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

11001011 10010111
01101000 00100110
00001001 01000100
11100010 11000101
00110111 01011101

Figure 2

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

00101101 01010001 0
01110001 11101011 0
11001000 00010011 0
10100111 01010100 0
00011100 11110000 1
00101111 00000101 1

Figure 3

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

00011110 00010111 0
00010111 10011111 0
11100000 01011101 1
11100101 11001011 0
10111100 10100011 1
10100000 10110101 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:

11001011 10010111 0
01101000 00100110 0
00001001 01000100 0
11100010 11000101 0
00110111 01011101 0
01111111 01101101 0

1. The parity bits for the 16 columns is: 01111111 01101101

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 (12,0):

00101101 01010001 0
01110001 11101011 0
11001000 00010011 0
10100111 01010100 0
00011100 11110000 1
00101111 00000101 1

For figure 3, the bits that were flipped are (12,2) and (3,5):

00011110 00010111 0
00010111 10011111 0
11100000 01011101 1
11100101 11001011 0
10111100 10100011 1
10100000 10110101 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: 0111111101101101

Question 1 of 5

The answer was: 00000

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 12,0

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