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Lecture 2

Arrays

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Linear Data structures

Linear form a sequence Linear relationship b/w the elements

represented by means of sequentialmemory location

Link list and arrays are linear relationship

Non-Linear Structure are Trees and Graphs

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Operation performed by LinearStructure

Traversal: Processing each element in the list Search : Find the location of the element with a

given value or the record with a given

key Insertion : Adding new element to the list Deletion : Removing an element from the list

Sorting : Arranging the elements in some typeof order

Merging : Combining two list into single list

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Linear Array

List of finite number N of homogenousdata elements (i.e. data elements of sametype) such that The elements of the array are referenced

respectively by an index set consisting of Nconsecutive number

The elements of the array are storedrespectively in successive memory location

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Length of Array

N = length of array

Length = UB LB + 1

UB = Upper Bound or Largest Index

LB= Lower Bound or smallest Index

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Representation in Memory

Address of any element in Array =LOC(LA[k])=Base (LA) + w (k - LB)

LOC(LA[k]) =Address of element LA[k] of the Array LA

Base (LA) = Base Address of LA

w = No. of words per memory cell for the Array LA k = Any element of Array

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Multidimensional Arrays

Two Dimensional Arrays Matrices in Mathematics Tables in Business Applications

Matrix Arrays1 2 3

1 A[1,1] A[1,2] A[1,3]

2 A[2,1] A[2,2] A[2,3]3 A[3,1] A[3,2] A[3,3]

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Column and Row Major Order 2 Dimension

Loc(A[J,K])= Base(A) + w [M(K-LB)+(J-LB)] (Column Major Order)

Loc(A[J,K])= Base(A) + w [N(J-LB)+(K-LB)] (Row Major Order)

1,1

2,1

3,1

1,2

2,2

3,2

1,32,3

3,3

1,4

2,43,4

1,1

1,2

1,3

1,4

2,1

2,2

2,32,4

3,1

3,2

3,33,4

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Example: Row and Column Major Order, 2 Dimension Array.

AAA (4:13,-7:10)

How many elements can be stored in the array. Calculate the address of AAA(6,-6) by Row Major Order and Column Major Order Base address = 1500 and word size = 4

Length of first dimension = UB LB + 1 => 13 4 + 1 = 10 (M) Length of second dimension = UB LB + 1 => 10 (-7) + 1 = 18 (N)

Size = M * N => 10 * 18 = 180 elements can be stored in the array.

Address of (6,-6) by Row Major Order: Loc(A[J,K]) = Base(A) + w [N(J-LB)+(K-LB)] Loc(6,-6) = 1500 + 4 [18(6-4)+(-6-(-7))] => 1500 + 4(37) => 1500 + 148 = 1648

Address of (6,-6) by Column Major Order: Loc(A[J,K]) = Base(A) + w [M(K-LB)+(J-LB)] Loc(6,-6) = 1500 + 4 [10(-6-(-7))+(6-4)] => 1500 + 4(12) => 1500 + 48 = 1548

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Column and Row Major Order 3 Dimension

Length = Upper Bound Lower Bound +1

Effective Index = Index Number Lower Bound

1,1,1

2,1,1

1,2,1

2,2,1

1,3,1

2,3,1

1,4,1

2,4,1

1,4,32,4,3

1,1,1

1,1,2

1,1,3

1,2,1

1,2,2

1,2,3

1,3,1

1,3,2

2,4,22,4,3

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Column and Row Major Order 3 Dimension

Length = Upper Bound Lower Bound +1

Effective Index = Index Number Lower Bound

Row Major Order:Loc = Base + w(E1L2+E2)L3+E3)

Column Major Order:

Loc = Base + w(E3L2+E2)L1+E1)

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Column and Row Major Order 3 DimensionRow Major Order:

B ( 2 : 8 , - 4 : 1 , 6 : 10 )

L1 = 8 2 + 1 = 7L2 = 1 - ( - 4 ) + 1 = 6L3 = 10 - 6 + 1 = 5

E1 = 5 2 = 3E2 = - 1 - ( - 4 ) = 3E3 = 8 - 6 = 2

E1 . L2 = 3 . 6 = 18E1 . L2 + E2 = 18 + 3 = 21

( E1 . L2 + E2 ) . L3 = 21 . 5 = 105(E1.L2 + E2).L3 + E3 = 105+2 =107

Loc = Base + w(E1L2+E2)L3+E3)

Loc [5,-1,8] = 200 + 4 (107) = 628

Column Major Order:

B ( 1 : 8 , - 5 : 5 , - 10 : 5 )

L1 = 8 1 + 1 = 8L2 = 5 - ( - 5 ) + 1 = 11L3 = 5 - ( - 10 ) + 1 = 16

E1 = 3 1 = 2E2 = 3 - ( - 5 ) = 8E3 = 3 - ( - 10 ) = 13

E3 . L2 = 13 . 11 = 143E3 . L2 + E2 = 143 + 8 = 151

( E3 . L2 + E2 ) . L1 = 151 . 8 = 1208(E3.L2 + E2).L1 + E1 = 1208+2 =1210

Loc = Base + w(E3L2+E2)L1+E1)

Loc [3,3,3] = 400 + 4 (1210) = 5240

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Operations on Array

Traversing a Linear Array

TraverseArray (LA, LB, UB)Function: This algorithm traverse LA

applying an operation PROCESS

to each element of LAInput: LA is a Linear Array with Lower BoundLB and Upper bound UB

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Algorithm:1. [Initialize Counter] Set K:=LB2. Repeat Steps 3 and 4 while K UB 3. [Visit element] Apply PROCESS to

LA[K]4. [Increase counter] Set K:=K+1

[End of Step 2 loop]5. Exit

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Operations Cont.

Insert an element in Linear ArrayInsertElement (LA, ITEM, N, K)Function: This algorithm insert an element in

a Linear Array at required positionInput: LA is a Linear Array having N

elementsITEM is the element to be inserted atgiven position K

Precondition: K N where K is a +ve integer

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Algorithm:

1. [Initialize Counter] Set J:=N2. Repeat Steps 3 and 4 while J K 3. [Move Jth element downward]

Set LA[J+1] := LA[J]4. [Decrease counter] Set J:=J-1

[End of Step 2 loop]

5. [Insert element] Set LA[K]:=ITEM6. [Reset N] N:= N+17. Exit

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Operation Cont. Delete an element from a Linear Array

DeleteElement (LA, ITEM, N, K)Function: This algorithm delete an element

from a given position in Linear Array

Input: LA is a Linear Array having N elementsK is the position given from which ITEM

needs to be deletedOutput: ITEM is the element deleted fromthe given position K

Precondition: K N where K is a +ve integer

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Algorithm:

1. Set ITEM:=LA[K]2. Set J:= K3. Repeat Steps 4 and 5 while J