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

01010000 01001100
11000101 00000101
10101101 01100101
01111001 10101011
00110001 11100101

Figure 2

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

11111000 10100110 1
10110011 01100011 1
11111000 10111000 1
10111100 10000000 0
00100001 11010110 0
00001110 00101011 1

Figure 3

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

00010000 01100110 1
01010101 11101000 0
00010000 11100100 1
11010100 10100110 0
00001111 00000111 0
10001110 11001010 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:

01010000 01001100 1
11000101 00000101 0
10101101 01100101 1
01111001 10101011 0
00110001 11100101 0
01110000 01100010 0

1. The parity bits for the 16 columns is: 01110000 01100010

2. The parity bits for the 5 rows is: 10100

3. The parity bit for the parity row is: 0

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

11111000 10100110 1
10110011 01100011 1
11111000 10111000 1
10111100 10000000 0
00100001 11010110 0
00001110 00101011 1

For figure 3, the bits that were flipped are (15,5) and (16,4):

00010000 01100110 1
01010101 11101000 0
00010000 11100100 1
11010100 10100110 0
00001111 00000111 0
10001110 11001010 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: 0111000001100010

Question 1 of 5

The answer was: 10100

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 2,4

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