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

11100000 10000010
01111110 00011101
01111010 11101010
11000101 00100100
00010101 01111111

Figure 2

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

00111001 00101000 0
11010010 10000111 1
10001111 11000111 0
00110000 01011111 0
01100010 00000010 0
00110110 10110101 1

Figure 3

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

11111001 10100111 1
10101011 10110011 0
00011110 00110001 1
11001011 00010010 1
00110000 01101001 1
10010111 11011110 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:

11100000 10000010 1
01111110 00011101 0
01111010 11101010 0
11000101 00100100 0
00010101 01111111 0
00110100 00101110 1

1. The parity bits for the 16 columns is: 00110100 00101110

2. The parity bits for the 5 rows is: 10000

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

4. The bit that was flipped in figure 2 is (8,1):

00111001 00101000 0
11010010 10000111 1
10001111 11000111 0
00110000 01011111 0
01100010 00000010 0
00110110 10110101 1

For figure 3, the bits that were flipped are (8,5) and (2,4):

11111001 10100111 1
10101011 10110011 0
00011110 00110001 1
11001011 00010010 1
00110000 01101001 1
10010111 11011110 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: 0011010000101110

Question 1 of 5

The answer was: 10000

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 8,1

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