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

01110101 11110100
00101001 01110001
01101000 00100000
11111100 10110100
11010000 10100100

Figure 2

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

00101000 00001111 0
10110011 11110011 1
10000110 00111000 0
11000111 11010010 1
11110100 00011100 0
00101110 00001010 1

Figure 3

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

00001000 10000100 1
10000011 01100101 1
01011000 11101101 1
10110101 11111001 1
10010010 10100100 1
11110101 00010001 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:

01110101 11110100 0
00101001 01110001 1
01101000 00100000 0
11111100 10110100 0
11010000 10100100 0
00011000 10110101 1

1. The parity bits for the 16 columns is: 00011000 10110101

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 (16,5):

00101000 00001111 0
10110011 11110011 1
10000110 00111000 0
11000111 11010010 1
11110100 00011100 0
00101110 00001010 1

For figure 3, the bits that were flipped are (9,4) and (7,5):

00001000 10000100 1
10000011 01100101 1
01011000 11101101 1
10110101 11111001 1
10010010 10100100 1
11110101 00010001 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: 0001100010110101

Question 1 of 5

The answer was: 01000

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 16,5

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