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

01001010 10010010
10110011 01111010
10000110 11011011
00011110 01110101
01110010 01110101

Figure 2

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

10101100 01101100 0
01110000 11011010 0
01000000 00110111 0
11110011 00111001 1
01001100 01000100 1
00100010 11111100 0

Figure 3

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

00000111 01111110 1
00100001 10000011 0
11010100 00011100 1
00110100 11010001 1
01101001 01011111 1
10101011 11101111 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:

01001010 10010010 0
10110011 01111010 0
10000110 11011011 1
00011110 01110101 1
01110010 01110101 1
00010011 00110011 1

1. The parity bits for the 16 columns is: 00010011 00110011

2. The parity bits for the 5 rows is: 00111

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

4. The bit that was flipped in figure 2 is (7,3):

10101100 01101100 0
01110000 11011010 0
01000000 00110111 0
11110011 00111001 1
01001100 01000100 1
00100010 11111100 0

For figure 3, the bits that were flipped are (5,1) and (8,4):

00000111 01111110 1
00100001 10000011 0
11010100 00011100 1
00110100 11010001 1
01101001 01011111 1
10101011 11101111 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: 0001001100110011

Question 1 of 5

The answer was: 00111

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 7,3

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