Vertex Cover, Clique, and Independent Set:
VCDP: Given a graph G=(V,E) and a positive integer k<=|V| is
there a set of vertices V' contained in V such that |V'|<=k and such
that for every edge <u,v> either u is in V' or v is in V'.
CDP: Given a graph G=(V,E) and a positive integer k<=|V| is
there a set of vertices V' contained in V such that |V'|>=k and such
that if u and v are in V' then the edge <u,v> is in E. (In other
words, the complete graph on V' is a subgraph of G. In fact, a CLIQUE
is a complete subgraph.)
INDEPENDENT SET: Given a graph G=(V,E) and a positive integer
k<=|V|, is there a set of vertices V' contained in V such that if u
and v are in V' then the edge <u,v> is NOT in E and such that
|V'|>=k? (In other words, no two vertices of V' are connected by an
edge in G.)
The complement graph: If G=(V,E) is a graph, then the complement
graph of G, Gc=(G,Ec) where Ec
contains edge <u,v> exactly when <u,v> is NOT an edge for E.
Lemma: V' is a vertex cover for G <=> V-V' is an independent
set for G <=> V-V' is a clique for Gc.
Proof: Suppose V' is a vertex covder. Then every edge <u,v> is
such that either u or v is in V'. Let W=V-V'. Then if u and v are in W,
<u,v> cannot be an edge since either u or v must be in V'. Thus,
vertex cover for G => independent set for G.
Now suppose V-V' is an independent set for G. Then no edges between
vertices in V-V' is in E. Buth, then, every edge between vertices in
V-V' must be in Ec. Thus, V-V' is a clique in Gc.
Thus we have V-V' is an independent set for G => V-V' is a clique
for Gc.
Now suppose V-V' is a clique in Gc. Then every edge
<u,v> with u and v in V-V' is in Ec. Then it is the
case that there are no edges <u,v> in E with u and v in V-V'.
Thus every edge <u,v> in E must be such that either u or v are in
V'. Thus, V-V' is a clique for Gc => V' is a vertex cover
for G.
QED
Now if we suppose CDP is NP-hard, we
can show INDEPDENDENT SET is NP-hard as follows:
Let G=(V,E) and number k be an instance of CDP. Then construct Gc=(V,Ec)
and use the number |V|-k. Then if Gc has an independent set
V' of size at least |V|-k we can conclude from our lemma that G
contains a clique of size at most |V|-k. This is clearly a
polynomial-time transformation. Even if G has no edges, the complement
has at most n(n-1)/2 edges where n=|V|.
QED
Now if we suppose INDEPENDENT SET is
NP-hard we can show VCDP is NP-hard as follows:
Let G=(V,E) and integer k be an instance of INDEPENDENT SET.
Then create the instance for VCDP as G=(V,E) and number |V|-k. If G has
a vertex cover of size |V|-k it means that there is a set V' contained
in V such that if <u,v> is an edge then either u or v is in V and
|V'|<=k. Then it must follow that |V-V'|>=k and by our lemma V-V'
is an independent set.
QED
Now is we suppose VCDP is NP-hard we
can show CDP is NP-hard as follows:
Let G=(V,E) and number k be an instance of VCDP. Let us construct an
instance of CDP consisting of Gc=(V,Ec) and
number |V|-k. Then suppose Gc has a clique V' of size
>=|V|-k. Then by our lemma, G has no edges between elements of V' so
V-V' must be a vertex cover since for every edge <u,v> either u
or v is not in V' and, therefore, is in V-V'. Also, since V' is size
>= |V|-k it mus be the case that the size of V-V'<=|V|-(|V|-k)=k.
QED
Proofs that all of these are in NP is a homework exercise.