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

11010111 00000000
01100000 01011101
00000101 11000001
10101110 01010011
10000011 11011101

Figure 2

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

10100001 01000100 1
10000000 11101110 1
10110000 01000001 1
10010011 01011001 0
11010111 11100101 0
11110101 01010111 1

Figure 3

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

11011110 01001101 1
10100011 10110000 1
00101101 01101011 0
10011010 01100101 0
01111100 00000000 1
10111100 11110011 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:

11010111 00000000 0
01100000 01011101 1
00000101 11000001 1
10101110 01010011 1
10000011 11011101 1
10011111 00010010 0

1. The parity bits for the 16 columns is: 10011111 00010010

2. The parity bits for the 5 rows is: 01111

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

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

10100001 01000100 1
10000000 11101110 1
10110000 01000001 1
10010011 01011001 0
11010111 11100101 0
11110101 01010111 1

For figure 3, the bits that were flipped are (4,0) and (6,2):

11011110 01001101 1
10100011 10110000 1
00101101 01101011 0
10011010 01100101 0
01111100 00000000 1
10111100 11110011 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: 1001111100010010

Question 1 of 5

The answer was: 01111

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 2,4

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