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

00011001 10011100
00111010 00010001
00010001 00000111
10111110 01100110
00110101 10001100

Figure 2

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

10001100 10111111 1
00001000 00100001 1
00011100 10101001 1
10011111 01100000 0
01011111 00100101 1
11011000 01110010 0

Figure 3

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

01010111 00100001 1
01000000 10110101 1
00010010 11000110 1
01011111 01101010 0
11111111 01100100 1
00100111 01011100 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:

00011001 10011100 1
00111010 00010001 0
00010001 00000111 1
10111110 01100110 0
00110101 10001100 1
10111001 01100000 1

1. The parity bits for the 16 columns is: 10111001 01100000

2. The parity bits for the 5 rows is: 10101

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

4. The bit that was flipped in figure 2 is (0,0):

10001100 10111111 1
00001000 00100001 1
00011100 10101001 1
10011111 01100000 0
01011111 00100101 1
11011000 01110010 0

For figure 3, the bits that were flipped are (0,2) and (6,1):

01010111 00100001 1
01000000 10110101 1
00010010 11000110 1
01011111 01101010 0
11111111 01100100 1
00100111 01011100 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: 1011100101100000

Question 1 of 5

The answer was: 10101

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 0,0

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