2})$, this translates to $O(n \log n)$. We also have a lower bound of $\Omega(n \log (2+k/n))$. Thus, we know a linear bound is impossible.

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    $\begingroup$ I took the liberty to fix the bound. The original implied that sorting can be done with $O(k)$ comparisons, which is easily seen to be impossible for $1\le k\ll n$. $\endgroup$ Commented Oct 8, 2015 at 13:37

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