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 01110001
01000001 01100101
00111010 11010111
00000110 10100100
00010110 10111100

Figure 2

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

01001011 00001001 0
10011001 00111111 0
10111010 01100000 1
00010100 10110100 0
11000001 00010011 0
10111101 11110101 1

Figure 3

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

10010011 01011001 1
11111001 01101000 1
10111001 00000110 1
10010001 11010111 1
01111000 10010011 1
00011110 01110011 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:

01111111 01110001 1
01000001 01100101 0
00111010 11010111 0
00000110 10100100 1
00010110 10111100 0
00010100 11011011 0

1. The parity bits for the 16 columns is: 00010100 11011011

2. The parity bits for the 5 rows is: 10010

3. The parity bit for the parity row is: 0

4. The bit that was flipped in figure 2 is (13,5):

01001011 00001001 0
10011001 00111111 0
10111010 01100000 1
00010100 10110100 0
11000001 00010011 0
10111101 11110101 1

For figure 3, the bits that were flipped are (5,0) and (2,4):

10010011 01011001 1
11111001 01101000 1
10111001 00000110 1
10010001 11010111 1
01111000 10010011 1
00011110 01110011 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: 0001010011011011

Question 1 of 5

The answer was: 10010

Question 2 of 5

The answer was: 0

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