NP Hardness Reductions

Reductions to Prove NP Hardness

Nabil Mustafa

Computational Complexity

1 / 121 Cook-Levin: All L NP can be reduced to SAT ∈ I polynomial g( ) such that x L g(x) SAT ∃ · ∈ ⇐⇒ ∈ I Very important to remember the implication goes both ways

Gave a reduction from general CNF SAT to 3-CNF SAT

I 2-CNF SAT is in P through a poly-time algorithm.

3-CNF SAT then becomes a very important problem. Also,

I An elementary problem, very simple to understand I Captures the ‘pure’ combinatorial choices of algorithms

Given 3-CNF SAT , can construct a host of other reductions.

Reductions

The class NP : languages with succinct membership certiﬁcates

2 / 121 I Very important to remember the implication goes both ways

Gave a reduction from general CNF SAT to 3-CNF SAT

I 2-CNF SAT is in P through a poly-time algorithm.

3-CNF SAT then becomes a very important problem. Also,

I An elementary problem, very simple to understand I Captures the ‘pure’ combinatorial choices of algorithms

Given 3-CNF SAT , can construct a host of other reductions.

Reductions

The class NP : languages with succinct membership certiﬁcates

Cook-Levin: All L NP can be reduced to SAT ∈ I polynomial g( ) such that x L g(x) SAT ∃ · ∈ ⇐⇒ ∈

3 / 121 Gave a reduction from general CNF SAT to 3-CNF SAT

I 2-CNF SAT is in P through a poly-time algorithm.

3-CNF SAT then becomes a very important problem. Also,

I An elementary problem, very simple to understand I Captures the ‘pure’ combinatorial choices of algorithms

Given 3-CNF SAT , can construct a host of other reductions.

Reductions

The class NP : languages with succinct membership certiﬁcates

Cook-Levin: All L NP can be reduced to SAT ∈ I polynomial g( ) such that x L g(x) SAT ∃ · ∈ ⇐⇒ ∈ I Very important to remember the implication goes both ways

4 / 121 I 2-CNF SAT is in P through a poly-time algorithm.

3-CNF SAT then becomes a very important problem. Also,

I An elementary problem, very simple to understand I Captures the ‘pure’ combinatorial choices of algorithms

Given 3-CNF SAT , can construct a host of other reductions.

Reductions

The class NP : languages with succinct membership certiﬁcates

Cook-Levin: All L NP can be reduced to SAT ∈ I polynomial g( ) such that x L g(x) SAT ∃ · ∈ ⇐⇒ ∈ I Very important to remember the implication goes both ways

Gave a reduction from general CNF SAT to 3-CNF SAT

5 / 121 3-CNF SAT then becomes a very important problem. Also,

I An elementary problem, very simple to understand I Captures the ‘pure’ combinatorial choices of algorithms

Given 3-CNF SAT , can construct a host of other reductions.

Reductions

The class NP : languages with succinct membership certiﬁcates

Cook-Levin: All L NP can be reduced to SAT ∈ I polynomial g( ) such that x L g(x) SAT ∃ · ∈ ⇐⇒ ∈ I Very important to remember the implication goes both ways

Gave a reduction from general CNF SAT to 3-CNF SAT

I 2-CNF SAT is in P through a poly-time algorithm.

6 / 121 I An elementary problem, very simple to understand I Captures the ‘pure’ combinatorial choices of algorithms

Given 3-CNF SAT , can construct a host of other reductions.

Reductions

The class NP : languages with succinct membership certiﬁcates

Cook-Levin: All L NP can be reduced to SAT ∈ I polynomial g( ) such that x L g(x) SAT ∃ · ∈ ⇐⇒ ∈ I Very important to remember the implication goes both ways

Gave a reduction from general CNF SAT to 3-CNF SAT

I 2-CNF SAT is in P through a poly-time algorithm.

3-CNF SAT then becomes a very important problem. Also,

7 / 121 I Captures the ‘pure’ combinatorial choices of algorithms

Given 3-CNF SAT , can construct a host of other reductions.

Reductions

The class NP : languages with succinct membership certiﬁcates

Cook-Levin: All L NP can be reduced to SAT ∈ I polynomial g( ) such that x L g(x) SAT ∃ · ∈ ⇐⇒ ∈ I Very important to remember the implication goes both ways

Gave a reduction from general CNF SAT to 3-CNF SAT

I 2-CNF SAT is in P through a poly-time algorithm.

3-CNF SAT then becomes a very important problem. Also,

I An elementary problem, very simple to understand

8 / 121 Given 3-CNF SAT , can construct a host of other reductions.

Reductions

The class NP : languages with succinct membership certiﬁcates

Cook-Levin: All L NP can be reduced to SAT ∈ I polynomial g( ) such that x L g(x) SAT ∃ · ∈ ⇐⇒ ∈ I Very important to remember the implication goes both ways

Gave a reduction from general CNF SAT to 3-CNF SAT

I 2-CNF SAT is in P through a poly-time algorithm.

3-CNF SAT then becomes a very important problem. Also,

I An elementary problem, very simple to understand I Captures the ‘pure’ combinatorial choices of algorithms

9 / 121 Reductions

The class NP : languages with succinct membership certiﬁcates

Cook-Levin: All L NP can be reduced to SAT ∈ I polynomial g( ) such that x L g(x) SAT ∃ · ∈ ⇐⇒ ∈ I Very important to remember the implication goes both ways

Gave a reduction from general CNF SAT to 3-CNF SAT

I 2-CNF SAT is in P through a poly-time algorithm.

3-CNF SAT then becomes a very important problem. Also,

I An elementary problem, very simple to understand I Captures the ‘pure’ combinatorial choices of algorithms

Given 3-CNF SAT , can construct a host of other reductions.

10 / 121 i) It is in NP ii) Reduce a NP hard problem to it Show: If one can solve INDSET in polynomial time, then can solve 3-CNF SAT problem in polynomial time as well.

INDSET Claim INDSET : Given a graph G = (V , E) and a parameter k, does G have an independent set of size k? The independent set problem (INDSET ) is NP complete.

11 / 121 ii) Reduce a NP hard problem to it Show: If one can solve INDSET in polynomial time, then can solve 3-CNF SAT problem in polynomial time as well.

INDSET Claim INDSET : Given a graph G = (V , E) and a parameter k, does G have an independent set of size k? The independent set problem (INDSET ) is NP complete.

i) It is in NP

12 / 121 Show: If one can solve INDSET in polynomial time, then can solve 3-CNF SAT problem in polynomial time as well.

INDSET Claim INDSET : Given a graph G = (V , E) and a parameter k, does G have an independent set of size k? The independent set problem (INDSET ) is NP complete.

i) It is in NP ii) Reduce a NP hard problem to it

13 / 121 INDSET Claim INDSET : Given a graph G = (V , E) and a parameter k, does G have an independent set of size k? The independent set problem (INDSET ) is NP complete.

i) It is in NP ii) Reduce a NP hard problem to it Show: If one can solve INDSET in polynomial time, then can solve 3-CNF SAT problem in polynomial time as well.

14 / 121 Construct a graph G = (V , E) as follows: i i I Vertices: Each literal xj i corresponds to the vertex vj . i i 0 i∈ Ci 0 I Edges: (vj , vj0 ) E if xj , xj0 are the same variable negated. ∈ i i i I Add all possible edges between v1, v2 and v3 for all i. (x1 x2 x3) (x1 x3 x4) (x2 x4) ∨ ∨ ∧ ∨ ∨ ∧ ∨

i i i φ = 1 2 ... k , where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

15 / 121 i i I Vertices: Each literal xj i corresponds to the vertex vj . i i 0 i∈ Ci 0 I Edges: (vj , vj0 ) E if xj , xj0 are the same variable negated. ∈ i i i I Add all possible edges between v1, v2 and v3 for all i. (x1 x2 x3) (x1 x3 x4) (x2 x4) ∨ ∨ ∧ ∨ ∨ ∧ ∨

i i i φ = 1 2 ... k , where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3 Construct a graph G = (V , E) as follows:

16 / 121 i i 0 i i 0 I Edges: (vj , vj0 ) E if xj , xj0 are the same variable negated. ∈ i i i I Add all possible edges between v1, v2 and v3 for all i. (x1 x2 x3) (x1 x3 x4) (x2 x4) ∨ ∨ ∧ ∨ ∨ ∧ ∨

i i i φ = 1 2 ... k , where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3 Construct a graph G = (V , E) as follows: i i I Vertices: Each literal x i corresponds to the vertex v . j ∈ C j

17 / 121 i i i I Add all possible edges between v1, v2 and v3 for all i. (x1 x2 x3) (x1 x3 x4) (x2 x4) ∨ ∨ ∧ ∨ ∨ ∧ ∨

i i i φ = 1 2 ... k , where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3 Construct a graph G = (V , E) as follows: i i I Vertices: Each literal xj i corresponds to the vertex vj . i i 0 i∈ Ci 0 I Edges: (v , v 0 ) E if x , x 0 are the same variable negated. j j ∈ j j

18 / 121 (x1 x2 x3) (x1 x3 x4) (x2 x4) ∨ ∨ ∧ ∨ ∨ ∧ ∨

i i i φ = 1 2 ... k , where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3 Construct a graph G = (V , E) as follows: i i I Vertices: Each literal xj i corresponds to the vertex vj . i i 0 i∈ Ci 0 I Edges: (vj , vj0 ) E if xj , xj0 are the same variable negated. ∈ i i i I Add all possible edges between v1, v2 and v3 for all i.

19 / 121 i i i φ = 1 2 ... k , where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3 Construct a graph G = (V , E) as follows: i i I Vertices: Each literal xj i corresponds to the vertex vj . i i 0 i∈ Ci 0 I Edges: (vj , vj0 ) E if xj , xj0 are the same variable negated. ∈ i i i I Add all possible edges between v1, v2 and v3 for all i. (x1 x2 x3) (x1 x3 x4) (x2 x4) ∨ ∨ ∧ ∨ ∨ ∧ ∨

20 / 121 i i i φ = 1 2 ... k , where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3 Construct a graph G = (V , E) as follows: i i I Vertices: Each literal xj i corresponds to the vertex vj . i i 0 i∈ Ci 0 I Edges: (vj , vj0 ) E if xj , xj0 are the same variable negated. ∈ i i i I Add all possible edges between v1, v2 and v3 for all i. (x1 x2 x3) (x1 x3 x4) (x2 x4) ∨ ∨ ∧ ∨ ∨ ∧ ∨

x1 x3 x4

x3 x1

21 / 121 i i i φ = 1 2 ... k , where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3 Construct a graph G = (V , E) as follows: i i I Vertices: Each literal xj i corresponds to the vertex vj . i i 0 i∈ Ci 0 I Edges: (vj , vj0 ) E if xj , xj0 are the same variable negated. ∈ i i i I Add all possible edges between v1, v2 and v3 for all i. (x1 x2 x3) (x1 x3 x4) (x2 x4) ∨ ∨ ∧ ∨ ∨ ∧ ∨

x1 x3 x4

x3 x1

22 / 121 i i i φ = 1 2 ... k , where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3 Construct a graph G = (V , E) as follows: i i I Vertices: Each literal xj i corresponds to the vertex vj . i i 0 i∈ Ci 0 I Edges: (vj , vj0 ) E if xj , xj0 are the same variable negated. ∈ i i i I Add all possible edges between v1, v2 and v3 for all i. (x1 x2 x3) (x1 x3 x4) (x2 x4) ∨ ∨ ∧ ∨ ∨ ∧ ∨

x1 x3 x4

x3 x1

23 / 121 0 I The independent set, say V , cannot have two vertices from the same clause 0 I Therefore, V has exactly one vertex from each of the k clauses

I No pair maps to literals x and x i i i I If vj corresponds to literal xj = x, set xj = 1 i i i I If vj corresponds to literal xj = x, set xj = 0

I Each clause satisﬁed, and no variable-setting conﬂicts between clauses

Claim: If φ is satisﬁable, G has an independent set of size k

Reduction for INDSET

Claim: If G has an independent set of size k, φ is satisﬁable

24 / 121 0 I Therefore, V has exactly one vertex from each of the k clauses

I No pair maps to literals x and x i i i I If vj corresponds to literal xj = x, set xj = 1 i i i I If vj corresponds to literal xj = x, set xj = 0

I Each clause satisﬁed, and no variable-setting conﬂicts between clauses

Claim: If φ is satisﬁable, G has an independent set of size k

Reduction for INDSET

Claim: If G has an independent set of size k, φ is satisﬁable

0 I The independent set, say V , cannot have two vertices from the same clause

25 / 121 I No pair maps to literals x and x i i i I If vj corresponds to literal xj = x, set xj = 1 i i i I If vj corresponds to literal xj = x, set xj = 0

I Each clause satisﬁed, and no variable-setting conﬂicts between clauses

Claim: If φ is satisﬁable, G has an independent set of size k

Reduction for INDSET

Claim: If G has an independent set of size k, φ is satisﬁable

0 I The independent set, say V , cannot have two vertices from the same clause 0 I Therefore, V has exactly one vertex from each of the k clauses

26 / 121 i i i I If vj corresponds to literal xj = x, set xj = 1 i i i I If vj corresponds to literal xj = x, set xj = 0

I Each clause satisﬁed, and no variable-setting conﬂicts between clauses

Claim: If φ is satisﬁable, G has an independent set of size k

Reduction for INDSET

Claim: If G has an independent set of size k, φ is satisﬁable

0 I The independent set, say V , cannot have two vertices from the same clause 0 I Therefore, V has exactly one vertex from each of the k clauses

I No pair maps to literals x and x

27 / 121 i i i I If vj corresponds to literal xj = x, set xj = 0

I Each clause satisﬁed, and no variable-setting conﬂicts between clauses

Claim: If φ is satisﬁable, G has an independent set of size k

Reduction for INDSET

Claim: If G has an independent set of size k, φ is satisﬁable

0 I The independent set, say V , cannot have two vertices from the same clause 0 I Therefore, V has exactly one vertex from each of the k clauses

I No pair maps to literals x and x i i i I If vj corresponds to literal xj = x, set xj = 1

28 / 121 I Each clause satisﬁed, and no variable-setting conﬂicts between clauses

Claim: If φ is satisﬁable, G has an independent set of size k

Reduction for INDSET

Claim: If G has an independent set of size k, φ is satisﬁable

0 I The independent set, say V , cannot have two vertices from the same clause 0 I Therefore, V has exactly one vertex from each of the k clauses

I No pair maps to literals x and x i i i I If vj corresponds to literal xj = x, set xj = 1 i i i I If vj corresponds to literal xj = x, set xj = 0

29 / 121 Claim: If φ is satisﬁable, G has an independent set of size k

Reduction for INDSET

Claim: If G has an independent set of size k, φ is satisﬁable

0 I The independent set, say V , cannot have two vertices from the same clause 0 I Therefore, V has exactly one vertex from each of the k clauses

I No pair maps to literals x and x i i i I If vj corresponds to literal xj = x, set xj = 1 i i i I If vj corresponds to literal xj = x, set xj = 0

I Each clause satisﬁed, and no variable-setting conﬂicts between clauses

30 / 121 Reduction for INDSET

Claim: If G has an independent set of size k, φ is satisﬁable

0 I The independent set, say V , cannot have two vertices from the same clause 0 I Therefore, V has exactly one vertex from each of the k clauses

I No pair maps to literals x and x i i i I If vj corresponds to literal xj = x, set xj = 1 i i i I If vj corresponds to literal xj = x, set xj = 0

I Each clause satisﬁed, and no variable-setting conﬂicts between clauses

Claim: If φ is satisﬁable, G has an independent set of size k

31 / 121 Reduction from INDSET Given G, consider the complement graph G Claim: G has an indset of size k iﬀ G has clique of size k

v3 v3

v2 v2

v4 v4

v1 v1

v5 v5

CLIQUE Claim CLIQUE : Given a graph G = (V , E) and a parameter k, does G have a clique of size k? The clique problem (CLIQUE ) is NP complete.

32 / 121 Given G, consider the complement graph G Claim: G has an indset of size k iﬀ G has clique of size k

v3 v3

v2 v2

v4 v4

v1 v1

v5 v5

CLIQUE Claim CLIQUE : Given a graph G = (V , E) and a parameter k, does G have a clique of size k? The clique problem (CLIQUE ) is NP complete.

Reduction from INDSET

33 / 121 Claim: G has an indset of size k iﬀ G has clique of size k

v3 v3

v2 v2

v4 v4

v1 v1

v5 v5

CLIQUE Claim CLIQUE : Given a graph G = (V , E) and a parameter k, does G have a clique of size k? The clique problem (CLIQUE ) is NP complete.

Reduction from INDSET Given G, consider the complement graph G

34 / 121 v3 v3

v2 v2

v4 v4

v1 v1

v5 v5

CLIQUE Claim CLIQUE : Given a graph G = (V , E) and a parameter k, does G have a clique of size k? The clique problem (CLIQUE ) is NP complete.

Reduction from INDSET Given G, consider the complement graph G Claim: G has an indset of size k iﬀ G has clique of size k

35 / 121 CLIQUE Claim CLIQUE : Given a graph G = (V , E) and a parameter k, does G have a clique of size k? The clique problem (CLIQUE ) is NP complete.

Reduction from INDSET Given G, consider the complement graph G Claim: G has an indset of size k iﬀ G has clique of size k

v3 v3

v2 v2

v4 v4

v1 v1

v5 v5

36 / 121 Construct a graph G = (V , E) as follows:

I Vertices: Each literal corresponds to a vertex.

I Edges: All vertices are connected with an edge except the vertices of the same clause and vertices with negated literals.

F xi , xj ∈ Ck does not have an edge.

F xi ∈ Ck , x¯i ∈ Cj does not have an edge.

Clique of size k in G iﬀ 3-CNF satisﬁable.

Same CLIQUE Reduction ‘Unrolled’

Reduce 3-CNF SAT to CLIQUE

i i i φ = C1 C2 ... Ck , where Ci = (x x x ) ∧ ∧ ∧ 1 ∨ 2 ∨ 3

37 / 121 I Vertices: Each literal corresponds to a vertex.

I Edges: All vertices are connected with an edge except the vertices of the same clause and vertices with negated literals.

F xi , xj ∈ Ck does not have an edge.

F xi ∈ Ck , x¯i ∈ Cj does not have an edge.

Clique of size k in G iﬀ 3-CNF satisﬁable.

Same CLIQUE Reduction ‘Unrolled’

Reduce 3-CNF SAT to CLIQUE

i i i φ = C1 C2 ... Ck , where Ci = (x x x ) ∧ ∧ ∧ 1 ∨ 2 ∨ 3

Construct a graph G = (V , E) as follows:

38 / 121 I Edges: All vertices are connected with an edge except the vertices of the same clause and vertices with negated literals.

F xi , xj ∈ Ck does not have an edge.

F xi ∈ Ck , x¯i ∈ Cj does not have an edge.

Clique of size k in G iﬀ 3-CNF satisﬁable.

Same CLIQUE Reduction ‘Unrolled’

Reduce 3-CNF SAT to CLIQUE

i i i φ = C1 C2 ... Ck , where Ci = (x x x ) ∧ ∧ ∧ 1 ∨ 2 ∨ 3

Construct a graph G = (V , E) as follows:

I Vertices: Each literal corresponds to a vertex.

39 / 121 F xi , xj ∈ Ck does not have an edge.

F xi ∈ Ck , x¯i ∈ Cj does not have an edge.

Clique of size k in G iﬀ 3-CNF satisﬁable.

Same CLIQUE Reduction ‘Unrolled’

Reduce 3-CNF SAT to CLIQUE

i i i φ = C1 C2 ... Ck , where Ci = (x x x ) ∧ ∧ ∧ 1 ∨ 2 ∨ 3

Construct a graph G = (V , E) as follows:

I Vertices: Each literal corresponds to a vertex.

I Edges: All vertices are connected with an edge except the vertices of the same clause and vertices with negated literals.

40 / 121 F xi ∈ Ck , x¯i ∈ Cj does not have an edge.

Clique of size k in G iﬀ 3-CNF satisﬁable.

Same CLIQUE Reduction ‘Unrolled’

Reduce 3-CNF SAT to CLIQUE

i i i φ = C1 C2 ... Ck , where Ci = (x x x ) ∧ ∧ ∧ 1 ∨ 2 ∨ 3

Construct a graph G = (V , E) as follows:

I Vertices: Each literal corresponds to a vertex.

I Edges: All vertices are connected with an edge except the vertices of the same clause and vertices with negated literals.

F xi , xj ∈ Ck does not have an edge.

41 / 121 Clique of size k in G iﬀ 3-CNF satisﬁable.

Same CLIQUE Reduction ‘Unrolled’

Reduce 3-CNF SAT to CLIQUE

i i i φ = C1 C2 ... Ck , where Ci = (x x x ) ∧ ∧ ∧ 1 ∨ 2 ∨ 3

Construct a graph G = (V , E) as follows:

I Vertices: Each literal corresponds to a vertex.

I Edges: All vertices are connected with an edge except the vertices of the same clause and vertices with negated literals.

F xi , xj ∈ Ck does not have an edge.

F xi ∈ Ck , x¯i ∈ Cj does not have an edge.

42 / 121 Same CLIQUE Reduction ‘Unrolled’

Reduce 3-CNF SAT to CLIQUE

i i i φ = C1 C2 ... Ck , where Ci = (x x x ) ∧ ∧ ∧ 1 ∨ 2 ∨ 3

Construct a graph G = (V , E) as follows:

I Vertices: Each literal corresponds to a vertex.

I Edges: All vertices are connected with an edge except the vertices of the same clause and vertices with negated literals.

F xi , xj ∈ Ck does not have an edge.

F xi ∈ Ck , x¯i ∈ Cj does not have an edge.

Clique of size k in G iﬀ 3-CNF satisﬁable.

43 / 121 Example Clique Reduction

φ = (x x x ) (x x x ) (x x ) 1 ∨ 2 ∨ 3 ∧ 1 ∨ 3 ∨ 4 ∧ 2 ∨ 4

x1 x2 x3

x3 x4

x2 x4

44 / 121 Vertex Cover Deﬁnition A Vertex Cover of an undirected graph G = (V , E) is a subset V 0 of vertex set V of the vertices of G which contains at least one of the two end points of each edge.

0 0 0 V V : (vi , vj ) E : vi V vj V ⊆ ∀ ∈ ∈ ∨ ∈

45 / 121 Note that VERTEX-COVER is in NP

We reduce 3-CNF SAT to VERTEX-COVER

VERTEX-COVER

Claim VERTEX-COVER : Given a graph G = (V , E) and a parameter k, does G have a vertex cover of size k? The vertex cover problem (VERTEX-COVER ) is NP complete.

46 / 121 We reduce 3-CNF SAT to VERTEX-COVER

VERTEX-COVER

Claim VERTEX-COVER : Given a graph G = (V , E) and a parameter k, does G have a vertex cover of size k? The vertex cover problem (VERTEX-COVER ) is NP complete.

Note that VERTEX-COVER is in NP

47 / 121 VERTEX-COVER

Claim VERTEX-COVER : Given a graph G = (V , E) and a parameter k, does G have a vertex cover of size k? The vertex cover problem (VERTEX-COVER ) is NP complete.

Note that VERTEX-COVER is in NP

We reduce 3-CNF SAT to VERTEX-COVER

48 / 121 Construct a graph G = (V , E) as follows:

1 0 I Variable vertices: Each variable xi creates two vertices vi , vi

I Edge vertices: Each clause j nj creates a set V ( j ) of nj vertices C C 1 0 I Edges: For all xi , (v , v ) E i i ∈ I Edges: Add all possible edges between each set V ( j ) C I Edges: For each literal xi j , connect its vertex in j to the corresponding literal vertex.∈ C C

I k = n + 2m

Question: Does G have a vertex-cover of size k?

Reduction for VERTEX-COVER

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

49 / 121 1 0 I Variable vertices: Each variable xi creates two vertices vi , vi

I k = n + 2m

Question: Does G have a vertex-cover of size k?

Reduction for VERTEX-COVER

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Construct a graph G = (V , E) as follows:

50 / 121 I Edge vertices: Each clause j nj creates a set V ( j ) of nj vertices C C 1 0 I Edges: For all xi , (v , v ) E i i ∈ I Edges: Add all possible edges between each set V ( j ) C I Edges: For each literal xi j , connect its vertex in j to the corresponding literal vertex.∈ C C

I k = n + 2m

Question: Does G have a vertex-cover of size k?

Reduction for VERTEX-COVER

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Construct a graph G = (V , E) as follows:

1 0 I Variable vertices: Each variable xi creates two vertices vi , vi

51 / 121 1 0 I Edges: For all xi , (v , v ) E i i ∈ I Edges: Add all possible edges between each set V ( j ) C I Edges: For each literal xi j , connect its vertex in j to the corresponding literal vertex.∈ C C

I k = n + 2m

Question: Does G have a vertex-cover of size k?

Reduction for VERTEX-COVER

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Construct a graph G = (V , E) as follows:

1 0 I Variable vertices: Each variable xi creates two vertices vi , vi

I Edge vertices: Each clause j nj creates a set V ( j ) of nj vertices C C

52 / 121 I Edges: Add all possible edges between each set V ( j ) C I Edges: For each literal xi j , connect its vertex in j to the corresponding literal vertex.∈ C C

I k = n + 2m

Question: Does G have a vertex-cover of size k?

Reduction for VERTEX-COVER

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Construct a graph G = (V , E) as follows:

1 0 I Variable vertices: Each variable xi creates two vertices vi , vi

I Edge vertices: Each clause j nj creates a set V ( j ) of nj vertices C C 1 0 I Edges: For all xi , (v , v ) E i i ∈

53 / 121 I Edges: For each literal xi j , connect its vertex in j to the corresponding literal vertex.∈ C C

I k = n + 2m

Question: Does G have a vertex-cover of size k?

Reduction for VERTEX-COVER

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Construct a graph G = (V , E) as follows:

1 0 I Variable vertices: Each variable xi creates two vertices vi , vi

I Edge vertices: Each clause j nj creates a set V ( j ) of nj vertices C C 1 0 I Edges: For all xi , (v , v ) E i i ∈ I Edges: Add all possible edges between each set V ( j ) C

54 / 121 I k = n + 2m

Question: Does G have a vertex-cover of size k?

Reduction for VERTEX-COVER

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Construct a graph G = (V , E) as follows:

1 0 I Variable vertices: Each variable xi creates two vertices vi , vi

55 / 121 Question: Does G have a vertex-cover of size k?

Reduction for VERTEX-COVER

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Construct a graph G = (V , E) as follows:

1 0 I Variable vertices: Each variable xi creates two vertices vi , vi

I k = n + 2m

56 / 121 Reduction for VERTEX-COVER

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Construct a graph G = (V , E) as follows:

1 0 I Variable vertices: Each variable xi creates two vertices vi , vi

I k = n + 2m

Question: Does G have a vertex-cover of size k?

57 / 121 Reduction for VERTEX-COVER

(x1 x2 x3) (x1 x3 x4) (x2 x4 x1), n = 4, m = 3 ∨ ∨ ∧ ∨ ∨ ∧ ∨ ∨

58 / 121 Reduction for VERTEX-COVER

(x1 x2 x3) (x1 x3 x4) (x2 x4 x1), n = 4, m = 3 ∨ ∨ ∧ ∨ ∨ ∧ ∨ ∨

v1 v1 v2 v2 v3 v3 v4 v4

59 / 121 Reduction for VERTEX-COVER

(x1 x2 x3) (x1 x3 x4) (x2 x4 x1), n = 4, m = 3 ∨ ∨ ∧ ∨ ∨ ∧ ∨ ∨

v1 v1 v2 v2 v3 v3 v4 v4

x2 x4 x3 x1 x1 x2 x3 x1 x4

60 / 121 Reduction for VERTEX-COVER

(x1 x2 x3) (x1 x3 x4) (x2 x4 x1), n = 4, m = 3 ∨ ∨ ∧ ∨ ∨ ∧ ∨ ∨

v1 v1 v2 v2 v3 v3 v4 v4

x2 x4 x3 x1 x1 x2 x3 x1 x4

61 / 121 0 If variable xi = 0, pick vi to be in the vertex cover 1 If variable xi = 1, pick vi to be in the vertex cover j Each clause j has at least one literal, say x2, set to true. C j I Don’t pick edge vertex of xi , pick other two edge vertices Claim: The picked vertices cover all the edges of G. Clearly, the number of vertices picked is n + 2m. An edge between two variable vertices is covered

I For each i, exactly one of the variable vertices picked.

For each j , edges to the picked vertices covered. C j Crucial: What about the edges incident upon x2? I Edges within its clause covered by the other two picked vertices j 1 j I Remaining edge of the form: (xi , vi ) if xi is un-negated. 1 I We know that vi is picked, since vi = 1.

Reduction for VERTEX-COVER Claim: If φ is satisﬁable, then a vertex cover of size k = n + 2m ∃

62 / 121 1 If variable xi = 1, pick vi to be in the vertex cover j Each clause j has at least one literal, say x2, set to true. C j I Don’t pick edge vertex of xi , pick other two edge vertices Claim: The picked vertices cover all the edges of G. Clearly, the number of vertices picked is n + 2m. An edge between two variable vertices is covered

I For each i, exactly one of the variable vertices picked.

Reduction for VERTEX-COVER Claim: If φ is satisﬁable, then a vertex cover of size k = n + 2m 0 ∃ If variable xi = 0, pick vi to be in the vertex cover

63 / 121 j Each clause j has at least one literal, say x2, set to true. C j I Don’t pick edge vertex of xi , pick other two edge vertices Claim: The picked vertices cover all the edges of G. Clearly, the number of vertices picked is n + 2m. An edge between two variable vertices is covered

I For each i, exactly one of the variable vertices picked.

64 / 121 j I Don’t pick edge vertex of xi , pick other two edge vertices Claim: The picked vertices cover all the edges of G. Clearly, the number of vertices picked is n + 2m. An edge between two variable vertices is covered

I For each i, exactly one of the variable vertices picked.

65 / 121 Claim: The picked vertices cover all the edges of G. Clearly, the number of vertices picked is n + 2m. An edge between two variable vertices is covered

I For each i, exactly one of the variable vertices picked.

66 / 121 Clearly, the number of vertices picked is n + 2m. An edge between two variable vertices is covered

I For each i, exactly one of the variable vertices picked.

Reduction for VERTEX-COVER Claim: If φ is satisﬁable, then a vertex cover of size k = n + 2m 0 ∃ If variable xi = 0, pick vi to be in the vertex cover 1 If variable xi = 1, pick vi to be in the vertex cover j Each clause j has at least one literal, say x2, set to true. C j I Don’t pick edge vertex of xi , pick other two edge vertices Claim: The picked vertices cover all the edges of G.

67 / 121 An edge between two variable vertices is covered

I For each i, exactly one of the variable vertices picked.

68 / 121 I For each i, exactly one of the variable vertices picked.

69 / 121 For each j , edges to the picked vertices covered. C j Crucial: What about the edges incident upon x2? I Edges within its clause covered by the other two picked vertices j 1 j I Remaining edge of the form: (xi , vi ) if xi is un-negated. 1 I We know that vi is picked, since vi = 1.

I For each i, exactly one of the variable vertices picked.

70 / 121 j Crucial: What about the edges incident upon x2? I Edges within its clause covered by the other two picked vertices j 1 j I Remaining edge of the form: (xi , vi ) if xi is un-negated. 1 I We know that vi is picked, since vi = 1.

I For each i, exactly one of the variable vertices picked.

For each j , edges to the picked vertices covered. C

71 / 121 I Edges within its clause covered by the other two picked vertices j 1 j I Remaining edge of the form: (xi , vi ) if xi is un-negated. 1 I We know that vi is picked, since vi = 1.

I For each i, exactly one of the variable vertices picked.

For each j , edges to the picked vertices covered. C j Crucial: What about the edges incident upon x2?

72 / 121 j 1 j I Remaining edge of the form: (xi , vi ) if xi is un-negated. 1 I We know that vi is picked, since vi = 1.

I For each i, exactly one of the variable vertices picked.

For each j , edges to the picked vertices covered. C j Crucial: What about the edges incident upon x2? I Edges within its clause covered by the other two picked vertices

73 / 121 1 I We know that vi is picked, since vi = 1.

I For each i, exactly one of the variable vertices picked.

74 / 121 Reduction for VERTEX-COVER Claim: If φ is satisﬁable, then a vertex cover of size k = n + 2m 0 ∃ If variable xi = 0, pick vi to be in the vertex cover 1 If variable xi = 1, pick vi to be in the vertex cover j Each clause j has at least one literal, say x2, set to true. C j I Don’t pick edge vertex of xi , pick other two edge vertices Claim: The picked vertices cover all the edges of G. Clearly, the number of vertices picked is n + 2m. An edge between two variable vertices is covered

I For each i, exactly one of the variable vertices picked.

75 / 121 For each i, at least one of the variable vertices in vertex cover.

In each set V (Cj ), at least 2 vertices in vertex cover This already makes n + 2m vertices! Exactly one vertex not picked in each clause

Consider the clause (x1 x2 x3) ∨ ∨ Lets say x1 and x2 were picked in vertex cover

The edge (x3, v 3) must be covered

Therefore v 3 must be in the vertex cover.

Then set the boolean variable x3 = 0 Similarly, whichever vertex not picked in cover, set corresponding literal to 1 Thus, setting variables as above, each clause satisﬁed

Reduction for VERTEX-COVER

Claim: If a vertex cover of size k = n + 2m, then φ satisﬁable ∃

76 / 121 In each set V (Cj ), at least 2 vertices in vertex cover This already makes n + 2m vertices! Exactly one vertex not picked in each clause

Consider the clause (x1 x2 x3) ∨ ∨ Lets say x1 and x2 were picked in vertex cover

The edge (x3, v 3) must be covered

Therefore v 3 must be in the vertex cover.

Then set the boolean variable x3 = 0 Similarly, whichever vertex not picked in cover, set corresponding literal to 1 Thus, setting variables as above, each clause satisﬁed

Reduction for VERTEX-COVER

Claim: If a vertex cover of size k = n + 2m, then φ satisﬁable ∃ For each i, at least one of the variable vertices in vertex cover.

77 / 121 This already makes n + 2m vertices! Exactly one vertex not picked in each clause

Consider the clause (x1 x2 x3) ∨ ∨ Lets say x1 and x2 were picked in vertex cover

The edge (x3, v 3) must be covered

Therefore v 3 must be in the vertex cover.

Then set the boolean variable x3 = 0 Similarly, whichever vertex not picked in cover, set corresponding literal to 1 Thus, setting variables as above, each clause satisﬁed

Reduction for VERTEX-COVER

Claim: If a vertex cover of size k = n + 2m, then φ satisﬁable ∃ For each i, at least one of the variable vertices in vertex cover.

In each set V (Cj ), at least 2 vertices in vertex cover

78 / 121 Exactly one vertex not picked in each clause

Consider the clause (x1 x2 x3) ∨ ∨ Lets say x1 and x2 were picked in vertex cover

The edge (x3, v 3) must be covered

Therefore v 3 must be in the vertex cover.

Then set the boolean variable x3 = 0 Similarly, whichever vertex not picked in cover, set corresponding literal to 1 Thus, setting variables as above, each clause satisﬁed

Reduction for VERTEX-COVER

Claim: If a vertex cover of size k = n + 2m, then φ satisﬁable ∃ For each i, at least one of the variable vertices in vertex cover.

In each set V (Cj ), at least 2 vertices in vertex cover This already makes n + 2m vertices!

79 / 121 Consider the clause (x1 x2 x3) ∨ ∨ Lets say x1 and x2 were picked in vertex cover

The edge (x3, v 3) must be covered

Therefore v 3 must be in the vertex cover.

Then set the boolean variable x3 = 0 Similarly, whichever vertex not picked in cover, set corresponding literal to 1 Thus, setting variables as above, each clause satisﬁed

Reduction for VERTEX-COVER

Claim: If a vertex cover of size k = n + 2m, then φ satisﬁable ∃ For each i, at least one of the variable vertices in vertex cover.

In each set V (Cj ), at least 2 vertices in vertex cover This already makes n + 2m vertices! Exactly one vertex not picked in each clause

80 / 121 Lets say x1 and x2 were picked in vertex cover

The edge (x3, v 3) must be covered

Therefore v 3 must be in the vertex cover.

Then set the boolean variable x3 = 0 Similarly, whichever vertex not picked in cover, set corresponding literal to 1 Thus, setting variables as above, each clause satisﬁed

Reduction for VERTEX-COVER

Claim: If a vertex cover of size k = n + 2m, then φ satisﬁable ∃ For each i, at least one of the variable vertices in vertex cover.

In each set V (Cj ), at least 2 vertices in vertex cover This already makes n + 2m vertices! Exactly one vertex not picked in each clause

Consider the clause (x1 x2 x3) ∨ ∨

81 / 121 The edge (x3, v 3) must be covered

Therefore v 3 must be in the vertex cover.

Then set the boolean variable x3 = 0 Similarly, whichever vertex not picked in cover, set corresponding literal to 1 Thus, setting variables as above, each clause satisﬁed

Reduction for VERTEX-COVER

Claim: If a vertex cover of size k = n + 2m, then φ satisﬁable ∃ For each i, at least one of the variable vertices in vertex cover.

In each set V (Cj ), at least 2 vertices in vertex cover This already makes n + 2m vertices! Exactly one vertex not picked in each clause

Consider the clause (x1 x2 x3) ∨ ∨ Lets say x1 and x2 were picked in vertex cover

82 / 121 Therefore v 3 must be in the vertex cover.

Then set the boolean variable x3 = 0 Similarly, whichever vertex not picked in cover, set corresponding literal to 1 Thus, setting variables as above, each clause satisﬁed

Reduction for VERTEX-COVER

Claim: If a vertex cover of size k = n + 2m, then φ satisﬁable ∃ For each i, at least one of the variable vertices in vertex cover.

In each set V (Cj ), at least 2 vertices in vertex cover This already makes n + 2m vertices! Exactly one vertex not picked in each clause

Consider the clause (x1 x2 x3) ∨ ∨ Lets say x1 and x2 were picked in vertex cover

The edge (x3, v 3) must be covered

83 / 121 Then set the boolean variable x3 = 0 Similarly, whichever vertex not picked in cover, set corresponding literal to 1 Thus, setting variables as above, each clause satisﬁed

Reduction for VERTEX-COVER

Claim: If a vertex cover of size k = n + 2m, then φ satisﬁable ∃ For each i, at least one of the variable vertices in vertex cover.

In each set V (Cj ), at least 2 vertices in vertex cover This already makes n + 2m vertices! Exactly one vertex not picked in each clause

Consider the clause (x1 x2 x3) ∨ ∨ Lets say x1 and x2 were picked in vertex cover

The edge (x3, v 3) must be covered

Therefore v 3 must be in the vertex cover.

84 / 121 Similarly, whichever vertex not picked in cover, set corresponding literal to 1 Thus, setting variables as above, each clause satisﬁed

Reduction for VERTEX-COVER

Claim: If a vertex cover of size k = n + 2m, then φ satisﬁable ∃ For each i, at least one of the variable vertices in vertex cover.

In each set V (Cj ), at least 2 vertices in vertex cover This already makes n + 2m vertices! Exactly one vertex not picked in each clause

Consider the clause (x1 x2 x3) ∨ ∨ Lets say x1 and x2 were picked in vertex cover

The edge (x3, v 3) must be covered

Therefore v 3 must be in the vertex cover.

Then set the boolean variable x3 = 0

85 / 121 Thus, setting variables as above, each clause satisﬁed

Reduction for VERTEX-COVER

Claim: If a vertex cover of size k = n + 2m, then φ satisﬁable ∃ For each i, at least one of the variable vertices in vertex cover.

In each set V (Cj ), at least 2 vertices in vertex cover This already makes n + 2m vertices! Exactly one vertex not picked in each clause

Consider the clause (x1 x2 x3) ∨ ∨ Lets say x1 and x2 were picked in vertex cover

The edge (x3, v 3) must be covered

Therefore v 3 must be in the vertex cover.

Then set the boolean variable x3 = 0 Similarly, whichever vertex not picked in cover, set corresponding literal to 1

86 / 121 Reduction for VERTEX-COVER

Claim: If a vertex cover of size k = n + 2m, then φ satisﬁable ∃ For each i, at least one of the variable vertices in vertex cover.

In each set V (Cj ), at least 2 vertices in vertex cover This already makes n + 2m vertices! Exactly one vertex not picked in each clause

Consider the clause (x1 x2 x3) ∨ ∨ Lets say x1 and x2 were picked in vertex cover

The edge (x3, v 3) must be covered

Therefore v 3 must be in the vertex cover.

Then set the boolean variable x3 = 0 Similarly, whichever vertex not picked in cover, set corresponding literal to 1 Thus, setting variables as above, each clause satisﬁed

87 / 121 Let I = v1, v2,..., vn−k be INDSET of size n k { } − I There is no edge in G with both endpoints in I

I So if we pick the remaining k vertices, all edges covered

0 V = v1, v2,..., vk is a vertex cover of size k { } 0 I Each edge in G has at least one endpoint in V

I So the remaining n k vertices form an independent set −

Another reduction for VERTEX-COVER

Claim V 0 is a vertex cover of size k in G = (V , E) iﬀ V V 0 of size (n k) is an independent set. \ −

88 / 121 I There is no edge in G with both endpoints in I

I So if we pick the remaining k vertices, all edges covered

0 V = v1, v2,..., vk is a vertex cover of size k { } 0 I Each edge in G has at least one endpoint in V

I So the remaining n k vertices form an independent set −

Another reduction for VERTEX-COVER

Claim V 0 is a vertex cover of size k in G = (V , E) iﬀ V V 0 of size (n k) is an independent set. \ −

Let I = v1, v2,..., vn−k be INDSET of size n k { } −

89 / 121 I So if we pick the remaining k vertices, all edges covered

0 V = v1, v2,..., vk is a vertex cover of size k { } 0 I Each edge in G has at least one endpoint in V

I So the remaining n k vertices form an independent set −

Another reduction for VERTEX-COVER

Claim V 0 is a vertex cover of size k in G = (V , E) iﬀ V V 0 of size (n k) is an independent set. \ −

Let I = v1, v2,..., vn−k be INDSET of size n k { } − I There is no edge in G with both endpoints in I

90 / 121 0 V = v1, v2,..., vk is a vertex cover of size k { } 0 I Each edge in G has at least one endpoint in V

I So the remaining n k vertices form an independent set −

Another reduction for VERTEX-COVER

Claim V 0 is a vertex cover of size k in G = (V , E) iﬀ V V 0 of size (n k) is an independent set. \ −

Let I = v1, v2,..., vn−k be INDSET of size n k { } − I There is no edge in G with both endpoints in I

I So if we pick the remaining k vertices, all edges covered

91 / 121 0 I Each edge in G has at least one endpoint in V

I So the remaining n k vertices form an independent set −

Another reduction for VERTEX-COVER

Claim V 0 is a vertex cover of size k in G = (V , E) iﬀ V V 0 of size (n k) is an independent set. \ −

Let I = v1, v2,..., vn−k be INDSET of size n k { } − I There is no edge in G with both endpoints in I

I So if we pick the remaining k vertices, all edges covered

0 V = v1, v2,..., vk is a vertex cover of size k { }

92 / 121 I So the remaining n k vertices form an independent set −

Another reduction for VERTEX-COVER

Claim V 0 is a vertex cover of size k in G = (V , E) iﬀ V V 0 of size (n k) is an independent set. \ −

Let I = v1, v2,..., vn−k be INDSET of size n k { } − I There is no edge in G with both endpoints in I

I So if we pick the remaining k vertices, all edges covered

0 V = v1, v2,..., vk is a vertex cover of size k { } 0 I Each edge in G has at least one endpoint in V

93 / 121 Another reduction for VERTEX-COVER

Claim V 0 is a vertex cover of size k in G = (V , E) iﬀ V V 0 of size (n k) is an independent set. \ −

Let I = v1, v2,..., vn−k be INDSET of size n k { } − I There is no edge in G with both endpoints in I

I So if we pick the remaining k vertices, all edges covered

0 V = v1, v2,..., vk is a vertex cover of size k { } 0 I Each edge in G has at least one endpoint in V

I So the remaining n k vertices form an independent set −

94 / 121 S1 S4 8 1 H1 S2 12 7 15

6 9 3 S3 14

10 H2

Reduction from? Consider each set having size two!

HITTING-SET

Claim HITTING-SET : A collection C of subsets of a set V , and parameter k, ﬁnd a hitting set V 0 V of size k. ⊆ The hitting set problem (HITTING-SET ) is NP complete.

95 / 121 Reduction from? Consider each set having size two!

HITTING-SET

Claim HITTING-SET : A collection C of subsets of a set V , and parameter k, ﬁnd a hitting set V 0 V of size k. ⊆ The hitting set problem (HITTING-SET ) is NP complete.

S1 S4 8 1 H1 S2 12 7 15

6 9 3 S3 14

10 H2

96 / 121 Consider each set having size two!

HITTING-SET

Claim HITTING-SET : A collection C of subsets of a set V , and parameter k, ﬁnd a hitting set V 0 V of size k. ⊆ The hitting set problem (HITTING-SET ) is NP complete.

S1 S4 8 1 H1 S2 12 7 15

6 9 3 S3 14

10 H2

Reduction from?

97 / 121 HITTING-SET

Claim HITTING-SET : A collection C of subsets of a set V , and parameter k, ﬁnd a hitting set V 0 V of size k. ⊆ The hitting set problem (HITTING-SET ) is NP complete.

S1 S4 8 1 H1 S2 12 7 15

6 9 3 S3 14

10 H2

Reduction from? Consider each set having size two!

98 / 121 u1 + 4u2 32u3 34 − ≥ 2u2 2u4 + 7u3 239 − ≤ 43u4 2u1 + 17u2 17 − ≥

Note that INTEGER-PROGRAMMING is in NP Reduction from 3-CNF SAT

INTEGER-PROGRAMMING

Claim INTEGER-PROGRAMMING : Given a set of linear inequalities over variables v1,..., vn, does there exist an satisfying assignment of vi to positive integers The integer programming problem is NP complete.

99 / 121 2u2 2u4 + 7u3 239 − ≤ 43u4 2u1 + 17u2 17 − ≥

Note that INTEGER-PROGRAMMING is in NP Reduction from 3-CNF SAT

INTEGER-PROGRAMMING

u1 + 4u2 32u3 34 − ≥

100 / 121 43u4 2u1 + 17u2 17 − ≥

Note that INTEGER-PROGRAMMING is in NP Reduction from 3-CNF SAT

INTEGER-PROGRAMMING

u1 + 4u2 32u3 34 − ≥ 2u2 2u4 + 7u3 239 − ≤

101 / 121 Note that INTEGER-PROGRAMMING is in NP Reduction from 3-CNF SAT

INTEGER-PROGRAMMING

u1 + 4u2 32u3 34 − ≥ 2u2 2u4 + 7u3 239 − ≤ 43u4 2u1 + 17u2 17 − ≥

102 / 121 Reduction from 3-CNF SAT

INTEGER-PROGRAMMING

u1 + 4u2 32u3 34 − ≥ 2u2 2u4 + 7u3 239 − ≤ 43u4 2u1 + 17u2 17 − ≥

Note that INTEGER-PROGRAMMING is in NP

103 / 121 INTEGER-PROGRAMMING

u1 + 4u2 32u3 34 − ≥ 2u2 2u4 + 7u3 239 − ≤ 43u4 2u1 + 17u2 17 − ≥

Note that INTEGER-PROGRAMMING is in NP Reduction from 3-CNF SAT

104 / 121 Given φ, construct the set of inequalities as following:

I Each boolean variable xi becomes integer variable vi I Each clause becomes an equation set to be greater than 1:

xi vi , xi (1 vi ), + → → − ∨ →

0 v1 1, 0 v2 1, 0 v3 1, 0 v4 1 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤

Reduction for INTEGER-PROGRAMMING

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

105 / 121 I Each boolean variable xi becomes integer variable vi I Each clause becomes an equation set to be greater than 1:

xi vi , xi (1 vi ), + → → − ∨ →

0 v1 1, 0 v2 1, 0 v3 1, 0 v4 1 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤

Reduction for INTEGER-PROGRAMMING

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Given φ, construct the set of inequalities as following:

106 / 121 I Each clause becomes an equation set to be greater than 1:

xi vi , xi (1 vi ), + → → − ∨ →

0 v1 1, 0 v2 1, 0 v3 1, 0 v4 1 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤

Reduction for INTEGER-PROGRAMMING

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Given φ, construct the set of inequalities as following:

I Each boolean variable xi becomes integer variable vi

107 / 121 xi vi , xi (1 vi ), + → → − ∨ →

0 v1 1, 0 v2 1, 0 v3 1, 0 v4 1 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤

Reduction for INTEGER-PROGRAMMING

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Given φ, construct the set of inequalities as following:

I Each boolean variable xi becomes integer variable vi I Each clause becomes an equation set to be greater than 1:

108 / 121 0 v1 1, 0 v2 1, 0 v3 1, 0 v4 1 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤

Reduction for INTEGER-PROGRAMMING

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Given φ, construct the set of inequalities as following:

I Each boolean variable xi becomes integer variable vi I Each clause becomes an equation set to be greater than 1:

xi vi , xi (1 vi ), + → → − ∨ →

109 / 121 0 v1 1, 0 v2 1, 0 v3 1, 0 v4 1 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤

Reduction for INTEGER-PROGRAMMING

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Given φ, construct the set of inequalities as following:

I Each boolean variable xi becomes integer variable vi I Each clause becomes an equation set to be greater than 1:

xi vi , xi (1 vi ), + → → − ∨ →

(x1 x2 x3) ∨ ∨

110 / 121 0 v1 1, 0 v2 1, 0 v3 1, 0 v4 1 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤

Reduction for INTEGER-PROGRAMMING

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Given φ, construct the set of inequalities as following:

I Each boolean variable xi becomes integer variable vi I Each clause becomes an equation set to be greater than 1:

xi vi , xi (1 vi ), + → → − ∨ →

(x1 x2 x3) ∨ ∨ v1 + v2 + (1 v3) 1 − ≥

111 / 121 0 v1 1, 0 v2 1, 0 v3 1, 0 v4 1 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤

Reduction for INTEGER-PROGRAMMING

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Given φ, construct the set of inequalities as following:

I Each boolean variable xi becomes integer variable vi I Each clause becomes an equation set to be greater than 1:

xi vi , xi (1 vi ), + → → − ∨ →

(x1 x2 x3) (x1 x3 x4) ∨ ∨ ∧ ∨ ∨

112 / 121 0 v1 1, 0 v2 1, 0 v3 1, 0 v4 1 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤

Reduction for INTEGER-PROGRAMMING

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Given φ, construct the set of inequalities as following:

I Each boolean variable xi becomes integer variable vi I Each clause becomes an equation set to be greater than 1:

xi vi , xi (1 vi ), + → → − ∨ →

(x1 x2 x3) (x1 x3 x4) ∨ ∨ ∧ ∨ ∨ v1 + v2 + (1 v3) 1, (1 v1) + v3 + (1 v4) 1 − ≥ − − ≥

113 / 121 0 v1 1, 0 v2 1, 0 v3 1, 0 v4 1 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤

Reduction for INTEGER-PROGRAMMING

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Given φ, construct the set of inequalities as following:

I Each boolean variable xi becomes integer variable vi I Each clause becomes an equation set to be greater than 1:

xi vi , xi (1 vi ), + → → − ∨ →

(x1 x2 x3) (x1 x3 x4) (x2 x4 x1) ∨ ∨ ∧ ∨ ∨ ∧ ∨ ∨

114 / 121 0 v1 1, 0 v2 1, 0 v3 1, 0 v4 1 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤

Reduction for INTEGER-PROGRAMMING

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Given φ, construct the set of inequalities as following:

I Each boolean variable xi becomes integer variable vi I Each clause becomes an equation set to be greater than 1:

xi vi , xi (1 vi ), + → → − ∨ →

(x1 x2 x3) (x1 x3 x4) (x2 x4 x1) ∨ ∨ ∧ ∨ ∨ ∧ ∨ ∨

v1 + v2 + (1 v3) 1, (1 v1) + v3 + (1 v4) 1, (1 v2) + v4 + v1 1 − ≥ − − ≥ − ≥

115 / 121 0 v1 1, 0 v2 1, 0 v3 1, 0 v4 1 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤

Reduction for INTEGER-PROGRAMMING

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Given φ, construct the set of inequalities as following:

I Each boolean variable xi becomes integer variable vi I Each clause becomes an equation set to be greater than 1:

xi vi , xi (1 vi ), + → → − ∨ →

(x1 x2 x3) (x1 x3 x4) (x2 x4 x1) ∨ ∨ ∧ ∨ ∨ ∧ ∨ ∨ v1 + v2 + (1 v3) 1, (1 v1) + v3 + (1 v4) 1, (1 v2) + v4 + v1 1 − ≥ − − ≥ − ≥

116 / 121 Reduction for INTEGER-PROGRAMMING

i i i φ = 1 2 ... m, where i = (x x x ) C ∧ C ∧ ∧ C C 1 ∨ 2 ∨ 3

Given φ, construct the set of inequalities as following:

I Each boolean variable xi becomes integer variable vi I Each clause becomes an equation set to be greater than 1:

xi vi , xi (1 vi ), + → → − ∨ →

(x1 x2 x3) (x1 x3 x4) (x2 x4 x1) ∨ ∨ ∧ ∨ ∨ ∧ ∨ ∨ v1 + v2 + (1 v3) 1, (1 v1) + v3 + (1 v4) 1, (1 v2) + v4 + v1 1 − ≥ − − ≥ − ≥

0 v1 1, 0 v2 1, 0 v3 1, 0 v4 1 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤

117 / 121 Each vi can be either 0 or 1

A clause Cj satisﬁed corresponding inequality satisﬁed ⇐⇒

(x1 x2 x3) = 1 v1 + v2 + (1 v3) 1 ∨ ∨ ⇐⇒ − ≥

Likewise in the other direction, set vi = xi .

Reduction for INTEGER-PROGRAMMING

Claim: Solve INTEGER-PROGRAMMING to get a valid solution. Set boolean variable xi = vi . This satisﬁes φ.

118 / 121 A clause Cj satisﬁed corresponding inequality satisﬁed ⇐⇒

(x1 x2 x3) = 1 v1 + v2 + (1 v3) 1 ∨ ∨ ⇐⇒ − ≥

Likewise in the other direction, set vi = xi .

Reduction for INTEGER-PROGRAMMING

Claim: Solve INTEGER-PROGRAMMING to get a valid solution. Set boolean variable xi = vi . This satisﬁes φ.

Each vi can be either 0 or 1

119 / 121 Likewise in the other direction, set vi = xi .

Reduction for INTEGER-PROGRAMMING

Claim: Solve INTEGER-PROGRAMMING to get a valid solution. Set boolean variable xi = vi . This satisﬁes φ.

Each vi can be either 0 or 1

A clause Cj satisﬁed corresponding inequality satisﬁed ⇐⇒

(x1 x2 x3) = 1 v1 + v2 + (1 v3) 1 ∨ ∨ ⇐⇒ − ≥

120 / 121 Reduction for INTEGER-PROGRAMMING

Claim: Solve INTEGER-PROGRAMMING to get a valid solution. Set boolean variable xi = vi . This satisﬁes φ.

Each vi can be either 0 or 1

A clause Cj satisﬁed corresponding inequality satisﬁed ⇐⇒

(x1 x2 x3) = 1 v1 + v2 + (1 v3) 1 ∨ ∨ ⇐⇒ − ≥

Likewise in the other direction, set vi = xi .

121 / 121