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Let, $x_{1} ⊕ x_{2} ⊕ x_{3} ⊕ x_{4}= 0$ where $x_{1}, x_{2}, x_{3}, x_{4}$ are Boolean variables, and $⊕$ is the XOR operator.

Which one of the following must always be TRUE?

  1. $x_{1}x_{2}x_{3}x_{4} = 0$
  2. $x_{1}x_{3} + x_{2} = 0$
  3. $\bar{x}_{1} ⊕ \bar{x}_{3} = \bar{x}_{2} ⊕ \bar{x}_{4}$
  4. $x_{1} + x_{2} + x_{3} + x_{4} = 0$
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10 Answers

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A XOR B is simply the addition of  bits of A and B without carry.

So, out of 4 variables given here x1, x2, x3, x4 , given their XOR is 0 ,  

  1. they all should be 0  or
  2. all should be 1 or
  3. any two of them should be 1.  

The reason why C is possible : 

LHS can have combination of x1 and x3 as 

  1. (1,1), when all of them are 1 and so RHS will also have (1,1) for (x2,x4).
  2. (0,0),when all of them are 0 and so RHS will also have (0,0) for (x2,x4).
  3. (1,0) or (1,1), when two them are 1 and so RHS will have (1,0) or (0,0) for (x2,x4).
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1 votes
EXOR gate gives output as 1 when we have odd number of 1’s in the input.
(EXNOR gate gives output as 1 when we have even number of 0’s in the input.)

Now, given x1⊕x2⊕x3⊕x4=0 which means number of 1’s in the input is not odd (but even).
Possible input values of x1,x2,x3,x4 are: Zero 1’s, two 1’s or all four 1’s.

Now let’s check the options:
    A) x1x2x3x4=0. False when we have all 1’s in the input.
    B) x1x3+x2=0. False when two 1’s occur for the inputs x1,x3.
    C) x¯1⊕x¯3=x¯2⊕x¯4. This is same as x1⊕x3=x2⊕x4
       Now, since the number of 1’s in the input is even,
       if number of 1’s in x1,x3 is even then number of 1’s in x2,x4 is also even and if number of 1’s in x1,x3 is odd then number of 1’s in x2,x4 is also odd. Hence this is always true.
    D) x1+x2+x3+x4=0.  False when we have zero 1’s in the input.
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1 votes

XOR is associative and commutative.

$x_1$ or $x_2$ or any other $x$'s order doesn't matter here.

 

$x_1⊕x_2⊕x_3⊕x_4= x_2⊕x_1⊕x_4⊕x_3$ or any other jumbled order.

 

The given function is true when we have two or four $1$'s

 

Since we can have four 1's; Options A and D are eliminated.

Since we can have two 1's, we can assign those 1's to $x_1$ and $x_3$ and Option B is eliminated.

It is a child's play to verify that Option C is the correct answer.

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0 votes
Answer c

Note that XOR is symmetric operation.For it to be zero the parity of number of 1s of mutually disjoint and exclusive sets should be same.
Answer:

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