Simplify & K-Map Homework Statement

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Discussion Overview

The discussion revolves around the process of transforming a given Boolean expression into a Karnaugh Map (K-Map). Participants explore methods for simplifying the transformation, particularly focusing on whether complete expansion is necessary or if shortcuts exist. The conversation includes aspects of homework-related problem-solving in the context of digital logic design.

Discussion Character

  • Homework-related
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant questions if the only way to convert a Boolean expression into a K-Map is through complete expansion, seeking potential shortcuts.
  • Another participant explains that with three variables, there are only eight states, and emphasizes that if any factor in the product of sums is zero, the result is zero.
  • Some participants express concern that manually plugging in values for each combination could be time-consuming during exams and seek more efficient methods.
  • A later reply suggests that for a product of sums, one can quickly determine where to place zeros in the K-Map by identifying the conditions that make each factor zero.
  • There is a discussion about the method of labeling K-Map rows and columns using reflected Gray code, which could simplify the process of filling in the K-Map without needing to convert to binary each time.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method for transforming the Boolean expression into a K-Map. While some suggest that manual methods may be the fastest, others propose alternative techniques that could save time.

Contextual Notes

Participants mention various methods for filling out the K-Map, including the use of binary conversion and Gray code labeling, but do not resolve which method is superior or more efficient in practice.

DorumonSg
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Homework Statement



Okay, I have question I was pondering about for a while...

If I had the equation below and wanted to transform it into a K-Map, is the only way I can do it to expand it completely? Are there any shortcuts?

Homework Equations



This is just a random equation out of my mind :

(A+B+C)(A+B'+C)(A+B+C')(A'+B+C)(A'+B')

So I been taught when handling K-Map if I get a SOP, I would make all of portions similar to include all 3 variables eg.(A+B+C) then convert them to binary, invert them and take their sum to map it into the K-Map as a short cut.

Example for (A + B + C) I would take 111, invert it 000, the decimal will be 0, and then I map this as a 0(Maxterm or was it minterm @_@) into the 0 portion of the K-Map.

But I was thinking is there anyway to convert the portions of my above equation to a (A+B+C) form easily or do I have to expand them to make it a POS then map it into a K-Map.
 
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When you only have three variables A, B, and C, there are only the 8 states:

ABC
000
001
011
010
110
111
101
100

If any of your factors is 0 in the product of sums, the result is 0. And in any individual factor in any term is 1 the sum is 1. It only takes a minute or so to just plug them into fill out the K-table.
 
You are telling me to pluck in the individual values 1 by 1 in the equation than sort it out in a K-Map which would cost a lot of time in an exam...

I am looking for shortcuts to save time like the one I posted above...

I am asking if besides sorting it out manually and expanding the above equation to a POS than map it. Is there anyway to map the above into K-map quick or change the portions to include all 3 A, B, C using any techniques?

Yeah, but i c ur point, it seems quickest in this case.
 
DorumonSg said:
You are telling me to pluck in the individual values 1 by 1 in the equation than sort it out in a K-Map which would cost a lot of time in an exam...

I am looking for shortcuts to save time like the one I posted above...

I am asking if besides sorting it out manually and expanding the above equation to a POS than map it. Is there anyway to map the above into K-map quick or change the portions to include all 3 A, B, C using any techniques?

Yeah, but i c ur point, it seems quickest in this case.

Well, if time was of the essence, I might do it this way for a product of sums. Here's your example:

(A+B+C)(A+B'+C)(A+B+C')(A'+B+C)(A'+B')

This will be 0 if and only if one of the factors is zero, and each factor will be zero only if each term is zero. So for each factor, note what makes it zero:

(A+B+C) 0,0,0
(A+B'+C) 0,1,0
(A+B+C') 0,0,1
(A'+B+C) 1.0.0
(A'+B') 1,1,X (here it doesn't matter what X is)

The above can be read off immediately from your formula. Put in 0's accordingly in your K map and 1's everywhere else. Should be fast enough for an exam.
 
LCKurtz said:
Well, if time was of the essence, I might do it this way for a product of sums. Here's your example:

(A+B+C)(A+B'+C)(A+B+C')(A'+B+C)(A'+B')

This will be 0 if and only if one of the factors is zero, and each factor will be zero only if each term is zero. So for each factor, note what makes it zero:

(A+B+C) 0,0,0
(A+B'+C) 0,1,0
(A+B+C') 0,0,1
(A'+B+C) 1.0.0
(A'+B') 1,1,X (here it doesn't matter what X is)

The above can be read off immediately from your formula. Put in 0's accordingly in your K map and 1's everywhere else. Should be fast enough for an exam.

I see what you mean, but I can't map the above... see the way I been taugh a K-map can be divided into portions of numbers.

0 1 3 2
4 5 7 6
12 13 15 16
8 9 11 10

So by adding up the binary numbers of the portions I can immediately know where to map it.. example (A'+B+C) 1.0.0 gives me 4... so it goes in 4.

Yep, so I suppose from the equation I gave, doing it manually maybe the fastest way.
 
DorumonSg said:
I see what you mean, but I can't map the above... see the way I been taugh a K-map can be divided into portions of numbers.

0 1 3 2
4 5 7 6
12 13 15 16
8 9 11 10

So by adding up the binary numbers of the portions I can immediately know where to map it.. example (A'+B+C) 1.0.0 gives me 4... so it goes in 4.

Yep, so I suppose from the equation I gave, doing it manually maybe the fastest way.

But you don't have to do the binary conversion to fill in your k-table. Don't you label the top and left side with the reflected Gray code, like this?:

--- 00 01 11 10
---------------------
00|
01| 0
11|
10|

Your example only has three bits, so you would only need two rows. But the point is that you can read the binary right off the table and put the 0's in. For example, I have put the 0 in the 100 spot; I don't have to think about the fact that is 4.
 

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