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

01111101 01101110
10111101 10001011
10111100 01110100
11101011 00010110
00001100 10000100

Figure 2

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

10111111 10000011 0
11100101 10001110 1
00100100 11111110 1
00100001 11101001 1
00100101 11110000 1
01111110 11101010 0

Figure 3

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

10110011 01111101 1
11001001 00000101 1
00010001 10010010 0
11101001 00100101 0
10100100 01000000 0
00000110 10001101 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:

01111101 01101110 1
10111101 10001011 0
10111100 01110100 1
11101011 00010110 1
00001100 10000100 0
10011011 00000011 1

1. The parity bits for the 16 columns is: 10011011 00000011

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 (5,5):

10111111 10000011 0
11100101 10001110 1
00100100 11111110 1
00100001 11101001 1
00100101 11110000 1
01111110 11101010 0

For figure 3, the bits that were flipped are (14,2) and (2,1):

10110011 01111101 1
11001001 00000101 1
00010001 10010010 0
11101001 00100101 0
10100100 01000000 0
00000110 10001101 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: 1001101100000011

Question 1 of 5

The answer was: 10110

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 5,5

Question 4 of 5

The answer was: No

Question 5 of 5

Try Another Problem

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