February 26, 2008

* Circuit Value Problem is P-complete
* Circuit complexity and time complexity
* Circuit-SAT is NP-complete (alternative proof of Cook-Levin)

CIRCUIT VALUE IS P-COMPLETE

A Boolean circuit consists of a collection of 2-input AND, 2-input OR,
1-input NOT gates, and inputs, connected together by wires in an
acyclic manner.  A Boolean circuit computes an output value from a
setting of inputs.  Note that a Boolean circuit can be written out as
a SAT formula and vice versa.

The Circuit Value Problem (CVP) is to calculate the value of a given Boolean
circuit on a given input setting.  

Theorem: CVP is P-complete.

Proof: We first note that CVP is in P.

Now, we reduce every language in P to CVP.  The proof is similar to
Cook-Levin's theorem.  Let M be a deterministic Turing machine running
in t(n) = O(n^k) time.  We construct a tableau that lists the
configurations of M on given input, one in each row.  The tableau is
t(n) x t(n).

We assume that M accepts only when its head is on the leftmost cell
and the tape is blank.  Also, once M halts the remaining
configurations are the same.

P_{i,j,a} = Symbol occupying tape cell j at time i is a

Q_{i,j,q} = The machine is in state q scanning cell j at time t

P_{i,j,b} = [OR_{d(p,a) = (q,b,d)} (Q_{i-1,j,p} AND P_{i-1,j,a})]
            OR [P_{i-1,j,b} AND (AND_{p in Q} NOT(Q_{i-1,j,p})

That is, the symbol occupying tape cell j at time i is a if either the
machine is in state p scanning cell j at time t and the machine writes
a after appropriate transition or the cell j had at time i-1 and the
machine was not scanning the cell j at time t-1.

Q_{i,j,q} = [OR_{d(p,a) = (q,b,R)} (Q_{i-1,j-1,p} AND P_{i-1,j-1,a})]
            OR [OR_{d(p,a) = (q,b,L)} (Q_{i-1,j+1,p} AND P_{i-1,j+1,a})]

Special case for j = 0 and j = t(n).

All that remains to take care are the input and the output.  The input
is simply the specification of the start configuration.

P_{0,0,#} = 1
P_{0,0,b} = 0 for all b in Gamma
P_{0,j,w_j} = 1
P_{0,j,x} = 0 if x not= w_j
P_{0,j,u} = 1 for j >= n+1
P_{0,j,x} = 0 if x not= u for j >= n+1
Q_{0,0,q_0} = 1 where q_0 is start state
Q_{0,0,q} = 0 where q not= q_0

The output is Q_{t(n),0,q_f} where q_f is accept state of M.

Note that each variable is a gate.  The total number of variables is
O(t(n)^2).

End Proof

The above theorem can be easily generalized to establishing the
following between time complexity and circuit complexity.  A language
L over {0,1} is said to have circuit complexity C(n) if for every n,
there exists a circuit of size C(n) that evaluates to true for
precisely the set of n-length strings that are in L.

CIRCUIT COMPLEXITY AND TIME COMPLEXITY

Theorem: Every language in TIME(t(n)) has a circuit of size O(t(n)^2)
if t(n) >= n.

Proof: The proof is identical to that of the above theorem except for
the fact that we do not specify the inputs.  The inputs simply
correspond to the string w that is being considered for membership in L.

P_{0,0,#} = 1
P_{0,0,b} = 0 for all b in Gamma
P_{0,j} = w_j
P_{0,j,u} = 1 for j >= n+1
P_{0,j,x} = 0 if x not= u for j >= n+1
Q_{0,0,q_0} = 1 where q_0 is start state
Q_{0,0,q} = 0 where q not= q_0

Thus, the only inputs are the P_{0,j}s.

End Proof

Circuits offer an alternative way of looking at languages.  Define
Circuit-SAT as follows.

Circuit-SAT = {<C> : C is a satisfiable Boolean circuit}

ALTERNATIVE PROOF OF COOK-LEVIN THEOREM

Theorem: Circuit-SAT is NP complete.

Proof: Clearly Circuit-SAT is in NP.  For the reduction, given any
language L in NP, we need a poly-time computable f such that 

w is in L iff f(w) is a satisfiable Boolean circuit.

L is in NP so for every w, there is a poly-size certificate c and a
deterministic TM M (the verifier) that accepts <w,c>.  Use the proof
of the theorem connecting circuit complexity and time complexity to
construct the circuit.  Call this circuit C; we specify the input bits
corresponding to w, and c is unspecified.

Clearly C is satisfiable iff there is an assignment to c that
satisfies C.  The size of the circuit is poly in the size of w and c,
which is poly in the size of w.

End Proof