GATE IT 2004 | Question: 84

Question

Consider a parity check code with three data bits and four parity check bits. Three of the Code Words are $0101011, 1001101$ and $1110001.$ Which of the following are also code words?

1. $0010111$
2. $0110110$
3. $1011010$
4. $0111010$

• A. I and III
• B. I, II and III
• C. II and IV
• D. I, II, III and IV

Comments

can any one have reference for this

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How cn we sure which bits to be considered as parity and data bits??

And what method?

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Refer this https://www.geeksforgeeks.org/gate-gate-it-2004-question-84/

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Even though Ayush Upadhyaya's answer is top class, then too in exam we can do even fast.

Then the total number of codewords is

                                                                   2^k

Here Data bits (3), hence k = 3   and total = 8

0000000 (always present) 

C1        ....given
C2        ....given
C3        ....given 

C1 ⊕ C2            
C1 ⊕ C3
C2 ⊕ C3
C1 ⊕ C2 ⊕ C3    ...............find these 4 values and match with options.

Answer 1

Let $X_1, X_2$ and $X_3$ are data bits, and $C_1, C_2, C_3$ and $C_4$ are parity check bits.

Given transmitted codewords are
$$\begin{array}{|c|c|c|c|c|c|c|} \hline \textbf {$X _1$} & \textbf {$X_2$} &\text {$X_3$} & \textbf {$C_1$}& \textbf {$C_2$}& \textbf {$C_3$}& \textbf{$C_4$} \\\hline \text{0 }& \text{1} & \text{0} & \text{1} & \text{0} & \text{1}& \text{1} \\\hline \text{1 }& \text{0} & \text{0} & \text{1} & \text{1} & \text{0}& \text{1} \\\hline  \text{1 }& \text{1} & \text{1} & \text{0} & \text{0} & \text{0}& \text{1} \\\hline \end{array}$$
By inspection, we can find the rule for generating each of the parity bits –
$$\begin{array}{|c|c|c|} \hline \textbf {$X _1$} & \textbf {$X_2$} &\text {$X_3$} & \textbf {$X_1$} \oplus  \textbf {$X_2$}&\textbf {$X_1$} \oplus  \textbf {$X_3$}& \textbf {$X_2$} \oplus  \textbf {$X_3$}& \textbf {$X_1$} \oplus  \textbf {$X_2$} \oplus \textbf{$X_3$} \\\hline \color{blue}{\text{0}}& \color{blue}{\text{1}} &\color{blue}{\text{0}} & \text{1} & \text{0} & \text{1}& \text{1} \\\hline 
\color{blue}{\text{1}} & \color{blue}{\text{0}} &\color{blue}{\text{0}} & \text{1} & \text{1} & \text{0}& \text{1} \\\hline  
\color{blue}{\text{1}} & \color{blue}{\text{1}} & \color{blue}{\text{1}} & \text{0} & \text{0} & \text{0}& \text{1} \\\hline \end{array}$$

Now we can not only eliminate options, we can also find the set of all eight code words using $X_1, X_2$ and $X_3$.

$$\begin{array}{|c|c|c|} \hline \textbf {$X _1$} & \textbf {$X_2$} &\text {$X_3$} & \textbf {$X_1$} \oplus  \textbf {$X_2$}&\textbf {$X_1$} \oplus  \textbf {$X_3$}& \textbf {$X_2$} \oplus  \textbf {$X_3$}& \textbf {$X_1$} \oplus  \textbf {$X_2$} \oplus \textbf{$X_3$} \\\hline \text{0 }& \text{0} & \text{0}  \\\hline \text{0 }& \text{0} & \text{1}  \\\hline\text{0 }& \text{1} & \text{0} & \text{1} & \text{0} & \text{1}& \text{1} \\\hline  \text{0 }& \text{1} & \text{1}  \\\hline \text{1 }& \text{0} & \text{0} & \text{1} & \text{1} & \text{0}& \text{1} \\\hline \text{1 }& \text{0} & \text{1}   \\\hline  \text{1 }& \text{1} & \text{0}   \\\hline  \text{1 }& \text{1} & \text{1} & \text{0} & \text{0} & \text{0}& \text{1} \\\hline \end{array}$$

We can fill all remaining entries above :)

Now I am directly writing answer, Option A is correct choice !

Comments

Made Easy copied this answer in their Previous Year Book or it may be other way round

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WAS THIS TAUGHT IN CLASS?
IS THIS EVEN IN SYLLABUS ANY MORE

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yes it is in syllabus

Answer 2

I had tried a new method hope this is less intuitive and more calculative.

In all valid codewords, the hamming distance between any two valid codewords remains constant.

Example say if we have a set of 16 valid codewords, then between any two valid codewords in this set of 16, the hamming distance would be same.

The given codewords are

0101011( Let's Say A), 1001101 (let's say B) and 1110001 ( say C)

For finding the hamming distance between any two codewords, perform XOR operation of both and find the number of 1's present in the result.
 A⊕B = 1100110 (Hamming distance is 4)

B⊕C=0111100 (Hamming distance is 4)

A⊕C=1011010(Hamming distance is 4)

So, in our given code system, all codewords are at a hamming distance of 4. 

Any new valid codeword must also have a hamming distance of 4 from these 3(A, B and C) valid codewords.

Now considering each option one by one

(I) 0010111 (Let's call it D)

A⊕ D=111100 (Hamming distance =4)

B⊕D=1011010 (Hamming distance =4)

C⊕D = 1100110 (Hamming distance =4)

This new codeword (D) has same hamming distance with the present codewords. Hence, this is a valid codeword.

(II)0110110 (Assume is as E)

  A⊕E=0011101 (Hamming distance = 4)

B⊕E = 1111011 (Hamming distance = 6)

Now, Stop here as this codeword E has a hamming distance of 6 with codeword B. But it should have same hamming distance with all the valid codewords.

So, Choice (II) is surely Invalid.

(III) 1011010 (Let's call it F)

A⊕F = 1110001 (Hamming distance =4)

B⊕F= 0010111 (Hamming distance =4)

C⊕F=0101011 (Hamming distance =4)

Yes, this is a valid codeword.

(IV)0111010 (Let's call it G)

A⊕G=0010011 (Hamming distance =3)

This is an invalid codeword.

So, clearly, our answer is option A

Comments

90+ upvotes v 80+ upvotes and this is not best answer. Update your Algorithms gateoverflow

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Great explaination Ayush Bhaiya :)

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Thank you sir .

Answer 3

The simplest way to solve this is to use XOR property of codewords which says that XOR of two codewords is itself a codeword.
Upon XORing 1st and 3rd codeword we get another codeword 1011010, which is III .
And on XORing this new generated codeword with 2nd codeword given we get 0010111, which is I.

Hence Answer = A
 

Comments

but this property is only used for linear codes right?

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We can not just apply XOR, bcoz this is not linear block code.

There has to some rule for generating each of the parity checks.

Similar to question 2 here or question 1 here

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@Sachin Mittal 1

http://www.seas.upenn.edu/~tcom501/Other/t501_midterm_spring00.pdf link is not working.. Can you provide working link..

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@Sachin Mittal 1 they have given in question parity check code. Parity check codes are also one type of linear block codes right?

Answer 4

This is nice Question. I'm giving probable answer, which seems best to me Though do comment if you feel there is any mistake.

010 1 0 1 1 => Here 1 is Even parity for 01 (starting 2 char of 010)

                         Here 0  is Even parity for 010 (First & last characters of 010)

                         Here 1 is Even parity for 010 (Last 2 character of 010)

Finally last 1 is Even parity for total string 010 1 0 1 => 1.

Same way you can do for 1001101 and 1110001.

If you try for using similar test for

I. 0010111             II. 0110110         III. 1011010             IV. 0111010

Then I & III pass in test. So answer is A .

Comments

please  explain in simple manner

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i also answered with same idea :)

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I am not getting this pls explain

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please  explain in simple way 

@Akash Kanase

Answer 5

Answer is option A.

 

Since in question it has given 3 code words so clearly it's indicating there is some pattern there so,

When you clearly see

P1: 0101011

P2:1001101

P3:1110001

So when you'll calculate hamming distance (no. Of unmatched bits) between p1 and p2 it come out to be 4. So does in p2 and p3 and p1 and p3 also.

Similar in option 1 and 3 there hamming distance is also 4. So option A is correct.

Answer 6

Here is another approach to look at the question.

3 data bits and 4 parity check bits. Isn't that overkill for error detection? We can pretty much send the same 3 bits again with an additional parity bit to establish which 3 data bits had an error.

So could this be a self-correcting hamming code?

For a self correcting hamming code, with 3 data bits (up to 4 data bits), we need an additional 3 parity bits. That still comes out to be 3+3=6. However, the problem with hamming codes is that they cannot tell if a double-bit error caused the modification or a single bit error. It assumes that it was a single bit error and the auto-correct mechanism gives us wrong data (in case more bits were modified). Adding 1 extra parity bit solves that to some extent. 3+3+1 = 7! https://en.wikipedia.org/wiki/Hamming_code
 

Just casually looking at the last bits of the codewords will tell you that they are even parity bits. Let us call them [Pa] as they are a single additional parity bit at the end.
[D3], [D5] and [D6] should be data bits. Remaining:  [P1],[P2] and [P4] should be the parity bits used for correction.

The codewords should therefore look like: [P1][P2][D3][P4][D5][D6][Pa].
The hamming code part should only be [P1][P2][D3][P4][D5][D6]

For even parities (which seems to be the norm here): 
[P1] = [D3]⊕[D5]  (D7 does not exist!)
[P2] = [D3]⊕[D6]  (D7 does not exist!)
[P4] = [D5]⊕[D6]  (D7 does not exist!)

This holds for all the codewords provided. And it holds for I and III only. The correct answer should therefore be A.