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

10001100 11000000
01101000 11010011
00101011 10111001
11101001 10110000
01101100 10110010

Figure 2

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

11110010 01101110 0
10000001 01111001 1
11100100 00111011 1
10100111 01100010 1
01100000 00100101 1
01010000 01111011 0

Figure 3

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

00111011 11001101 0
10101111 11010101 1
11010010 00010001 1
01110111 00100110 1
00000011 11000011 0
00110010 00101100 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:

10001100 11000000 1
01101000 11010011 0
00101011 10111001 1
11101001 10110000 0
01101100 10110010 0
01001010 10101000 0

1. The parity bits for the 16 columns is: 01001010 10101000

2. The parity bits for the 5 rows is: 10100

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

4. The bit that was flipped in figure 2 is (11,3):

11110010 01101110 0
10000001 01111001 1
11100100 00111011 1
10100111 01100010 1
01100000 00100101 1
01010000 01111011 0

For figure 3, the bits that were flipped are (9,5) and (8,2):

00111011 11001101 0
10101111 11010101 1
11010010 00010001 1
01110111 00100110 1
00000011 11000011 0
00110010 00101100 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: 0100101010101000

Question 1 of 5

The answer was: 10100

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 11,3

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