Smallest number of operators for a given output of a 3-variable truth table For a $3$-variable truth table, there are $8$ combinations of variable states that must be examined. The output can be any of the $256$ ($2^8$) possible truth table values.
I am looking to find the most compact way to represent each possible value of the truth table (i.e. the fewest number of operators as possible).
I know the upper bound for the given output is 5 operators: $(A \land (B \,\text{op}_1 C)) \lor (\lnot A \land (B \,\text{op}_2 \, C))$. This function joins the two cases of where $A$ is true, and where $A$ is false, with any of the binary operators that produce every form of $B$ and $C$ on their own. If you count the "not" in front of $A$, then we actually have $6$ operators.
I also know there are some that cannot be derived from $3$ variables and two operators (where each variable appears exactly once). $(A \leftrightarrow B) \land (B \leftrightarrow C)$ is the most compact way to represent all $3$ variables being the same value.
 A: At the moment I have only a partial solution. I count the $\lnot$ as a binary operator and also the constants (0-ary operators) $T$ and $F$. I did some calculations with python and if I did not made an error it is not possible to express all possible ternary operators by three binary operators. My program was only able to find such representations of three binary operators for only 232 of the 256 ternary operators. One of the ternary operator that I could not express by three binary operators was
 n a b c *
 1 0 0 0 0
 2 0 0 1 0
 3 0 1 0 0
 4 0 1 1 1
 5 1 0 0 0
 6 1 0 1 1
 7 1 1 0 1
 8 1 1 1 0

Her n is the row number of the table, a,b,c are the variables and * is the value of the operator with arguments a,b,c.
Bu I am not sure that my program is working correctly. But what I can show here is that ternary operator cannot be expressed by two binary operations.
Proof:
Assume that * can be expressed by (a op1 b) op2 c. Then from line 1 and 7 follows
(0 op1 0) op2 0 = 0
(1 op1 1) op2 0 = 1

Form this follows that
(0 op1 0) != (1 op1 1)   (1)

From line 2 and 4 follows
(0 op1 0) != (0 op1 1)    (2)

and from 4 and 8
( 0 op1 1) != (1 op1 1)    (3) 

From the last two inequalities follows
(0 op1 0) = (1 op1 1)     (4)

But (4) contradicts (1). so we cannot find two binary operators such that
$$( a\ \text{op}_1\ b)\ \text{op}_2\ c = *(a,b,c)$$
One also has to check if
$$ a\ \text{op}_1\ (b\ \text{op}_2\ c) = *(a,b,c)$$
but we will get a similar contradiction.
