CODEWITHSUNDEEP
boolean-logicalgebraboolean-expressionboolean-operations
user476033
user476033

Reputation: 4727

Basic boolean minimization

I am trying to simplify the following piece of boolean algebra so I can construct the circuit :

A'.B'.C.D  +  A'.B.C.D'  +  A'.B.C.D  +  A.B'.C'.D +  A.B'.C.D  +  A.B.C'.D  +  A.B.C.D' + A.B.C.D

So far I have gotten it to :

(C.D) + (B.C) + (A.C'.D)

Is this correct?

I want to get the best possible minimization.

The steps I have went through so far are :

A'.B'.C.D  +  A'.B.C.D'  +  A'.B.C.D  +  A+B'+C'+D +  A.B'+C+D  +  A.B.C'.D  +  A.B.C.D' + A.B.C.D 
= A.A'(B'.C.D)  +  A.A'(B.C.D')  +  A.A'(B.C.D)  +  B.B'(A.C'.D)
= (B.C.D) + (B'.C.D) + (B.C.D) + (B.C.D') + (A.C'.D)
= (C.D) + (B.C) + (A.C'.D)

Can I do any more?

Upvotes: 2

Views: 2945

Answers (4)

Achilles Ram Nakirekanti
Achilles Ram Nakirekanti

Reputation: 4011

for simplifying boolean expressions use karnaugh maps. i think it is very much useful if we less number of variables. but if we have more variables then we can follow methods because this method is not that preferable.

Upvotes: 0

Paul R
Paul R

Reputation: 212979

Assuming your equation is actually:

X = (A'.B'.C.D) + (A'.B.C.D') + (A'.B.C.D) + (A+B'+C'+D) + (A.B'+C+D) + (A.B.C'.D) + (A.B.C.D') + (A.B.C.D);

I just ran this through Logic Friday and it factored it down to:

X = 1;

So you might want to check your simplification work and/or check that you've given the correct equation.

However I suspect there may be typos in the original equation above, and perhaps it should be:

X = (A'.B'.C.D) + (A'.B.C.D') + (A'.B.C.D) + (A.B'.C'.D) + (A.B'.C.D) + (A.B.C'.D) + (A.B.C.D') + (A.B.C.D);

?

In which case Logic Friday simplifies it to:

X = B.C + A.D + C.D;

Upvotes: 1

500 - Internal Server Error
500 - Internal Server Error

Reputation: 28839

Here's another solution (found by brute force):

(a+c).(b+d).(c+d)

Upvotes: 1

Vivian River
Vivian River

Reputation: 32390

The only thing I can see that you could possibly do is distribute the "C" in the left two terms:

(C).(B+D)+(A.C'.D)

Or you could distribute the "D":

(C+A.C').D + (B.C)

Response to Comment: The distributive law is described here: http://www.ee.surrey.ac.uk/Projects/Labview/boolalgebra/. See the information under heading "T3"

Upvotes: 1

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