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

00010111 11010110
10001111 11101110
11111110 11001111
10000011 00101011
11001111 11011011

Figure 2

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

00111001 00000101 0
00011101 11100010 1
01001000 00111100 0
11100111 10011011 1
01011111 11011000 0
11010100 10011100 0

Figure 3

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

10001101 00000010 0
01111001 11110111 0
01110001 11100000 0
10101010 00010101 1
11110100 10001100 0
11011011 10000100 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:

00010111 11010110 1
10001111 11101110 1
11111110 11001111 1
10000011 00101011 1
11001111 11011011 0
00101010 00000111 0

1. The parity bits for the 16 columns is: 00101010 00000111

2. The parity bits for the 5 rows is: 11110

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

4. The bit that was flipped in figure 2 is (13,1):

00111001 00000101 0
00011101 11100010 1
01001000 00111100 0
11100111 10011011 1
01011111 11011000 0
11010100 10011100 0

For figure 3, the bits that were flipped are (16,2) and (12,0):

10001101 00000010 0
01111001 11110111 0
01110001 11100000 0
10101010 00010101 1
11110100 10001100 0
11011011 10000100 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: 0010101000000111

Question 1 of 5

The answer was: 11110

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 13,1

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