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

11110111 10100100
11111000 00011010
10110011 00001001
00110010 11111001
00111110 11110011

Figure 2

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

01111001 11110001 1
10001010 11011100 0
11000010 10111111 0
11000011 11001001 0
01101000 01000110 0
10011010 00011111 1

Figure 3

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

11111011 00000011 1
11100010 10000110 0
10001101 10000100 0
10100110 10010101 0
10101000 01111111 1
10011010 10101011 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:

11110111 10100100 0
11111000 00011010 0
10110011 00001001 1
00110010 11111001 1
00111110 11110011 1
10110000 10111101 1

1. The parity bits for the 16 columns is: 10110000 10111101

2. The parity bits for the 5 rows is: 00111

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

4. The bit that was flipped in figure 2 is (14,0):

01111001 11110001 1
10001010 11011100 0
11000010 10111111 0
11000011 11001001 0
01101000 01000110 0
10011010 00011111 1

For figure 3, the bits that were flipped are (9,1) and (16,4):

11111011 00000011 1
11100010 10000110 0
10001101 10000100 0
10100110 10010101 0
10101000 01111111 1
10011010 10101011 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: 1011000010111101

Question 1 of 5

The answer was: 00111

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 14,0

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