2.2k views

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 | 2.2k views
+1
can somebody remove that cursor from option A.
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Noooooo

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

Alternatively,

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

Here, $x$ means do not care ($x$ takes value either $0$ or $1$)

$P : X' Y' Z' + W' X Y' + W Y' Z + X Z$
$\qquad = x X' Y' Z' + W' X Y' x + W x Y' Z + x X x Z$
$\qquad= x 000 + 010x + 1 x 01 + x 1 x 1$
$\qquad= 0000 + 1000 + 0100 + 0101 + 1001 + 1101 + 0101+0111+1101+1111$
$\qquad= 0 + 8 + 4 + 5 + 9+13 + 5+7+13+15$
$\qquad= ∑ m(0,4,5,7,8,9,13,15)$

$Q : W' Y' Z' + W X' Y' + XZ$
$\qquad= W' x Y' Z' + W X' Y' x + x X x Z$
$\qquad= 0 x 0 0 + 100 x + x 1 x 1$
$\qquad= 0000+0100 + 1000+1001 + 0101 + 0111+1101+1111$
$\qquad= 0+4 + 8+9 + 5+7+13+15$
$\qquad= ∑ m(0,4,5,7,8,9,13,15)$

$R : W' Y' Z' + W X' Y' + XYZ + X Y' Z$
$\qquad= W' x Y' Z' + W X' Y' x + x X Y Z + x X Y' Z$
$\qquad= 0 x 0 0 + 100 x + x 111 + x 101$
$\qquad= 0000+ 0100 + 1000+1001 + 0111+1111 + 0101 + 1101$
$\qquad= 0+4 + 8+9 + 7+15 + 5+13$
$\qquad= ∑ m(0,4,5,7,8,9,13,15)$

$S : X' Y' Z' + W X' Y' + W' Y$
$\qquad= x X' Y' Z' + W X' Y' x + W' x Y x$
$\qquad= x 000 + 100 x + 0 x 1 x$
$\qquad= 0000+1000 + 1000+1001 + 0010+0011+0110+0111$
$\qquad= 0+8 + 8+9 + 2+3+6+7$
$\qquad= ∑ m(0,2,3,6,7,8,9)$

Correct Answer: $D$

edited
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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?
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@Hirak

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thank you sir
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P : X' Y' Z' + W' X' Y' + W Y' Z + X Z

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

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@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.

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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

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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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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)
+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.
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xz term is compulsory  but s does not fulfill it

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