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

1. 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 *(P* N * (i-x) + P*(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

2. 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 A[i][j][k]=B+W×(M×P×(k−z)+M×(j−y)+(i−x))

Solution:
Address of arr[3][3][3] = 400 + 4 * ((11*16*(3-1)+11*(3-(-10)+(3-(-5)))
                                    = 400 + 4 * ((176*2 + 11*13 + 8)
                                   = 400 + 4 * (503)
                                  = 400 + 2012
                                 = 2412

This article focuses on calculating the address of any element in a 1-Dimensional, 2-Dimensional, and 3-Dimensional array in Row major order and Column major order.