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

00011111 00100110
01010110 11011001
10010000 11111111
10000000 10101110
01100011 11111111

Figure 2

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

11100100 01010100 1
11000111 01101010 1
10011010 01010010 1
11111101 11111010 1
01111011 00101001 0
00111111 10011111 0

Figure 3

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

00001011 01011000 0
11000010 10111111 0
01110001 10011101 0
00011111 01000001 0
11111011 10110000 0
01011100 10011001 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:

00011111 00100110 0
01010110 11011001 1
10010000 11111111 0
10000000 10101110 0
01100011 11111111 0
00111010 01010001 1

1. The parity bits for the 16 columns is: 00111010 01010001

2. The parity bits for the 5 rows is: 01000

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

4. The bit that was flipped in figure 2 is (10,4):

11100100 01010100 1
11000111 01101010 1
10011010 01010010 1
11111101 11111010 1
01111011 00101001 0
00111111 10011111 0

For figure 3, the bits that were flipped are (14,3) and (11,2):

00001011 01011000 0
11000010 10111111 0
01110001 10011101 0
00011111 01000001 0
11111011 10110000 0
01011100 10011001 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: 0011101001010001

Question 1 of 5

The answer was: 01000

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 10,4

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