You may only use two input gates, AND, OR and NOT gates are only gates accepted.
I'm new to this stuff so I know its probably really simple but anyways I have a truth table, a representation (sum-of-products and product-of-sum), and I ended up making the gate circuit for the given representation.
Heres the representation: ~(A*B*C) + (~A*B*C) + (A*~B*C) + (A*B*~C), the * is AND, the + is OR.
I'm suppose to reduce this using only two input gates instead of the 5 that i'm using now. The representation I have is the one I came up with based on the truth table, not a given one on the homework so it might be different then what something else.
Anyways I've tried reducing this representation and I've come to the conclusion theres no way I can? I need the ~(A*B*C) because 1 1 1 = 0 and 0 0 0 = 1 and if you try pulling out letters and reducing it you just end up with something dumb like B(~AC + A~C), A(~BC + ~CB) etc..
Any ideas?
I'm new to this stuff so I know its probably really simple but anyways I have a truth table, a representation (sum-of-products and product-of-sum), and I ended up making the gate circuit for the given representation.
Heres the representation: ~(A*B*C) + (~A*B*C) + (A*~B*C) + (A*B*~C), the * is AND, the + is OR.
I'm suppose to reduce this using only two input gates instead of the 5 that i'm using now. The representation I have is the one I came up with based on the truth table, not a given one on the homework so it might be different then what something else.
Anyways I've tried reducing this representation and I've come to the conclusion theres no way I can? I need the ~(A*B*C) because 1 1 1 = 0 and 0 0 0 = 1 and if you try pulling out letters and reducing it you just end up with something dumb like B(~AC + A~C), A(~BC + ~CB) etc..
Any ideas?