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I have this truth table:

   CD | 00 | 01 | 11 | 10
AB    |    |    |    |
------+----+----+----+----
00    |  1 |  1 |  0 |  0
------+----+----+----+----
01    |  0 |  1 |  1 |  0
------+----+----+----+----
11    |  0 |  1 |  1 |  0
------+----+----+----+----
10    |  0 |  1 |  1 |  0
------+----+----+----+----

Now I need the f function... and I already have the result:

f=¬A¬B¬C+BD+AD

...but I really can't get there. I checked my calculations for hours and I can't get that result. Can you help me in understanding the logic? Thans a lot in advance!

Cholesky
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1 Answers1

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I assume you are currently studying Karnaugh Maps (K-Maps). If you have not been introduced to them, I strongly suggest you do a search for them and look at some examples and explanations. Anywa, the idea here is to identify groups or blocks of 1's that represent a simple expression. Thus, for example, in your diagram the four 1's in the very middle corresponds to the term BD. And the two 1's in row 3 together with the two 1's in row four combine to the term AD. Finally, the two 1's in row 1 is your A'B'C' term. Add them together, and you have your function.

Addendum

OK, so you got to:

$$A'B'C' + BD + AB'D$$

So far so good! But you can continue:

$$= A'B'C' + (B + AB')D = A'B'D' + (B +A)D = A'B'C'+BD+AD$$

That step where I go from $B +AB'$ to $B+A$ is called Reduction ... Few texts mention it (because you can derive it from more basic principles) but it's a very common pattern when doing boolean algebra, so add it to your toolbox!

Bram28
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  • Thanks a lot for your answer... but there's something I can't understand: you are using the two 1's in the third row TWICE... is this possibile?

    This is my calculation step by step: f=A'B'C'D'+A'B'C'D+A'BC'D+A'BCD+ABC'D+ABCD+AB'C'D+AB'CD f=A'B'C'(D'+D)+A'BD(C'+C)+ABD(C+C')+AB'D(C'+C) f=A'B'C'+A'BD+ABD+AB'D............................ f=A'B'C'+BD(A'+A)+AB'D........................................................... f=A'B'C'+BD+AB'D...............................................

    (instead of f=A'B'C'+BD+AD, that should be the correct result)

    Where am I wrong?

     – Cholesky Mar 14 '17 at 21:07
  • @Cholesky yes, when you're working with K-Maps you can use 1's in any number of groupings. As far as your specific derivation goes, let me add something to my answer – Bram28 Mar 14 '17 at 21:46
  • Oh my God! Thanks you a lot!!! It was NOT simple at all... I couldn't find that formula anywhere! Thanks you, now I'll remember it for sure! :) – Cholesky Mar 14 '17 at 22:28
  • @Cholesky Cool! Glad I could help! :) – Bram28 Mar 14 '17 at 22:29