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Simplify Boolean Expression Using Karnot Map

I have the following problem:

Express the following boolean expressions as the sum of products and simplify as much as possible using a Karnot map.

enter image description here

I drew a Karnaugh map and then put my values ​​in the table as true (First, B does not mean D means 10, and not B and D means 01). We then have the following values: 0100,0110,1100,1110 (since A and C can be either 0 or 1). So we get:

enter image description here

We notice that we only have one group (which is surrounded by blue) and then we have:

0100
0110
1100
1110

We see that the only variables that do not change their values ​​are B and D, and therefore we get the following simplified version:

B non D

But this answer is only for parenthesized expression without minus. Any ideas how I can solve this if I have a minus in front of the expression? How does this change my expression?

My second question is how am I supposed to solve it when I have double negation like this enter image description here

When matching, the first one means 1111 and the others are 0101, 1101, 0101 and then I solve it the same way? Any ideas? Thank!

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2 answers


enter image description hereFor the first question, an expression without negation can be called (B XOR D), so XOR with negation is basically XNOR. it can be represented in sums of works as (BD + B'D ')



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(AC! (! B! D)) + (! AC! (! B! D)) + (A! C! (! B! D)) + (! A! C! (! B. D!)) Steps



  • = C.! (! B.! D). (A +! A) +! C.! (! B! D). (A +! A)
  • = C.! (! B.! D) .1 +! C.! (! B.! D) .1
  • = C.! (! B.! D) +! C.! (! B.! D)
  • =! (! B.! D). (C +! C)
  • =! (! B.! D)
  • =! B +! D
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