Reduce boolean expression to 3 literals

[success2be]

Reduce the following to 3 literals
A'C' + ABC + AC'
C'(A'+A) + ABC
C' + ABC

or
A'C' + ABC + AC'
A'C' + A(BC + C')
A'C' + A(BC + C'(B + B'))
A'C' + A(BC + BC' + B'C'))

I can't go any further. What am I doing wrong?
The book has the answer as AB + C'

[success2be]

Figured it out. Man this is long.

From the second solution

A'C' + A (BC + BC' + B'C')
A'C' + A (B (C+C') + B'C')
A'C' + A ( B + B'C' )
A'C' + A ( (B + B') (B + C') )
A'C' + A ( B + C' )
A'C' + AB + AC'
C' (A' + A) + AB
C' + AB

I wonder is it always a trial and error process in getting the solution? Wonder if I'll ever have enough time during an exam to complete.

[Hurkyl]

Are you familiar with Karnaugh maps? (I hope I spelled that right)

[success2be]

Are you familiar with Karnaugh maps? (I hope I spelled that right)

The professor have talked about the K map. The chapter I'm working on have not discussed it officially yet. It'll be the next chapter.

I'm wondering if I'll be forced to not use the K map to solve a problem on an exam. Don't know if they only care as long as the solution is correct or that the student must know every method to the solution.

I think I'll just move on to the next chapter and come back to the previous chapter question when I understand K map. Thanks.

[Kenneth Mann]

I believe that the point of this is to give some understanding of the underlying principles and axioms. As convenient and time-saving as K-maps are, they don't really readily show why they work. In addition, most people can only apply K-maps to minimization problems of up to four or six variables. (In reality, if ordered K-maps are used, they can solve problems of as many variables as the user can practically map onto paper. This, however means the possibility of having to draw very large maps. As examples, a four-variable minimization problem requires a 16-cell map, but a ten-variable problem would require a 1024-cell map. Imagine how big that would be (but it can be done).

I'd suggest that you learn this approach well before worrying about mapping.

KM

[abdo375]

Reduce the following to 3 literals
A'C' + ABC + AC'
C'(A'+A) + ABC
C' + ABC

or
A'C' + ABC + AC'
A'C' + A(BC + C')
A'C' + A(BC + C'(B + B'))
A'C' + A(BC + BC' + B'C'))

I can't go any further. What am I doing wrong?
The book has the answer as AB + C'

I am continuing on your steps but first you need to START looking at the 'and' , 'or' signs as 'plus' and 'multiply' and to start solving the problems by means of both Boolean algebra and regular algebra


A'C' + ABC + AC'
C'(A'+A) + ABC
C' + ABC
(C'+A)*(C'+B)*(C+C')
since (C+C')=1
(C'+A)*(C'+B)
(C'*C')+(C'*B)+(A*C')+(A*B)
(C'*(1+B))+(AC')+(AB)
C'+AC'+AB
C'(A+1)+AB
AB+C'

[abdo375]

I wonder is it always a trial and error process in getting the solution? Wonder if I'll ever have enough time during an exam to complete.

No i don't think that this is the way to think about i'd rather think about it as simplifying it as possible
I can tell that this is your first course on this,
I took this course last year and I solved your problem by just looking at it ,so don't worry you'll get better at it soon.......

[simple_taurean]

Reduce the following to 3 literals
A'C' + ABC + AC'
C'(A'+A) + ABC
C' + ABC

or
A'C' + ABC + AC'
A'C' + A(BC + C')
A'C' + A(BC + C'(B + B'))
A'C' + A(BC + BC' + B'C'))

I can't go any further. What am I doing wrong?
The book has the answer as AB + C'


hey u wer solving it wrnd
di like this
ur eq. wz A'C'+ABC+AC'
it will be reduced like this
C'(A'+A)+ABC
C'+ABC
NOW CONSIDER AB AS 1 TERM AND C AS SECOND
THEN (AB+C')(C+C')
C+C'=1
SO UR ANS IS AB+C'
okieeeeeeeeeee
all the bst

[anollipian]

that's our assignment too :)
i guess these problems are nice .. much better than electronics :biggrin: