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CSC 363                 Lecture Summary for Week  9               Fall 2009
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Last week's tutorial had two unanswered questions...

3.  Prove or disprove:  A_TM (- NP.

    A_TM is undecidable but all languages in NP are decidable, by
    definition.  So A_TM does not belong to NP.

4.  Show that P is closed under complementation, i.e., for all languages
    L (- P, ~L (- P.

    Suppose L (- P.  By definition, there exists a decider D for L and some
    constant k (- N such that for all inputs w (- \Sigma*, D accepts w if
    w (- L and D rejects w if w !(- L, all in time O(|w|^k).
    Consider the decider D' = D with q_A and q_R interchanged.  Then D'
    runs in time O(|w|^k) for all input strings w (- \Sigma*, and D'
    accepts w iff D rejects w iff w !(- L, i.e., D' decides ~L in time
    O(n^k).  Hence, ~L (- P.

Examples of languages in NP.

 -  Definition: a propositional formula is in Conjunctive Normal Form (CNF)
    if it is written as a conjunction ("and") of one or more "clauses":
        C_1 /\ C_2 /\ ...  /\ C_r
    where each clause is a disjunction ("or") of one or more "literals":
        C_j = (a_{j,1} \/ a_{j,2} \/ ... \/ a_{j,s_j})
    where each literal is either a propositional variable or the negation
    of a propositional variable.
    For example, (p \/ ~q) /\ q is in CNF, but ~(p /\ q) /\ (r \/ ~p) is
    NOT in CNF (it is equivalent to some CNF formula, but the property of
    being in CNF is all about the way the formula is written, *not* what it
    is equivalent to).

    The proof of the Cook-Levin Theorem actually shows that CNF-SAT is NPc
    (where CNF-SAT = { <F> : F is a formula in CNF that is satisfiable }),
    because it's possible to construct the formula F_x to be in CNF.

 -  Definition: a propositional formula is in 3-CNF if it is in CNF where
    each clause contains exactly 3 literals, e.g.,
        (p \/ q \/ ~r) /\ (p \/ p \/ ~q)

    3SAT = { <F> : F is a formula in 3-CNF that is satisfiable } is NPc:
    3SAT (- NP because it's a special case of SAT (same verifier works).
    3SAT is NP-hard because CNF-SAT <=p 3SAT.

    Note:  Careful with directions!  Trivially, 3SAT <=p CNF-SAT (3SAT is
    special case of CNF-SAT).  But we need other direction to conclude 3SAT
    is NP-hard, transforming instances of general problem into instances of
    restricted problem (trickier).

    Given <F> (a CNF formula), construct <F'> (a 3-CNF formula) such that F
    is satisfiable iff F' is satisfiable, as follows.  Note that it is not
    necessary to make F and F' logically equivalent in order to achieve
    this.
    For each clause C of F:

     -  If C = (a1), then replace C with (a1 \/ a1 \/ a1).

     -  If C = (a1 \/ a2), then replace C with (a1 \/ a1 \/ a2).

     -  If C = (a1 \/ a2 \/ a3), then leave C the same.

     -  If C = (a1 \/ a2 \/ ... \/ ar) where r > 3, then replace C with
        (a1 \/ a2 \/ z1) /\ (~z1 \/ a3 \/ z2) /\ (~z2 \/ a4 \/ z3) /\ ...
         /\ (~z{r-4} \/ a{r-2} \/ z{r-3}) /\ (~z{r-3} \/ a{r-1} \/ ar),
        where z1, z2, ..., z{r-3} are new variables (not in F).

    For example, if
        F = (x1 \/ x2) /\ (~x1) /\ (x2 \/ ~x3 \/ x3 \/ x5 \/ ~x4)
    then
        F' = (x1 \/ x1 \/ x2) /\ (~x1 \/ ~x1 \/ ~x1) /\
             (x2 \/ ~x3 \/ z3) /\ (~z3 \/ x3 \/ z4) /\ (~z4 \/ x5 \/ ~x4)

    Clearly, this transformation can be carried out in polytime: at most,
    each clause of length r gets replaced with O(r) 3-clauses using O(r)
    new variables.

    Also, if F is satisfiable, then there is an assignment of truth values
    to the variables of F that makes at least one literal true in each
    clause of F.  This assignment can be extended to include values for the
    new variables of F' that will make each clause of F' true:

     -  For 1-/2-/3-clauses of F with at least one true literal, the
        corresponding clause in F' is also true because it contains the
        same literals, at least one of which is true.

     -  For r-clauses of F with at least one true literal, say the original
        clause is (a1 \/ a2 \/ ... \/ ar) and the true literal is ai.  Then
        pick values for the new variables as follows:
         .  if i=1 or i=2, then (a1 \/ a2 \/ z1) is satisfied so pick
            z1 = z2 = ... = z{r-3} = false to satisfy every other clause;
         .  if i=r-1 or i=r, then (~z{r-3} \/ a{r-1} \/ ar) is satisfied so
            pick z1 = z2 = ... = z{r-3} = true to satisfy every other
            clause;
         .  if 2 < i < r-1, then (~z{i-2} \/ ai \/ z{i-1}) is satisfied so
            pick z1 = z2 = ... = z{i-2} = true to satisfy the first i-2
            clauses and pick z{i-1} = zi = ... = z{r-3} = false to satisfy
            the last r-i-1 clauses.
        For example, if x3 = true satisfies (x2 \/ ~x3 \/ x3 \/ x5 \/ ~x4),
        then pick z1 = true and z2 = false to satisfy
        (x2 \/ ~x3 \/ z1) /\ (~z1 \/ x3 \/ z2) /\ (~z2 \/ x5 \/ ~x4)
        (the first clause is satisfied by z1 = true, the second clause is
        satisfied by x3 = true, the last clause is satisfied by z2 =
        false).

    Finally, if F' is satisfiable, then the assignment of values to the
    variables of F' must include values to the variables of F that satisfy
    F:

     -  If the new clause (a1 \/ a2 \/ a3) is satisfied, then the original
        clause is also satisfied because it's the same, and similarly for
        the new clauses (a1 \/ a1 \/ a2) and (a1 \/ a1 \/ a1), because they
        are logically equivalent to the original clauses.

     -  If the new clauses
        (a1 \/ a2 \/ z1) /\ (~z1 \/ a3 \/ z2) /\ (~z2 \/ a4 \/ z3) /\ ...
         /\ (~z{r-4} \/ a{r-2} \/ z{r-3}) /\ (~z{r-3} \/ a{r-1} \/ ar)
        are satisfied, then:
         .  if none of the new variables z1, ..., z{r-3} is set to false,
            then (~z{r-3} \/ a{r-1} \/ ar) can only be satisfied by setting
            a{r-1} = true or ar = true;
         .  else, if z1 is set to false, then (a1 \/ a2 \/ z1) can only be
            satisfied by setting a1 = true or a2 = true;
         .  else let zi be the first new variable set to false (so
            z1 = z2 = ... = z{i-1} = true): (~z{i-1} \/ a{i+1} \/ zi) can
            only be satisfied by setting a{i+1} = true;
        in all cases, one of the original literals must be set to true so
        the original clause (a1 \/ a2 \/ ... \/ ar) is also satisfied.

    We have shown that any CNF formula F can be transformed in polytime to
    a 3-CNF formula F' such that F is satisfiable iff F' is satisfiable;
    this completes the polytime reduction from CNF-SAT to 3SAT.

 -  VERTEX-COVER (VC for short) = { <G,k> : G is a graph that contains a
    vertex cover of size k, i.e., a set C of k vertices such that each edge
    of G has at least one endpoint in C }
     .  VC (- NP:  certificate = vertex cover of size k.
     .  VC is NP-hard:  3SAT <=p VC.
        Given F = (a1 \/ b1 \/ c1) /\ ... /\ (ar \/ br \/ cr), where
        ai,bi,ci in {x1,~x1,x2,~x2,...,xs,~xs}, construct G=(V,E) and k
        such that F satisfiable iff G contains vertex cover of size k, as
        follows:
            k = s + 2r
            V = { a1,b1,c1, ..., ar,br,cr, x1,~x1, ..., xs,~xs }
            E = { (xi,~xi) : 1 <= i <= s } U
                { (ai,bi),(bi,ci),(ci,ai) : 1 <= i <= r } U
                { (l,x) : l = ai or bi or ci, and x = xj or ~xj
                          corresponding to l }
        For example, if
        F = (x1 \/ ~x2 \/ ~x4) /\ (x2 \/ ~x3 \/ x1) /\ (~x3 \/ x4 \/ ~x2),
        then a1=x1, b1=~x2, c1=~x4, a2=x2, b2=~x3, c2=x1, a3=~x3, b3=x4,
        c3=~x2 so
        k = 4 + 2*3 = 10
        V = {a1,b1,c1, a2,b2,c2, a3,b3,c3, x1,~x1, x2,~x2, x3,~x3, x4,~x4}
        E = { (x1,~x1), (x2,~x2), (x3,~x3), (x4,~x4),
              (a1,b1), (b1,c1), (c1,a1), (a1,x1), (b1,~x2), (c1,~x4),
              (a2,b2), (b2,c2), (c2,a2), (a2,x2), (b2,~x3), (c2,x1),
              (a3,b3), (b3,c3), (c3,a3), (a3,~x3), (b3,x4), (c3,~x2) }

        Clearly, construction can be done in polytime (with one scan of F).