ISRO CSE 2014 | Question: 48

Question

A frame buffer array is addressed in row major order for a monitor with pixel locations starting from $(0,0)$ and ending with $(100,100).$ What is address of the pixel $(6,10)?$ Assume one bit storage per pixel and starting pixel location is at $0.$

• A. $1016$
• B. $1006$
• C. $610$
• D. $616$

Comments

@Gaurav Raj 2 isn't that column-major order for pixels? As I understand it row-major for pixels corresponds to the technique we use as column-major for arrays

---

if i solve by row major order answer i am getting is 610 like why people are solving by column major order ?

---

they have not mentioned that (rows =6, column = 10). in computer graphics the standard convention is (column,row).

(row * #columns) + column = 10 * 101 + 6 = 1016.

Answer 1

Address of pixel (6,10) in row major order is,

= 0 + 1((6 - 0) +101 (10 - 0))

= 1016

(For pixel calculation address in row major order , use logic of how to find address of array element in column major order .)

Comments

Why have you taken 101?

---

they have explicitly mentioned row-major order.

---

u are using coloumn major order formula...if we use row major order answer is 616......even if u count manually answer is 616...ISRO should have said coloumn major order for 1016....there is a lot of ambiguity in these ISRO questions

Answer 2

Let’s imagine the frame buffer to be a matrix of 100x100, so we define this as 2d array in C(or any language for that matter), note that the size of the array should be actually array[101][101] since we need to store 100, this is from the fact that if you need to store a single element in a array you need to declare it as array[1] and not array[0] which would not make sense, right?

Now we have maximum rows = maximum colums = 101. So we can get the address of element by using row-major order.

 

Case 1: Row Major Order

Address = Base Adress + Size of Single Element * (Row * Maximum Row Value) + Column

So, we have,

0 + 1 * (6 * 101) + 10 = 616

 

Case 2: Column Major Order

Address = Base Address + Size of Single Element * (Column * Maximum Column Value) + Row

So, we have,

0 + 1 * (10 * 101) + 6 = 1016

I have no clue what is the answer, some website says it’s 616 and some say it’s 1016. If anyone have any ideas, please comment below. I really would like to know.

Comments

The confusion usually stems from the difference between mathematical array indexing and screen coordinate conventions.

While it is true that in programming languages like C, we often write a[row][column], in the context of computer graphics and monitor screens, a point (x, y) follows the standard Cartesian convention where:

x is the horizontal distance (Column).

y is the vertical distance (Row).

Why the answer is 1016 (Standard Graphics Convention)

In most computer graphics problems (including the ISRO 2014 exam where this specific question originated), the notation (6, 10) is treated as (x, y).

Interpret the coordinates: (x, y) = (6, 10) means we are at column 6 and row 10.

Row-Major Logic: To get to row 10, we must skip all pixels in rows 0 through 9.

The Math: * There are 101 pixels per row (from 0 to 100).

10 full rows skipped * 101 pixels per row = 1010 pixels.

Add the 6 pixels in the current row: 1010 + 6 = 1016.

Why you got 616 (Programming Array Convention)

Your calculation of 616 is mathematically correct if the input (6, 10) is interpreted as a[row][column]:

Row index = 6

Column index = 10

Calculation: (6 * 101) + 10 = 606 + 10 = 616.

Answer 3

Column Major System:

The address of a location in Column Major System is calculated using the following formula:

Address of A [ I ][ J ] = B + W * [  ( I – Lr ) + M*( J – Lc ) ]

A[6][10] = 0 + 1[(6-0) +100*10] = 1016

Answer is A

Comments

@ dattasai Please share revised answer key of ISRO

---

https://www.isro.gov.in/sites/default/files/pdf/recruitment/icrb2014-01/Computer%20Science-BE003-key-2014-revised.pdf

 

---

The correct address for the pixel $\mathbf{(6, 10)}$ is 1016.

 

🔍 Calculation

 

$$\text{Address} = (\text{Row Index}) \times (\text{Width}) + (\text{Column Index})$$

• Pixel Coordinates: $(\mathbf{x}, \mathbf{y}) = (6, 10)$.

  • In screen graphics, $\mathbf{y}$ is the Row Index (10).

  • $\mathbf{x}$ is the Column Index (6).

• Monitor Width (W): From 0 to 100 is $100 - 0 + 1 = \mathbf{101}$ pixels.

$$\text{Address} = (\mathbf{10}) \times (101) + (\mathbf{6})$$

$$\text{Address} = 1010 + 6 = \mathbf{1016}$$

---

 

💡 Why This Convention?

 

The formula prioritizes $\mathbf{y}$ (the row) because the storage is Row-Major Order (row-by-row).

1. $(\mathbf{y}) \times (\text{Width})$: This term counts all the pixels in the $\mathbf{10}$ full rows that precede the current row.

2. $+(\mathbf{x})$: This term counts the $\mathbf{6}$ pixels that precede the target pixel in its own row.

$$\text{Total Preceding Pixels} = \text{Address} = \mathbf{1016}$$

---

Note on 616: The option 616 results from incorrectly swapping the coordinates: $(\mathbf{6}) \times 101 + (\mathbf{10})$. While common in certain matrix contexts, it contradicts the standard $(\text{Column}, \text{Row})$ convention for screen graphics.

I got this from internet but it clears my confusion

Answer 4

Hope the answer is A