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Let $f(w, x, y, z) = \sum {\left(0,4,5,7,8,9,13,15\right)}$. Which of the following expressions are NOT equivalent to $f$?

P: $x'y'z' + w'xy' + wy'z + xz$

Q: $w'y'z' + wx'y' + xz$

R: $w'y'z' + wx'y' + xyz+xy'z$

S: $x'y'z' + wx'y'+ w'y$

1. P only
2. Q and S
3. R and S
4. S only
edited | 1.9k views
+1
can somebody remove that cursor from option A.

K-map
w'x' w'x wx wx'
y'z' 1 1   1
y'z   1 1 1
yz   1 1
yz'

So, minimized expression will be

$xz + w'y'z' + wx'y'$ which is Q. From the K-map, we can also get P and R. So, only S is NOT equivalent to $f$.

http://www.eecs.berkeley.edu/~newton/Classes/CS150sp98/lectures/week4_2/sld011.htm

Go with Minterm representation of each option, (Note that order of W,X,Y,Z is should be preserved. )

P : X' Y' Z' + W' X Y' + W Y' Z + X Z

= () X' Y' Z' + W' X Y' () + W () Y' Z + () X () Z

= () 000                  +             010()          +           1 () 01          +                        () 1 () 1

= 0000 + 1000        +      0100 + 0101      +       1001 + 1101      +    0101+0111+1101+1111

= 0 + 8                  +         4 + 5              +          9+13              +      5+7+13+15

= ∑ m(0,4,5,7,8,9,13,15)

Q : W' Y' Z'   + W X' Y'  + XZ

= W' () Y' Z'   + W X' Y' () + () X () Z

= 0 () 0 0             +        100 ()       +           () 1 () 1

= 0000+0100       +    1000+1001   +    0101 + 0111+1101+1111

= 0+4                 +     8+9            +       5+7+13+15

= ∑ m(0,4,5,7,8,9,13,15)

R : W' Y' Z'  +  W X' Y'  + XYZ + X Y' Z

= W' () Y' Z'     +     W X' Y' ()     +    () X Y Z    +    () X Y' Z

= 0 () 0 0           +  100 ()         +    () 111   +    () 101

=  0000+ 0100   + 1000+1001  +  0111+1111   + 0101 + 1101

= 0+4              +     8+9          +   7+15         +     5+13

= ∑ m(0,4,5,7,8,9,13,15)

S : X' Y' Z'  +  W X' Y'  + W' Y

= () X' Y' Z'  +  W X' Y' ()  + W' () Y ()

= () 000            +     100 ()          +      0 () 1 ()

=  0000+1000   +   1000+1001    +    0010+0011+0110+0111

= 0+8              +     8+9             +   2+3+6+7

= ∑ m(0,2,3,6,7,8,9)

answered by Veteran (369k points)
edited
0
I am not understanding how to check P and R. one way is that we can draw the 4bit truth table but that is time consuming. How to derive it(P and R) using k-map?
0

@Hirak

NOW, CHECK THE ANSWER

0

thank you sir
0

P : X' Y' Z' + W' X' Y' + W Y' Z + X Z

P: X'Y'Z' + W'XY' + WY'Z + XZ

0

@register_user_19

on the first line only, i wrote X', actually it is X only.... from second line onwards it is taken as X only... everything is correct in the derivation of statement P.

i will update it... thanks for pointing out the mistake.

0
nice exp...

Let me show u a very simple method

Let w =1 ,x =1 ,y=1 ,z=1 then the value of f is 1

consider each statement

x'y'z' + w' x y' +w y' z +x z = 0.0.0 + 0.1.0 + 1.0.1 + 1.1 =1

w' y' z' +w x' z' y' +x z = 0.0.0 +1.0.0 +1.1  =1

w' y' z' +w x' y' +x y z +x y' z =0.0.0 +1.0.0 +1.1.1 +1.0.1 =1

x' y' z' +w x' y' + w' y = 0.0.0 + 1.0.0 + 0.1 =0

So statement (d) is false because w=1 x=1 y=1 z=1 the value of f is 0 .

(d) does not contain the essential Minterms .

xyz+wxy+wyz+xz

xyz+wxy+wyz+xz

answered by Boss (45.4k points)
0
Is this method correct for generalisation? If yes,why?
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is this a perfect method to solve such question, sometimes I got the error  by this method ?

need help @shekhar_chauhan
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@arjun sir confirm it please
0
Here it works by only putting 0 and 1 values. But this is not a general way to solve this problem. For e.g. for $x.z$ we get $1.1=1$ though it's not a correct representation of function.
Minterms of P, Q and R are same as f.

So, S is not equivalent to f

Ans (D)
answered by Boss (13.5k points)
+1 vote
putting x=y=z=w=1

s is false.

So option A is eliminated as S is not there.

Out of B and C ,

Q=R , so either both B and C correct or both wrong.

As this ques can have only one answer , so B and C are incorrect.

Only S ie D is the correct answer.
answered by (227 points)
0
xz term is compulsory  but s does not fulfill it

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