NP-completeness and Cook's Theorem.
A is reducible to B if there is a polynomial-time function f such that
- x in A implies f(x) in B
- x not in A implies f(x) not in B
L is NP-complete if
- L is in NP, and
- For all A in NP, A is reducible to L
We will show that CNF-SAT is NP-complete. Let A be a language in NP accepted by a nondeterministic Turing machine M. Fix an input x. We will create a 3CNF formula φ that will be satisfiable if and only if there is a proper tableau for M and x.
Let m be the running time of M on x. m is bounded by a polynomial in |x| since A is in NP. m is also a bound on the size of the configurations of M(x).
We will create a set of Boolean variables to describe a tableau and a set of clauses that will all be true if and only if the tableau is proper. The variables are as follows.
- qij: true if confi is in state j.
- hik: true if the head in confi is at location k.
- tikr: true if the tape cell in location k of confi has element r.
We create four clause groups to check that the tableau is proper.
- Every configuration has exactly one state, head location and each tape cell has one element.
- conf0 is the initial configuration.
- confm is accepting.
- For each i≤m, confi+1 follows from confi in one step.
1. We will just do this for states. The others are similar. Suppose we have u possible states.
Each configuration in at least one state. For each i we have
qi0 OR qi1 OR ... OR qiu
Each configuration in at most one state. For each i and possible states v and w, v≠w
(NOT qiv) OR (NOT qiw)
2. Let x=x1...xn, b the blank character and state 0 the initial state. We have the following single variable clauses,
q00
h01
t0ixi for i≤n
t0ib for i>n
3. Let a be the accept state. We need only one single variable clause.
qma
4. We need two parts. First if the head is not over a tape location then it should not change.
((NOT hik) AND tikr)→ ti(k+1)r
Now this doesn't look like an OR of literals. We now apply the facts that P→Q is the same as (NOT P) OR Q and NOT(R AND S) is equivalent to (NOT R) OR (NOT S) to get
hik OR (NOT tikr) OR t(i+1)kr
At this point none of the formula has been dependent on the machine M. Our last set of clauses will take care of this. Recall the program of a Turing machine is a mapping from current state and tape character under the head to a new state, a possibly new character under the head and a movement of the tape head one space right or left. A nondeterministic machine may allow for several of these possibilities. We add clauses to prevent the wrong operations.
Suppose that the following is NOT a legitimate transition of M: In state j and tape symbol r, will write s, move left and go to state v. We prevent this possibility with the following set of clauses (for all appropriate i and k).
(qij AND hik AND tikr)→ NOT(t(i+1)ks AND hi(k-1) AND q(i+1)v)
which is equivalent to
(NOT qij) OR (NOT hik) OR (NOT tikr) OR (NOT t(i+1)ks) OR (NOT hi(k-1)) OR (NOT q(i+1)v)
We do this for every possible illegitimate transition. Finally we just need to make sure the head must go one square right or left in each turn.
(NOT hik) OR h(i+1)(k-1) OR h(i+1)(k+1)
Just to note in the above formula we need special care to handle the boundary conditions (where k is 1 or m) but this is straightforward.
