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

01101110 10011010
00000101 00100011
10011001 00111011
01010101 11101001
11001100 10110110

Figure 2

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

11001001 01001110 0
01011100 11001001 0
11001101 10110010 1
10011000 01110110 0
11111010 10101000 0
00111010 11101011 0

Figure 3

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

01011110 00000110 0
11001110 01110110 0
10000111 10011101 1
01000011 00011111 1
11010111 00011000 0
10000111 11001010 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:

01101110 10011010 1
00000101 00100011 1
10011001 00111011 1
01010101 11101001 1
11001100 10110110 1
01101011 11011101 1

1. The parity bits for the 16 columns is: 01101011 11011101

2. The parity bits for the 5 rows is: 11111

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

4. The bit that was flipped in figure 2 is (16,4):

11001001 01001110 0
01011100 11001001 0
11001101 10110010 1
10011000 01110110 0
11111010 10101000 0
00111010 11101011 0

For figure 3, the bits that were flipped are (10,3) and (5,0):

01011110 00000110 0
11001110 01110110 0
10000111 10011101 1
01000011 00011111 1
11010111 00011000 0
10000111 11001010 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: 0110101111011101

Question 1 of 5

The answer was: 11111

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 16,4

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