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

01100101 11111101
01001111 00111110
11101000 10101000
11001101 11110110
00110111 11111111

Figure 2

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

10011100 11011101 0
01001100 00010101 0
01000011 11100010 0
10100001 01010100 0
01100011 10110000 1
01011001 11001110 1

Figure 3

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

01100000 11101110 1
00101110 01001110 1
00010001 00100100 0
01010011 00100111 0
10100100 11010110 0
10101010 01110001 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:

01100101 11111101 1
01001111 00111110 0
11101000 10101000 1
11001101 11110110 1
00110111 11111111 1
00111000 01100010 0

1. The parity bits for the 16 columns is: 00111000 01100010

2. The parity bits for the 5 rows is: 10111

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

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

10011100 11011101 0
01001100 00010101 0
01000011 11100010 0
10100001 01010100 0
01100011 10110000 1
01011001 11001110 1

For figure 3, the bits that were flipped are (13,0) and (6,1):

01100000 11101110 1
00101110 01001110 1
00010001 00100100 0
01010011 00100111 0
10100100 11010110 0
10101010 01110001 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: 0011100001100010

Question 1 of 5

The answer was: 10111

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 4,2

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