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

11110010 01110100
10011010 00010000
11011011 00111010
11011000 00010010
01011101 10000101

Figure 2

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

01000000 00110101 0
00000000 11111000 1
00111000 11011110 1
11011111 10100101 1
10110011 10001000 1
00010100 01111110 0

Figure 3

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

00000110 11001001 0
11011101 00101011 0
10101000 11101010 0
10001101 01110110 0
00001101 11110101 0
10100011 10001011 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:

11110010 01110100 1
10011010 00010000 1
11011011 00111010 0
11011000 00010010 0
01011101 10000101 0
00110110 11001001 0

1. The parity bits for the 16 columns is: 00110110 11001001

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 (9,0):

01000000 00110101 0
00000000 11111000 1
00111000 11011110 1
11011111 10100101 1
10110011 10001000 1
00010100 01111110 0

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

00000110 11001001 0
11011101 00101011 0
10101000 11101010 0
10001101 01110110 0
00001101 11110101 0
10100011 10001011 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: 0011011011001001

Question 1 of 5

The answer was: 11000

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 9,0

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