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Given I am still learning how to do these Boolean Algebra, I was wondering if someone could tell me if I am on the right track.

All work is pictured. I figured that the circuit drawn is the actual circuit and not simplified using the Karnaugh map. Also does that have 3 gates technically or just two?

enter image description here

Any help is appreciated! EDIT: I know this is wrong but I am not sure where I went wrong. Has to be in the Karnaugh map but I dont understand why or where.

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  • \$\begingroup\$ Technically 5 depends on your definition of a gate \$\endgroup\$ – Voltage Spike Apr 20 '17 at 7:13
  • \$\begingroup\$ In your Kmap, you should change 111 to X, where X can be 1 or 0 as required to simplify the solution. In your case, X = 1, but your instructor probably mark it wrong as workings. \$\endgroup\$ – StainlessSteelRat Apr 20 '17 at 13:50
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What constitutes one single gate is open to interpretation. Other than that, use the algebra right there:

AB + BC = F

Factor B to give:

F = B (A + C)

This equates to one AND gate and one OR gate.

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  • \$\begingroup\$ So was mine right then? F=AB+BC? Cause he has F=AB+C but I can't seem to get that \$\endgroup\$ – noreturn Apr 20 '17 at 3:27
  • \$\begingroup\$ Correct - F=AB+C is wrong. Looking at the original table, you can see this: when C is 1, F is 1 according to the formula - but the table shows that F is not always 1 when C is 1 - and there are other examples. \$\endgroup\$ – CL22 Apr 20 '17 at 3:32
  • \$\begingroup\$ Thanks makes sense why I couldn't get what he had. Is there a reason we prefer f=B(A+B)? \$\endgroup\$ – noreturn Apr 20 '17 at 3:37
  • \$\begingroup\$ It removes the repeated instance of "B". It just is easier to read and manipulate algebraically \$\endgroup\$ – CL22 Apr 20 '17 at 8:41

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