![Dr. Pradyumansinh U. Jadeja #3130702 (DS) Unit 2 – Linear Data Structure (Array) 3
One Dimensional Array
Simplest data structure that makes use of computed address to locate its elements is the one-
dimensional array or vector.
Number of memory locations is sequentially allocated to the vector.
A vector size is fixed and therefore requires a fixed number of memory locations.
Vector A with subscript lower bound of “one” is represented as below….
• L0 is the address of the first word allocated to the first element of vector A
• C words is size of each element or node
• The address of element Ai is
Loc(Ai) = L0 + (C*(i-1))
• Let’s consider the more general case of a vector A with lower bound for it’s
subscript is given by some variable b.
• The address of element Ai is
Loc(Ai) = L0 + (C*(i-b))
A[i]
L0
L0+(i-1)C
i-1](https://image.slidesharecdn.com/array-221223114246-47a7c2e3/85/array-pptx-3-320.jpg)
![Dr. Pradyumansinh U. Jadeja #3130702 (DS) Unit 2 – Linear Data Structure (Array) 4
Two Dimensional Array
Two dimensional arrays are also called table or matrix
Two dimensional arrays have two subscripts
Column major order matrix: Two dimensional array in which elements are stored column by
column is called as column major matrix
Two dimensional array consisting of two rows and four columns is stored sequentially by
columns : A[1,1], A[2,1], A[1,2], A[2,2], A[1,3], A[2,3], A[1,4], A[2,4]
[1,1] [1,2] [1,3] [1,4]
[2,1] [2,2] [2,3] [2,4]
Row 1
Row 2
Col-1 Col-2 Col-3 Col-4](https://image.slidesharecdn.com/array-221223114246-47a7c2e3/85/array-pptx-4-320.jpg)
![Dr. Pradyumansinh U. Jadeja #3130702 (DS) Unit 2 – Linear Data Structure (Array) 5
Column major order matrix
The address of element A [ i , j ] can be obtained by expression
Loc (A [ i , j ]) = L0 + (j-1)*2 + (i-1)
Loc (A [2, 3]) = L0 + (3-1)*2 + (2-1) = L0 + 5
In general for two dimensional array consisting of n rows and m columns the address element A [
i , j ] is given by
Loc (A [ i , j ]) = L0 + (j-1)*n + (i – 1)
[1,1] [1,2] [1,3] [1,4]
[2,1] [2,2] [2,3] [2,4]
Row 1
Row 2
Col-1 Col-2 Col-3 Col-4
[1,1]
[2,1]
[1,2]
[2,2]
[1,3]](https://image.slidesharecdn.com/array-221223114246-47a7c2e3/85/array-pptx-5-320.jpg)
![Dr. Pradyumansinh U. Jadeja #3130702 (DS) Unit 2 – Linear Data Structure (Array) 6
Row major order matrix
Row major order matrix: Two dimensional array in which elements are stored row by row is called
as row major matrix
• The address element A [ i , j ] is given by
Loc (A [ i , j ]) = L0 + (i-1)*m + (j – 1)
n = no of rows, m = no of columns
b1 = lower bound subscript of row
u1 = upper bound subscript of row
n = u1 – b1 + 1
b2 = lower bound subscript of column
u2 = upper bound subscript of column
m = u2 – b2 + 1
• The address element A [ i , j ] is given by
[1,1] [1,2] [1,3] [1,m]
[2,1] [2,2] [2,3] [2,m]
[n,1] [n,2] [n,3] [n,m]
nxm
b1
u1
b2 u2
Loc (A [ i , j ]) = L0 + (i-b1)*(u2-b2+1) + (j – b2)](https://image.slidesharecdn.com/array-221223114246-47a7c2e3/85/array-pptx-6-320.jpg)
array.pptx
- 1. Darshan Institute of Engineering & Technology, Rajkot Unit-2 Linear Data Structure Array pradyuman.jadeja@darshan.ac.in +91 9879461848 Computer Engineering Department Data Structures (DS) GTU # 3130702
- 2. Looping Outline Array • Representation of arrays • One dimensional array • Two dimensional array Applications of arrays • Symbol Manipulation (matrix representation of polynomial equation) • Sparse matrix Sparse matrix and its representation
- 3. Dr. Pradyumansinh U. Jadeja #3130702 (DS) Unit 2 – Linear Data Structure (Array) 3 One Dimensional Array Simplest data structure that makes use of computed address to locate its elements is the one- dimensional array or vector. Number of memory locations is sequentially allocated to the vector. A vector size is fixed and therefore requires a fixed number of memory locations. Vector A with subscript lower bound of “one” is represented as below…. • L0 is the address of the first word allocated to the first element of vector A • C words is size of each element or node • The address of element Ai is Loc(Ai) = L0 + (C*(i-1)) • Let’s consider the more general case of a vector A with lower bound for it’s subscript is given by some variable b. • The address of element Ai is Loc(Ai) = L0 + (C*(i-b)) A[i] L0 L0+(i-1)C i-1
- 4. Dr. Pradyumansinh U. Jadeja #3130702 (DS) Unit 2 – Linear Data Structure (Array) 4 Two Dimensional Array Two dimensional arrays are also called table or matrix Two dimensional arrays have two subscripts Column major order matrix: Two dimensional array in which elements are stored column by column is called as column major matrix Two dimensional array consisting of two rows and four columns is stored sequentially by columns : A[1,1], A[2,1], A[1,2], A[2,2], A[1,3], A[2,3], A[1,4], A[2,4] [1,1] [1,2] [1,3] [1,4] [2,1] [2,2] [2,3] [2,4] Row 1 Row 2 Col-1 Col-2 Col-3 Col-4
- 5. Dr. Pradyumansinh U. Jadeja #3130702 (DS) Unit 2 – Linear Data Structure (Array) 5 Column major order matrix The address of element A [ i , j ] can be obtained by expression Loc (A [ i , j ]) = L0 + (j-1)*2 + (i-1) Loc (A [2, 3]) = L0 + (3-1)*2 + (2-1) = L0 + 5 In general for two dimensional array consisting of n rows and m columns the address element A [ i , j ] is given by Loc (A [ i , j ]) = L0 + (j-1)*n + (i – 1) [1,1] [1,2] [1,3] [1,4] [2,1] [2,2] [2,3] [2,4] Row 1 Row 2 Col-1 Col-2 Col-3 Col-4 [1,1] [2,1] [1,2] [2,2] [1,3]
- 6. Dr. Pradyumansinh U. Jadeja #3130702 (DS) Unit 2 – Linear Data Structure (Array) 6 Row major order matrix Row major order matrix: Two dimensional array in which elements are stored row by row is called as row major matrix • The address element A [ i , j ] is given by Loc (A [ i , j ]) = L0 + (i-1)*m + (j – 1) n = no of rows, m = no of columns b1 = lower bound subscript of row u1 = upper bound subscript of row n = u1 – b1 + 1 b2 = lower bound subscript of column u2 = upper bound subscript of column m = u2 – b2 + 1 • The address element A [ i , j ] is given by [1,1] [1,2] [1,3] [1,m] [2,1] [2,2] [2,3] [2,m] [n,1] [n,2] [n,3] [n,m] nxm b1 u1 b2 u2 Loc (A [ i , j ]) = L0 + (i-b1)*(u2-b2+1) + (j – b2)
- 7. Dr. Pradyumansinh U. Jadeja #3130702 (DS) Unit 2 – Linear Data Structure (Array) 7 Applications of Array 1. Symbol Manipulation (matrix representation of polynomial equation) 2. Sparse Matrix Matrix representation of polynomial equation We can use array for different kind of operations in polynomial equation such as addition, subtraction, division, differentiation etc… We are interested in finding suitable representation for polynomial so that different operations like addition, subtraction etc… can be performed in efficient manner. Array can be used to represent Polynomial equation.
- 8. Dr. Pradyumansinh U. Jadeja #3130702 (DS) Unit 2 – Linear Data Structure (Array) 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 X X2 X3 X4 Y Y2 Y3 Y4 Representation of Polynomial equation 2X2 + 5XY + Y2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 X X2 X3 X4 Y Y2 Y3 Y4 2 5 1 X2 + 3XY + Y2+Y-X 1 3 1 1 -1 Y Y2 Y3 Y4 X XY XY2 XY3 XY4 X2 X3Y X2Y2 X2Y3 X2Y4 X3 X3Y X3Y2 X3Y3 X3Y4 X4 X4Y X4Y2 X4Y3 X4Y4
- 9. Dr. Pradyumansinh U. Jadeja #3130702 (DS) Unit 2 – Linear Data Structure (Array) 9 Sparse matrix An m x n matrix is said to be sparse if “many” of its elements are zero. A matrix that is not sparse is called a dense matrix. We can device a simple representation scheme whose space requirement equals the size of the non-zero elements. 0 0 0 2 0 0 1 0 0 6 0 0 7 0 0 3 0 0 0 9 0 8 0 0 0 4 5 0 0 0 0 0 Terms Row Column Value 4x8 Row - 1 Row - 2 Row - 3 Row - 4 Column - 1 Column - 2 Column - 3 Column - 4 Column - 5 Column - 6 Column - 7 Column - 8 Linear Representation of given matrix 0 1 4 2 1 1 7 1 2 2 2 6 3 2 5 7 4 2 8 3 5 3 4 9 6 3 6 8 7 4 2 4 8 4 3 5
- 10. Dr. Pradyumansinh U. Jadeja #3130702 (DS) Unit 2 – Linear Data Structure (Array) 10 Sparse matrix Cont… To construct matrix structure from liner representation we need to record. Original row and columns of each non zero entries. Number of rows and columns in the matrix. So each element of the array into which the sparse matrix is mapped need to have three fields: row, column and value
- 11. Dr. Pradyumansinh U. Jadeja #3130702 (DS) Unit 2 – Linear Data Structure (Array) 11 Sparse matrix Cont… 0 0 6 0 9 0 0 2 0 0 7 8 0 4 10 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 5 Row Column A 6x7 1 2 3 4 5 6 1 2 3 4 5 6 7 Memory Space required to store 6x7 matrix 42 x 2 = 84 bytes Memory Space required to store Linear Representation 30 x 2 = 60 bytes Linear representation of Matrix A = 1 3 6 1 5 9 2 1 2 2 4 7 2 5 8 2 7 4 3 1 10 4 3 12 6 4 3 6 7 5 Space Saved = 84 – 60 = 24 bytes
- 12. Dr. Pradyumansinh U. Jadeja #3130702 (DS) Unit 2 – Linear Data Structure (Array) 12 Sparse matrix Cont… Row 1 1 2 2 2 2 3 4 6 6 Column 3 5 1 4 5 7 1 3 4 7 A 6 9 2 7 8 4 10 12 3 5 Linear Representation of Matrix Column 3 5 1 4 5 7 1 3 4 7 A 6 9 2 7 8 4 10 12 3 5 Linear Representation of Matrix Row 1 3 7 8 0 9 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 Memory Space required to store Liner Representation = 26 x 2 = 42 bytes
- 13. Darshan Institute of Engineering & Technology, Rajkot Thank You pradyuman.jadeja@darshan.ac.in +91 9879461848 Computer Engineering Department Data Structures (DS) GTU # 3130702