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

01100011 11110101
10110010 01110000
11100000 00111001
10100110 01110000
01110000 00011111

Figure 2

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

00001011 01001101 0
10000000 00101111 0
11111100 10111101 0
00100100 01101100 0
01101100 10110000 1
00111101 00000011 1

Figure 3

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

00001110 00000011 1
10101011 10101101 1
00111101 00011001 0
11000000 01001110 0
11100111 10001001 0
11110111 01110000 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:

01100011 11110101 0
10110010 01110000 1
11100000 00111001 1
10100110 01110000 1
01110000 00011111 0
11100111 11010011 1

1. The parity bits for the 16 columns is: 11100111 11010011

2. The parity bits for the 5 rows is: 01110

3. The parity bit for the parity row is: 1

4. The bit that was flipped in figure 2 is (6,0):

00001011 01001101 0
10000000 00101111 0
11111100 10111101 0
00100100 01101100 0
01101100 10110000 1
00111101 00000011 1

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

00001110 00000011 1
10101011 10101101 1
00111101 00011001 0
11000000 01001110 0
11100111 10001001 0
11110111 01110000 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: 1110011111010011

Question 1 of 5

The answer was: 01110

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 6,0

Question 4 of 5

The answer was: No

Question 5 of 5

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