The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+9 votes
991 views

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. $(X-Y)\cup (Y-X)$
asked in Algorithms by Active (3.7k points) | 991 views
+7
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'.

2 Answers

+14 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 

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

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

answered by Active (4.8k 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)
+10 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)..
answered by Active (5k points)
Answer:

Related questions



Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

41,055 questions
47,653 answers
147,217 comments
62,380 users