CompSci 100E29.1Sorting: From Theory to Practiceÿ Why do we study sorting? Because we have to Because sorting is beautiful Example of algorithm analysis in a simple, useful settingÿ There are n sorting algorithms, how many should we study? O(n), O(log n), … Why do we study more than one algorithm?o Some are good, some are bad, some are very, very sado Paradigms of trade-offs and algorithmic design Which sorting algorithm is best? Which sort should you call from code you write?CompSci 100E29.2Sorting out sortsÿ Simple, O(n2) sorts --- for sorting n elements Selection sort --- n2comparisons, n swaps, easy to code Insertion sort --- n2comparisons, n2moves, stable, fast on nearly sorted vectors: O(n) Bubble sort --- n2everything, slowerÿ Divide and conquer faster sorts: O(n log n) for n elements Quick sort: fast in practice, O(n2) worst case Merge sort: good worst case, great for linked lists, stable, uses extra storage for vectors/arraysÿ Other sorts: Heap sort, basically priority queue sorting Radix sort: doesn’t compare keys, uses digits/characters Shell sort: quasi-insertion, fast in practice, non-recursiveCompSci 100E29.3Selection sort: summaryÿ Simple to code n2sort: n2comparisons, n swapsvoid selectSort(String[] a){for(int k=0; k < a.length; k++){int minIndex = findMin(a,k);swap(a,k,minIndex);}}ÿ # comparisons: Swaps? Invariant:ΣΣΣΣk=1nk=1+2+…+n=n(n+1)/2=O(n2)Sorted, won’t movefinal position?????CompSci 100E29.4Insertion Sort: summaryÿ Stable sort, O(n2) -- ( O(n) on nearly sorted vectors!) Stable sorts maintain order of equal keys Good for sorting on two criteria: name, then agevoid insertSort(String[] a){int k, loc; string elt;for(k=1; k < a.length; k++) {elt = a[k];loc = k;// shift until spot for elt is foundwhile (0 < loc && elt.compareTo(a[loc-1]) < 0) {a[loc] = a[loc-1]; // shift rightloc=loc-1;}a[loc] = elt;}}Sorted relative toeach other?????CompSci 100E29.5Bubble sort: summary of a dogÿ For completeness you should know about this sort Few, if any, redeeming features. Really slow, really Can code to recognize already sorted vector (see insertion)o Not worth it for bubble sort, much slower than insertionvoid bubbleSort(String[] a) {for(int j=a.length-1; j >= 0; j--) {for(int k=0; k < j; k++) {if (a[k] > a[k+1])swap(a,k,k+1);}}}ÿ “bubble” elements down the vector/arraySorted, in finalposition?????CompSci 100E29.6Summary of simple sortsÿ Selection sort has n swaps, good for “heavy” data moving objects with lots of state, e.g., …o In C or C++ this is an issueo In Java everything is a pointer/reference, so swapping is fast since it's pointer assignmentÿ Insertion sort is good on nearly sorted data, it’s stable, it’s fast Also foundation for Shell sort, very fast non-recursive More complicated to code, but relatively simple, and fastÿ Bubble sort is a travesty? But it's fast to code if you know it! Can be parallelized, but on one machine don’t go near it (see quotes at end of slides)CompSci 100E29.7Quicksort: fast in practiceÿ Invented in 1962 by C.A.R. Hoare, didn’t understand recursion Worst case is O(n2), but avoidable in nearly all cases In 1997 Introsort published (Musser, introspective sort)o Like quicksort in practice, but recognizes when it will be bad and changes to heapsortvoid quick(String[], int left, int right) {if (left < right) {int pivot = partition(a, left, right);quick(a, left, pivot-1);quick(a, pivot+1, right);}}ÿ Recurrence?<= X>XXpivot indexCompSci 100E29.8Partition codefor quicksortleftÿEasy to develop partitionint partition(String[] a,int left, int right){String pivot = a[left];int k, pIndex = left;for(k=left+1, k <= right; k++){if (a[k].compareTo(pivot) <= 0){pIndex++;swap(a,k,pIndex);}}swap(a,left,pIndex);return pIndex;}ÿ Loop invariant: statement true each time loop test is evaluated, used to verify correctness of loopÿ Can swap into a[left] before loop Nearly sorted data still ok??????????????<=>???<= pivot > pivotpIndexleftrightrightleft rightwhat we wantwhat we start withinvariantpIndexkCompSci 100E29.9Analysis of Quicksortÿ Average case and worst case analysis Recurrence for worst case: T(n) = What about average?ÿ Reason informally: Two calls vector size n/2 Four calls vector size n/4 … How many calls? Work done on each call?ÿ Partition: typically find middle of left, middle, right, swap, go Avoid bad performance on nearly sorted dataÿ In practice: remove some (all?) recursion, avoid lots of “clones”T(n-1) + T(1) + O(n)T(n) = 2T(n/2) + O(n)CompSci 100E29.10Tail recursion eliminationÿ If the last statement is a recursive call, recursion can be replaced with iteration Call cannot be part of an expression Some compilers do this automaticallyvoid foo(int n) void foo2(int n){{if(0<n){ while(0<n){System.out.println(n); System.out.println(n);foo(n-1); n = n-1;}}}}ÿ What if print and recursive call switched?ÿ What about recursive factorial? return n*factorial(n-1);CompSci 100E29.11Merge sort: worst case O(n log n)ÿ Divide and conquer --- recursive sort Divide list/vector into two halveso Sort each halfo Merge sorted halves together What is complexity of merging two sorted lists? What is recurrence relation for merge sort as described?T(n) =ÿ What is advantage of array over linked-list for merge sort? What about merging, advantage of linked list? Array requires auxiliary storage (or very fancy coding)T(n) = 2T(n/2) + O(n)CompSci 100E29.12Merge sort: lists or vectorsÿ Mergesort for vectorsvoid mergesort(String[] a, int left, int right) {if (left < right) {int mid = (right+left)/2;mergesort(a, left, mid);mergesort(a, mid+1, right);merge(a,left,mid,right);}}ÿ What’s different when linked lists used? Do differences affect complexity? Why?ÿ How does merge work?CompSci 100E29.13Mergesort continuedÿ Array code for merge isn’t pretty, but it’s not hard Mergesort itself is elegantvoid merge(String[] a,int left, int middle, int right)// pre: left <= middle <= right,// a[left] <= … <= a[middle],// a[middle+1] <= … <= a[right]// post: a[left] <= … <= a[right]ÿ Why is this prototype potentially simpler for linked lists? What will prototype be? What is complexity?CompSci 100E29.14Mergesort continuedvoid merge(String[] a, int left, int middle, int right) {String[] b = new String[right - left + 1];int k = 0, kl = left, kr = middle + 1;for (; kl <= middle
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