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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. $(X-Y)\cup (Y-X)$
| 1.3k views
+8
Apply simple set theory to it.
(X ∧ ~Y) ∨ (~X ∧ Y)

Draw venn diagram. Answer is D.
0
Hi Guys,

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

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

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

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

by Active (4.7k points)
edited by
0
This can also be solve using some example and even by using ven  diagram which resuts in exor operation at end hence d)