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

10101000 10010100
00100110 11101111
11101111 00110100
01110100 00001000
11011111 10100001

Figure 2

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

00101101 11010100 0
10001111 01111000 0
00011001 10001110 1
01000101 01000110 0
01000101 10001110 1
10111011 11001010 0

Figure 3

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

11010001 01111011 1
11011110 10100110 0
01011000 00001011 0
11100001 10011111 1
00001111 10001111 1
10111001 10000010 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:

10101000 10010100 0
00100110 11101111 0
11101111 00110100 0
01110100 00001000 1
11011111 10100001 0
11001010 11100110 1

1. The parity bits for the 16 columns is: 11001010 11100110

2. The parity bits for the 5 rows is: 00010

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

4. The bit that was flipped in figure 2 is (10,1):

00101101 11010100 0
10001111 01111000 0
00011001 10001110 1
01000101 01000110 0
01000101 10001110 1
10111011 11001010 0

For figure 3, the bits that were flipped are (13,3) and (9,0):

11010001 01111011 1
11011110 10100110 0
01011000 00001011 0
11100001 10011111 1
00001111 10001111 1
10111001 10000010 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: 1100101011100110

Question 1 of 5

The answer was: 00010

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 10,1

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