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

01011000 10110010
10110100 11011101
10111010 00111000
11101001 11010011
00001011 11110111

Figure 2

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

11100101 01000010 0
00100011 01001111 0
00100110 11000110 1
00010010 11001011 1
11000111 11011101 1
00110100 11011101 1

Figure 3

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

10010110 01110110 1
01100101 11100101 1
00101101 10010110 1
10110101 10001010 0
11010010 00101011 0
10111001 11000100 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:

01011000 10110010 1
10110100 11011101 0
10111010 00111000 0
11101001 11010011 0
00001011 11110111 0
10110100 01110011 1

1. The parity bits for the 16 columns is: 10110100 01110011

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 (7,0):

11100101 01000010 0
00100011 01001111 0
00100110 11000110 1
00010010 11001011 1
11000111 11011101 1
00110100 11011101 1

For figure 3, the bits that were flipped are (10,2) and (9,5):

10010110 01110110 1
01100101 11100101 1
00101101 10010110 1
10110101 10001010 0
11010010 00101011 0
10111001 11000100 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: 1011010001110011

Question 1 of 5

The answer was: 10000

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 7,0

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