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

00000101 00001011
01100111 11101111
01111110 11000011
11001101 01111011
10111101 11001011

Figure 2

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

10101011 10011011 0
11101100 00111001 1
01101111 00110010 1
11001001 01011111 0
01100101 10011100 1
10000100 01110011 1

Figure 3

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

10011001 10010010 1
11010011 00110101 0
01010010 01001111 0
11101000 10010100 0
10100111 01111011 1
01110011 00000111 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:

00000101 00001011 1
01100111 11101111 0
01111110 11000011 0
11001101 01111011 1
10111101 11001011 1
01101100 10010111 1

1. The parity bits for the 16 columns is: 01101100 10010111

2. The parity bits for the 5 rows is: 10011

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

4. The bit that was flipped in figure 2 is (10,4):

10101011 10011011 0
11101100 00111001 1
01101111 00110010 1
11001001 01011111 0
01100101 10011100 1
10000100 01110011 1

For figure 3, the bits that were flipped are (5,3) and (2,1):

10011001 10010010 1
11010011 00110101 0
01010010 01001111 0
11101000 10010100 0
10100111 01111011 1
01110011 00000111 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: 0110110010010111

Question 1 of 5

The answer was: 10011

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 10,4

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