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Now we can telescope:
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If we add all the equations in Equation 99, we have
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= O(l0g N )
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We use the fact that the Nth harmonic number is O(logN)
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The last line in Equation 910 follows from Theorem 65 When we multiply both sides by N + 1 , we obtain the final result: T ( N ) = O(N log N )
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(911)
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963 Picking the Pivot
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Now that we have established that quicksort will run in O(N log N ) time on average, our primary concern is to ensure that the worst case does not occur By performing a complex analysis, we can compute the standard deviation of quicksort's running time The result is that, if a single random permutation is presented, the running time used to sort it will almost certainly be close to the average Thus we must see to it that degenerate inputs do not result in bad running times Degenerate inputs include data that have already been sorted and data that contain only N completely identical elements Sometimes it is the easy cases that give algorithms trouble
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A Wrong Way
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The popular, uninformed choice is to use the first element (ie, the element in position low) as the pivot This selection is acceptable if the input is random, but if the input has been presorted or is in reverse order, the pivot provides a poor partition because it is an extreme element Moreover, this behavior will continue recursively As we demonstrated earlier in the chapter, we would end up with quadratic running time to do absolutely nothing Needless to say, that would be embarrassing Never use the first element as the pivot Another popular alternative is to choose the larger of the first two distinct keys4 as the pivot, but this selection has the same bad effects as choosing the first key Stay away from any strategy that looks only at some key near the front or end of the input group
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Picking the pivot is crucial to good performance Never choose the first element as pivot
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A Safe Choice
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A perfectly reasonable choice for the pivot is the middle element (ie, the element in array cell (low+high)/2) When the input has already been sorted, this selection gives the perfect pivot in each recursive call Of course, we could construct an input sequence that forces quadratic behavior for this strategy (see Exercise 98) However, the chances of randomly running into a case that took even twice as long as the average case is extremely small
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The middle element is a reasonable but passive choice
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Median-of-Three Partitioning
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Choosing the middle element as the pivot avoids the degenerate cases that arise from nonrandom inputs Note that this is a passive choice, however That is, we do not attempt to choose a good pivot Instead, we merely try to avoid picking a bad pivot Median-of-three partitioning is an attempt to pick a better than average pivot In median-of-three partitioning, the median of the first, middle, and last elements is used as the pivot The median of a group of N numbers is the r N / 2 1 th smallest number The best choice for the pivot clearly is the median because it guarantees an even split of the elements Unfortunately, the median is hard to calculate, which would slow quicksort considerably So, we want to get a good estimate of the median without spending too much time doing so We can obtain such an estimate by sampling-the classic method used in opinion polls That is, we pick a subset of these numbers and find their median The larger the sample, the more accurate is the estimate However the larger sample takes longer to evaluate A sample size of 3 gives a small improvement in the average running time of quicksort and also simplifies the resulting partitioning code by
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4 In a complex object, the key is usually the part of the object on which the comparison is
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In median-of-three partitioning, the median of the first, middle, and last elements is used as the pivotThis approach simplifies the partitioning stage of quicksort
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eliminating some special cases Large sample sizes d o not significantly improve performance and thus are not worth using The three elements used in the sample are the first middle, and last elements For instance with input 8 1 4 9 6 3, 5 2 7, 0 the leftmost element is 8 the rightmost element is 0 and the center element is 6: thus the pi\,ot would be 6 Note that for already sorted items, we keep the middle element as the pivot and in this case the pivot is the median