How can I simplify this circuit using boolean algebra for XOR and XNOR gates?

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Discussion Overview

The discussion revolves around simplifying a boolean expression involving XOR and XNOR gates. Participants explore various methods, including K-maps and alternative logic gate configurations, to achieve a simpler circuit representation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks help to simplify the expression M'(A'B'C+ABC')+M(AB'C'+A'BC) using boolean algebra.
  • Another participant suggests using K-maps but believes the terms cannot be simplified further using only AND, OR, and NOT gates.
  • A participant mentions that they initially derived the expression using K-maps and expresses a desire to simplify it to a form involving XOR, specifically (AB Oplus C), but finds it challenging.
  • There is a question raised about whether the term (A'B'C+ABC') can reduce to (A Oplus B Oplus C).
  • One participant discusses the possibility of using a single Programmable Logic Array (PLA) for simplification, noting the need for a burner.
  • A participant shares their experience of trying to express the equation using XOR/XNOR and mentions discovering a pattern that aids in isolation.
  • Another participant proposes using a multiplexer as a potential solution, highlighting its simplicity and commonality.
  • A claim is made regarding a specific expression (A xnor C) nor (B xor M), questioning its validity and discussing its implications for circuit simplicity.

Areas of Agreement / Disagreement

Participants express differing opinions on the simplification methods and the potential forms of the boolean expression. No consensus is reached on the simplest form or the best approach to take.

Contextual Notes

Participants mention various assumptions and conditions, such as the roles of variables M and B, and A and C, which may affect the simplification process. There are also references to the complexity of implementing certain logic gates in terms of physical components.

speck
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I want to simplfy M'(A'B'C+ABC')+M(AB'C'+A'BC) to as simple a circuit as possible.

I don't know the boolean algebra to simplfy the ABC terms. Help please, Speck
 
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Tried K-maps?

From quick Venn inspection of (A'B'C+ABC') and (AB'C'+A'BC), I don't think you can simplify them further using AND, OR, NOT only
 
K-map is how I initial got the Eq. , right, it won't simplify with AND, OR, NOT. I want to use XOR with XNOR gates. I would really like it to simplify to something like (AB Oplus C) using XOR, but it does not. Thks, Speck
 
Does anyone think that the (A'B'C+ABC') part of the Eq. will reduce to (A Oplus B Oplus C)?
 
By simple, do you mean the least number of packages? It's trivial with a single PLA, but you'd need a burner...
 
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I tried to put it into XOR/XNOR but I really couldn't find any way.

P.S. (I learned this stuff few weeks ago, so all I know is that there should be checkboard pattern)

Now that I said that I realized that there is infact a pattern and it is easier to isolate it when you look at it. You got to approach it differently.
See K-Map When A = 0 and C = 1
A = 1 C = 0

I get something like

A!C!(B XOR M) + A!C (M XOR B)

So far, I look at K-Map and try to isolate 2 literal K-Maps that look like XOR and "and" it with conditions like A = 1 and C = 0 .. It works so far
 
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Another PLA-type cheater's answer: use a multiplexer. Input ABCM as the addresses, hardwire the 16 inputs to 1 or 0 to synthetise the desired logic function. The 4067 is such a 16-to-1 mux-demux and seems to be still relatively common (hey, I just feel younger!). One single package, no programming needed.

For a non-cheater answer, you'll have to wait a bit more. M and B have similar roles, as do A and C, so combining these pairs first could bring something.
 
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(A xnor C) nor (B xor M)
please check!

It's not "the simplest" form for the number of packages on a breadboard.

In a chip, I guess it's not the minimum number of Mos neither, as an xor or an xnor needs two inverters and 8 transistors, and a 16-to-1 mux also needs one inverter per input and this function consumes 16 N-channels and 16 P-channels.

Well, this form must be the simplest in the mind of some teacher at least.
 
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