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

01100110 00111100
11101011 10101010
01101010 00110011
10011011 01100001
11111011 00001101

Figure 2

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

00100011 10000111 1
01011011 00100100 1
11111111 11110100 1
11110111 10011100 0
10110011 00000101 1
11000001 11001110 0

Figure 3

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

00010001 10101011 1
10101111 01011001 1
00110101 11011010 0
10010011 00010001 0
10011110 01111101 1
10010010 01000100 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:

01100110 00111100 0
11101011 10101010 0
01101010 00110011 0
10011011 01100001 0
11111011 00001101 0
10000111 11001001 0

1. The parity bits for the 16 columns is: 10000111 11001001

2. The parity bits for the 5 rows is: 00000

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

4. The bit that was flipped in figure 2 is (6,3):

00100011 10000111 1
01011011 00100100 1
11111111 11110100 1
11110111 10011100 0
10110011 00000101 1
11000001 11001110 0

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

00010001 10101011 1
10101111 01011001 1
00110101 11011010 0
10010011 00010001 0
10011110 01111101 1
10010010 01000100 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: 1000011111001001

Question 1 of 5

The answer was: 00000

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 6,3

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