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

00111000 01111110
10010111 11101100
11010101 10110001
10101111 11000101
00000110 11011101

Figure 2

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

00001110 01010101 1
01000000 11001010 1
00101100 10110011 1
00010010 11100110 1
11111010 01011011 1
10001010 00010001 1

Figure 3

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

10101100 11101110 0
00101101 01111000 0
11100100 01010101 0
11001011 11110010 1
10111111 00011111 0
00110001 00101010 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:

00111000 01111110 1
10010111 11101100 0
11010101 10110001 1
10101111 11000101 0
00000110 11011101 0
11010011 00111011 0

1. The parity bits for the 16 columns is: 11010011 00111011

2. The parity bits for the 5 rows is: 10100

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

4. The bit that was flipped in figure 2 is (8,2):

00001110 01010101 1
01000000 11001010 1
00101100 10110011 1
00010010 11100110 1
11111010 01011011 1
10001010 00010001 1

For figure 3, the bits that were flipped are (2,3) and (13,5):

10101100 11101110 0
00101101 01111000 0
11100100 01010101 0
11001011 11110010 1
10111111 00011111 0
00110001 00101010 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: 1101001100111011

Question 1 of 5

The answer was: 10100

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 8,2

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