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

11001110 01111100
00101100 01110010
00111000 11101110
11101110 10001101
00101010 01001001

Figure 2

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

00111100 11111110 1
11000101 10000110 1
10101001 11101001 1
10100000 00001001 0
00110101 11101100 1
11000101 01110000 0

Figure 3

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

00110101 01100010 0
01101111 01101011 1
10111011 11011010 1
00100010 11110001 1
01110000 11010011 0
10100011 11110000 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:

11001110 01111100 0
00101100 01110010 1
00111000 11101110 1
11101110 10001101 0
00101010 01001001 0
00011110 00100100 0

1. The parity bits for the 16 columns is: 00011110 00100100

2. The parity bits for the 5 rows is: 01100

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

4. The bit that was flipped in figure 2 is (13,5):

00111100 11111110 1
11000101 10000110 1
10101001 11101001 1
10100000 00001001 0
00110101 11101100 1
11000101 01110000 0

For figure 3, the bits that were flipped are (3,5) and (15,0):

00110101 01100010 0
01101111 01101011 1
10111011 11011010 1
00100010 11110001 1
01110000 11010011 0
10100011 11110000 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: 0001111000100100

Question 1 of 5

The answer was: 01100

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 13,5

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