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

01111111 00101110
11000101 00101110
01011001 11111110
00101010 00000110
01100100 01101110

Figure 2

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

11011100 11110111 0
01111001 11011000 1
10110101 11111001 1
00111000 11000001 0
10110010 00100001 0
10111010 00110110 0

Figure 3

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

00000001 10000100 0
00011001 10101110 1
10011110 00000100 0
10010101 10101010 0
00000000 00100010 0
00010011 10110110 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:

01111111 00101110 1
11000101 00101110 0
01011001 11111110 1
00101010 00000110 1
01100100 01101110 0
10101101 10010110 1

1. The parity bits for the 16 columns is: 10101101 10010110

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 (2,5):

11011100 11110111 0
01111001 11011000 1
10110101 11111001 1
00111000 11000001 0
10110010 00100001 0
10111010 00110110 0

For figure 3, the bits that were flipped are (11,1) and (16,0):

00000001 10000100 0
00011001 10101110 1
10011110 00000100 0
10010101 10101010 0
00000000 00100010 0
00010011 10110110 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: 1010110110010110

Question 1 of 5

The answer was: 10110

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 2,5

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