cs223 advanced data structures and algorithms 1 sorting and master method neil tang 01/21/2009

CS223 Advanced Data Structures and Algorithms 1

Sorting and Master Method Sorting and Master Method

Neil TangNeil Tang01/21/200901/21/2009


Class OverviewClass Overview

Review of sorting algorithms: Insertion, merge and quick

The master method


Insertion SortInsertion Sort

Start from the second element and select an element (key) in each step.

Insert the key to the current sorted subsequence such that the new subsequence is in the sorted order.



3 5 2 8

5

ji

key

3

3 5 2 8

2

ji

key

3

2 3 5 8 3

2 3 5 8

8

ji

key

3

2 3 5 8

3

ji

key

3

2 3 3 5 8

CS223 Advanced Data Structures and Algorithms


Insertion-Sort(A[1..n])1 for j := 2 to n do 2 key := A[j]

i := j -15 while i > 0 and A[i] > key do 6 A[i+1] := A[i]7 i := i - 18 A[i+1] := key

5



Best case: the array is sorted

each while loop only takes constant time, which is repeated (n-1) times, so time complexity is Θ(n) .

Worst case: the array is in reverse order

each while loop takes cj time. So time complexity is

)(2

)1)(2( 2

2

nnnc

jcn

j


Merge SortMerge Sort

Divide: Divide the N-element sequence into 2 subsequences of N/2 each.

Conquer: Sort each subsequence recursively using merge sort.

Combine: Merge two sorted subsequences to produce a single sorted sequence.

The time complexity is O(NlogN)


Merge SortMerge Sort

5, 3, 6, 8, 12, 3, 1, 4, 7, 9, 13, 5, 6, 8, 12 1, 3, 4, 7, 9, 14

1, 3, 3, 4, 5, 6, 7, 8, 9, 12, 14


Quick SortQuick Sort

Divide: Divide the sequence into 2 subsequences, s.t. each element in the 1st subsequence is less than or equal to each element in the 2nd subsequence.

Conquer: Sort each subsequence recursively using quick sort.

Combine: no work is needed.


Quick SortQuick Sort

age25

Peopleage < 25

peopleage 25

age23

peopleage < 23

age30

peopleage 30

peopleage 23

peopleage < 30


Master MethodMaster Method

Recurrence in general form: T(N) = aT(N/b) + f(N), a 1 and b > 1.

Case 1: If f(N) = O(Nlogb

a - ) for some constant > 0, then T(N) = (Nlog

ba) .

Case 2: If f(N) = (Nlogb

a), then T(N) = (Nlogb

a logN).

Case 3: If f(N) = (Nlogb

a + ) for some constant > 0, and a*f(N/b)c*f(N) for some constant c < 1 and all sufficiently large N, then T(N) = (f(N)) .


Example: Binary Search Example: Binary Search

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

a=1, b=2, f(N)=1

log21 = 0

So we have case 2, T(N) = (logN).


Example: Merge Sort Example: Merge Sort

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

a=2, b=2, f(N)=N

log22 = 1

So we have case 2, T(N) = (NlogN).


Example: Matrix Multiplication Example: Matrix Multiplication

T(N) = 8T(N/2) + N2

a=8, b=2, f(N)=N2

log28 = 3

= 0.5

So we have case 1, T(N) = (N3).


More ExampleMore Example

T(N) = 3T(N/4) + NlogN

a=3, b=4, f(N) = NlogN

log43 = 0.7925

=0.2, c=3/4

So we have case 3, T(N) = (NlogN).


LimitationsLimitations

Does the master method work all the time?

Quick sort: T(N) = T(N-1) + N

Fibonacci number: T(N) = T(N-1) + T(N-2) + 1

Does it work for all the cases in the format of

T(N) =aT(N/b) + f(N)?

cs223 advanced data structures and algorithms 1 sorting and master method neil tang 01/21/2009

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