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

10101100 10000001
10110100 10001100
00100110 11011101
11100010 11110010
11010010 00010101

Figure 2

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

10001001 10101001 1
10000111 10001010 1
10011100 00101100 1
01000001 11110010 0
11010011 01010111 0
00000100 10101010 1

Figure 3

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

10111101 10001000 0
10010000 01001011 0
00100011 00001000 1
00111001 01101110 0
00110111 01100010 0
00000100 11000101 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:

10101100 10000001 0
10110100 10001100 1
00100110 11011101 1
11100010 11110010 1
11010010 00010101 1
00001110 00110111 0

1. The parity bits for the 16 columns is: 00001110 00110111

2. The parity bits for the 5 rows is: 01111

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

4. The bit that was flipped in figure 2 is (5,3):

10001001 10101001 1
10000111 10001010 1
10011100 00101100 1
01000001 11110010 0
11010011 01010111 0
00000100 10101010 1

For figure 3, the bits that were flipped are (5,2) and (14,3):

10111101 10001000 0
10010000 01001011 0
00100011 00001000 1
00111001 01101110 0
00110111 01100010 0
00000100 11000101 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: 0000111000110111

Question 1 of 5

The answer was: 01111

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 5,3

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