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

10000100 01110101
11000000 11000111
00011001 00111101
00010011 10101011
00001011 11110100

Figure 2

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

01101010 00000100 1
00010101 10010110 1
11111100 11011100 1
01001110 11101000 1
11101010 00101010 0
00100111 10101100 0

Figure 3

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

01000011 10110101 0
11111110 01110101 0
10000001 00000000 1
10011001 10011001 1
11100110 11101011 1
01100111 10110010 1


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:

10000100 01110101 1
11000000 11000111 1
00011001 00111101 0
00010011 10101011 0
00001011 11110100 0
01000101 11010000 0

1. The parity bits for the 16 columns is: 01000101 11010000

2. The parity bits for the 5 rows is: 11000

3. The parity bit for the parity row is: 0

4. The bit that was flipped in figure 2 is (10,3):

01101010 00000100 1
00010101 10010110 1
11111100 11011100 1
01001110 11101000 1
11101010 00101010 0
00100111 10101100 0

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

01000011 10110101 0
11111110 01110101 0
10000001 00000000 1
10011001 10011001 1
11100110 11101011 1
01100111 10110010 1

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: 0100010111010000

Question 1 of 5

The answer was: 11000

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 10,3

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