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

10111011 10000001
00110100 10111101
00111010 00111101
10000001 01101110
11000000 01001101

Figure 2

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

01010111 00001000 0
01010110 11010101 1
11010110 11111110 0
10010100 01111100 0
01110100 01100101 0
00110011 00111010 1

Figure 3

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

11111011 00000000 0
11100100 11010100 1
01100111 01011000 0
01101010 01010100 1
11010010 10010110 0
11010000 11001110 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:

10111011 10000001 0
00110100 10111101 1
00111010 00111101 1
10000001 01101110 1
11000000 01001101 0
11110100 00100010 1

1. The parity bits for the 16 columns is: 11110100 00100010

2. The parity bits for the 5 rows is: 01110

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

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

01010111 00001000 0
01010110 11010101 1
11010110 11111110 0
10010100 01111100 0
01110100 01100101 0
00110011 00111010 1

For figure 3, the bits that were flipped are (3,0) and (8,1):

11111011 00000000 0
11100100 11010100 1
01100111 01011000 0
01101010 01010100 1
11010010 10010110 0
11010000 11001110 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: 1111010000100010

Question 1 of 5

The answer was: 01110

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 5,5

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