CSC – 332 Data StructuresSorting

Dr. Curry Guinn

Today’s Class

•Don’t forget Homework 7•All the sorts

– Insertion sort– Shell sort– Heap sort– Merge sort– Quick sort

•What’s the best we can do?•In class quiz on Tuesday (See last slide for questions)

Sorting

• Most “easy” algorithms are O(N2)

• Any algorithm that sorts by swapping adjacent elements will be O(N2)

Insertion sort

Take an array of numbers of size n.

1) Initially p = 1

2) Let the first p elements be sorted.

3) Insert the (p+1)th element properly in the list so that now p+1 elements are sorted.

4) Increment p and go to step (3)

Insertion Sort...

Consists of N - 1 passes For pass p = 1 through N - 1, ensures that the

elements in positions 0 through p are in sorted order elements in positions 0 through p - 1 are already sorted move the element in position p left until its correct place is

found among the first p + 1 elements

http://www.cis.upenn.edu/~matuszek/cse121-2003/Applets/Chap03/Insertion/InsertSort.html

Analysis: worst-case running time

• Inner loop is executed p times, for each p=1..N-1

Overall: 1 + 2 + 3 + . . . + N-1 = …= O(N2)

Analysis• The bound is tight (N2)• That is, there exists some input which actually

uses (N2) time

Average case analysis

• Given an array A of elements, an inversion is an ordered pair (i, j) such that i < j, but A[i] > A[j]. (out of order elements)

• Assume no duplicate elements.

• Theorem: The average number of inversions in an array of n distinct elements is n (n - 1) / 4.

• Corollary: Any algorithm that sorts by exchanging adjacent elements requires O(n2) time on average.

Shellsort

• What’s the concept?

• Increment sequence

• Shell’s sequence {1, 2, 4, 8, …} O(N2)

• Hibbard’s sequence {1, 3, 7, 15, …}O(N3/2)

• best sequence known { 1, 5, 19, 41, 109, …} O(N7/6) – (either 9*4i -9*2i + 1 or 4i - 3*2i + 1)

Shell Sort

0 1 2 3 4 5 6 7 8

Example: an array of int, length n = 9

24 344 27 10 4147 2531

gap = n/2 = 4

for (int i = gap; i < A.length; i++) {

int temp = A[i];

for ( j = i; j >= gap && temp < A[j-gap]; j -= gap)

A[j] = A[j – gap];

A[j] = temp; }

A

i

2410

i i

4441

i

273

i

gap = max(1, (int) gap /2) = 2

i i

3 31

i

4124

i i i

27 473127

i

44254125

1

3 10 2725 4744

Heapsort

• What’s the concept?

Performance of Heapsort

MergesortBased on divide-and-conquer strategy

• Divide the list into two smaller lists of about equal sizes• Sort each smaller list recursively• Merge the two sorted lists to get one sorted list

How do we divide the list? How much time needed?

How do we merge the two sorted lists? How much time needed?

Dividing

• If the input list is a linked list, dividing takes (N) time– We scan the linked list, stop at the N/2 th entry and

cut the link• If the input list is an array A[0..N-1]: dividing takes O(1)

time– we can represent a sublist by two integers left and right: to divide A[left..right], we compute center=(left+right)/2 and obtain A[left..center] and A[center+1..right]

– Try left=0, right = 50, center=?

Mergesort

• Divide-and-conquer strategy– recursively mergesort the first half and the

second half– merge the two sorted halves together

http://www.cosc.canterbury.ac.nz/people/mukundan/dsal/MSort.html

How to merge?

• Input: two sorted array A and B• Output: an output sorted array C• Three counters: Actr, Bctr, and Cctr

– initially set to the beginning of their respective arrays

(1)   The smaller of A[Actr] and B[Bctr] is copied to the next entry in C, and the appropriate counters are advanced

(2)   When either input list is exhausted, the remainder of the other list is copied to C

Example: Merge

Example: Merge...

Running time analysis: Clearly, merge takes O(m1 + m2) where m1 and m2 are

the sizes of the two sublists.

Space requirement:merging two sorted lists requires linear extra memoryadditional work to copy to the temporary array and back

Mergesortvoid MergeSort(int arr[], int temp[], int left,

int right)

{

if (left < right) {

int center = (left + right)/ 2;

// sort left half

MergeSort(arr, temp, left, center);

// sort right half

MergeSort(arr, temp, center+1, right);

// merge left and right halves

Merge(arr, temp, left, center+1, right);

}

}

void Merge(int arr[], int temp[], int curLeft, int curRight, int endRight){ int endLeft = curRight – 1; int curTemp = curLeft; int numElems = endRight – startLeft + 1; // Main loop while (curLeft <= endLeft && curRight <= endRight) if ( arr[curLeft] <= arr[curRight]) temp[curTemp++] = arr[curLeft++]; else temp[curTemp++] = arr[curRight++];

while (curLeft <= endLeft) // finish left temp[curTemp++] = arr[curLeft++]; while (curRight <= endRight) // finish right temp[curTemp++] = arr[curRight++]; // copy back to arr for (int = i = 0; i < numElems; i++, endRight--) arr[endRight] = tempArray[endRight];}

Analysis of mergesort Let T(N) denote the worst-case running time of mergesort

to sort N numbers.

Assume that N is a power of 2.

• Divide step: O(1) time• Conquer step: 2 T(N/2) time• Combine step: O(N) time Recurrence equation:

T(1) = 1 T(N) = 2T(N/2) + N

Analysis of mergesort Let T(N) denote the worst-case running time of

mergesort to sort N numbers.

Assume that N is a power of 2.

• Divide step: O(1) time• Conquer step: 2 T(N/2) time• Combine step: O(N) time

Solving the Recurrence Relation

• See board

• Other important recurrence relations– T(n) = T(n – 1) + 1

• O(n)

– T(n) = T(n/2) + 1• O(log n)

– T(n) = T(n – 1) + n• O(n2)

Master Theorem

• Let T(n) be a monotonically increasing function that satisfies

T(n) = a T(n/b) + f(n)

T(1) = c

where a 1, b 2, c>0. If f(n) is (nd) where d 0 then

if a < bd

T(n) = If a = bd

if a > bd

Quicksort

• Fastest known sorting algorithm in practice for many data types

• Average case: O(N log N)

• Worst case: O(N2)– But, the worst case seldom happens.

• Another divide-and-conquer recursive algorithm like mergesort

Quicksort

• Divide step: – Pick any element (pivot) v in S – Partition S – {v} into two disjoint groups S1 = {x S – {v} | x v} S2 = {x S – {v} | x v}

• Conquer step: recursively sort S1 and S2

• Combine step: combine the sorted S1, followed by v, followed by the sorted S2

v

v

S1 S2

S

Example: Quicksort

Example: Quicksort...

Pseudocode

Input: an array A[p, r]

Quicksort (A, p, r) {

if (p < r) {

q = Partition (A, p, r) //q is the position of the pivot element

Quicksort (A, p, q-1)

Quicksort (A, q+1, r)

}

}

Partitioning• Partitioning

– Key step of quicksort algorithm– Goal: given the picked pivot, partition the

remaining elements into two smaller sets– Many ways to implement

• We will learn an easy and efficient partitioning strategy here.

• How to pick a pivot will be discussed later

Partitioning Strategy

• Want to partition an array A[left .. right]• First, get the pivot element out of the way by

swapping it with the last element. (Swap pivot and A[right])

• Let i start at the first element and j start at the next-to-last element (i = left, j = right – 1)

pivot i j

5 6 4 6 3 12 19 5 6 4 63 1219

swap

Partitioning Strategy• Want to have

– A[p] <= pivot, for p < i– A[p] >= pivot, for p > j

• When i < j– Move i right, skipping over elements smaller than the pivot– Move j left, skipping over elements greater than the pivot– When both i and j have stopped

• A[i] >= pivot• A[j] <= pivot

i j

5 6 4 63 1219

i j

5 6 4 63 1219

Partitioning Strategy• When i and j have stopped and i is to the left of j

– Swap A[i] and A[j]• The large element is pushed to the right and the small element

is pushed to the left

– After swapping• A[i] <= pivot• A[j] >= pivot

– Repeat the process until i and j cross

swap

i j

5 6 4 63 1219

i j

5 3 4 66 1219

Partitioning Strategy

• When i and j have crossed– Swap A[i] and pivot

• Result:– A[p] <= pivot, for p < i– A[p] >= pivot, for p > i

i j

5 3 4 66 1219

ij

5 3 4 66 1219

ij

5 3 4 6 6 12 19

Small arrays

• For very small arrays, quicksort does not perform as well as insertion sort– How small depends on many factors, such as

the time spent making a recursive call, the compiler, etc

• Do not use quicksort recursively for small arrays– Instead, use a sorting algorithm that is efficient

for small arrays, such as insertion sort

Picking the Pivot

• Use the first element as pivot– If the input is random, ok– If the input is presorted (or in reverse order)

• All the elements go into S2 (or S1)• This happens consistently throughout the recursive calls• Results in O(n2) behavior (Analyze this case later)

• Choose the pivot randomly– Generally safe– Random number generation can be expensive

Picking the Pivot

• Use the median of the array– Partitioning always cuts the array into roughly half– An optimal quicksort (O(N log N))– However, hard to find the exact median

• e.g., sort an array to pick the value in the middle

Pivot: median of three• We will use median of three

– Compare just three elements: the leftmost, rightmost and center– Swap these elements if necessary so that

• A[left] = Smallest• A[right] = Largest• A[center] = Median of three

– Pick A[center] as the pivot– Swap A[center] and A[right – 1] so that pivot is at second last position (why?)

Pivot: median of three

pivot

5 6 4

6

3 12 192 13 6

5 6 4 3 12 192 6 13

A[left] = 2, A[center] = 13, A[right] = 6

Swap A[center] and A[right]

5 6 4 3 12 192 13

pivot

65 6 4 3 12192 13

Choose A[center] as pivot

Swap pivot and A[right – 1]

Note we only need to partition A[left + 1, …, right – 2]. Why?

Main Quicksort Routine

For small arrays

Recursion

Choose pivot

Partitioning

Analysis• Assumptions:

– A random pivot (no median-of-three partitioning– No cutoff for small arrays

• Running time– pivot selection: constant time O(1)– partitioning: linear time O(N)– running time of the two recursive calls

• T(N)=T(i)+T(N-i-1)+cN where c is a constant– i: number of elements in S1

Worst-Case Analysis

• What will be the worst case?– The pivot is the smallest element, all the

time– Partition is always unbalanced

Best-case Analysis• What will be the best case?

– Partition is perfectly balanced.– Pivot is always in the middle (median of the

array)

Average-Case Analysis

• Assume– Each of the sizes for S1 is equally likely

• This assumption is valid for our pivoting (median-of-three) and partitioning strategy

• On average, the running time is O(N log N)

Quicksort Faster than Mergesort• Both quicksort and mergesort take O(N log N) in the

average case.• Why is quicksort faster than mergesort?

– The inner loop consists of an increment/decrement (by 1, which is fast), a test and a jump.

– There is no extra juggling as in mergesort.

inner loop

Is N Log N the best we can do?

• For comparison-based sorting, yes.

Lower Bound for Sorting

• Mergesort and heapsort– worst-case running time is O(N log N)

• Are there better algorithms?

• Goal: Prove that any sorting algorithm based on only comparisons takes (N log N) comparisons in the worst case (worse-case input) to sort N elements.

Lower Bound for Sorting

• Suppose we want to sort N distinct elements

• How many possible orderings do we have for N elements?

• We can have N! possible orderings (e.g., the sorted output for a,b,c can be a b c, b a c, a c b, c a b, c b a, b c a.)

Lower Bound for Sorting

• Any comparison-based sorting process can be represented as a binary decision tree.– Each node represents a set of possible

orderings, consistent with all the comparisons that have been made

– The tree edges are results of the comparisons

Decision tree for

Algorithm X for sorting

three elements a, b, c

Lower Bound for Sorting• The worst-case number of comparisons used by the

sorting algorithm is equal to the depth of the deepest leaf– The average number of comparisons used is equal to the

average depth of the leaves

• A decision tree to sort N elements must have N! leaves– a binary tree of depth d has at most 2d leaves

the tree must have depth at least log2 (N!)

• Therefore, any sorting algorithm based on only comparisons between elements requires at least

log2(N!) comparisons in the worst case.

Lower Bound for Sorting

• Any sorting algorithm based on comparisons between elements requires (N log N) comparisons.

Linear time sorting

• Can we do better (linear time algorithm) if the input has special structure (e.g., uniformly distributed, every numbers can be represented by d digits)? Yes.

• Bucket Sort (Counting Sort)

Bucket Sort• Assume N integers to be sorted, each is in the range 1 to M.• Define an array B[1..M], initialize all to 0 O(M)• Scan through the input list A[i], insert A[i] into B[A[i]] O(N)• Scan B once, read out the nonzero integers O(M)

Total time: O(M + N)– if M is O(N), then total time is O(N)– Can be bad if range is very big, e.g. M=O(N2)

N=7, M = 9,

Want to sort 8 1 9 5 2 6 3

1 2 5 8 9

Output: 1 2 3 5 6 8 9

3 6

For Next Class, Tuesday

• Homework 7 due Monday, 03/24

• For Tuesday– More Sorting– Quiz (See next slide)

Quiz Questions• Weiss, Chapter 7

1. Shell’s original increments result in an algorithm that is O(N2). What is the problem with Shell’s increments as described at the bottom of page 252 (2nd Edition) or page 276 (3rd Edition) (7.4.1)?

2. What is the worst case analysis of heapsort, given on page 256 (2nd Edition) or page 279-80 (3rd Edition) (7.5.1)?

3. Mergesort uses the minimum number of comparisons of any sort but it does use extra memory. Quicksort might use more comparisons, but it does not use extra memory. For objects that are large, which sort is better according to the discussion on page 263 (2nd edition) or page 288 (3rd edition) (7.6.1). Does it depend on the language of implementation?

4. Why is choosing the first element as the pivot “an absolutely horrible idea” according to Weiss?

• GEB, Chapter 6:5. Definition of Genotype and Phenotype6. Definition of Aleatoric music7. What were the 3 languages encoded on the Rosetta Stone?

Quiz is open notes and must be taken in class. No electronic submission.