Logic Gates!

Logic Level 4

What is the minimum number of basic one/two input gates needed to implement the following Boolean Function?

F ( w , x , y , z ) = w x y z + w x y z + x y z + w x z + w x z + x y z + w x z F(w, x, y, z) = w'x'y'z' + wx'yz' + x'y'z' + w'x'z' + wxz + xy'z + w'xz

Clarifications:

  • Basic Gates include AND, OR, and NOT only. XOR, NAND and NOR gates do not count in this problem.
  • Each AND/OR gate must only accept two inputs. Each NOT gate must only have one output.
  • Available inputs are only w, x, y, and z.

As an explicit example,

For the function F ( x , y , z ) = x y z + x y z + x y F (x,y,z) = xy'z + xy'z' + x'y' , this can be simplified to be F ( x , y , z ) = y F (x,y,z) = y' , which means it only needs one NOT gate to have the function implemented.


The answer is 5.

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1 solution

Farhabi Mojib
Oct 29, 2015

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