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

00010100 10010011
00100111 01001000
11111101 10110110
11001010 01000001
10111100 01110110

Figure 2

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

11001110 01101111 0
01110011 00110000 1
10010001 01010100 0
10101101 10100100 0
10101010 00011101 0
10101011 10110010 1

Figure 3

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

10101110 01000110 0
01111000 01111111 1
00101011 10100000 1
11001111 11111001 0
10110010 11001111 1
10000000 10101110 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:

00010100 10010011 0
00100111 01001000 0
11111101 10110110 0
11001010 01000001 0
10111100 01110110 0
10111000 01011010 0

1. The parity bits for the 16 columns is: 10111000 01011010

2. The parity bits for the 5 rows is: 00000

3. The parity bit for the parity row is: 0

4. The bit that was flipped in figure 2 is (0,0):

11001110 01101111 0
01110011 00110000 1
10010001 01010100 0
10101101 10100100 0
10101010 00011101 0
10101011 10110010 1

For figure 3, the bits that were flipped are (15,2) and (16,4):

10101110 01000110 0
01111000 01111111 1
00101011 10100000 1
11001111 11111001 0
10110010 11001111 1
10000000 10101110 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: 1011100001011010

Question 1 of 5

The answer was: 00000

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 0,0

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