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

11100011 10011111
01001011 00001110
00011010 11101001
10001111 10110100
01000000 10100101

Figure 2

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

11000110 00011010 1
00000000 01101101 1
11010110 00010101 0
10001101 11011011 0
10001011 11000100 1
00010010 01111101 1

Figure 3

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

00110010 10101110 1
11110011 11011000 0
10011011 00000110 1
10000111 00110101 0
01110000 10100111 1
10001101 01100010 1


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:

11100011 10011111 1
01001011 00001110 1
00011010 11101001 0
10001111 10110100 1
01000000 10100101 1
01111101 01101001 0

1. The parity bits for the 16 columns is: 01111101 01101001

2. The parity bits for the 5 rows is: 11011

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

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

11000110 00011010 1
00000000 01101101 1
11010110 00010101 0
10001101 11011011 0
10001011 11000100 1
00010010 01111101 1

For figure 3, the bits that were flipped are (2,0) and (8,4):

00110010 10101110 1
11110011 11011000 0
10011011 00000110 1
10000111 00110101 0
01110000 10100111 1
10001101 01100010 1

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: 0111110101101001

Question 1 of 5

The answer was: 11011

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 5,5

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