1

Sorting

• We have actually seen already two efficient ways to sort:

2

A kind of “insertion” sort

• Insert the elements into a red-black tree one by one

• Traverse the tree in in-order and collect the keys

• Takes O(nlog(n)) time

3

Heapsort (Willians, Floyd, 1964)

• Put the elements in an array• Make the array into a heap• Do a deletemin and put the

deleted element at the last position of the array

4

Put the elements in the heap

65

79

26

3323

2924 1519

79 65 26 24 19 15 29 23 33 40 7

740

Q

5

65

79

26

3323

2924 1519

79 65 26 24 19 15 29 23 33 40 7

740

Q

Make the elements into a heap

6

Make the elements into a heap

65

79

26

3323

2924 1519

79 65 26 24 19 15 29 23 33 40 7

740

Q

Heapify-down(Q,4)

7

65

79

26

3323

2924 157

79 65 26 24 7 15 29 23 33 40 19

1940

Q

Heapify-down(Q,4)

8

65

79

26

3323

2924 157

79 65 26 24 7 15 29 23 33 40 19

1940

Q

Heapify-down(Q,3)

9

65

79

26

3324

2923 157

79 65 26 23 7 15 29 24 33 40 19

1940

Q

Heapify-down(Q,3)

10

65

79

26

3324

2923 157

79 65 26 23 7 15 29 24 33 40 19

1940

Q

Heapify-down(Q,2)

11

65

79

15

3324

2923 267

79 65 15 23 7 26 29 24 33 40 19

1940

Q

Heapify-down(Q,2)

12

65

79

15

3324

2923 267

79 65 15 23 7 26 29 24 33 40 19

1940

Q

Heapify-down(Q,1)

13

7

79

15

3324

2923 2665

79 7 15 23 65 26 29 24 33 40 19

1940

Q

Heapify-down(Q,1)

14

7

79

15

3324

2923 2619

79 7 15 23 19 26 29 24 33 40 65

6540

Q

Heapify-down(Q,1)

15

7

79

15

3324

2923 2619

79 7 15 23 19 26 29 24 33 40 65

6540

Q

Heapify-down(Q,0)

16

79

7

15

3324

2923 2619

7 79 15 23 19 26 29 24 33 40 65

6540

Q

Heapify-down(Q,0)

17

19

7

15

3324

2923 2679

7 19 15 23 79 26 29 24 33 40 65

6540

Q

Heapify-down(Q,0)

18

19

7

15

3324

2923 2640

7 19 15 23 40 26 29 24 33 79 65

6579

Q

Heapify-down(Q,0)

20

Summery

• We can build the heap in linear time (we already did this analysis)

• We still have to deletemin the elements one by one in order to sort that will take O(nlog(n))

21

Quicksort (Hoare 1961)

22

quicksort

Input: an array A[p, r]

Quicksort (A, p, r) if (p < r)

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

Quicksort (A, p, q-1) Quicksort (A, q+1, r)

23

2 8 7 1 3 5 6 4

i j

2 8 7 1 3 5 6 4

i j

2 8 7 1 3 5 6 4

i j

2 8 7 1 3 5 6 4

i j

2 1 7 8 3 5 6 4

i j

p r

24

2 1 7 8 3 5 6 4

i j

2 1 3 8 7 5 6 4

i j

2 1 3 8 7 5 6 4

i j

2 1 3 8 7 5 6 4

i j

2 1 3 4 7 5 6 8

i j

25

2 8 7 1 3 5 6 4p r

Partition(A, p, r) x ←A[r]

i ← p-1 for j ← p to r-1

do if A[j] ≤ x then i ← i+1 exchange A[i] ↔ A[j] exchange A[i+1] ↔A[r] return i+1

26

Analysis

• Running time is proportional to the number of comparisons

• Each pair is compared at most once O(n2)

• In fact for each n there is an input of size n on which quicksort takes cn2 Ω(n2)

27

But

• Assume that the split is even in each iteration

28

T(n) = 2T(n/2) + bn

How do we solve linear recurrences like this ? (read Chapter 4)

29

Recurrence tree

T(n/2)

bn

T(n/2)

30

Recurrence tree

bn/2

bn

bn/2

T(n/4)T(n/4)T(n/4)T(n/4)

31

Recurrence tree

bn/2

bn

bn/2

T(n/4)T(n/4)T(n/4)T(n/4)logn

In every level we do bn comparisonsSo the total number of comparisons is O(nlogn)

34

Observations

• We can’t guarantee good splits

• But intuitively on random inputs we will get good splits

35

Randomized quicksort

• Use randomized-partition rather than partition

Randomized-partition (A, p, r) i ← random(p,r)

exchange A[r] ↔ A[i] return partition(A,p,r)

36

• On the same input we will get a different running time in each run !

• Look at the average for one particular input of all these running times

37

Expected # of comparisons

Let X be the expected # of comparisons

This is a random variable

Want to know E(X)

38

Expected # of comparisons

Let z1,z2,.....,zn the elements in sorted order

Let Xij = 1 if zi is compared to zj and 0 otherwise

So,

1n

1i

n

1ijijXX

39

n 1 n n 1 n

ij iji 1 j i 1 i 1 j i 1

E X E X E X

by linearity of expectation

n 1 n

i ji 1 j i 1

Pr{z is compared to z }

40

n 1 n n 1 n

ij iji 1 j i 1 i 1 j i 1

E X E X E X

by linearity of expectation

n 1 n

i 1 j ii j

1

Pr{z is compared to z }

41

Consider zi,zi+1,.......,zj ≡ Zij

Claim: zi and zj are compared either zi or zj is the first chosen in Zij

Proof: 3 cases:– {zi, …, zj} Compared on this

partition, and never again.

– {zi, …, zj} the same

– {zi, …, zk, …, zj} Not compared on this partition. Partition separates them, so no future partition uses both.

42

= 1/(j-i+1) + 1/(j-i+1)= 2/(j-i+1)

Pr{zi is compared to zj}

= Pr{zi or zj is first pivot chosen from Zij} just explained

= Pr{zi is first pivot chosen from Zij} +

Pr{zj is first pivot chosen from Zij}

mutually exclusivepossibilities

43

1n

1i

n

1ij 1ij

2XE

n 1 n i+1

i 1 k 2

2

kSimplify with a change of variable, k=j-i+1.

1n

1i

n

1k k

2Simplify and overestimate, by adding terms.

1n

1i

n lgO

n) lg O(n

44

Lower bound for sorting in the comparison model

45

A lower bound

• Comparison model: We assume that the operation from which we deduce order among keys are comparisons

• Then we prove that we need Ω(nlogn) comparisons on the worst case

46

Model the algorithm as a decision tree

12 1

2

1 2

2

12

3

3

2

2

1

3

3

1

1

21

3

3

2

2

3

1

1

1

1

2

3

3

1

1

3

2

2

47

Important Observations

• Every algorithm can be represented as a (binary) tree like this

• Each path corresponds to a run on some input

• The worst case # of comparisons corresponds to the longest path

48

The lower bound

Let d be the length of the longest path

#leaves ≤ 2dn! ≤

log2(n!) ≤d

49

Lower Bound for Sorting

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

50

Beating the lower bound

• We can beat the lower bound if we can deduce order relations between keys not by comparisons

Examples:• Count sort• Radix sort

51

Linear time sorting

• Or assume something about the input: random, “almost sorted”

52

Sorting an almost sorted input

• Suppose we know that the input is “almost” sorted

• Let I be the number of “inversions” in the input: The number of pairs ai,aj such that i<j and ai>aj

53

Example

1, 4 , 5 , 8 , 3

I=3

8, 7 , 5 , 3 , 1 I=10

54

• Think of “insertion sort” using a list

• When we insert the next item ak, how deep it gets into the list?

• As the number of inversions ai,ak for i < k lets call this Ik

55

Analysis

The running time is:

1

n

jj

I n I n

56

Thoughts

• When I=Ω(n2) the running time is Ω(n2)

• But we would like it to be O(nlog(n)) for any input, and faster whan I is small

57

Finger red black trees

58

Finger treeTake a regular search tree and reverse the direction of the pointers on the rightmost spine

We go up from the last leaf until we find the subtree containing the item and we descend into it

59

Finger treesSay we search for a position at distance d from the end

Then we go up to height O(log(d))

Insertions and deletions still take O(log n) worst case time but O(log(d)) amortized time

So search for the dth position takes O(log(d)) time

60

Back to sorting

• Suppose we implement the insertion sort using a finger search tree

• When we insert item k then d=O(Ik) and it take O(log(Ik)) time

61

Analysis

The running time is:

1

( log( ) )n

jj

O I n

Since ∑Ij = I this is at most

logI

O n nn

62

Selection

Find the kth element

63

Randomized selection

Randomized-select (A, p, r,k) if p=r then return A[p]

q←randomized-partition(A,p,r) j ← q-p+1 if j=k then return A[q] else if k < j then return randomized-select(A,p,q-

1,k) else return randomized-select(A,q+1,r,k-j)

65

Expected running time

With probability 1/n, A[p,q] contains exactly k elements, for k=1,2,…,n

1

1( ( )) ( ) ( (max( 1, )))

n

k

E T n O n E T k n kn

66

Assume n is even

1

1( ( )) ( ) ( (max( 1, )))

n

k

E T n O n E T k n kn

( ( 1)) ( ( 2)) ....

1( ( )) ( )

2 2

1 ...... ( ( 1))2

E T n E T n

n nE T n O n E T E T

n

nE T E T n

67

In general

1

1( ( )) ( ) ( (max( 1, )))

n

k

E T n O n E T k n kn

1

/ 2

2( ( )) ( ) ( ( ))

n

k n

E T n O n E T kn

68

Solve by “substitution”

1

/ 2

2( ( )) ( ) ( ( ))

n

k n

E T n O n E T kn

Assume T(k) ≤ ck for k < n, and prove T(n) ≤ cn

1

/ 2

2( ( ))

n

k n

E T n an ckn

69

Solve by “substitution”1

/ 2

2( ( ))

n

k n

E T n an ckn

/ 2 11

1 1

2 nn

k k

can k k

n

/ 2 ( / 2 1)2 ( 1)

2 2

n nc n nan

n

70

( / 2 1) / 22 ( 1)( ( ))

2 2

n nc n nE T n an

n

2 ( 1) ( / 2 2)( / 2 1)

2 2

c n n n nan

n

232

4 2

c n nan

n

3

4

cnan

4

cncn an

Choose c ≥4a

cn

71

Selection in linear worst case time

Blum, Floyd, Pratt, Rivest, and Tarjan (1973)

72

5-tuples

6

2

9

5

1

73

Sort the tuples

9

6

5

2

1

74

Recursively find the median of the medians

9

6

5

2

1

75

Recursively find the median of the medians

9

6

5

2

1

7 10 1 3 2 11

76

Recursively find the median of the medians

9

6

5

2

1

7 10 1 3 2 11

78

Partition around the median of the medians

5

Continue recursively with the side that contains the kth element

79

Neither side can be large

5

≤ ¾n

≤ ¾n

80

The reason

1

3 2

9

6

5

2

1

7 10 11

≥

81

The reason

1

3 2

9

6

5

2

1

7 10 11

≤

82

Analysis

3 1( ) ( )

4 5T n O n T n T n

( ) ( )T n O n

83

Order statistics, a dynamic version

rank and select

84

The dictionary ADT

• Insert(x,D)• Delete(x,D)• Find(x,D): Returns a pointer to x if x ∊ D, and

a pointer to the successor or predecessor of x if x is not in D

85

Suppose we want to add to the dictionary ADT

• Select(k,D): Returns the kth

element in the dictionary:

An element x such that k-1 elements are smaller than x

86

Select(5,D)

9089

7773

70673426

2120194

87

Select(5,D)

9089

7773

70673426

2120194

88

9089777370673426

2120194

Can we still use a red-black tree ?

89

For each node v store # of leaves in the subtree of v

9089777370673426

2120194

22

4

2 2

4 4

2 2

8

12

90

Select(7,T)

9089777370673426

2120194

22

4

2 2

4 4

2 2

8

12

91

Select(7,T)

9089777370673426

2120194

22

4

2 2

4 4

2 2

8

12Select(3, )

92

Select(7,T)

9089777370673426

2120194

22

4

2 2

4 4

2 2

8

12

Select(3, )

93

Select(1,)

Select(7,T)

9089777370673426

2120194

22

4

2 2

4 4

2 2

8

12

94

Select(i,T)

Select(i,T): Select(i,root(T))

Select(k,v): if k = 1 then return v.left if k = 2 then return v.right if k ≤ (v.left).size

then return Select(k,v.left) else return Select(k – (v.left).size),v.right)

O(logn) worst case time

95

Rank(x,T)

• Return the index of x in T

96

Rank(x,T)

xNeed to return 9

97

9089777370673426

2120194

22

4

2 2

4 4

2 2

8

12

xSum up the sizes of the subtrees to the left of the path

98

Rank(x,T)

• Write the p-code

99

Insertion and deletions

• Consider insertion, deletion is similar

100

Insert

4

2

8

12

101

Insert (cont)

5

3

9

13

2

102

Easy to maintain through rotations

x

y

B

C

y

Ax

B C

<===>

A

size(x) ← size(B) + size(C)

size(y) ← size(A) + size(x)

103

Summary

• Insertion and deletion and other dictionary operations still take O(log n) time