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

01000110 11011011
11011101 00011101
01001000 00001111
10100111 01101010
01111010 00110111

Figure 2

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

00101110 00011010 1
00110110 10010111 0
00011000 10001101 0
00000010 00101110 1
10011110 00010001 1
10011100 00011111 1

Figure 3

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

11111100 10100101 0
11011000 00110100 0
01111010 01001111 0
11100010 00010011 1
11000100 10010011 0
11111001 01011110 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:

01000110 11011011 1
11011101 00011101 0
01001000 00001111 0
10100111 01101010 1
01111010 00110111 0
00001110 10010100 0

1. The parity bits for the 16 columns is: 00001110 10010100

2. The parity bits for the 5 rows is: 10010

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

4. The bit that was flipped in figure 2 is (10,1):

00101110 00011010 1
00110110 10010111 0
00011000 10001101 0
00000010 00101110 1
10011110 00010001 1
10011100 00011111 1

For figure 3, the bits that were flipped are (7,4) and (0,1):

11111100 10100101 0
11011000 00110100 0
01111010 01001111 0
11100010 00010011 1
11000100 10010011 0
11111001 01011110 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: 0000111010010100

Question 1 of 5

The answer was: 10010

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 10,1

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