Sorting

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## Sorting

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**Sorting**What makes it hard? Chapter 7 in DS&AA Chapter 8 in DS&PS**Insertion Sort**• Algorithm • Conceptually, incremental add element to sorted array or list, starting with an empty array (list). • Incremental or batch algorithm. • Analysis • In best case, input is sorted: time is O(N) • In worst case, input is reverse sorted: time is O(N2). • Average case is (loose argument) is O(N2) • Inversion: elements out of order • critical variable for determining algorithm time-cost • each swap removes exactly 1 inversion**Inversions**• What is average number of inversions, over all inputs? • Let A be any array of integers • Let revA be the reverse of A • Note: if (i,j) are in order in A they are out of order in revA. And vice versa. • Total number of pairs (i,j) is N*(N-1)/2 so average number of inversions is N*(N-1)/4 which is O(N2) • Corollary: any algorithm that only removes a single inversion at a time will take time at least O(N2)! • To do better, we need to remove more than one inversion at a time.**BubbleSort**• Most frequently used sorting algorithm • Algorithm: for j=n-1 to 1 …. O(n) for i=0 to j ….. O(j) if A[i] and A[i+1] are out of order, swap them (that’s the bubble) …. O(1) • Analysis • Bubblesort is O(n^2) • Appropriate for small arrays • Appropriate for nearly sorted arrays • Comparision versus swaps ?**Shell Sort: 1959 by Shell**• Motivated by inversion result - need to move far elements • Still quadratic • Only in text books • Historical interest and theoretical interest - not fully understood. • Algorithm: (with schedule 1, 3, 5) • bubble sort things spaced 5 apart • bubble sort things 3 apart • bubble sort things 1 apart • Faster than insertion sort, but still O(N^2) • No one knows the best schedule**Divide and Conquer: Merge Sort**• Let A be array of integers of length n • define Sort (A) recursively via auxSort(A,0,N) where • Define array[] Sort(A,low, high) • if (low == high) return • Else • mid = (low+high)/2 • temp1 = sort(A,low,mid) • temp2 = sort(A,mid,high) • temp3 = merge(temp1,temp2)**Merge**• Int[] Merge(int[] temp1, int[] temp2) • int[] temp = new int[ temp1.length+temp2.length] • int i,j,k • repeat • if (temp1[i]<temp2[j]) temp[k++]=temp1[i++] • else temp[k++] = temp2[j++] • for all appropriate i, j. • Analysis of Merge: • time: O( temp1.length+temp2.length) • memory: O(temp1.length+temp2.length)**Analysis of Merge Sort**• Time • Let N be number of elements • Number of levels is O(logN) • At each level, O(N) work • Total is O(N*logN) • This is best possible for sorting. • Space • At each level, O(N) temporary space • Space can be freed, but calls to new costly • Needs O(N) space • Bad - better to have an in place sort • Quick Sort (chapter 8) is the sort of choice.**Quicksort: Algorithm**• QuickSort - fastest algorithm • QuickSort(S) • 1. If size of S is 0 or 1, return S • 2. Pick element v in S (pivot) • 3. Construct L = all elements less than v and R = all elements greater than v. • 4. Return QuickSort(L), then v, then QuickSort(R) • Algorithm can be done in situ (in place). • On average runs in O(NlogN), but can take O(N2) time • depends on choice of pivot.**Quicksort: Analysis**• Worst Case: • T(N) = worst case sorting time • T(1) = 1 • if bad pivot, T(N) = T(N-1)+N • Via Telescope argument (expand and add) • T(N) = O(N^2) • Average Case (text argument) • Assume equally likely subproblem sizes • Note: chance of picking ith is 1/N • T(N) average cost to sort**Analysis continued**• T(left branch) = T(right branch) (average) so • T(N) = 2* ( T(0)+T(1)….T(N-1) )/N + N, where N is cost of partitioning • Multiply by N: • NT(N) = 2(T(0)+…+T(N-1)) +N^2 (*) • Subtract N-1 case of (*) • NT(N) - (N-1)T(N-1) = 2T(N-1) +2N-1 • Rearrange and drop -1 • NT(N) = (N+1)T(N-1) + 2N -1 • Divide by N(N+1) • T(N)/(N+1) = T(N-1) + 2/(N+1)**Last Step**• Substitute N-1, N-2,... 3 for N • T(N-1)/N = T(N-2)/(N-1) + 2/N • … • T(2)/3 = T(1)/2 +2/3 • Add • T(N)/(N+1) = T(1)/2+ 2(1/3+1/4 + ..+1/(N+1) • = 2( 1+1/2 +…) -5/2 since T(1) = 0 • = O(logN) • Hence T(N) = N logN • In literature, more accurate proof. • For better results, choose pivot as median of 3 random values.**Quickselect: Algorithm**• Problem: find the kth smallest item • Algorithm: modify Quicksort • let |S| be the number of elements in S. • QuickSelect(S, k) • if |S| = 1, return element in S • Pick element p in S (the pivot) • Partition S via p as in QuickSort into L and R • if k < |L| return QuickSelect(L,k) • if k = |L|+1, return pivot • otherwise return QuickSelect(R, k - |L|-1)**Quickselect: Analysis**• Worst Case is O(N^2) • Average Case: analysis similar to quicksort’s. • Here T(N) = 1*(T(0)+T(1)+…+T(N-1))/N + N • Multiply by N • NT(N) = T(0)+T(1) +T(N-1) + N^2 • Substitute with N = N-1 and subtract: • NT(N) -(N-1)T(N-1) = T(N-1) + 2N -1 • Rearrange and divide by N • T(N)= T(N-1)+2 • T(N) = T(N-2) + 4….. = T(1)+2*N = O(N) • Average Case: Linear.**Bucket Sort**• A linear time sort algorithm! • Need to know the possible values. • Example 1: to sort N integers less than M. • Make array A of size M • Read each integer i and update, A[i]++ • Example 2: 200 names • make array of size 26*26 = 676 • Using first 2 letters of each name, put it in [char-char] bucket (usually a short ordered linked list) • Collect them up**Radix Sorting (card sorting)**• Uses linked lists • Idea: Multiple passes of Bucket Sort • Trick: Iteratively sort by last index, next to last, etc. • Example ed ca xa cd xd bd pass1: a:{ca, xa} d:{ed, cd, xd, bd} ca xa ed cd xd bd pass 2: b{bd} c: {ca, cd} e: {ed} x:{xa, xd} bd ca cd ed xa xd • Complexity: O(N* number of passes) • number of passes = length of key**External Sorting (Tape or CD)**• Idea: merge sort (2-way) • Suppose memory size is M (enough to sort internally) • Ta1, Ta2, Tb1, Tb2 are tape drives • Data on Ta1 (initially) • Pass 1: • read M records • sort and write to Tb1, Tb2 alternatively (each run of M records on Tb1, Tb2 is sorted) • Pass 2: • merge sort Tb1 and Tb2 onto Ta1 and Ta2 • Note this takes O(1) memory • Each run of 2*M records is sorted**External Sorting**• Continuing merging, alternating writing to ta1, ta2. • Number of passes is log(N/M) • Time comlexity is O( N/M *log(M)) for first pass • O(N) for subsequent passes • Total: O(max(N log(N/M), N/M*log(M)) • With more tapes, can reduce time by doing k-way merge rather than 2-way merge • Replace Log base 2 with log base k • A trickier algorithm (Polyphase) can do it with fewer tapes. • Who uses tapes? Algorithm works for CDs**Lower Bound for Sorting**• Theorem: if you sort by comparisons, then must use at least log(N!) comparisons. Hence N logN algorithm. • Proof: • N items can be rearranged in N! ways. • Consider a decision tree where each internal node is a comparison. • Each possible array goes down one path • Number of leaves N! • minimum depth of a decision tree is log(N!) • log(N!) = log1+log2+…+log(N) is O(N logN) • Proof: use partition trick • sum log(N/2) + log(N/2+1)….log(N) >N/2*log(N/2)**Summary**• For online sorting, use heapsort. • Online : get elements one at at time • Offline or Batch: have all elements available • For small collections, bubble sort is fine • For large collections, use quicksort • You may hybridize the algorithms, e.g • use quicksort until the size is below some k • then use bubble sort • Sorting is important and well-studied and often inefficiently done. • Libraries often contain sorting routines, but beware: the quicksort routine in Visual C++ seems to run in quadratic time. Java sorts in Collections are fine.