I need clarification on the arbitrary decision that is made to decide when light will be on.
According to the explanation in the book by Kenneth Rosen on Discreet Mathematics and Its Applications:
Question:
Sometimes light fixtures are controlled by more than one switch. Circuits need to be designed so that flipping any one of the switches for the fixture turns the light on when it is off and turns the light off when it is on. Design circuits that accomplish this when there are two switches and when there are three switches.
Explanation:
Let x = 1 when the first switch is closed and x = 0 when it is open, and let y = 1 when the second switch is closed and y = 0 when it is open. Let F(x, y) = 1 when the light is on and F(x, y) = 0 when it is off. We can arbitrarily decide that the light will be on when both switches are closed, so that F(1, 1) = 1. This determines all the other values of F. When one of the two switches is opened, the light goes off, so F(1, 0) = F(0, 1) = 0. When the other switch is also opened, the light goes on, so F(0, 0) = 1. Note that F(x,y) = xy + $\bar x$$\bar y$
As far as I understood, the values that the function F(x,y) will take depends on the arbitrary decision that is made to decide when light will be on.
Is it also correct to make an arbitrary decision about when the light will be off and then follow the same logic as given in the explanation in the book to design a circuit?
ATTEMPTS:
Considering the case with only on switch:
If we consider a negative logic and a light fixture controlled by only one switch, we can let x = 0 when the switch is closed and x = 1 when the switch is open. Also, let F(x) = 1 when the light is on and F(x) = 0 when it is off. We can arbitrarily decide that the light will be on when switch is closed, which happens when x = 0. Therefore the light goes on when F(0) = 1. This determines the other value of F. When switch is opened the light goes off, so F(1) = 0.
I suppose this logic will work.
Considering the case with two switches:
Can we also arbitrarily decide that the light will be off when both switches are open so that F(0,0) = 0? If Yes, this determines all other values of F. When one of the two switches is closed, the light goes on, so F(1,0) = F(0,1) = 1. When the other switch is also closed, the light goes off, so F(1,1) = 0.
Note that F(x,y) = $\bar x$y + x$\bar y$.
The circuit in this case will be different from the one given in the explanation in the book because the boolean expressions are different in the two cases, i.e, this boolean expression is different from the one in the explanation given in the book for two switch case. Will this circuit function as required?
Can we also arbitrarily decide that the light will be on when one switch is off and the other is on and then follow the same logic as given in the explanation in the book to design a circuit?