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

01000011 00111111
10001010 10111110
11010000 10101000
01100101 11010010
11001110 00000001

Figure 2

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

11100000 01011011 0
11111010 01011010 0
11011010 10000101 1
10101010 10001111 1
11100011 00010010 1
10000001 00011001 1

Figure 3

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

11100100 00010110 1
10000111 01011110 1
10011010 01010110 0
01001001 11101100 0
01011110 01101111 0
11111110 10011100 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:

01000011 00111111 1
10001010 10111110 1
11010000 10101000 0
01100101 11010010 0
11001110 00000001 0
10110010 11111010 0

1. The parity bits for the 16 columns is: 10110010 11111010

2. The parity bits for the 5 rows is: 11000

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

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

11100000 01011011 0
11111010 01011010 0
11011010 10000101 1
10101010 10001111 1
11100011 00010010 1
10000001 00011001 1

For figure 3, the bits that were flipped are (3,5) and (15,4):

11100100 00010110 1
10000111 01011110 1
10011010 01010110 0
01001001 11101100 0
01011110 01101111 0
11111110 10011100 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: 1011001011111010

Question 1 of 5

The answer was: 11000

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 4,2

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