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In the partition step every element except the pivot is placed in of two groups
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t Quicksort small items
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array cell p The recursive calls would then be on the parts from low to p-1 and then p + l to h i g h Because recursion allows us to take the giant leap of faith, the correctness of the algorithm is guaranteed as follows The group of small elements is sorted, by virtue of the recursion
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The largest element in the group of small elements is not larger than the pivot, by virtue of the partition The pivot is not larger than the smallest element in the group of large elements, by virtue of the partition The group of large elements is sorted, by virtue of the recursion Although the correctness of the algorithm is easily established, why it is faster than mergesort is not clear Like mergesort, it recursively solves two subproblems and requires linear additional work (in the form of the partitioning step) Unlike mergesort, however, quicksort subproblems are not guaranteed to be of equal size, which is bad for performance However, quicksort is faster than mergesort because the partitioning step can be performed significantly faster than the merging step can In particular, the partitioning step can be performed without using an extra array, and the code to implement it is very compact and efficient This advantage makes up for the lack of equally sized subproblems
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Quicksort is fast because the partitioning step can be performed quickly and in place
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The algorithm description leaves several questions unanswered: How do we choose the pivot How do we perform the partition What do we do if we see an element that is equal to the pivot All these questions can dramatically affect the running time of the algorithm We perform an analysis to help us decide how to implement the unspecified steps in quicksort
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The best case for quicksort is that the pivot partitions the set into two equally sized subsets and that this partioning happens at each stage of the recursion We then have two half-sized recursive calls plus linear overhead, which matches the performance of mergesort The running time for this case is O(N log N) (We have not actually proved that this is the best case Although such a proof is possible, we omit the details here)
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The best case occurs when the partition always splits into equal subsets The running time is O(N log N)
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Since equally sized subsets are good for quicksort, you might expect that unequally sized subsets are bad That indeed is the case Let us suppose that, in each step of the recursion, the pivot happens to be the smallest element Then the set of small elements L will be empty, and the set of large elements R will have all the elements except the pivot We then have to recursively call quicksort on subset R Suppose also that T(N) is the running time to quicksort N elements and we assume that the time to sort 0 or I element is just 1 time unit
The worst case occurs when the partition repeatedly generates an empty subset The running time is O(N2)
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Suppose further that we charge N units to partition a set that contains N elements Then for N > 1, we obtain a running time that satisfies
In other words, Equation 91 states that the time required to quicksort N items equals the time to sort recursively the N - 1 items in the subset of larger elements plus the N units of cost to perform the partition This assumes that in each step of the iteration we are unfortunate enough to pick the smallest element as the pivot To simplify the analysis, we normalize by throwing out constant factors and solve this recurrence by telescoping Equation 9 l repeatedly: