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

00001111 11000110
10001100 11111001
11011110 01101010
11100001 10000001
01001101 11111101

Figure 2

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

00011110 00101100 1
10000010 01011001 1
11110001 11111010 1
00011110 11001110 1
00110010 11101100 0
00000001 10101101 0

Figure 3

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

01100000 00110010 1
11001010 00011011 1
10011110 11000001 0
10011000 10110110 0
11000111 01101101 1
01001111 00110011 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:

00001111 11000110 0
10001100 11111001 1
11011110 01101010 0
11100001 10000001 0
01001101 11111101 1
11110001 00101001 0

1. The parity bits for the 16 columns is: 11110001 00101001

2. The parity bits for the 5 rows is: 01001

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

4. The bit that was flipped in figure 2 is (1,1):

00011110 00101100 1
10000010 01011001 1
11110001 11111010 1
00011110 11001110 1
00110010 11101100 0
00000001 10101101 0

For figure 3, the bits that were flipped are (5,1) and (2,4):

01100000 00110010 1
11001010 00011011 1
10011110 11000001 0
10011000 10110110 0
11000111 01101101 1
01001111 00110011 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: 1111000100101001

Question 1 of 5

The answer was: 01001

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 1,1

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