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

00010101 11110000
00100000 10000000
10111010 10000100
11101110 10110000
10000010 11011110

Figure 2

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

10011011 01101001 0
10010101 11010001 0
01100001 01100101 1
00100111 01000000 1
11111100 00100011 1
10110100 10111010 1

Figure 3

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

11001111 01100111 0
10001000 01111111 1
10011000 00100000 0
01101010 00100001 1
00000011 01001101 0
10110111 01010101 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:

00010101 11110000 1
00100000 10000000 0
10111010 10000100 1
11101110 10110000 1
10000010 11011110 0
11100011 10011010 1

1. The parity bits for the 16 columns is: 11100011 10011010

2. The parity bits for the 5 rows is: 10110

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

4. The bit that was flipped in figure 2 is (13,0):

10011011 01101001 0
10010101 11010001 0
01100001 01100101 1
00100111 01000000 1
11111100 00100011 1
10110100 10111010 1

For figure 3, the bits that were flipped are (7,0) and (15,3):

11001111 01100111 0
10001000 01111111 1
10011000 00100000 0
01101010 00100001 1
00000011 01001101 0
10110111 01010101 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: 1110001110011010

Question 1 of 5

The answer was: 10110

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 13,0

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