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Simplification of Boolean Expression

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blade88

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Hello,

I am trying to simplify the following expression:

F=A'BC'D+A'BCD'+AB'C'D+AB'CD'+ABCD

However, the Karnaugh Map doesn't show any pairs of ones.

Are there any suggestions?

Thanks
 

not in terms of not-and-or logic. Maybe there is something using xor.

For FPGA applications, it is a 4LUT -- K-maps don't make sense for FPGA applications. (you should still know them for interviews)
 
Hi,

It's been a long time since I posted here.

I solved it:

F=A'BC'D+A'BCD'+AB'C'D+AB'CD'+ABCD=
=(A'B+AB')C'D+(A'B+AB')CD'+ABCD=
=(A'B+AB')(C'D+CD')+ABCD=
=(A XOR B)(C XOR D)+ABCD

Using Karnaugh Map, I should find F' using zeros in the table and I could find the same result.

Thank you
 

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