step 3: Swap the in position w ~ t h pivot the

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S o r t i n g Algorithms

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Partitioning algorithm: i stops at large element 9; j stops at small element 3

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Partitioning algorithm: Swap pivot and element in position i

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Keys Equal to the Pivot

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Counters i and j when they must encounter an item equal to the pivot to guarantee good performance

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One important detail that we must consider is how to handle keys that are equal to the pivot Should i stop when it encounters a key equal to the pivot, and should j stop when it encounters a key equal to the pivot Counters i and j should do the same thing; otherwise, the partitioning step is biased For instance, if i stops and j does not, all keys that are equal to the pivot wind up on the right-hand side Let us consider the case in which all elements in the array are identical If both i and j stop, many swaps will occur between identical elements Although these actions seem useless, the positive effect is that i and j cross in the middle, so when the pivot is replaced the partition creates two nearly equal subsets Thus the best-case analysis applies, and the running time is O(N log N) If neither i nor j stops, then i winds up at the last position (assuming of course that it does stop at the boundary), and no swaps are performed This result seems great until we realize that the pivot is then placed as the last element because that is the last cell that i touches The result is widely uneven subsets and a running time that matches the worst-case bound of O(N2)The effect is the same as using the first element as a pivot for presorted input: It takes quadratic time to do nothing We conclude that doing the unnecessary swaps and creating even subsets is better than risking widely uneven subsets Therefore we have both i and j stop if they encounter an element equal to the pivot This action turns out to be the only one of the four possibilities that does not take quadratic time for this input At first glance, worrying about an array of identical elements may seem silly After all, why would anyone want to sort 5000 identical elements However, recall that quicksort is recursive Suppose that there are 100,000 elements, of which 5000 are identical Eventually quicksort could make the recursive call on only the 5000 identical elements Then, ensuring that 5000 identical elements can be sorted efficiently really is important

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966 Median-of-Three Partitioning

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When we do median-of-three partitioning, we can do a simple optimization that saves a few comparisons and also greatly simplifies the code Figure 917 shows the original array Recall that median-of-three partitioning requires that we find the median of the first middle and last elements The easiest wav to do so is to sort them in the array The result is shown in Figure 918 Note the resulting shading: The element that winds up in the first position is guaranteed to be smaller than (or equal to) the pivot, and the element in the last position is guaranteed to be larger than (or equal to) the pivot This outcome tells us four things We should not swap the pivot with the element in the last position Instead, we should swap it with the element in the next-to-last position, as shown in Figure 919 Wecanstart i at low+l and j at high-2 We are guaranteed that whenever i searches for a large element, it will stop because in the worst case it will encounter the pivot (and we stop on equality) We are guaranteed that, whenever j searches for a small element, it will stop because in the worst case it will encounter the first element (and we stop on equality) All of these optimizations are incorporated into the final C++ code

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Computing the involves sorting three elements Hence we can give the Partitioning step a head start and also never about running off the end of the array

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