Interactive end-of-chapter exercises


Error Detection and Correction: Two Dimensional Parity

Suppose that a packet’s payload consists of 10 eight-bit values (e.g., representing ten ASCII-encoded characters) shown below. (Here, we have arranged the ten eight-bit values as five sixteen-bit values):

Figure 1

00000110 01011111
10110011 11111111
00000010 01001010
01110000 00011100
10100001 00101001

Figure 2

Both the payload and parity bits are shown. One of these bits is flipped.

11000100 10000100 1
00000011 00110001 1
11101011 00011010 1
10010001 11110100 0
10000111 10000100 0
00111010 11011110 1

Figure 3

Both the payload and parity bits are shown; Either one or two of the bits have been flipped.

01001000 01111000 0
11000000 10101000 1
11111000 11110010 1
10101011 01010010 1
10110110 01100011 1
00001101 00010011 0


Question List


1. For figure 1, compute the two-dimensional parity bits for the 16 columns. Combine the bits into one string

2. For figure 1, compute the two-dimensional parity bits for the 5 rows (starting from the top). Combine the bits into one string

3. For figure 1, compute the parity bit for the parity bit row from question 1. Assume that the result should be even.

4. For figure 2, indicate the row and column with the flipped bit (format as: x,y), assuming the top-left bit is 0,0

5. For figure 3, is it possible to detect and correct the bit flips? Yes or No




Solution


The full solution for figure 1 is shown below:

00000110 01011111 0
10110011 11111111 1
00000010 01001010 0
01110000 00011100 0
10100001 00101001 0
01100110 11011111 1

1. The parity bits for the 16 columns is: 01100110 11011111

2. The parity bits for the 5 rows is: 01000

3. The parity bit for the parity row is: 1

4. The bit that was flipped in figure 2 is (15,5):

11000100 10000100 1
00000011 00110001 1
11101011 00011010 1
10010001 11110100 0
10000111 10000100 0
00111010 11011110 1

For figure 3, the bits that were flipped are (2,2) and (1,3):

01001000 01111000 0
11000000 10101000 1
11111000 11110010 1
10101011 01010010 1
10110110 01100011 1
00001101 00010011 0

5. No, with 2D parity, you can detect the presence of two flipped bits, but you can't know their exact locations in order to correct them.



That's incorrect

That's correct

The answer was: 0110011011011111

Question 1 of 5

The answer was: 01000

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 15,5

Question 4 of 5

The answer was: No

Question 5 of 5

Try Another Problem

We greatly appreciate the work of John Broderick (UMass '21) in helping to develop these interactive problems.

Copyright © 2010-2025 J.F. Kurose, K.W. Ross
Comments welcome and appreciated: kurose@cs.umass.edu