Engineering Is it Possible to Solve This Problem - Equations and Building Circuits

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Building the functions F, G, and H using only four chips is challenging due to the limited number of available logic gates and the complexity of the equations. The user has outlined their input configuration with eight switches and specified the chips they can use, which include various 7400 series logic gates. Despite spending several hours on the problem, they believe it may not be possible to achieve the desired outcomes within the constraints. Other participants are encouraged to share their attempts and solutions to help find a feasible approach. The discussion highlights the difficulty of circuit design with strict limitations on components.
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# -> means XOR
' => means NOT
AB => means A and B
A+B => means A or B

Is it possible to build the following 3 functions
F=(D#B)+(C#A)
G=DCA'+DB'+CB'A'
H=DC'B+D'C'A+D'BA+C'BA

Using only 4 chips (Build all three functions using only 4 chips not 4 for each function)

I have 8 switches for the inputs and was going to use
Switch 1 for A
Switch 2 for B
Switch 3 for C
Switch 4 for D
Switch 5 for A'
Switch 6 for B'
Switch 7 for C'
Switch 8 for D'

I don't have any other switches. I'm just using a switch and flipping switch 8 to the opposite of switch 4 to get the NOT of D.

I can only use the following chips. At the end of the day there can only be 4 chips on my protoboard so if I use the same chip say twice than this counts as two chips.

7400, 7402, 7404, 7408, 7410, 7411, 7420, 7421, 7425, 7427,
7432, and 7486.

I'm getting that it's not possible and have spent several hours trying to do it. Thanks for any help that you can provide me!
 
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GreenPrint said:
# -> means XOR
' => means NOT
AB => means A and B
A+B => means A or B

Is it possible to build the following 3 functions
F=(D#B)+(C#A)
G=DCA'+DB'+CB'A'
H=DC'B+D'C'A+D'BA+C'BA

Using only 4 chips (Build all three functions using only 4 chips not 4 for each function)

I have 8 switches for the inputs and was going to use
Switch 1 for A
Switch 2 for B
Switch 3 for C
Switch 4 for D
Switch 5 for A'
Switch 6 for B'
Switch 7 for C'
Switch 8 for D'

I don't have any other switches. I'm just using a switch and flipping switch 8 to the opposite of switch 4 to get the NOT of D.

I can only use the following chips. At the end of the day there can only be 4 chips on my protoboard so if I use the same chip say twice than this counts as two chips.

7400, 7402, 7404, 7408, 7410, 7411, 7420, 7421, 7425, 7427,
7432, and 7486.

I'm getting that it's not possible and have spent several hours trying to do it. Thanks for any help that you can provide me!

What is the closest that you've gotten? Can you post what you have tried?
 
GreenPrint said:
I have 8 switches for the inputs and was going to use
Switch 1 for A
Switch 2 for B
Switch 3 for C
Switch 4 for D
Switch 5 for A'
Switch 6 for B'
Switch 7 for C'
Switch 8 for D'
Normally you would use a switch to provide D, and use a gate to generate D'.
 

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