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

00010000 10001101
11010000 00010001
10100000 00111111
01100101 01000001
01000010 01001001

Figure 2

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

00010101 01110110 1
11001000 00111011 0
10100110 01001010 1
00101101 00110100 1
00001110 01010011 1
01001000 01100000 0

Figure 3

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

01100000 11101100 0
01110001 10001000 0
00001111 10010101 0
10100111 11100000 1
00011001 11010111 1
00100000 11000010 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:

00010000 10001101 1
11010000 00010001 1
10100000 00111111 0
01100101 01000001 0
01000010 01001001 1
01000111 10101011 1

1. The parity bits for the 16 columns is: 01000111 10101011

2. The parity bits for the 5 rows is: 11001

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

4. The bit that was flipped in figure 2 is (3,0):

00010101 01110110 1
11001000 00111011 0
10100110 01001010 1
00101101 00110100 1
00001110 01010011 1
01001000 01100000 0

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

01100000 11101100 0
01110001 10001000 0
00001111 10010101 0
10100111 11100000 1
00011001 11010111 1
00100000 11000010 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: 0100011110101011

Question 1 of 5

The answer was: 11001

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 3,0

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