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

00001001 01010011
00110110 10001101
11001001 01001101
00011100 00001101
01000111 10010111

Figure 2

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

10100101 00100100 0
10111110 00100100 0
11100110 00010010 1
11100110 11000000 1
11110011 10100110 1
11101000 01110100 0

Figure 3

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

11001101 01001100 0
00111011 00101111 1
00100100 10111110 1
01010011 11110011 0
00111010 01000011 1
10111001 01101111 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:

00001001 01010011 0
00110110 10001101 0
11001001 01001101 0
00011100 00001101 0
01000111 10010111 1
10101101 00001001 1

1. The parity bits for the 16 columns is: 10101101 00001001

2. The parity bits for the 5 rows is: 00001

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

4. The bit that was flipped in figure 2 is (16,4):

10100101 00100100 0
10111110 00100100 0
11100110 00010010 1
11100110 11000000 1
11110011 10100110 1
11101000 01110100 0

For figure 3, the bits that were flipped are (14,1) and (6,2):

11001101 01001100 0
00111011 00101111 1
00100100 10111110 1
01010011 11110011 0
00111010 01000011 1
10111001 01101111 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: 1010110100001001

Question 1 of 5

The answer was: 00001

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 16,4

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.

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