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CSC 363                 Lecture Summary for Week 10             Winter 2008
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Examples of NP-completeness:

 -  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).

    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: reduction was written up and
    posted along with summary of lecture notes for last week.

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

 -  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).

        Also, if F is satisfiable, then there is an assignment of truth
        values that make at least one literal in each clause true.  Pick a
        cover C as follows: for each variable, C contains xi or ~xi,
        whichever is true under the truth assignment; for each clause, C
        contains every literal except one that's true (pick arbitrarily if
        more than one true literal).  C contains exactly s+2r vertices and
        is a cover: all edges (xi,~xi) are covered; all edges in clause
        triangles are covered (because we picked two vertices from each
        triangle); all edges between "clauses" and "variables" are covered
        (two from inside triangle, one from true literal for that clause).

        Finally if G contains a cover C of size k=s+2r, C must contain at
        least one of xi or ~xi for each i (because of edges (xi,~xi)) and
        at least two of ai,bi,ci for each i (because of triangle), so only
        way for C to have size s+2r is to contain exactly one of xi or ~xi
        and exactly two of ai,bi,ci, for each i.  Since C covers all edges
        with only two vertices per triangle, the third vertex in each
        triangle must have its "outside" edge covered because of xi or ~xi.
        If we set literals according to choices of xi or ~xi in C, this
        will make formula F true: at least one literal will be true in each
        clause (because at least one edge from "variables" to "clauses" is
        covered by the variable in C).

 -  INDEPENDENT-SET = { <G,k> | G is an undirected graph that contains
            an independent set of size k -- a subset of vertices with NO
            edge between any two of them }

    INDEPENDENT-SET (IS) is NPc:
     .  in NP: certificate = independent set
     .  NP-hard: VC <=p IS:  Given <G,k>, construct <G',k'> as
        follows: G' = G, k' = n-k.  Clearly this can be done in polytime.
        Also, if G contains a vertex cover of size k, the vertices outside
        the cover form an independent set of size n-k (no edge can have
        both endpoint outside the cover).  Finally, if G' contains an
        independent set of size n-k, the vertices outside the independent
        set form a vertex cover of size k (every edge has at least one
        endpoint in the cover).

 -  CLIQUE is NPc:  Already known in NP.  For NP-hardness, use reduction
    from INDEPENDENT-SET:
    On input <G,k>, construct <G',k'> as follows:
        G' = complement of G (same vertices, edge in G' iff edge not in G)
        k' = k
    This construction can obviously be carried out in polytime.
    Moreover, if G contains an I.S. of size k, then the same set in G' will
    be a clique of size k; conversely, if G' contains a clique of size k,
    the same set in G will be an I.S. of size k.
    Note:  textbook uses different reduction from 3SAT.

 -  SUBSET-SUM is NPc:  Already known in NP.  For NP-hardness, show
    3SAT <=p SS:
    Given formula F = (a1 \/ b1 \/ c1) /\ ... /\ (ar \/ br \/ cr) where
    ai,bi,ci in {x1,~x1,...,xs,~xs}, construct numbers as follows:
     .  For j = 1,...,s,
        number xj = 1 followed by s-j 0s followed by r digits where
        k-th next digit equals 1 if xj appears in clause C_k, 0 otherwise;
        number ~xj = 1 followed by s-j 0s followed by r digits where
        k-th next digit equals 1 if ~xj appears in clause C_k, 0 otherwise.
     .  For j = 1,...,r,
        number Cj = 1 followed by r-j 0s and
        number C'j = 2 followed by r-j 0s.
     .  Target t = s 1s followed by r 4s.
    Clearly, this can be constructed in polytime.

    Example of reduction for
    F = (x1 \/ ~x2 \/ ~x4) /\ (x2 \/ ~x3 \/ x1) /\ (~x3 \/ x4 \/ ~x2):
    S = {  x1 = 1000110,
          ~x1 = 1000000,
           x2 = 0100010,
          ~x2 = 0100101,
           x3 = 0010000,
          ~x3 = 0010011,
           x4 = 0001001,
          ~x4 = 0001100,
           C1 = 0000100,
          C'1 = 0000200,
           C2 = 0000010,
          C'2 = 0000020,
           C3 = 0000001,
          C'3 = 0000002 }
            t = 1111444
    Note: slightly different from book to ensure S contains all distinct
    numbers (book's reduction constructs S with repeated numbers, so S not
    really a set).

    If F is satisfiable, then there is a setting of variables such that
    each clause of F contains at least one true literal.  Consider the
    subset S' = {numbers that correspond to true literals}.  By
    construction, \sum_{x (- S'} x = s 1s followed by r digits, each one of
    which is either 1, 2, or 3 (because each clause contains at least one
    true literal).  This means it is possible to add suitable numbers from
    {C1,C'1,...,Cr,C'r} so that the last r digits of the sum are equal to
    4, i.e., there is a subset S' such that \sum_{x (- S'} x = t.

    If there is a subset S' of S such that \sum_{x (- S'} x = t, then S'
    must contain exactly one of {xj,~xj} for j = 1,...,n, because that is
    the only way for the numbers in S' to add to the target (with a 1 in
    the first s digits).  Then, F is satisfied by setting each variable
    according to the numbers in S': for each clause j, the corresponding
    digit in the target is equal to 4 but the numbers Cj and C'j together
    only add up to 3 in that digit; this means that the selection of
    numbers in S' must include some literal with a 1 in that digit, i.e.,
    clause j contains at least one true literal.