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

11110011 11011001
11011010 11110111
01110011 10100001
11101010 11100001
00001100 00100101

Figure 2

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

01001010 01000100 1
11101111 01100011 1
11001110 00001011 0
10010101 10010011 0
10001100 10110100 0
01110010 01001011 0

Figure 3

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

11101001 00101011 1
00101110 11110000 0
00110110 00110000 1
11101110 00001110 0
10110011 01100000 1
01101100 10000101 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:

11110011 11011001 1
11011010 11110111 0
01110011 10100001 0
11101010 11100001 1
00001100 00100101 1
10111100 01001011 1

1. The parity bits for the 16 columns is: 10111100 01001011

2. The parity bits for the 5 rows is: 10011

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

4. The bit that was flipped in figure 2 is (9,4):

01001010 01000100 1
11101111 01100011 1
11001110 00001011 0
10010101 10010011 0
10001100 10110100 0
01110010 01001011 0

For figure 3, the bits that were flipped are (1,3) and (0,2):

11101001 00101011 1
00101110 11110000 0
00110110 00110000 1
11101110 00001110 0
10110011 01100000 1
01101100 10000101 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: 1011110001001011

Question 1 of 5

The answer was: 10011

Question 2 of 5

The answer was: 1

Question 3 of 5

The answer was: 9,4

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