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

01101110 10100100
11001110 01100100
11000000 10001000
00101011 10011011
00111001 00010101

Figure 2

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

11010100 11110110 0
10111101 11110111 0
00100110 11111110 0
01000110 10011011 0
11100100 00100010 0
11101101 11000110 0

Figure 3

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

00111101 00100101 1
01111110 11111011 1
00000001 00001110 0
11111110 01001110 1
10100000 10101101 0
00001100 10110011 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:

01101110 10100100 0
11001110 01100100 0
11000000 10001000 0
00101011 10011011 1
00111001 00010101 1
01110010 11000110 0

1. The parity bits for the 16 columns is: 01110010 11000110

2. The parity bits for the 5 rows is: 00011

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

4. The bit that was flipped in figure 2 is (8,1):

11010100 11110110 0
10111101 11110111 0
00100110 11111110 0
01000110 10011011 0
11100100 00100010 0
11101101 11000110 0

For figure 3, the bits that were flipped are (8,0) and (3,4):

00111101 00100101 1
01111110 11111011 1
00000001 00001110 0
11111110 01001110 1
10100000 10101101 0
00001100 10110011 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: 0111001011000110

Question 1 of 5

The answer was: 00011

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 8,1

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