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

10001011 10000001
00011010 01001000
11010100 01011011
11100100 01101111
00100111 00000101

Figure 2

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

11000101 10000110 1
11011011 01001111 1
00101011 01001101 0
00100110 11111011 0
10110000 01010000 0
10100010 00101111 0

Figure 3

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

01110010 00010101 1
01001011 00000111 0
10011011 01100110 1
00110001 01110010 0
10100011 00010111 0
00100000 00011001 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:

10001011 10000001 0
00011010 01001000 1
11010100 01011011 1
11100100 01101111 0
00100111 00000101 0
10000110 11111000 0

1. The parity bits for the 16 columns is: 10000110 11111000

2. The parity bits for the 5 rows is: 01100

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

4. The bit that was flipped in figure 2 is (7,4):

11000101 10000110 1
11011011 01001111 1
00101011 01001101 0
00100110 11111011 0
10110000 01010000 0
10100010 00101111 0

For figure 3, the bits that were flipped are (12,3) and (3,1):

01110010 00010101 1
01001011 00000111 0
10011011 01100110 1
00110001 01110010 0
10100011 00010111 0
00100000 00011001 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: 1000011011111000

Question 1 of 5

The answer was: 01100

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 7,4

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