CLIQUE(G,k):  Given the graph G = (V, E) and an integer k,
does the graph G have a clique of at least size k?

Theorem: CLIQUE is NP-complete.
 
===============

Proof (with allowable fudges on the sizes of input and certificate
strings s and t):

Claim 1: CLIQUE is in NP

===============
Pf Claim 1: 

CLIQUE(G, k) is clearly a decision problem. 
  
Moreover, let G = (V, E) and k be specified by the input string s.
 
*MARKING POLICY*: In this course it is sufficient to ignore the 
difference between the input string s, and the items that s specifies,
in this case (G, k), where G = (V, E).  We will take |s| to 
denote some reasonable measure of the size of (G,k), say 
|s| = |V| + |E|.  For homework and exam problems, you need
to say what you take s to specify (typically the input), and 
also state precisely how you define the size of s. (The text for the
course doesn't take care with this.)

We will argue that a suitable certificate t for the CLIQUE(G, k)
problem is a string t which specifies a subset W of V, for which W 
forms a clique and has size |W| >= k.
 
*MARKING POLICY*: Again, it is sufficient in this course to fudge 
the difference between the string t, and the items that t specifies 
(in this case W).  Set |t| to be any reasonable measure of the
size of the set that t specifies.  For example, here we take 
|t| = |W|. (For homework and exam problems, you need to state 
exactly what t specifies, and also what you use for |t|.)
 
With these choices and fudges, |t| <= p(|s|), so the certificate t
is of polynomial size in |s| (e.g., we can use the polynomial p(x) = x).  

We are left with showing that there is a poly-time certifier, that is
an algorithm C((G,k), W) such that, for any graph G and integer k:
    CLIQUE(G,k) is true
  iff (*)  (here * is simply a bookmark)
    there exists a W such that C((G,k), W) is true.
(Here we have omitted the constraints on the size of the 
certificate t from the definition of NP, which we have argued
above are satisfied by W.) 

Suppose C((G, k), W) is the straight-forward algorithm which 
checks the following conditions:
  a) W is a subset of V, 
  b) |W| >= k, 
and
  c) for every pair u, v in W, with u not equal v, 
     the edge (u,v) is in E.  
Here C((G, k), W) returns true iff all these conditions are satisfied.
 
We need to confirm the logical equivalence at the bookmark (*) above.

First, consider the implication, CLIQUE(G,k) implies there exists
a W s.t. C((G,k), W) is true.

Suppose CLIQUE(G,k) is true.  Then, by defn of CLIQUE, there must
exist a clique of G with size >= k.  Let W (a subset of V) be such
a clique with |W| >= k.  Then, by the definition of clique it follows 
W must satisfy that each of the conditions in the algorithm C((G, k), W),
and therefore C((G, k), W) returns true.

We also need to consider the reverse implicaiton, namely that 
the existence of a W s.t. C((G,k), W) is true implies CLIQUE(G,k).
This again follows since the algorithm C((G,k), W) explicitly
checks that W satisfies the definition of a clique, and that
|W| >= k.  Therefore CLIQUE(G,k) must be true.
 
Finally, we need to show that this algorithm C((G, k), W) 
runs in polynomial time, as a function of the size of the input 
|s| + |t|.  The brute force implementation that was indicated
runs in time O(|V| + |V|^2 |E|), which is bounded by O(|s|^3) 
(since we defined |s| = |V| + |E|, and |t| = |W| <= |V|.)
 
Therefore C((G, k), W) is a poly-time certifier for CLIQUE(G, k).  
This completes the proof of Claim 1.
 
====================

At this point we have established that CLIQUE(G,k) is in NP.

We next what to show that CLIQUE is NP-complete.  To do that
we show that there is a polynomial reduction from an problem
X that is known to be NP-complete to CLIQUE, i.e., X <=_p CLIQUE.

One convenient choice for X is VERTEX-COVER, another is INDEPENDENT_SET.
Below we use 3-SAT for X.

Claim 2: 3-SAT <=_p CLIQUE (i.e., 3-SAT polynomially reduces to CLIQUE).

Since we know 3-SAT is NP-complete, it will follow 
from Claim 2 that CLIQUE(G, k) is also NP-complete, thereby
proving the desired Theorem.
 
====================
Pf Claim 2:

Consider any instance of the 3-SAT problem, say 3-SAT(s).  Specifically,
let PHI = C_1 And C_2 And ...  And C_m be a 3-CNF formula, where
each clause C_k is the disjunction of three literals (i.e. x_n or 
(not x_n)). (Moreover, a variable cannot appear more than once in
the same clause.  Although we don't use this constraint on PHI
here.)
 
We again need to measure the size of the input string s which 
specifies PHI.  In particular, we need to argue that the reduction
described below can be done in polynomial time (and with only
polynomially many calls to CLIQUE) where these polynomals are
in terms of the size of the input |s|.  Here we take |s| = m, 
the number of clauses in PHI.
 
Given PHI define a set of vertices V as follows.  For each clause C_k 
define three vertices v(k,j) for j = 1,2,3.  Here the vertex v(k,j) 
corresponds to the j^th literal (textually from the left) in the 
k^th clause, i.e., L(k, j) is either x_n or (not x_n) for some n.

So V = {v(k,j) | 1<= k <= m, 1<= j < 3}, and |V| = 3m, where m 
equals the number of clauses in PHI.

We define the set of edges E as follows.  For every pair k, i of clauses,
with k not equal to i, we include the edge (v(k,p), v(i,q)) in E iff
v(k,p) and v(i,q) do not refer to a single literal and its negation.
That is, if v(k,p) refers to L(k,p) = (not x_n) and v(i,q) refers 
to L(i,q) = (x_n), or vice versa, for some fixed n, then there 
is no edge between these vertices.  Otherwise we include
the edge (v(k,p), v(i,q)) in E.  

Note there are no edges between vertices associated with the same 
clause, so (v(k,p), v(k,q)) is not in E, for any k, p, q.

Concretely, write out an example of this graph on the board for 

PHI =
(x or (not y) or (not z)) and  (not x or (not y) or z ) and (x or y or (not z))

     k= 1              2             3             
j=1:    x             not x          x
 
j=2:  not y           not y          y
  
j=3:  not z             z           not z

Draw edges between all vertices in different columns, except when
the endpoints correspond to a literal and it's negation.  So 
(v(k=1,j=1), v(3,1)), (v(1,1), v(2,2)), (v(1,1), v(2,3)) are all in E.
But (v(1,1), v(2,1)) and (v(1,2), v(3,2)) are not in E (since the
former would link x with (not x), and the later would link (not y) with y.

This graph can be constructed in O(m^2) time (note the number of
distinct variables is at most 3m).  Since we set |s| = m, this
is polynomial in the size |s|.

With the above graph G = (V, E) constructed in poly-time, 
we claim that the solution of the original 3-SAT(PHI) problem 
is given by the result of one call to CLIQUE(G,m).  That is,
the proof of the reduction will be complete if we can show:
 
Claim 3: 3-SAT(PHI) is true iff CLIQUE(G, m) is true. 

Note Claim 3 will show that the value of CLIQUE(G, m) is
the correct answer for the orginal decision problem
3-SAT(PHI).  Therefore the correctness of the polynomial reduction 
(and Claim 2) follows from the proof of Claim 3 given below.
 
=================
Proof of Claim 3:

We first prove that CLIQUE(G, m) implies 3-SAT(PHI). 
 
Suppose CLIQUE(G, m) is true, with G = (V,E) as constructed above.  
Then there must exist a clique W of size >= m (W a subset of V). 
Since there are no edges between vertices for the same clause in E,
and there are m clauses, |W| must equal m.  

Why?  The maximum clique size in G is <= m.
To show this, use contradiction. Suppose there is a clique with size > m.  
Then there must be at least one clause k such that  v(k,j) and v(k,i) must be
in the clique, with i not equal j.  But there is no edge between vertices
arising from the same clause, therefore (v(k,j), v(k,i)) is not in E.  But
this contradicts definition of a clique (every pair of vertices in the
clique must be connected by an edge).

Moreover, W must contain exactly one vertex from each clause, i.e. for
some choices for p_k, k = 1,...,m,  we have
  W = { v(k, p_k) for k = 1, ... m }.
 
Why? Otherwise, since there are exactly m clauses, one of the clauses
would have to have more than one vertex in W.  But this is impossible
as there are no edges between such a pair of vertices and W is a clique.
 
For each such v(k, j_k) in W, we show we can set the corresponding literal 
to be true (i.e., either x_n or (not x_n)).  By the clique property,
every pair of vertices in W is connected by an edge.  And by
construction of the edge set E, such vertices cannot be associated
with both x_n and (not x_n).  Therefore, the set of all literals
associated with vertices in W cannot include both x_n and (not x_n)
for any n.  Therefore we can consistently set all these literals
to true (i.e., either x_n is true and (not x_n) is false, or vice versa).
 
Any other variable x_p that appears in PHI (either in
the literal x_p or (not x_p) or both), can simply be set to false.
 
This then defines consistent settings for all the variables x_n in PHI. 
Moreover, since at least one literal in each clause has
been set to true (i.e., the literal associated with v(k,j_k) in clause k),
each clause must be true.  Therefore, these settings of the variable
{x_n}  satisfy PHI itself.
 
This completes the argument showing that CLIQUE(G, m) implies 3-SAT(PHI).
 
To complete the proof of Claim 3 we also need to show that
3-SAT(PHI) implies CLIQUE(G, m).

Suppose 3-SAT(PHI) is true. Here PHI = C_1 And ... And C_m, 
and the associated graph G = (V, E)  are as defined above.
  
Since 3-SAT(PHI) is true, there exists a satisfying boolean assignment 
of the all the variables, x_n, n = 1, ..., N, in PHI.  Moreover, for 
each clause C_k, there must be at least one literal that is true.  
(Otherwise PHI itself would be false.)  Select exactly one literal
that is true in each clause, and form W from the vertices in G
that are associated with these literals.  In the construction,
the literals appearing at particular positions in PHI are in a 
1-1 relationship with vertices in V.  (For example, the 2nd literal in 
the k-th clause is vertex v(k,2) in V.) So this selection of
literals, one per clause, uniquely corresponds to exactly m vertices 
in V.  Let W be this set of m vertices.

We will show that W is a clique. Suppose a,b are any pair 
of vertices in W  with b not equal to a. It will follow 
that W is a clique if we can show that there is necessarily 
an edge (a,b) in E.  First notice that, from the way W was 
constructed, a and b must be associated with different clauses.  
So a = v(k, p) and b = v(i, q) with i not equal to k.
In addition, suppose L(k, p) and L(i, q) are the literals associated 
with v(k, p) and v(i, q), respectively (i.e. the p-th literal from
the left in the k-th clause, and the q-th literal in the i-th clause).
 
Moreover, given the manner in which the literals were chosen, 
both these literals L(k, p) and L(i, q) must be true in the current 
boolean assignment.  Thus it cannot be the case that literal 
L(k, p) is simply the negation of L(i, q).  That is, one cannot be 
x_n while the other is (not x_n).  Since these literals cannot be 
negations of each other, and we have established that they appear 
in different clauses (i.e., k is not equal to i), it follows from 
the construction of E that (a,b) is a edge in E.

Therefore we have shown that an arbitrary pair of points a, b in W
are connected by an edge in E.  Therefore W is a clique.  We
showed above that |W| = m, so therefore CLIQUE(G, m) must be true.

This proves 3-SAT(PHI) implies CLIQUE(G,m), and completes the
proof of Claim 3.