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Find the minimal representation for:

$f(w,x,y,z)$ = summation $m(0,5,6,8,13.14)+d(4,9,11,12)$

I was a little confused what to do with the don't cares but I used all of them.

Based on the Karnaugh map I made $4$ groups:

$w'y'z'+y'z+wx'+yz'$

Then to simplify I factored out $y'$. So I got $y'(wz'+z)+wx'+yz'=y(w')+wx'+yz'$

Is this the minimal representation or can more simplification be done?

JKnecht
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Lil
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    I have no idea what "f(w,x,y,z)= summation m(0,5,6,8,13.14)+d(4,9,11,12)" is supposed to mean... – 5xum Oct 27 '14 at 10:21
  • what part of the question is confusing? Sorry for not asking it clearly. – Lil Oct 27 '14 at 10:25
  • The definition of $f$. The part I quoted. I have no idea what that means. What is the domain of $f$, and what is the definition of $f$? – 5xum Oct 27 '14 at 10:27
  • that was all my textbook provided."f" represents the function we are trying to find the minimal representation for. Does that help? Since there are 4 variables and the karnaugh map goes up to 15.. so I would assume that is our domain? – Lil Oct 27 '14 at 10:30
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    What textbook is that? – Git Gud Oct 27 '14 at 10:34
  • Discrete and Combinatorical Mathematics by Grimaldi – Lil Oct 27 '14 at 10:34
  • @Lil What do the numbers $0, 5,7,8,13.14$ mean?! – 5xum Oct 27 '14 at 11:03
  • they correspond to locations on the karnaugh map – Lil Oct 27 '14 at 11:08
  • $\begin{matrix} 0&1&3&2\ 4&5&7&6\ 12&13&15&14\ 8&9&11&10\ \end{matrix}

    \begin{matrix} m&m&·&·\ d&m&·&m\ d&m&·&m\ m&d&d&·\ \end{matrix}$

     – Senex Ægypti Parvi Nov 03 '14 at 00:55
  • $\begin{matrix} 0&1&3&2\ 4&5&7&6\ 12&13&15&14\ 8&9&11&10\ \end{matrix}

    \begin{matrix} m&·&·&·\ d&m&·&m\ d&m&·&m\ m&d&d&·\ \end{matrix}$

     – Senex Ægypti Parvi Nov 03 '14 at 02:26

1 Answers1

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I assume that $d(4,9,11,12)$ means that we don't care about these outputs. You don't have to use all the don't cares.

If $f(A,B,C,D)$ is the given function with A the Most Significant Bit then the minimal representation is:
$$C'D' + BD' + BC'$$

Joker
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  • yes d is the don't cares. So will using the cares such as I did prevent me form getting the minimal representation? I'm a little confused how to know if my solution is the simplest form. – Lil Oct 27 '14 at 10:34
  • So after creating the Karnaugh map I got xy'z+xyz'+w'x'y'z'+wx'y'z from there what simplification was done?

    Can I do: x(yz'+z')+x'y'(w'z+wz) ?

     – Lil Oct 27 '14 at 10:40
  • @Joker \ Are you sure? I got ${\overline{C}\overline{D}}+{B\overline{D}}+{B\overline{C}}$. – Senex Ægypti Parvi Oct 27 '14 at 12:01
  • @SenexÆgyptiParvi I went through the problem again on my own using the quine mccluskey algorithm but I got wx'z+y'z'+xz'+xy'z where did I go wrong? – Lil Oct 27 '14 at 18:02
  • @Joker I went through the problem again on my own using the quine mccluskey algorithm but I got wx'z+y'z'+xz'+xy'z where did I go wrong? – Lil Oct 27 '14 at 18:02
  • $\begin{matrix} 0&1&3&2\ 4&5&7&6\ 12&13&15&14\ 8&9&11&10\ \end{matrix}

    \begin{matrix} m&·&·&·\ d&m&·&m\ d&m&·&m\ m&d&d&·\ \end{matrix}$

     – Senex Ægypti Parvi Nov 03 '14 at 02:27