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

00010010 11011001
01100001 11101000
01110000 00000101
11111001 11001111
11001110 10001111

Figure 2

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

11111001 10010111 1
11101010 01001101 0
00010011 10100111 0
00010011 00010000 0
11110010 01110011 0
11101001 00011110 1

Figure 3

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

10000000 00001010 1
01111001 10100010 1
11011001 00111010 0
10100011 10000001 0
11100101 10111100 0
01100100 10111111 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:

00010010 11011001 1
01100001 11101000 1
01110000 00000101 1
11111001 11001111 0
11001110 10001111 0
00110100 01110100 1

1. The parity bits for the 16 columns is: 00110100 01110100

2. The parity bits for the 5 rows is: 11100

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

4. The bit that was flipped in figure 2 is (4,1):

11111001 10010111 1
11101010 01001101 0
00010011 10100111 0
00010011 00010000 0
11110010 01110011 0
11101001 00011110 1

For figure 3, the bits that were flipped are (11,1) and (6,2):

10000000 00001010 1
01111001 10100010 1
11011001 00111010 0
10100011 10000001 0
11100101 10111100 0
01100100 10111111 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: 0011010001110100

Question 1 of 5

The answer was: 11100

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 4,1

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