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

00010110 01100011
00001011 00111011
01000111 00010000
11110011 10101101
01110100 00101100

Figure 2

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

00000100 01010100 0
00111111 00010111 0
10100100 11100100 0
11001010 11010101 1
00011010 01001000 1
01001111 00111000 0

Figure 3

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

01011110 11101110 1
10111000 10010010 0
10101101 01100011 1
00110001 01100110 0
11010101 10101010 1
10101111 01011011 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:

00010110 01100011 1
00001011 00111011 0
01000111 00010000 1
11110011 10101101 1
01110100 00101100 1
11011101 11001001 0

1. The parity bits for the 16 columns is: 11011101 11001001

2. The parity bits for the 5 rows is: 10111

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

4. The bit that was flipped in figure 2 is (14,2):

00000100 01010100 0
00111111 00010111 0
10100100 11100100 0
11001010 11010101 1
00011010 01001000 1
01001111 00111000 0

For figure 3, the bits that were flipped are (8,1) and (12,3):

01011110 11101110 1
10111000 10010010 0
10101101 01100011 1
00110001 01100110 0
11010101 10101010 1
10101111 01011011 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: 1101110111001001

Question 1 of 5

The answer was: 10111

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 14,2

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