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Sorting – Part II CS 367 – Introduction to Data Structures Better Sorting • The problem with all previous examples is the O(n2) performance – this may be acceptable for small data sets, but not large ones • Theoretically, O(n log n) is possible – see proof in Section 9.2 of the book Heap Sort • Major problem with selection sort – it has to search entire back end of array on every search for next smallest item – what if we could make this search faster? • A heap always keeps the largest element at the top – it only takes O(log n) to remove the top – O(log n) is much better than O(n) search time of selection sort Heap Sort • Basic procedure – build a heap – swap the root with the last element – rebuild the heap excluding the last element • the last element is where it is supposed to be – repeat until only one item left in the heap Heap Sort - Conceptually Z 0 1 2 3 4 5 6 X M queue Z T N J L X 0 1 2 3 4 5 6 T M queue X Z L N J T 0 1 2 3 4 5 6 N M queue T X Z L J Heap Sort - Implementation 0 1 2 3 4 5 6 Z X M T N J L swap 0 and 6, rebuild 0 1 2 3 4 5 6 X T M L N J Z 0 1 2 3 4 5 6 J L M N T X Z swap 0 and 5, rebuild swap 0 and 1, done 0 1 2 3 4 5 6 0 1 2 3 4 5 6 T N M L J X Z L J M N T X Z swap 0 and 4, rebuild swap 0 and 2, rebuild 0 1 2 3 4 5 6 0 1 2 3 4 5 6 N L M J T X Z M L J N T X Z swap 0 and 3, rebuild Building the Heap • The heap will be build within the array – no extra data structures will be needed • Basic idea – start at the last non-terminal node – restore heap for tree rooted at this node • simply swap this node with it’s largest child if the child is larger – repeat this process for all non-terminal nodes Building the Heap 0 1 2 3 4 5 6 M J Z T X L N compare Z with its children (no move made) 0 1 2 3 4 5 6 M J Z T X L N compare J with its children (swap it with X) 0 1 2 3 4 5 6 M X Z T J L N compare M with its children (swap it with Z and then N) 0 1 2 3 4 5 6 Z X N T J L M Valid Heap Building the Heap • Code to re-build the heap void moveDown(Object[ ] data, int first, int last) { int child = 2 * first + 1; while(child <= last) { if((child < last) && ((child + 1) <= last)) { if(data[child] < data[child + 1]) { child++; } if(data[first] < data[child]) { swap(first, child); first = child; child = 2 * child + 1; } else { break; } } } Heap Sort • Code to build the heap and sort it void heapSort(Object[ ] data) { // build the heap out of the data for(int i=data.length / 2; i >= 0; i--) moveDown(data, i, data.length – 1); // now sort it for(int i = data.length – 1; i < 0; i--) { swap(0, i); moveDown(data, 0, i – 1); } } Heap Sort • Time to build the heap in worst case – O(n) – proof can be found in Section 6.9.2 of book • Number of swaps to perform – always (n – 1) • Performance to rebuild the heap – O(n log n) • Overall performance – O(n) + (n-1) + O(n log n) = O(n log n) Quicksort • Basic procedure – divide the initial array into two parts • all of the elements in the left side must be smaller than all of the elements in the right side – sort the two arrays separately and put them back together • we now have a completely sorted array – however, before sorting the two arrays, divided them each into two more arrays • we now have a total of 4 arrays • smallest elements in far left and largest in far right – repeat this process until only 1 element arrays remain • put them all together and the overall array is sorted Quicksort 0 1 2 3 4 5 6 M J Z X T L N break into two parts 0 1 2 3 0 1 2 M J L N Z X T break into four parts 0 1 0 1 0 1 0 J L M N X T Z break into 7 parts 0 0 0 0 0 0 0 J L M N T X Z Quicksort - Implementing • Steps 1. move the largest value to the highest spot – this prevents some array overflow problems 2. pick an upper bound for the left sub-array – pick the value in the center of the array – move this to first element so it doesn’t get moved 3. move all elements less than this to left side 4. move all elements greater to the right side 5. bound will now be in its final position 6. repeat with the two new arrays – from 0 to index(bound) – 1 – from index(bound) + 1 to array.length - 1 Quicksort - Implementing void quickSort(Object[ ] data) { if(data.length < 2) { return; } int max = 0; // find the highest value and put it in top spot for(int i=1; i<data.length; i++) if(data[i] > data[max]) { max = i; } swap(max, data.length – 1); // start the real algorithm quickSort(data, 0, data.length – 2); } Quicksort - Implementing void quickSort(Object[ ] data, int first, int last) { int lower = first + 1, upper = last; swap(first, (first + last) / 2); // find the bound Comparable bound = data[first]; while(lower <= upper) { // divides the array in half while(data[lower] < bound) { lower++; } // lowers that are right while(data[upper] > bound) { upper--; } // uppers that are right if(lower < upper) { swap(lower++, upper--); } else { lower++; } // arrays are already split } swap(upper, first); // puts bound in its final location if(first < upper – 1) { quickSort(data, first, upper – 1); } if(upper + 1 < last) { quickSort(data, upper + 1, last); } } Quicksort Performance • Worst case – consider selecting the smallest (or largest) number as the bound – then all of the numbers end up on one “side” – consider the sorting the following array • [5 3 2 1 4 6 8] • 1 will be the first bound and end up in its proper location • however, there will still be n – 1 elements to sort • this will happen on each iteration – the result is an O(n2) algorithm Quicksort Performance • So what’s the average case? – the answer is O(n log n) • In practice, quicksort is usually the best sorting algorithm – the closer the bound is to the median, the better it is – beware, for arrays under 30 elements, insertion sort is more efficient • can you think how quicksort and insertion sort could be combined? Mergesort • One of the first ever sorting algorithms used on a computer • It works on a principle similar to quicksort – each array is broken into two parts and then sorted separately – this partition and sort method continues until only single element arrays exist – then all of the arrays are put back together to form a sorted array Mergesort • Big difference from quicksort is that the arrays are always broken into equal partitions – or in the case of an odd sized array, as close as possible to even • There is no bound selected • To put the arrays back together, simply select the smallest element from either array and make it next Mergesort 0 1 2 3 4 5 6 Z M J R X T V break into 7 parts 0 0 0 0 0 0 0 Z M J R X T V 0 1 0 1 0 1 M Z J R T X 0 1 2 3 0 1 2 J M R Z T V X 0 1 2 3 4 5 6 J M R T V X Z Merging • The most sophisticated part of mergesort is recombining (or merging) two separate arrays • Just go through each array selecting the smallest remaining element from each array – add it to the new array Merging • Pseudo-code merge(array, first, last) { mid = (first + last) / 2; i1= 0; i2 = first; i3 = mid + 1; while( // both left and right sub-arrays contain elements ) { if(array[i2] < array[i3]) { tmp[i1++] = array[i2++]; } else { tmp[i1++ = array[i3++]; } } // load into temp array remaining elements of array // copy elements in temp back into array } Mergesort • Once the merge code is done, the code for mergesort is easy • Psuedo-code mergeSort(data, first, last) { if(first < last) { mid = (first + last) / 2; mergeSort(data, first, mid); mergeSort(data, mid + 1, last); merge(data, first, last); } } Mergesort Performance • Mergesort produces a lot of copying in memory • It also requires extra storage space for the temporary array – this can be prohibitive for very large data sets

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posted: | 9/10/2011 |

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