1. NP-complete Problems SAT 3SAT Independent Set Hamiltonian Cycle
Stephen Cook 1971
SAT
Richard Karp 1972
3SAT
Independent Set Hamiltonian Cycle
Vertex Cover CLIQUE Traveling-Salesman Problem(TSP)
Subset-Sum …… ……
About 1000 NP-complete problems have been discovered since.

2. Independent Set
Let G be an undirected graph.
A  V(G) is an independent set if no two vertices in A share an edge.
 u, v  A, (u, v)  E(G)
Ex. b d g f c a e e d a g
A = { a, d, e, g } is an independent set.
B = { a, c, f } is not

3. The Independent-Set Problem
IND-SET
Input: Graph G, integer K  1.
Q: Does G have an independent set of size  K ?
Theorem IND-SET is NP-complete.
Proof  IND-SET  NP. Below is a non-deterministic algorithm:
a) Pick K vertices from V(G) to form a subset A  V.
non-deterministic O(K) time
certificate
b) Check if  u, v  A, (u, v)  E(G).
If so answer Yes; otherwise, answer No.
deterministic O(K ) time 2

4. 3SAT ≤ IND SET P Therefore IND-SET is NP-complete. Proof (cont’d)
 We show that 3SAT is polynomial-time reducible to IND-SET.
Construct (G , K) such that   3SAT  (G , K)  IND-SET.
 
This will follow from Lemma 1.
Therefore IND-SET is NP-complete.

5. Constructing G  x  x  One vertex for each literal.
Ex.  = ( x  x   x )  ( x  x  x )  ( x  x   x ) x
 x 1 2 3 4
 One vertex for each literal.
 Literals in the same clause form a triangle.
 Opposite literals share edges.

6. Satisfiability vs. Independent Set
Lemma 1  satisfiable  G has an independent set of size K
(= #clauses in ).
Proof (  ) Suppose  is satisfiable.
Then at least one literal from every clause is true. Pick exactly
one such literal from each clause and pick its corresponding
vertex.
 K vertices are picked
 For two of these vertices to share an edge, the
corresponding literals would be
 either in the same clause
Impossible given the way these vertices are picked
 or opposite literals.
Impossible given that  is satisfiable.
So none of the K vertices are adjacent.
Thus the K vertices form an independent set.

7. Cont’d (  ) Suppose G has an independent set of size K.
Then the vertices in the set
 must be in different triangles.
 do not include a pair of opposite literals.
Now we assign T to the corresponding literals in ,
which will be true under the induced truth value assignment
to the variables.

8. Vertex Cover G = K |C|  |V| – 1
Let G be an undirected graph.
C  V(G) is a vertex cover if for every edge (u, v)  E(G)
either u  C or v  C a c d b e a
G = K n
|C|  |V| – 1
complete graph
with n vertices d
{ a, d } is a vertex cover.

9. The Vertex Cover Problem
Vertex Cover (VC)
Input: Graph G = (V, E), positive integer K  | V |.
Q: Does G have a vertex cover of size  K ?
Lemma C is a vertex cover  V – C is an independent set. a c d b e
Proof () Let C be a vertex cover.
Suppose V – C is not an independent set.
Then there exists two vertices u, v  V – C
such that (u, v)  E.
{a, d}: vertex cover
So edge (u, v) has both vertices not in C
and C is not a vertex cover. a c d b e
( ) Similarly.
{b, c, e}: independent set

10. VC is NP-complete Corollary IND-SET  VC
It is easy to show that VC  NP.
Theorem VC  NPC.

11. CLIQUE A clique Q is a subset of vertices such that
(u, v) is an edge for every u, v  Q. b a d e c f g
cliques: { a, b, c, d }
{ e, f, g }
{ d, e }
{ g }, …
The subgraph induced by Q is a complete graph.

12. Complement of a Graph The complement G of a graph G = (V, E) has
 the same vertex set V
 edge set E such that
(u, v)  E  (u, v)  E
Complement:
Original graph: b a d e c f g b a d e c f g

13. The CLIQUE Problem
Input: Graph G = (V, E), positive integer K  | V |.
Q: Does G have a clique of size  K ?
Lemma Q is a clique in G 
Q is an independent set in the complement G .
IND-SET  CLIQUE P
Theorem CLIQUE  NPC.