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

11010111 10100001
10011011 11000000
11101111 01111001
01101000 01110011
10100000 11010110

Figure 2

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

01010111 00011101 1
01101010 00001111 0
11101011 10000111 0
11010001 10001001 1
11010111 00101100 1
11010000 00110100 1

Figure 3

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

00000010 11010100 1
00010011 11100010 0
10100110 01001100 1
01101010 11011010 1
01010110 00100110 0
10001001 10001110 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:

11010111 10100001 1
10011011 11000000 1
11101111 01111001 0
01101000 01110011 0
10100000 11010110 1
01101011 10111101 1

1. The parity bits for the 16 columns is: 01101011 10111101

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 (13,5):

01010111 00011101 1
01101010 00001111 0
11101011 10000111 0
11010001 10001001 1
11010111 00101100 1
11010000 00110100 1

For figure 3, the bits that were flipped are (6,4) and (12,1):

00000010 11010100 1
00010011 11100010 0
10100110 01001100 1
01101010 11011010 1
01010110 00100110 0
10001001 10001110 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: 0110101110111101

Question 1 of 5

The answer was: 11001

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 13,5

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