Build a 2-way universal logic processor using NAND logic gates

Question

A 2-way universal logic processor (2ULP) is a network of logic gates that takes two input wires A and B, as well as four other inputs L_, L_a, L_b, and L_ab, and produces a single output L(a, b) using the four L inputs as a truth table function:

• The 2ULP returns L_ if A and B are both 0.
• It returns L_a if A = 1 and B = 0.
• It returns L_b if A = 0 and B = 1.
• It returns L_ab if A and B are both 1.

For example, given the inputs L_ = 0, L_a = 1, L_b = 1, and L_ab = 0, then the output L(a, b) will be equal to A xor B.

Your task is to build a 2ULP using only NAND gates, using as few NAND gates as possible. To simplify things, you may use AND, OR, NOT, and XOR gates in your diagram, with the following corresponding scores:

• NOT: 1
• AND: 2
• OR: 3
• XOR: 4

Each of these scores corresponds to the number of NAND gates that it takes to construct the corresponding gate.

The logic circuit that uses the fewest NAND gates to produce a correct construction wins.

Answer 1

11 NANDs

Define the gate MUX (cost 4) as

MUX(P, Q, R) = (P NAND Q) NAND (NOT P NAND R)

with truth table

P Q R    (P NAND Q)   (NOT P NAND R)    MUX
0 0 0        1               1           0
0 0 1        1               0           1
0 1 0        1               1           0
0 1 1        1               0           1
1 0 0        1               1           0
1 0 1        1               1           0
1 1 0        0               1           1
1 1 1        0               1           1

Then this is the familiar ternary operator MUX(P, Q, R) = P ? Q : R

We simply have

2ULP = MUX(A, MUX(B, L_ab, L_a), MUX(B, L_b, L_))

for a cost 12, but there's a trivial one-gate saving by reusing the NOT B from the two inner MUXes.

Answer 2

cost: 4*4*14+4*(13)+13*3+3*3+24*1+4=352

I'm not a boolean man, this is my best in coding this things (I know this won't give me many immaginary internet points..).

# l1 is for a=F , b=F
# l2 is for a=F , b=T
# l3 is for a=T , b=F
# l4 is for a=T , b=T

2ULP(l1,l2,l3,l4,a,b) =
 (l1&l2&l3&l4)|             # always true
 (!l1&l2&l3&l4)&(a|b)|      # a=F,b=F->F; ee in T
 (l1&!l2&l3&l4)&(!a&b)|     # a=F,b=T->F; ee in T
 (!l1&!l2&l3&l4)&(a)|       # a=F,b=F->F; a=F,b=T->F; a=T,b=F->T; a=T,b=T->T; 
 (l1&l2&!l3&l4)&(a&!b)|     # a=T,b=F->F, ee in T
 (!l1&l2&!l3&l4)&(b)|       # a=T,b=F->F, ee in T
 (!l1&!l2&!l3&l4)&(a&b)|    # a=T,b=T->T, ee in F
 (l1&l2&l3&!l4)&(!(a|b))|   # a=T,b=T->F, ee in T
 (!l1&l2&l3&!l4)&(!(avb))|  # a=T,b=F->T, a=F,b=T->T, ee in T , note v is the exor.
 (l1&!l2&l3&!l4)&(!b)|      # T when b=T
 (!l1&!l2&l3&!l4)&(a&!b)|   # T when a=T,b=F
 (l1&l2&!l3&!l4)&(!a)|      # T when a=F
 (!l1&l2&!l3&!l4)&(!a&b)|   # T when a=F,B=T
 (l1&!l2&!l3&!l4)&(!(a|b))  # T when a=F,B=F

Answer 3

By using Wolfram language I can get a 13 gates formula:

logicfunc = Function[{a, b, Ln, La, Lb, Lab},
                   {a, b, Ln, La, Lb, Lab} -> Switch[{a, b},
                          {0, 0}, Ln, {1, 0}, La, {0, 1}, Lb, {1, 1}, Lab]
                  ];
trueRuleTable = Flatten[Outer[logicfunc, ##]] & @@ ConstantArray[{0, 1}, 6] /. {0 -> False, 1 -> True};
BooleanFunction[trueRuleTable, {a, b, Ln, La, Lb, Lab}] // BooleanMinimize[#, "NAND"] &

which outputs:

Here Ln, La, Lb and Lab are the L_, L_a, L_b and L_ab separately in OP.

sidenote: The results of the BooleanMinimize function in Wolfram language are restricted to two-level NAND and NOT when invoking as BooleanMinimize[(*blabla*), "NAND"], so it's not as good as Peter Taylor's four-level formula above.