Calculate the address of any element in the 3-D Array

A 3-Dimensional array is a collection of 2-Dimensional arrays. It is specified by using three subscripts:

1. Block size
2. Row size
3. Column size

More dimensions in an array mean more data can be stored in that array. 

Example:

To find the address of any element in 3-Dimensional arrays there are the following two ways:

• Row Major Order
• Column Major Order

Row Major Order:

To find the address of the element using row-major order, use the following formula:

Address of A[i][j][k] = B + W *(M * N * (i-x) + N *(j-y) + (k-z))

Here:

B = Base Address (start address)
W = Weight (storage size of one element stored in the array)
M = Row (total number of rows)
N = Column (total number of columns)
P = Width (total number of cells depth-wise)
x = Lower Bound of Row
y = Lower Bound of Column
z = Lower Bound of Width

Example: Given an array, arr[1:9, -4:1, 5:10] with a base value of 400 and the size of each element is 2 Bytes in memory find the address of element arr[5][-1][8] with the help of row-major order?

Solution:

Given:
Block Subset of an element whose address to be found I = 5
Row Subset of an element whose address to be found J = -1 
Column Subset of an element whose address to be found K = 8
Base address B = 400
Storage size of one element store in any array(in Byte) W = 2
Lower Limit of blocks in matrix x = 1
Lower Limit of row/start row index of matrix y = -4 
Lower Limit of column/start column index of matrix z = 5
M(row) = Upper Bound – Lower Bound + 1 = 1 – (-4) + 1 = 6
N(Column)= Upper Bound – Lower Bound + 1 = 10 – 5 + 1 = 6
 

Formula used:                                                
Address of[I][J][K] =B + W (M * N(i-x) + N *(j-y) + (k-z))

Solution:
Address of arr[5][-1][8] = 400 + 2 * {[6 * 6 * (5 – 1)] + 6 * [(-1 + 4)]} + [8 – 5]
                             = 400 + 2 * (6*6*4)+(6*3)+3
                            = 400 + 2 * (165)
                           = 730

Column Major Order:

To find the address of the element using column-major order, use the following formula:1

Address of A[i][j][k]= B + W(M * N(i – x) + M *(k – z) + (j – y))

Here:

B = Base Address (start address)
W = Weight (storage size of one element stored in the array)
M = Row (total number of rows)
N = Column (total number of columns)
P = Width (total number of cells depth-wise)
x = Lower Bound of block (first subscipt)
y = Lower Bound of Row
z = Lower Bound of Column

Example: Given an array arr[1:8, -5:5, -10:5] with a base value of 400 and the size of each element is 4 Bytes in memory find the address of element arr[3][3][3] with the help of column-major order ?

Solution:

Given:
Row Subset of an element whose address to be found I = 3
Column Subset of an element whose address to be found J = 3
Block Subset of an element whose address to be found K = 3
Base address B = 400
Storage size of one element store in any array(in Byte) W = 4
Lower Limit of blocks in matrix x = 1
Lower Limit of row/start row index of matrix y = -5
Lower Limit of column/start column index of matrix z = -10
M (row)= Upper Bound – Lower Bound + 1 = 5 +5 + 1 = 11
N (column)= Upper Bound – Lower Bound + 1 = 5 + 10 + 1 = 16

Formula used:
Address of[i][j][k] = B + W(M * N(i – x) + M * (j-y) + (k – z))

Solution:
Address of arr[3][3][3] = 400 + 4 * ((11*16*(3-1)+11*(3-(-5)+(3-(-10)))
                                    = 400 + 4 * ((176*2 + 11*8 + 13)
                                   = 400 + 4 * (453)
                                  = 400 + 1812
                                 = 2212

Arrays are fundamental data structures in computer science, and mastering them is crucial for success in the GATE exam. This article aims to provide concise yet comprehensive notes on arrays, covering essential concepts and strategies to help you tackle array-related questions in the GATE 2024 exam.