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

01011111 11011110
00110001 11010010
11011001 10111011
10110001 11011110
01010100 01111110

Figure 2

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

11001100 01010110 0
11011100 10011010 1
11101100 11100111 1
10101110 10010101 1
11100000 10101110 0
10110010 00010001 1

Figure 3

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

11111010 00011000 1
01110000 10100101 1
00011000 00001001 1
10001111 01010111 0
01001101 10101101 1
01010000 00001111 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:

01011111 11011110 0
00110001 11010010 1
11011001 10111011 1
10110001 11011110 0
01010100 01111110 1
01010010 00010111 1

1. The parity bits for the 16 columns is: 01010010 00010111

2. The parity bits for the 5 rows is: 01101

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

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

11001100 01010110 0
11011100 10011010 1
11101100 11100111 1
10101110 10010101 1
11100000 10101110 0
10110010 00010001 1

For figure 3, the bits that were flipped are (15,0) and (9,2):

11111010 00011000 1
01110000 10100101 1
00011000 00001001 1
10001111 01010111 0
01001101 10101101 1
01010000 00001111 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: 0101001000010111

Question 1 of 5

The answer was: 01101

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