daa ppt 28th jan

8/2/2019 Daa Ppt 28th Jan

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BITS PilaniPilani Campus

Design and Analysis of

AlgorithmsDr. Maheswari Karthikeyan

28/01/2012, Lecture2


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BITS PilaniPilani Campus

Bubble Sort AnalysisMerge Sort AnalysisGrowth of Functions

Asymptotic notation


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BITS Pilani, Pilani Campus

Idea: Repeatedly pass through the array

Swaps adjacent elements that are out of order

Easier to implement, but slower than Insertion sort

Bubble Sort

1 2 3 ni

1329648

j


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Example

1329648

i = 1 j

3129648

i = 1 j

3219648

i = 1 j

3291648

i = 1 j

3296148

i = 1 j

3296418

i = 1 j

3296481

i = 2 j

3964821

i = 3 j

9648321

i = 4 j

9684321

i = 5 j

9864321

i = 6 j

9864321

i = 7

j


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Alg.:BUBBLESORT(A)

fori 1tolength[A]

do forj length[A]downtoi + 1

do ifA[j] < A[j -1]then exchange A[j] A[j-1]

Bubble Sort

1329648

i = 1 j

i


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BITS Pilani, Pilani CampusT(n) =

(n

2

)

Bubble-Sort Running Time

222

)1()(

1 1

2

2

1

nnnnninin

n

i

n

i

n

i

T(n) = c1(n+1) +

n

i

in

1

)1(c2 c3

n

i

in

1

)( c4

n

i

in

1

)(

= (n) + (c2 + c2 + c4)

n

i

in1

)(

Alg.: BUBBLESORT(A)fori 1tolength[A]

do forj length[A]downtoi + 1

do ifA[j] < A[j -1]

then exchange A[j] A[j-1]

Exchanges: n2/2Comparisons n2/2

do if A[j] < A[j -1]


7/35BITS Pilani, Pilani Campus

Insertion sort Design approach:

Sorts in place:

Best case:

Worst case:

Bubble Sort

Design approach:

Sorts in place: Running time:

Sorting

Yes

(n)

(n2)

incremental

Yes(n2)

incremental



Selection sort Design approach:

Sorts in place:

Running time:

Merge Sort

Design approach:

Sorts in place:

Running time:

Sorting

Yes

(n2)

incremental

No

Lets see!!

divide and conquer



Divide the problem into a number of sub-problems

Similar sub-problems of smaller size

Conquer the sub-problems

Solve the sub-problems recursively

Sub-problem size small enough solve the problems in

straightforward manner

Combine the solutions to the sub-problems

Obtain the solution for the original problem

Divide-and-Conquer



To sort an array A[p . . r]:

Divide Divide the n-element sequence to be sorted into two subsequences of n/2

elements each

Conquer Sort the subsequences recursively using merge sort

When the size of the sequences is 1 there is nothing more to do

Combine

Merge the two sorted subsequences

Merge Sort Approach



Alg.: MERGE-SORT(A, p, r)

if p < r Check for base casethen q(p + r)/2 Divide

MERGE-SORT(A, p, q) ConquerMERGE-SORT(A, q + 1, r) Conquer

MERGE(A, p, q, r) Combine

Initial call:MERGE-SORT(A, 1, n)

Merge Sort

1 2 3 4 5 6 7 8

62317425

p rq



Examplen Power of 2

1 2 3 4 5 6 7 8

q = 462317425

1 2 3 4

7425

5 6 7 8

6231

1 2

25

3 4

74

5 6

31

7 8

62

1

5

2

2

3

4

4

7 1

6

3

7

2

8

6

5

Example
http://c/Documents%20and%20Settings/MCADEPARTMENT/Local%20Settings/Temporary%20Internet%20Files/Content.IE5/27PAKPBG/MergeSort.ppthttp://c/Documents%20and%20Settings/MCADEPARTMENT/Local%20Settings/Temporary%20Internet%20Files/Content.IE5/27PAKPBG/MergeSort.ppt



Examplen Power of 2

1

5

2

2

3

4

4

7 1

6

3

7

2

8

6

5

1 2 3 4 5 6 7 8

76543221

1 2 3 4

7542

5 6 7 8

6321

1 2

52

3 4

74

5 6

31

7 8

62



Examplen Not a Power of 2

62537416274

1 2 3 4 5 6 7 8 9 10 11

q = 6

416274

1 2 3 4 5 6

62537

7 8 9 10 11

q = 9q = 3

274

1 2 3

416

4 5 6

537

7 8 9

62

10 11

74

1 2

2

3

16

4 5

4

6

37

7 8

5

9

2

10

6

11

4

1

7

2

6

4

1

5

7

7

3

8



Examplen Not a Power of 2

77665443221

1 2 3 4 5 6 7 8 9 10 11

764421

1 2 3 4 5 6

76532

7 8 9 10 11

742

1 2 3

641

4 5 6

753

7 8 9

62

10 11

2

3

4

6

5

9

2

10

6

11

4

1

7

2

6

4

1

5

7

7

3

8

74

1 2

61

4 5

73

7 8



Input: Array Aand indices p, q, rsuch that p q < r Subarrays A[p . . q] and A[q + 1 . . r] are sorted

Output: One single sorted subarray A[p . . r]

Merging

1 2 3 4 5 6 7 8

63217542

p rq



Idea for merging:

Two piles of sorted cards

Choose the smaller of the two top cards

Remove it and place it in the output pile Repeat the process until one pile is empty

Take the remaining input pile and place it face-down onto the output pile

Merging



MERGE(A, 9, 12, 16)

p r



Example: MERGE(A, 9, 12, 16)



Example (cont.)



Example (cont.)

Done!


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Alg.:MERGE(A, p, q, r)

1. Compute n1and n2

2. Copy the first n1 elements into L[1 . . n1 + 1] and the next n2elements into R[1 . . n2 + 1]

3. L[n1 + 1]; R[n2 + 1]

4. i 1; j 1

5. for k pto r

6. do if L[ i ] R[ j ]7. then A[k] L[ i ]

8. ii + 1

9. else A[k] R[ j ]

10. j j + 1

Merge - Pseudocode

p q

7542

6321

rq + 1

L

R

1 2 3 4 5 6 7 8

63217542

p rq

n1 n2


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Initialization (copying into temporary arrays):

(n1 + n2) = (n)

Adding the elements to the final array (the last for loop):

n iterations, each taking constant time (n)

Total time for Merge:

(n)

Running Time of Merge


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The recurrence is based on the three steps of theparadigm: T(n) running time on a problem of size n

Divide the problem into a subproblems, each of size n/b: takes D(n)

Conquer (solve) the subproblems aT(n/b)

Combine the solutions C(n)

(1) if n c

T(n) = aT(n/b) + D(n) + C(n) otherwise

Analyzing Divide-and ConquerAlgorithms


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Divide: compute qas the average of pand r:D(n) = (1)

Conquer:

recursively solve 2 subproblems, each of size n/22T (n/2)

Combine:

MERGE on an n-element subarray takes (n) timeC(n) = (n)(1) if n =1

T(n) = 2T(n/2) + (n) if n > 1

MERGE-SORT Running Time


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T(n) = c if n = 12T(n/2) + cn if n > 1

Use Masters Theorem:

Compare n with f(n) = cn

Case 2: T(n) = (nlgn)

Solve the Recurrence


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A way to describe behavior of functions in the limit

How we indicate running times of algorithms

Describe the running time of an algorithm as n grows to

O notation: asymptotic less than: f(n) g(n)

notation: asymptotic greater than: f(n) g(n)

notation: asymptotic equality: f(n) = g(n)

Asymptotic Notations


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For each of the following pairs of functions, either f(n) isO(g(n)), f(n) is (g(n)), or f(n) is (g(n)). Determine

which relationship is correct. f(n) = log n2; g(n) = log n + 5

f(n) = n; g(n) = log n2

f(n) = log log n; g(n) = log n

f(n) = n; g(n) = log2 n

Asymptotic Notations -Examples

f(n) = (g(n))

f(n) = (g(n))

f(n) = O(g(n))

f(n) = (g(n))


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Intuitively: O(g(n)) = the set of

functions with a smaller or

same order of growth as g(n)

Asymptotic notations

O-notation


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3n + 2 = O(n) ; 3n + 2 = 2

3n + 3 = O(n) ; 3n + 3 = 3

100n + 6 = O(n) ; 100n + 6 = 6

Examples


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3n + 2 = O(n) ; 3n + 2 = 2

3n + 3 = O(n) ; 3n + 3 = 3

100n + 6 = O(n) ; 100n + 6 = 6

10 n 2 + 4n + 2

Examples

= O(n2)

10 n 2 + 4n + 2 < = 11 n2 for n >= 5


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- notation

Asymptotic notations (cont.)

Intuitively: (g(n)) = the set of

functions with a larger or same

order of growth as g(n)


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3n + 2 = ?

3n + 3 = ?

100n + 6 = ?

Examples

3n + 2 >= 3n for all n >= 1

3n + 3 >= 3n for all n >= 1

100n + 6 >= 100n for all n >= 1


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-notationAsymptotic notations (cont.)

Intuitively(g(n)) = the set offunctions with the same order of

growth as g(n)