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+11 votes

A set $X$ can be represented by an array $x[n]$ as follows: 

 $x\left [ i \right ]=\begin {cases} 1 & \text{if } i \in X \\ 0 & \text{otherwise}  \end{cases}$

Consider the following algorithm in which $x$, $y$, and $z$ are Boolean arrays of size $n$: 

algorithm zzz(x[], y[], z[]) {
    int i;
    for(i=0; i<n; ++i)
        z[i] = (x[i] ∧ ~y[i]) ∨ (~x[i] ∧ y[i]);

The set $Z$ computed by the algorithm is: 

  1. $(X\cup Y)$
  2. $(X\cap Y)$
  3. $(X-Y)\cap (Y-X)$
  4. $(X-Y)\cup (Y-X)$
in Algorithms by | 1.6k views
Apply simple set theory to it.
(X ∧ ~Y) ∨ (~X ∧ Y)

Draw venn diagram. Answer is D.
Hi Guys,

I think quickly answer could be obtained by 'Venn Diagram'.

2 Answers

+22 votes
Best answer

Option (D)

In the given algorithm the for loop contains a logical expression

 z[i] = (x[i] ∧ ~y[i]) ∨ (~x[i] ∧ y[i]);

The equivalent set representation of a given logical expression if we assume $z[i] = Z, x[i] = X, y[i] = Y$ then 

$Z = ( X \wedge \neg Y ) \vee ( \neg X \wedge  Y)$

$\implies Z = ( X - Y ) \cup ( Y - X )  [\because A \wedge \neg B = A - B]$

edited by
This can also be solve using some example and even by using ven  diagram which resuts in exor operation at end hence d)
+11 votes
experesion AB' + A'B is XOR.. XOR has property that either of A or B is 1 then output is 1 but not both...

this can be achieved by option (D)..

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