Notice that the logic equation appears to be reading the diagram from the right, then from the top to the bottom. The value of D is dependent on what the values of A, B, and C are, and we can show all possible values by creating a truth table.

The first step is to create a column for all the inputs and one for the output, and enough rows to hold all possible values.  The number of rows is calculated at 2^(number of inputs):

A	B	C	D
?	?	?	?
?	?	?	?
?	?	?	?
?	?	?	?
?	?	?	?
?	?	?	?
?	?	?	?
?	?	?	?

Once you have created your table, set out the possible combinations of inputs. The process of creating these is called grey code. Start with Column A – add in 0, then 1, then repeat a mirror of that until the column is full like this:

A	B	C	D
0	?	?	?
1	?	?	?
1	?	?	?
0	?	?	?
0	?	?	?
1	?	?	?
1	?	?	?
0	?	?	?

Next, fill in column B but this time use 2 zeros, then 2 ones like this:

A	B	C	D
0	0	?	?
1	0	?	?
1	1	?	?
0	1	?	?
0	1	?	?
1	1	?	?
1	0	?	?
0	0	?	?

Finally, complete Column C by using four zeros, then four ones (notice that we’re increasing by powers of two for each column). Like this:

A	B	C	D
0	0	0	?
1	0	0	?
1	1	0	?
0	1	0	?
0	1	1	?
1	1	1	?
1	0	1	?
0	0	1	?

Now our grey code is in place we can test each of the inputs against our logic circuit. Let’s start with the first line. Because A is part of the AND gate and it is false, then the output cannot be true.

A	B	C	D
0	0	0	0
1	0	0	?
1	1	0	?
0	1	0	?
0	1	1	?
1	1	1	?
1	0	1	?
0	0	1	?

Try the rest for yourself, then click the button below to reveal the answer. ​