1. Sorting
What makes it hard?
Chapter 7 in DS&AA
Chapter 8 in DS&PS

2. 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

3. 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.

4. 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 ?

5. 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

6. 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)

7. 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)

8. 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.

9. 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.

10. 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

11. 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)

12. 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.

13. 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)

14. 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.

15. 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

16. 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

17. 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

18. 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

19. 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)

20. 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.