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

10100001 10001110
10010010 10001110
11001000 11000110
01100111 01001111
10010011 00101100

Figure 2

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

01111101 10110000 0
01101110 11110011 1
11101100 01101001 1
00111011 00100111 1
01100111 10011100 1
10100010 10010001 0

Figure 3

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

00000110 10111100 1
00100011 01000010 0
10101000 10011000 1
01110101 10000001 1
00000000 01100001 1
11111100 10000110 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:

10100001 10001110 1
10010010 10001110 1
11001000 11000110 1
01100111 01001111 0
10010011 00101100 1
00001111 10100101 0

1. The parity bits for the 16 columns is: 00001111 10100101

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 (7,0):

01111101 10110000 0
01101110 11110011 1
11101100 01101001 1
00111011 00100111 1
01100111 10011100 1
10100010 10010001 0

For figure 3, the bits that were flipped are (5,2) and (16,1):

00000110 10111100 1
00100011 01000010 0
10101000 10011000 1
01110101 10000001 1
00000000 01100001 1
11111100 10000110 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: 0000111110100101

Question 1 of 5

The answer was: 11101

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 7,0

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