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

10110010 10110000
01010010 00110100
01000101 00101101
01010010 10000001
00100010 01000001

Figure 2

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

01011000 11001011 0
00001001 00111001 0
10101001 01111101 0
00001110 10011100 1
11101100 00011101 0
00011010 01001110 1

Figure 3

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

01000101 00001111 1
01101101 11101010 0
00010000 11100001 0
00011010 00101111 0
10101110 10111110 0
10101101 10010101 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:

10110010 10110000 1
01010010 00110100 0
01000101 00101101 1
01010010 10000001 1
00100010 01000001 0
11010101 01101001 1

1. The parity bits for the 16 columns is: 11010101 01101001

2. The parity bits for the 5 rows is: 10110

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

4. The bit that was flipped in figure 2 is (9,4):

01011000 11001011 0
00001001 00111001 0
10101001 01111101 0
00001110 10011100 1
11101100 00011101 0
00011010 01001110 1

For figure 3, the bits that were flipped are (2,4) and (7,2):

01000101 00001111 1
01101101 11101010 0
00010000 11100001 0
00011010 00101111 0
10101110 10111110 0
10101101 10010101 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: 1101010101101001

Question 1 of 5

The answer was: 10110

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 9,4

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