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

10001101 00110110
01011011 11110100
10010000 11001001
11000001 11100000
00001110 01110100

Figure 2

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

00011011 01111000 0
10101110 10100100 0
00110011 11010001 0
01111101 01001000 0
10111101 01111011 1
01000111 00111110 1

Figure 3

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

10010110 10100110 1
01110010 11011100 1
10010011 01110111 0
10111101 11011010 1
01001111 00110110 1
11010101 11100001 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:

10001101 00110110 0
01011011 11110100 0
10010000 11001001 0
11000001 11100000 0
00001110 01110100 1
10001001 10011111 1

1. The parity bits for the 16 columns is: 10001001 10011111

2. The parity bits for the 5 rows is: 00001

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

4. The bit that was flipped in figure 2 is (7,4):

00011011 01111000 0
10101110 10100100 0
00110011 11010001 0
01111101 01001000 0
10111101 01111011 1
01000111 00111110 1

For figure 3, the bits that were flipped are (3,0) and (1,5):

10010110 10100110 1
01110010 11011100 1
10010011 01110111 0
10111101 11011010 1
01001111 00110110 1
11010101 11100001 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: 1000100110011111

Question 1 of 5

The answer was: 00001

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 7,4

Question 4 of 5

The answer was: No

Question 5 of 5

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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