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

01011010 00000111
11111100 01101110
11100101 00000000
01000011 11001101
10101110 00000000

Figure 2

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

00101011 00111001 0
01010010 11101100 0
01110100 11010010 0
11111111 11101101 0
11100110 01100101 1
00010100 10000111 1

Figure 3

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

10010101 01011000 1
01001110 00011011 0
00111101 01010110 0
10001100 01100110 0
01001010 01111001 0
00100010 00001010 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:

01011010 00000111 1
11111100 01101110 1
11100101 00000000 1
01000011 11001101 0
10101110 00000000 1
10101110 10100100 0

1. The parity bits for the 16 columns is: 10101110 10100100

2. The parity bits for the 5 rows is: 11101

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

4. The bit that was flipped in figure 2 is (12,5):

00101011 00111001 0
01010010 11101100 0
01110100 11010010 0
11111111 11101101 0
11100110 01100101 1
00010100 10000111 1

For figure 3, the bits that were flipped are (6,2) and (16,3):

10010101 01011000 1
01001110 00011011 0
00111101 01010110 0
10001100 01100110 0
01001010 01111001 0
00100010 00001010 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: 1010111010100100

Question 1 of 5

The answer was: 11101

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 12,5

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.

Copyright © 2010-2025 J.F. Kurose, K.W. Ross
Comments welcome and appreciated: kurose@cs.umass.edu