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

10100100 01101010
00010110 10110101
11110111 11110100
10101011 00110001
01111101 11111000

Figure 2

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

10000011 11100111 1
00010100 10110111 0
10011000 10101011 0
11001101 10001011 0
10101011 10101001 1
01101001 11011101 0

Figure 3

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

01111010 11011010 0
11111010 00010000 0
10011110 00111110 0
11110001 01101011 0
11100010 00001010 1
10001101 10000101 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:

10100100 01101010 1
00010110 10110101 0
11110111 11110100 0
10101011 00110001 0
01111101 11111000 1
10010011 11100010 0

1. The parity bits for the 16 columns is: 10010011 11100010

2. The parity bits for the 5 rows is: 10001

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

4. The bit that was flipped in figure 2 is (13,3):

10000011 11100111 1
00010100 10110111 0
10011000 10101011 0
11001101 10001011 0
10101011 10101001 1
01101001 11011101 0

For figure 3, the bits that were flipped are (11,1) and (0,4):

01111010 11011010 0
11111010 00010000 0
10011110 00111110 0
11110001 01101011 0
11100010 00001010 1
10001101 10000101 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: 1001001111100010

Question 1 of 5

The answer was: 10001

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 13,3

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