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

10000011 10011001
11100010 11010101
10000101 11010010
00111111 00111001
01010101 01011011

Figure 2

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

10000100 01110001 0
10000010 01000011 1
11110011 11000000 1
01001111 00010001 1
10110011 01110110 0
00001001 11010101 1

Figure 3

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

01110011 01000010 1
11111101 00101010 1
10010000 11001001 1
11111011 10011100 1
10100111 11100110 0
01001010 10011011 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:

10000011 10011001 1
11100010 11010101 1
10000101 11010010 1
00111111 00111001 0
01010101 01011011 1
10001110 11111100 0

1. The parity bits for the 16 columns is: 10001110 11111100

2. The parity bits for the 5 rows is: 11101

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

4. The bit that was flipped in figure 2 is (9,2):

10000100 01110001 0
10000010 01000011 1
11110011 11000000 1
01001111 00010001 1
10110011 01110110 0
00001001 11010101 1

For figure 3, the bits that were flipped are (4,2) and (9,1):

01110011 01000010 1
11111101 00101010 1
10010000 11001001 1
11111011 10011100 1
10100111 11100110 0
01001010 10011011 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: 1000111011111100

Question 1 of 5

The answer was: 11101

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 9,2

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