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Truth Table, Canonical Sum and Minimizing the Canonical Sum

shamieh

Active member
Sep 13, 2013
539
Also would someone mind checking my work on these problems too? My answers are in BOLD

2a)Draw the truth table corresponding to $f$(X,Y,Z) = \(\displaystyle \pi\)M(2,4,6)

ANSWER:
  • x y z | f
  • 0 0 0| 1
  • 0 0 1| 1
  • 0 1 0| 0
  • 0 1 1| 1
  • 1 0 0| 0
  • 1 0 1| 1
  • 1 1 0| 0
  • 1 1 1| 1

2b) Write out the canonical product of sums (POS) expression for $f$(X,Y,Z) of 2a.

ANSWER:

x!yz + xy!z! + xyz!

2c) Minimize the expression of 2b.

ANSWER:
= x!yz + xy!z! + xyz!
= x(x!yz + y!z! + yz!)
= xy(x!z + y!z! + z!)

Yea okay I'm lost now. Help on this one please?
 
Last edited:

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Re: Drawing Truth Tables etc. Some clarification please?

2a)Draw the truth table corresponding to $f$(X,Y,Z) = \(\displaystyle \pi\)M(2,4,6)

ANSWER:
  • x y z | f
  • 0 0 0| 1
  • 0 0 1| 1
  • 0 1 0| 0
  • 0 1 1| 1
  • 1 0 0| 0
  • 1 0 1| 1
  • 1 1 0| 0
  • 1 1 1| 1
Correct.

2b) Write out the canonical product of sums (POS) expression for $f$(X,Y,Z) of 2a.

ANSWER:

x!yz + xy!z! + xyz!
What you wrote is a sum of products, not a product of sums. Also, see this info about indexing maxterms (i.e., the correspondence between maxterms and binary numbers).
 

shamieh

Active member
Sep 13, 2013
539
Re: Drawing Truth Tables etc. Some clarification please?

What you wrote is a sum of products, not a product of sums.
So what is the correct answer then? And what is the answer for minimizing it? Because I can't figure out how to do it...if you don't mind me asking(Wait)
 

shamieh

Active member
Sep 13, 2013
539
Re: Drawing Truth Tables etc. Some clarification please?

So what is the correct answer then? And what is the answer for minimizing it? Because I can't figure out how to do it...if you don't mind me asking(Wait)
Nevermind I figured it out. so I got xy!z + x!yz + x!y!z

but how do i Minimize that expression?
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Re: Drawing Truth Tables etc. Some clarification please?

Nevermind I figured it out. so I got xy!z + x!yz + x!y!z
You negated variables correctly this time, but don't you see that this is still a sum of products? Multiplication has a priority of addition, just as in ordinary arithmetic, so in your expression you first multiply x, y, !z and the other two triples and then you add the results: sum of products. In contrast, ΠM (capital M) means a product of sums (Greek Π stands for product and Greek Σ stands for sum in many areas of mathematics, and the capital M stands for maxterms, which are sums). The expression should be \[(x + y! + z)(x! + y + z)(x! + y! + z)\tag{1}\]

Minimizing it involves distributivity of addition over multiplication: \[(a + c)(b + c) = ab + c\tag{2}\] This law is less intuitive than distributivity of multiplication over addition because the former is not true in arithmetic, unlike the latter. Nevertheless, it is true in Boolean algebra. There are two ways to apply it to (1). The first one is to use the first and third sums of (1) and set $a=x$, $b=x!$ and $c=y!+z$ in (2). The right-hand side of (2) becomes $xx! + y! + z$, which equals $y!+z$ since $xx!=0$ and $0+d=d$ for any $d$. Thus, the whole (1) equals $(y!+z)(x! + y + z)$. The second way is to use the second and third sums of (1) and set $a=y$, $b=y!$ and $c=x!+z$ in (2). The right-hand side of (2) becomes $yy!+x!+z$, which equals $x!+z$. Thus, the whole (1) equals $(x+y!+z)(x!+z)$.
 

shamieh

Active member
Sep 13, 2013
539
Re: Drawing Truth Tables etc. Some clarification please?

Thus, the whole (1) equals $(x+y!+z)(x!+z)$.
Are you 100% positive that is the right answer? Looks like my teacher got it wrong then. He has (y! + z)(x! + z) for the final answer.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Re: Drawing Truth Tables etc. Some clarification please?

Are you 100% positive that is the right answer?
Why don't you check it using truth tables or another method?

My derivation to your teacher's answer seems a bit long, and I am not sure it's useful to post it. If you covered some methods for simplifying POS'es, you could post them and we can discuss how to apply them to this formula.