Here's the Karnaugh map:

alt text

The answer I should be getting from the Karnaugh should be:

T = R ∙ (CGM)'

I'm really not seeing how this was arrived at through any simplification methods I've learned thus far. I can see the answers that are intended are correct (I think), though.

From what I know, the best answer I can come up with to simplify (only) the Kargnaugh map is:


Help appreciated!

  • $\begingroup$ I edited this to make the problem more clear. Only thing (as of now) I'm still trying to figure is the truth table part. $\endgroup$ – Hamster Oct 19 '10 at 19:46

Using sum of products you should be able to derive:

RC' + RCG' + RCGM'

Substitute out the R:

R(C' + CG' + CGM')

Use the identity A + A'B = A + B

R(C' + G' + CGM')

That same identity works as a multivariable expression

R(C' + G' + M')

Then apply DeMorgans


  • $\begingroup$ So that identity A+A'B works even for just 2 of the terms in the formula? I was under the impression it had to be applied to each term... Well I'm glad I'm getting to wrap my brain around this now. Thanks! $\endgroup$ – Hamster Oct 18 '10 at 16:56
  • $\begingroup$ There are multiple ways of deriving an expression from a Karnaugh map. The one I chose was the entire row RC(10), and then I grouped (RC(11), GM(00)) and (RC(11), GM(01)) together. (RC(11), GM(10)) is by itself. $\endgroup$ – Unreasonable Sin Oct 19 '10 at 16:32
  • $\begingroup$ So is using RCGM' by itself (rather than trying to make a group out of it) the right way to do it? I was told it's acceptable to make a group out of it and another adjacent square, even if that other square is part of another group. $\endgroup$ – Hamster Oct 19 '10 at 16:44
  • $\begingroup$ Um, just one last question: How do I get that sum of products answer from the truth table, though? I just can't figure that one. $\endgroup$ – Hamster Oct 19 '10 at 17:02
  • $\begingroup$ There's really no right way to do Karnaugh map simplification. All of the different (correct) expressions you derive from a Karnaugh map are logically equivalent, even if on the surface they don't appear to be. It's important to understand the fundamentals, and be able to do simple Karnaugh maps by hand, but once you start designing logic systems you'll just have a computer do the simplification for you. $\endgroup$ – Unreasonable Sin Oct 19 '10 at 17:04

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