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

00000000 10001010
10010001 00010101
01111010 00010101
01101010 10110111
01011011 10110111

Figure 2

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

10000111 11010100 0
11000010 11010111 1
10000011 11111100 1
11101101 10101000 1
11110100 00011010 1
10011111 01001101 0

Figure 3

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

11000000 10001011 1
01001011 11100100 1
10110110 01110010 1
01110001 10111100 1
00100110 01011000 0
01001010 11101001 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:

00000000 10001010 1
10010001 00010101 0
01111010 00010101 0
01101010 10110111 0
01011011 10110111 1
11011010 10001010 0

1. The parity bits for the 16 columns is: 11011010 10001010

2. The parity bits for the 5 rows is: 10001

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

4. The bit that was flipped in figure 2 is (1,4):

10000111 11010100 0
11000010 11010111 1
10000011 11111100 1
11101101 10101000 1
11110100 00011010 1
10011111 01001101 0

For figure 3, the bits that were flipped are (2,1) and (11,0):

11000000 10001011 1
01001011 11100100 1
10110110 01110010 1
01110001 10111100 1
00100110 01011000 0
01001010 11101001 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: 1101101010001010

Question 1 of 5

The answer was: 10001

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 1,4

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