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

00111111 01111010
00000111 01110011
10101100 00110110
11111001 01010000
11101001 01111111

Figure 2

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

11000011 00011100 1
00001100 00001011 1
00010011 11111101 0
00001111 00110111 1
11011100 11001001 1
00001111 00010101 0

Figure 3

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

11111010 01010111 1
01001100 01110000 0
11000100 11111011 1
10101001 11001101 1
00001011 10110011 1
10010010 10100010 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:

00111111 01111010 1
00000111 01110011 0
10101100 00110110 0
11111001 01010000 0
11101001 01111111 0
10000100 00010000 1

1. The parity bits for the 16 columns is: 10000100 00010000

2. The parity bits for the 5 rows is: 10000

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

4. The bit that was flipped in figure 2 is (15,5):

11000011 00011100 1
00001100 00001011 1
00010011 11111101 0
00001111 00110111 1
11011100 11001001 1
00001111 00010101 0

For figure 3, the bits that were flipped are (6,4) and (1,2):

11111010 01010111 1
01001100 01110000 0
11000100 11111011 1
10101001 11001101 1
00001011 10110011 1
10010010 10100010 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: 1000010000010000

Question 1 of 5

The answer was: 10000

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 15,5

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