Sorting in data structure

8/13/2019 Sorting in data structure

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SORTING

Sorting is the process of arranging the elements insome logical order.

Sorting are classified into following categories:

External sorting:

deals with sorting of the data stored in data files.This method is used when the volume of data isvery large and cannot be held in computer mainmemory.

Internal sorting:deals with sorting the data held in memory of thecomputer


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SORTING METHODS

1. Bubble sort

2. Selection sort

3. Insertion sort

4. Bucket sort5. Merge sort

6. Quick sort

7. Heap sort8. Tree sort

9. Shell sort


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BUBBLE SORT

It requires n-1 passes to sort an array.

In each pass every element a[i] is compared witha[i+1], for i=0 to (n-k), where k is the passnumber and if they are out of order i. e. if

a[i]>a[i+1], they are swapped. This will cause the largest element move up or

bubble up.

Thus after the end of the first pass the largest

element in the array will be placed in the nthposition and on each successive pass, the nextlargest element is placed at position (n-1),(n-2).,2 respectively


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BUBBLE SORT

Pass1.

Step1. if a[0]>a[1] then swap a[0] and a[1].


Stepn-1. if a[n-2]>a[n-1] then swap a[n-2] and a[n-1].

Pass2.



Stepn-2. if a[n-3]>a[n-2] then swap a[n-3] and a[n-2].


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BUBBLE SORT

.

.

Pass k.

Step1. if a[0]>a[1] then swap a[0] and a[1].Step2. if a[1]>a[2] then swap a[1] and a[2].

Step n-k. if a[n-k+1]>a[n-k] then swap a[n-k+1] and

a[n-k].Pass n-1

Step 1 if a[0]>a[1] then swap a[0] and a[1].


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BUBBLE SORT

Example: 12 40 3 2 15

12

40

3

2

15

1240

3

215

123

40

215

123

2

4015

12

3

2

15

40

Given

array

Pass 1


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BUBBLE SORT

3

12

2

15

40

3

2

12

15

40

3

2

12

15

40

Pass 2


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BUBBLE SORT

Pass 3 pass 4

2

3

12

15

40

2

3

12

15

40

2

3

12

15

40


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ALGORITHM

Bubblesort(a,n)

for k=1 to (n-1) by 1 do

for j=0 to (n-k-1) by 1 do

if(a[j]>a[j+1]) then

set temp=[j]

set a[j]=a[j+1]

set a[j]=temp

endif

endfor

Endfor

end


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ANALYSIS OF BUBBLE SORT

First pass require n-1 comparison

Second pass requires n-2 comparison

Kth pass requires n-k comparisons

Last pass requires only one comparison

Therefore total comparisons are:

F(n)=(n-1)+(n-2)++(n-k)+3+2+1

=n(n-1)/2

=O(n2)


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SELECTION SORT

The selection sort also requires (n-1) passes

to sort an array.

In the first pass, find the smallest element

from elements a[0], a[1], a[2],.., a[n-1]

and swap with the first element, i.e. a[0].

In the second pass, find the smallest

element from elements a[1], a[2], a[3]..

a[n-1] and swap with a[1] and so on.


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SELECTION SORT

Pass1.

Find the location loc of the smallest element in the entire array, i.e.a[0],[1],a[2]a[n-1]

Interchange a[0] & a[loc]. Then a[0] is trivially sorted.

Pass2.

Find the location loc of the smallest element in the entire array, i.e.a[1],a[2]a[n-1]

Interchange a[1] & a[loc]. Then a[0], a[1] are sorted.

Passk.

Find the location loc of the smallest element in the entire array, i.e.a[k],a[k+1],a[k+2]a[n-1]

Interchange a[k] & a[loc]. Then a[0],a[1],a[2],a[k] are sorted.Passn-1.

Find the location loc of the smaller of the element a[n-2],a[n-1]

Interchange a[n-2] & a[loc]. Then elements a[0],a[1],a[2].a[n-1].


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EXAMPLE

Given array: 20 35 40 100 3 10 15

a[0] a[1] a[2] a[3] a[4] a[5] a[6]

20 35 40 100 3 10 15

Pass 1:

a[0] a[1] a[2] a[3] a[4] a[5] a[6]

20 35 40 100 3 10 15

Loc=4

Interchange elements a[0] & a[4] i.e. 20 and 3


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SELECTION SORT

a[0] a[1] a[2] a[3] a[4] a[5] a[6]

3 35 40 100 20 10 15

a[0] a[1] a[2] a[3] a[4] a[5] a[6]

3 10 40 100 20 35 15

Pass 2Loc=5


a[0] a[1] a[2] a[3] a[4] a[5] a[6]

3 10 15 100 20 35 15

Pass3

Pass 4



Loc=6

Loc=4


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SELECTION SORT

a[0] a[1] a[2] a[3] a[4] a[5] a[6]3 10 15 20 100 35 40

a[0] a[1] a[2] a[3] a[4] a[5] a[6]

3 10 15 20 35 100 40

a[0] a[1] a[2] a[3] a[4] a[5] a[6]

3 10 15 20 35 40 100

Pass 5

Pass 6

Loc=5

Loc=6




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ALGORITHM

Smallestelement(a,n,k,loc)

Here a is linear array of size n. this sub algorithm finds the location locof smallest element among a[k-1],a[k+1],a[k+2]a[n-1]. Temporaryvariable small is used to hold the current smllest element nd j isused loop control variable.

Begin

set small=a[k-1]

set loc=k-1

for j=k to (n-1) by 1 do

if(a[j]


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ALGORITHM

Selectionsort(a,n)

Here a is the linear array with n elements in memory. This algorithmsorts elements into ascending order. It uses a temporary variabletemp to facilitate the exchange of two values and variable I is usedloop control variable

Begin

for i=1 to (n-1) by 1 do

call smllest element(a,n,I,loc)

set temp=a[i-1]

set a[i-1]=a[loc]

set a[loc]=temp

endforend


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ANALYSIS OF SELECTION SORT

First pass require n-1 comparison to find thelocation loc of smallest element

Second pass requires n-2 comparison

Kth pass requires n-k comparisons Last pass requires only one comparison


F(n)=(n-1)+(n-2)++(n-k)+3+2+1

=n(n-1)/2

=O(n2)


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INSERTION SORT

This algorithm is very popular with bridge

players when they sort their cards. In this

procedure, we pick up a particular value

and then insert it at the appropriate placein the sorted sub list.

This algorithm also requires n-1 passes


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INSERTION SORT

Pass1: a[1] is inserted either before or after a[0] sothat a[0] and a[1] are sorted.

Pass2: a[2] is inserted either before a[0] or betweena[0] and a[1] or after a[1] so that the elements a[0], a[1],a[2] are sorted.

Pass3: a[3] is inserted either before a[0] or betweena[0] and a[1] or between a[1] and a[2] or after a[2] sothat the elements a[0], a[1], a[2], a[3] are sorted.

Passk: a[k] is inserted in proper place in the sortedsub array a[0], a[1], a[2],a[k-1] so that the elementsa[0], a[1], a[2],a[k-1],a[k] are sorted.

Passn-1: a[n-1] is inserted in proper place in the sortedsub array a[0], a[1], a[2],a[n-2] so that the elementsa[0], a[1], a[2],a[n-1] are sorted.


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EXAMPLE

Given array: 35 20 40 100 3 10 15

a[0] a[1] a[2] a[3] a[4] a[5] a[6]

35 20 40 100 3 10 15

Pass 1:

a[0] a[1] a[2] a[3] a[4] a[5] a[6]

35 20 40 100 3 10 15

Since a[1]< a[0] insert element a[1] before a[0]


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SELECTION SORT

a[0] a[1] a[2] a[3] a[4] a[5] a[6]

20 35 40 100 3 10 15

a[0] a[1] a[2] a[3] a[4] a[5] a[6]

20 35 40 100 3 10 15

Pass 2

Since a[2]>a[1] no action is performed

a[0] a[1] a[2] a[3] a[4] a[5] a[6]

20 35 40 100 3 10 15

Pass3

Pass 4

Since a[3]>a[2] no action is performed

Since a[4] is less than a[3], a[2], a[1] as well as a[0] thereforeinsert a[4] before a[0]


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SELECTION SORT

a[0] a[1] a[2] a[3] a[4] a[5] a[6]3 20 35 40 100 10 15

a[0] a[1] a[2] a[3] a[4] a[5] a[6]

3 10 20 35 40 100 15

a[0] a[1] a[2] a[3] a[4] a[5] a[6]

3 10 15 20 35 40 100

Pass 5

Pass 6

Since a[5] is less than a[4], a[3], a[2] as well as a[1]

therefore insert a[5] before a[1]

Since a[6] is less than a[5], a[4], a[3] as well as a[2] therefore

insert a[6] before a[2]


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ALGORITHM

insertionsort(a,n)

Here a is the linear array with n elements in memory. This algorithm sortselements into ascending order. It uses a temporary variable temp tofacilitate the exchange of two values and variable j and k are used loopcontrol variables.

Begin

for k=1 to (n-1) by 1 doset temp=a[k]

set a[j]=k-1

while((temp=0) do

set a[j+1]=a[j]

set j=j-1

endwhileset a[j+1]=temp

endfor

end


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ANALYSIS OF INSERTION SORT

The worst case performance occurs when the elementsof the input array are in descending order

First pass require 1 comparison to find the location loc ofsmallest element

Second pass requires 2 comparison Kth pass requires k-1 comparisons

Last pass requires (n-1) comparison


F(n)=1+2+3+..+(n-k)+.+(n-3)+(n-2)+(n-1)=n(n-1)/2

=O(n2)


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Bucket/Radix sort

This is used by most of the people when sorting a list ofnames in alphabetical order. The procedure is:

First, the names are grouped according to the first letter,thus the names are arranged in 26 classes, one for each

letter of alphabet. first class consists of those names thatbegin with letter A, the second class consists of thosenames that begins with letter B, and so on.

Next, the names are grouped according to the secondletter. After this step, the list of name will be sorted on

first two letter. This process is continued for number of times depending

on the length of the names with maximum letters.

B k t/R di t


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Bucket/Radix sort Since there are 26 letter of alphabet, we make

use of 26 buckets, one for each letter of thealphabet.

After grouping these names according to their

specific letter, we collect them according to order

of bucket. This new list becomes input for the next pass i.e

to separate them on the next letter from left.

To sort decimal number where base (radix) is10, we need 10 buckets are numbered from 0-9.

Unlike sorting names, decimal numbers are

sorted from right to left i.e. first on unit digits,

then on ten digit and so on.

example


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example321, 150, 235, 65, 573, 789, 928, 542

321

150

235

65

573

789

928

542

65150 321 542 573 235 928 789

0 1 2 3 4 5 6 7 8 9Input

Pass 1


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150

321

542

573

235

65

928

789

928

321 235 542 150 65 573 789

0 1 2 3 4 5 6 7 8 9Input

Pass 2


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321

928

235

542

150

65

573

789

573

65 150 235 321 542 789 928

0 1 2 3 4 5 6 7 8 9Input

Pass 3


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Bucket/Radix sort

After pass three, when the numbers are

collected, they re in following order

65, 150, 235, 321, 542, 573, 789, 928 thus

the numbers are sorted

Algorithm


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AlgorithmBucketsort(a,n)

Here a is linear array of integer with n elements, the variable digitcount is usedto store the number of digits in the largest number in order to control the

number of passes to be performed.Begin

find the largest number of the array

set digitcount=no. of digits of the largest no.

for pass=1 to digitcount by 1 do

initialize bucketsfor i=1 to n-1 by 1 do

set digit=obtain digit no. pass of a[i]

put a[i] in bucket no. digit

increment bucket count for bucket no. digit

endfor

collect all the numbers from buckets in order

endfor

end


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Analysis of bucket sort Let us suppose the number of digits in the

largest element of the given array is S.

the number of passes to be performed is n. Then , the umber of comparisons, f(n), needed

to sort the given array are

f(n)=


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Merge Sort


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Divide and Conquer

Divide-and-conquer method for algorithmdesign:

Divide: If the input size is too large to deal

with in a straightforward manner, divide theproblem into two or more disjoint subproblems

Conquer: Use divide and conquer recursivelyto solve the subproblems

Combine: Take the solutions to thesubproblems and merge these solutions intoa solution for the original problem


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Merge Sort Algorithm

Divide: If S has at least two elements (nothingneeds to be done if S has zero or one elements),remove all the elements from S and put theminto two sequences, S1and S2, each containing

about half of the elements of S. (i.e. S1containsthe firstn/2 elements and S2contains theremaining n/2 elements).

Conquer: Sort sequences S1and S2using

Merge Sort.

Combine: Put back the elements into S bymerging the sorted sequences S1and S2intoone sorted sequence


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Merge Sort: Algorithm

Merge-Sort(A, p, r)

if p < r then

q(p+r)/2Merge-Sort(A, p, q)

Merge-Sort(A, q+1, r)

Merge(A, p, q, r)

Merge(A, p, q, r)

Take the smallest of the two topmost elements of

sequences A[p..q] and A[q+1..r] and put into the

resulting sequence. Repeat this, until both sequences

are empty. Copy the resulting sequence into A[p..r].


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MergeSort (Example) - 1


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Merge Sort Revisited

To sort n numbers if n=1 done!

recursively sort 2 lists ofnumbers n/2and n/2elements

merge 2 sorted lists in Q(n)time

Strategy break problem into similar

(smaller) subproblems

recursively solvesubproblems

combine solutions to answer


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Analysis of merge Sort

The major work is done in the merge procedure, which isan O(n) operation.

Merge procedure is called from merge sort procedureafter the array is divided into two halves, and each

halves has been sorted. In each recursive calls, one for the left half and one for

the right half, array is divided into four segments.

At each level number of segments doubles, therefore thetotal division are log2n , hence total number of

comparisonsf(n)= n*log2n

=O(nlog2n)


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Quick-Sort


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Another divide-and-conquer sorting algorihm

To understand quick-sort, lets look at a high-level

description of the algorithm1)Divide: If the sequence S has 2 or more elements,

select an element x from S to be your pivot. Anyarbitrary element, like the last, will do. Remove all theelements of S and divide them into 3 sequences:

L, holds Ss elements less than x

E, holds Ss elements equal to x

G, holds Ss elements greater than x

2) Recurse: Recursively sort L and G3) Conquer: Finally, to put elements back into S in order,

first inserts the elements of L, then those of E, and thoseof G.

Here are some diagrams....

f Q S


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Idea of Quick Sort

1) Select: pick an element

2) Divide: rearrangeelements so that x goes

to itsfinal position E

3) Recurse and Conquer:recursively sort

I Pl Q i k S t


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In-Place Quick-Sort

Divide step: l scans the sequence from the left, and r from the right.

A swap is performed when l is at an element larger than the pivot and r is at onesmaller than the pivot.


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In Place Quick Sort (contd)

A final swap with the pivot completes the divide step


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Algorithm

quicksort(a , l, r)Begin

If(l


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Analysis of the quick sort

To find the location of the element that splits thearray into two sections is an O(n) operation,because every element in the array is comparedto the dividing element.

After the division each section is examinedseparately.

If the array is split approximately in half (which isno usually), then there will be log2n splits.

therefore the total comparisons are:

f(n)=n*log2n = O(nlog2n)


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Tree Sort

Tree sort method, in order to sort an array of

size n in ascending order, works in

following phases:

1. Build a binary search tree by using ncalls to insert operation

2. Print elements using inorder traversal.


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Tree Sort

Array : 50, 60, 40, 45, 31, 75, 53, 65

Solution:

When these elements are inserted in binary

search tree one by one, the final binary

search tree looks like in next slide;


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Given Array: 50, 60, 40, 45, 31, 75, 53, 65

50

40 60

75534531

65

Inorder traversal produces the following listing of elements:

31, 40, 45, 50, 53, 60, 65, 75


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Tree Sort v/s Quick Sort

In tree sort , the first item is inserted into the rootof the tree and all subsequent elements arepartitioned to the left or right depending on theirrelation to the first element. This is analogous to

quick sort, if the first element in the array is usedas the pivot element for partitioned.

Further in tree sort, second element becomesthe root of the sub tree. It becomes the pivot

element to partition all subsequent elements inthat subtree. This is analogous to quick sort forpartitioning of one of the sublist.


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Analysis of Tree Sort

All the comparisons for tree sort are done

during insert() calls. The insert() function

does the same number of comparisons as

quick sort. Therefore , tree sort has thesame running time as quick sort, i.e.

average case complexity is O(nlogn) and

worst case complexity is O(n2

)


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Tree Sort

It does not require that all the elements to

be present in the array at the beginning of

sorting

Elements can be added gradually as theybecome available

It also works on a linked structure that

allows easier insertions and deletions thana contiguous list

Shell sort


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Invented by Donald Shell in 1959, the shell sort is themost efficient of the O(n2) class of sorting algorithms. Ofcourse shell sort is also the most complex of the O(n2)

algorithms. This algorithm is similar to bubble sort in the sense it

also moves elements by exchanges.

It begins by comparing elements that are at a distance d.

With this the elements that are quite away from their

place will move rapidly than the simple bubble sort. In each pass, the value of d is reduced to half i.e.

di+1=(di+1)/2

In each pass, each element is compared with elementthat is located d position away from it, and exchange ismade if required.

The next iteration starts with new value of d.

The algorithm terminates when d=1

Example


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12,9,-10,22,2,35,40

Starting value of d=n/2=7/2=3

12

9

-10

22

2

35

40

12

9

-10

22

2

35

40

12

2

-10

22

9

35

40

12

2

-10

22

9

35

40

12

2

-10

22

9

35

40

Given

array Pass 1

12

9

-10

22

2

35

40

Example


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Starting value of di+1=(di+1)/2=(3+1)/2=2

12

2

-10

22

9

35

40

-10

2

12

22

9

35

40

-10

2

9

22

12

35

40

-10

2

9

22

12

35

40

-10

2

9

22

12

35

40

Pass 2

-10

2

12

22

9

35

40

Example


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Starting value of di+1=(di+1)/2=(2+1)/2=1

-10

2

9

22

12

35

40

-10

2

9

22

12

35

40

-10

2

9

22

12

35

40

-10

2

9

12

22

35

40

-10

2

9

22

12

35

40

Pass 3

-10

2

9

22

12

35

40

A l i f h ll


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Analysis of shell sort

It is difficult to predict the complexity of shellsort, as it is very difficult to show the effect of

one pass on another pass.

One thing is very clear that if the new distance dis computed using above formula, then the

number of passes will approximately be log2d

since d=1 will complete the sort.

Empirical studies have shown that the worstcase complexity of shell sort is O(n2)

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Sorting in data structure