GATE CSE 2000 | Question: 2.10

Question

The simultaneous equations on the Boolean variables $x, y, z$ and $w$,

• $x + y + z = 1 $
• $xy = 0$
• $xz + w = 1$
• $xy + \bar{z}\bar{w} = 0$

have the following solution for $x, y, z$ and $w,$ respectively:

• A. $0 \ 1 \ 0 \ 0$
• B. $1 \ 1 \ 0 \ 1$
• C. $1 \ 0 \ 1 \ 1$
• D. $1 \ 0 \ 0 \ 0$

Comments

Ok... Thanks for explaination bro..

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@Deepak Poonia sir...can we do it by taking the intersection of minterms of all inputs..?

f1 satisfies $\sum$(2,3,4,5.....15)

f2 satisfies $\sum$(2,3,4,5,6...11)

f3 satisfies $\sum$(1,3,5,7,9,10,11,12,13,15)

f4 satisfies $\sum$(1,2,3...11)

f1 $\cap$ f2 $\cap$ f3$\cap$ f4= $\sum$(3,5,7,9,10,11)

I Think this is the answer for your question sir..?

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@Deepak Poonia I think your question’s ans is 6, This can be easily done by Combinatorics. 

1. x+y+z=1 [total 2^3 -1(000 case) = 7 combinations]

2. xy=0 [3 combinations]

3. xz+w=1 [8 combinations]

4. xy+z’w’=0 => z’w’ = 0 => z+w=1 [only 3 combinations]

Now, we can create 2 cases w=0 or w=1,

Case 1:  w=0 :  i.e z=1 (from eq 4) , x=1 (from eq 3) , and x=z=1 means y=0 (from eq 1) 

So, Possible combination  = 1 only

Case 2: w=1   x=0,1   y=0,1  &  z=0,1  so  2c1*2c1*2c1 = 8 combinations but out of 8 i/p combinations, 3 are invalid (i.e x=1, y=1, z=1 & x=1, y=0, z=1 cz for x=y=1 violated eq 2 and x=y=z=0 violates eq 1 itself )

So, no possible combinations = 8-3 = 5

Total Combinations will be: 1+5 = $6$ :

x	y	z	w
1	0	1	0
0	1	0	1
0	0	1	1
0	1	1	1
1	0	0	1
1	0	1	1

 

Answer 1

option c is right.