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CSC 363H                Lecture Summary for Week 10             Summer 2006
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Template for proofs of NP-completeness:  To show A is NPc, prove that
    A in NP: Describe a polytime verifier for A.
        "Given <x,c>, check c has right format and properties..."
        Argue that verifier runs in polytime and that
        x in A iff verifier accepts <x,c> for some c.
    A is NP-hard: Show B <=p A for some NP-hard B.
        "Given y, construct x as follows: ..."
        Argue that construction can be carried out in polytime
        and that y in B iff x in A
        (often by showing y in B -> x in A and x in A -> y in B).

Examples:

    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.  Finally, if G'
        contains an independent set of size n-k, the vertices outside the
        independent set form a vertex cover of size k.

 -  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 = { <S,t> | S = {x_1,x_2,...,x_k} and there is a set
                           {y_1,y_2,...,y_j} such that SUM y_i = t }
    SUBSET-SUM is NPc:
     . in NP: verifier = "On input <S,t,c>:
	 1. check that c encodes a set of numbers;
	 2. check that c subset of S;
	 3. check that SUM_{y in c} y = t;
	 4. accept if all checks pass, reject otherwise."
       Clearly runs in polytime, and accepts for some c
       iff <S,t> in SUBSET-SUM.

     . NP-hardness:
       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).

    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 in 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 in S'} x = t.

    If there is a subset S' of S such that SUM_{x in 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.