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

00101001 00101000
11011010 10111010
00101001 00000010
00011111 11000010
11011110 00011100

Figure 2

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

10111100 11010010 1
00001110 01100100 1
00100010 00001000 1
01111100 01110110 0
01100011 10100001 1
10001101 01101001 0

Figure 3

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

11001011 10001111 0
10100101 10000010 0
10101110 10110000 1
00001101 00000011 1
00010010 00001111 0
01011111 10110000 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:

00101001 00101000 1
11011010 10111010 0
00101001 00000010 0
00011111 11000010 0
11011110 00011100 1
00011011 01001110 0

1. The parity bits for the 16 columns is: 00011011 01001110

2. The parity bits for the 5 rows is: 10001

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

4. The bit that was flipped in figure 2 is (6,1):

10111100 11010010 1
00001110 01100100 1
00100010 00001000 1
01111100 01110110 0
01100011 10100001 1
10001101 01101001 0

For figure 3, the bits that were flipped are (0,2) and (15,5):

11001011 10001111 0
10100101 10000010 0
10101110 10110000 1
00001101 00000011 1
00010010 00001111 0
01011111 10110000 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: 0001101101001110

Question 1 of 5

The answer was: 10001

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 6,1

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