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

00110110 10111111
01000010 10011100
00101100 01011010
11011011 01101100
01010000 01011000

Figure 2

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

10110010 11000101 0
10011110 00011111 0
11001110 11010100 1
10110000 00100010 0
10000011 10110101 0
11000001 10011001 1

Figure 3

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

00010101 00110111 0
00111011 11110110 0
11010010 01100101 0
00011100 10000001 1
10100110 10010100 1
01001100 10110001 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:

00110110 10111111 1
01000010 10011100 0
00101100 01011010 1
11011011 01101100 0
01010000 01011000 1
11010011 01001101 1

1. The parity bits for the 16 columns is: 11010011 01001101

2. The parity bits for the 5 rows is: 10101

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

4. The bit that was flipped in figure 2 is (3,3):

10110010 11000101 0
10011110 00011111 0
11001110 11010100 1
10110000 00100010 0
10000011 10110101 0
11000001 10011001 1

For figure 3, the bits that were flipped are (6,1) and (4,5):

00010101 00110111 0
00111011 11110110 0
11010010 01100101 0
00011100 10000001 1
10100110 10010100 1
01001100 10110001 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: 1101001101001101

Question 1 of 5

The answer was: 10101

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 3,3

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