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

01111101 00100101
10110101 01010111
00001110 00110111
10000011 11001001
01001101 10000110

Figure 2

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

01011000 11000001 0
11100100 01010111 1
00111010 00110010 1
10101010 10111010 1
01101010 01101100 0
01000110 01110000 1

Figure 3

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

00010100 11100101 0
10101100 01001111 0
00011110 11001000 1
11111100 10100110 0
01010111 01001110 1
00000101 10001110 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:

01111101 00100101 1
10110101 01010111 0
00001110 00110111 0
10000011 11001001 1
01001101 10000110 1
00001000 00001010 1

1. The parity bits for the 16 columns is: 00001000 00001010

2. The parity bits for the 5 rows is: 10011

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

4. The bit that was flipped in figure 2 is (14,5):

01011000 11000001 0
11100100 01010111 1
00111010 00110010 1
10101010 10111010 1
01101010 01101100 0
01000110 01110000 1

For figure 3, the bits that were flipped are (4,1) and (13,0):

00010100 11100101 0
10101100 01001111 0
00011110 11001000 1
11111100 10100110 0
01010111 01001110 1
00000101 10001110 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: 0000100000001010

Question 1 of 5

The answer was: 10011

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 14,5

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.

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Comments welcome and appreciated: kurose@cs.umass.edu