Truth Table, Canonical Sum and Minimizing the Canonical Sum

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Discussion Overview

The discussion revolves around constructing and minimizing a truth table and canonical product of sums (POS) expression for the function $f(X,Y,Z) = \pi M(2,4,6)$. Participants seek clarification on the correct formulation of the POS and the minimization process, involving both theoretical and practical aspects of Boolean algebra.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a truth table for the function, which is confirmed by another participant.
  • There is a challenge regarding the formulation of the canonical product of sums, with one participant asserting that the initial response is a sum of products rather than a product of sums.
  • Another participant expresses confusion about the minimization of the expression and seeks assistance.
  • A later reply provides a proposed correct expression for the product of sums and discusses the minimization process using Boolean algebra laws.
  • One participant questions the correctness of the derived expression and suggests verifying it through truth tables or other methods.

Areas of Agreement / Disagreement

Participants generally agree on the truth table's correctness but disagree on the formulation of the canonical product of sums and the subsequent minimization process. Multiple competing views on the final expression remain unresolved.

Contextual Notes

There are indications of missing assumptions regarding the definitions of terms in Boolean algebra, and the discussion reflects uncertainty in the application of minimization techniques.

shamieh
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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) = $$\pi$$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:
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Re: Drawing Truth Tables etc. Some clarification please?

shamieh said:
2a)Draw the truth table corresponding to $f$(X,Y,Z) = $$\pi$$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.

shamieh said:
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).
 
Re: Drawing Truth Tables etc. Some clarification please?

Evgeny.Makarov said:
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)
 
Re: Drawing Truth Tables etc. Some clarification please?

shamieh said:
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?
 
Re: Drawing Truth Tables etc. Some clarification please?

shamieh said:
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 $(x+y!+z)(x!+z)$.
 
Re: Drawing Truth Tables etc. Some clarification please?

Evgeny.Makarov said:
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.
 
Re: Drawing Truth Tables etc. Some clarification please?

shamieh said:
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.
 

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