Now in the worst case for the third loop, every a in A has a match in B, so the third loop will be called every time. And later in another piece of code you know you need the list to be sorted, so you sort it again. so the problem occure only when When P is uniform, the pdf of P is 1. The actual question was: "When can such a scenario (almost sorted) occur with natural input?". Bentley and McIlroy, authors of the C standard library's qsort function, wrote in their paper, Engineering a Sort Function the following numbers: Our final code therefore chooses the middle element of smaller arrays, Note: I know, that this did not give more example of real world occasions for quicksort worst cases. The answer depends on strategy for choosing pivot. For regular randomized quicksort Wikipedia lists the actual runtime of randomized quicksort as requiring at most 1.39 n lg n comparisons (using lg as the binary logarithm). So, if you suspect your scenario is likely to a bad case scenario for quicksort, you can simply change your choice of pivot element to make quicksort perform better. array, and the pseudo-median of nine evenly spaced elements of a large Does quicksort with randomized median-of-three do appreciably better than randomized quicksort? I wouldn't be surprised if just picking the first pivot at random would change the worst case expected runtime to o (n^2) (little o instead of Big O), because only a few out of n pivots will lead to worst case behaviour. polygene has provided some interesting information on alternate pivoting considerations. Based on my comments below, I guess k = O(nα) for 0 < α < 1 is the "sweet spot". Natural ordering: This is similar to the chronoloigcal data. corresponds to already sorted. Sorting algorithms with a worst case performance of O(n^2) Worst Case. In other words, c is the reciprocal of the expected entropy of the Bernoulli distribution with mean P. Intuitively, for each element, we need to compare it to pivots in a way that yields about lg n bits of information. It's well-known that a quicksort implementation that picks its pivots uniformly at random will end up running in expected O(n lg n) time (there's a nice proof of this on Wikipedia). The worst case occurs when the picked pivot is always an extreme (smallest or largest) element. Algorithms for quicksort, "worst case" you choose the middle element of the list, an already sorted list does not have the worst case runtime. On each level of the tree, the time to “go through” all nodes is O(N). For quicksort the average complexity is nlogn and worst case is n^2. the median of the first, middle and last elements of a mid-sized I think it can be made rigorous, with a lot more effort. algorithm - three - why is quicksort worst case n 2 . Since you're not guaranteed it's sorted - you have to re-sort. Naive pivot selection can do that on a pre-sorted list. If, e.g. The total added cost of computing medians of k elements, even O(n1 - ε) elements, is linear, and the constant for the n log n term decreases with the array being split more evenly. without using binary insertion). I'm hesitant to attempt your bolded question because I can answer your "underlying" question: asymptotically speaking, there is no "sweet spot". The constant is, When the pivot is a median of 3, the pdf of P is 6 x (1 - x). Let P be a continuous random variable with values in [0, 1]. 2) Array is already sorted in reverse order. Intuitively, P is the fraction of values less than the pivot. Certainly using QuickSort that pivots on the first element would exhibit O(n^2) performance. If we go even more generally, is there an expression for the constant factor on quicksort using a randomized median-of-k approach? Programmer error: Basically you land up sorting a list twice. More recently other modifications to Quicksort have been invented (e.g., Introsort, PDQ Sort) which prevent that O(N2) worst case. This pivot is the middle value and about half the values are less than the pivot and half are greater than it. Interleaved data: If you receive data from multiple sorted sources with overlapping keys, you could get keys resembling the following: 1 3 2 5 4 7 6 9 8 11 10 13 12 15 14 17 16 19 18. This is known to degenerate to O(n 2) in the worst-case (see this great paper on how to generate those worst-case inputs, for example). The motive for picking item by random would be that the deterministic method would give a result that is worse than the average. and Quicksort will avoid the worst case. Quicksort has its worst performance, if the pivot is likely to be either the smallest, or the largest element in the list (e.g. We have seen the analysis of the Quicksort algorithm and we know that the running time can go up to $\Theta(n^2)$ in the worst case. I believe that the worst case for quicksort depends on the choice of the pivot element at every step. When implemented well, it can be about two or three times faster than its main competitors, merge sort and heapsort. Although all the answers are dealing with "what causes worst case performance", none have covered "what causes data that meets the worst case performance scenario". The only example I came up with is re-indexing. [contradictory] The constant is. The catch is of course constants on the linear term that are spectacularly impractical, highlighting one of the drawbacks of asymptotic analysis. So, which ever method gives the best result depends on the input data, it can't be determined for every possible set. As a side-note: One of the usually worst performing sorting algorithms, actually does quite well with "almost-sorted" data. Algorithms for quicksort worst cases sorted list does not have the worst why is quicksort worst case n^2.... 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