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

01110110 00101101
00010111 00101001
10110000 10000100
10101001 00111100
10010101 00011111

Figure 2

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

01000010 01001001 1
01010010 01000110 0
00101001 10000101 0
10001001 10011000 0
11100100 01110010 1
00010100 01100000 0

Figure 3

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

11101100 01110100 1
10010000 10110010 1
00100000 01000000 1
01111110 00011011 0
11101110 10101011 1
11001010 00110110 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:

01110110 00101101 1
00010111 00101001 1
10110000 10000100 1
10101001 00111100 0
10010101 00011111 1
11101101 10100011 0

1. The parity bits for the 16 columns is: 11101101 10100011

2. The parity bits for the 5 rows is: 11101

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

4. The bit that was flipped in figure 2 is (1,4):

01000010 01001001 1
01010010 01000110 0
00101001 10000101 0
10001001 10011000 0
11100100 01110010 1
00010100 01100000 0

For figure 3, the bits that were flipped are (5,2) and (6,1):

11101100 01110100 1
10010000 10110010 1
00100000 01000000 1
01111110 00011011 0
11101110 10101011 1
11001010 00110110 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: 1110110110100011

Question 1 of 5

The answer was: 11101

Question 2 of 5

The answer was: 0

Question 3 of 5

The answer was: 1,4

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