•••• Proof Methods NP-Completeness : Proofs

NP-Completeness :

Proofs

• •

• • Proof Methods

A method to show a decision problem ΠΠΠΠ NP-complete is as follows.

(1) Show ΠΠΠΠ∈∈∈∈NP.

(2) Choose an NP-complete problem ΠΠΠΠ’.

(3) Show ΠΠ’ ∝ΠΠ ∝∝∝ ΠΠΠΠ.

A method to show an optimization problem ΨΨΨΨ NP-hard is as follows.

(1) Choose an NP-hard problem ΨΨΨ’Ψ (ΨΨΨΨ’ may be NP-complete).

(2) Show ΨΨΨΨ’ ∝∝∝∝ ΨΨΨΨ.

An alternative method to show ΨΨΨ NP-hard is to Ψ show the decision version of ΨΨΨ NP-complete. Ψ

• •

• • Two Simple Examples

Ex. Sum of Subsets

Instance: A finite set A of positive integers and a positive integer c.

Question: Is there a subset A’ of A whose elements sum to c?

For example, if A={7, 5, 19, 1, 12, 8, 14} and c=21, then the answer is yes (A’={7, 14}).

NP-completeness of Sum of Subsets is shown below.

♣

Exact Cover

Instance: A finite set S and k subsets S₁, S₂, …, S_k of S.

Question: Is there a subset of {S₁, S₂, …, S_k} that forms a partition of S?

For example, if S={7, 5, 19, 1, 12, 8, 14}, k=4, S₁ ={7, 19, 12, 14}, S₂ ={7, 5, 8}, S₃ ={5, 1, 8}, and S₄ ={19, 1, 8, 14}, then the answer is yes ({S₁, S₃} forms a partition of S).

♣

Let S={u₁, u₂, …, u_m} and S₁, S₂, …, S_k be an arbitrary instance of Exact Cover.

An instance of Sum of Subsets can be obtained in polynomial time as follows.

A={a₁, a₂, …, a_k} and c= ^m (

)

i

k i

−

=

∑

0

1 , where for 1≤≤≤≤j≤≤≤≤k,

a_j =

, (

)

m i j i

e k i

=

+ −

∑

1

1 1,

with e_j,i =1 if u_i ∈∈∈∈S_j and e_j,i =0 if u_i ∉∉∉∉S_j.

⇒ Sum of Subsets has the answer yes if and

For example, given the following instance of Exact Cover:

S={7, 5, 19, 1, 12, 8, 14}, k=4, S₁ ={7, 19, 12, 14}, S₂ ={7, 5, 8},

S₃ ={5, 1, 8}, and S₄ ={19, 1, 8, 4},

a matrix

e

7 5 19 1 12 8 14

1 2 3 4

1 0 1 0 1 0 1 1 1 0 0 0 1 0 0 1 0 1 0 1 0 0 0 1 1 0 1 1 S

S S S

 

 

 

 

 

 

An instance of Sum of Subsets is constructed as follows.

A={a₁, a₂, a₃, a₄} and c=5⁰+5¹+5²+ … +5⁶, where

a₁ =5⁰+5²+5⁴+5⁶ (1^st row of

e

a₂ =5⁰+5¹+5⁵ (2^nd row of

e

a₃ =5¹+5³+5⁵ (3^rd row of

e

a₄ =5²+5³+5⁵+5⁶ (4^th row of

e

The construction relates a_i with S_i and c with S.

It is not difficult to see

a₁ +a₃ =c ⇔⇔⇔⇔ S₁ ∪∪∪∪S₃ =S and S₁ ∩∩∩∩S₃ =∅∅. ∅∅

Ex. Partition

Instance: A multiset B={b₁, b₂, …, b_n} of positive integers.

Question: Is there a subset B’⊆⊆⊆⊆B such that

i

b B' i

∑

∈

∈

∈

∈

=

j

b B B' j

b

−

∑

∈

∈∈

∈

?

For example, when B={17, 53, 9, 35, 41, 32, 35}, then the answer is yes (B’={17, 53, 41}).

NP-completeness of Partition is shown below.

♣

♣

♣

Let A={a₁, a₂, …, a_m} and c be an arbitrary instance of Sum of Subsets.

An instance of Partition can be obtained in polynomial time as follows:

B = A ∪∪∪∪ {a_m+1, a_m+2},

where a_m+1 =c+1 and a_m+2 =1−−−−c+

i

a A i

∑

∈

∈∈

∈

.

Since a_m+1 +a_m+2 =

i

a A i

∑

∈

∈∈

∈

+2, we have {a_m+1, a_m+2}⊄⊄⊄⊄B’ and {a_m+1, a_m+2}⊄⊄⊄⊄B−−−−B’.

We show below that

i

a Α' i

∑

∈

∈

∈

∈

= c if and only if a +

∑

∑

(⇒⇒⇒⇒) Suppose

i

a Α' i

∑

∈

∈

∈

∈

= c.

a_m+2 +

i

a Α' i

∑

∈

∈∈

∈

= (1−−−−c+

i

a A i

∑

∈

∈

∈

∈

)+

i

a Α' i

∑

∈

∈

∈

∈

= 1+

i

a Α i

∑

∈

∈∈

∈

.

a_m+1 +

i

a Α Α' i

a

−

∑

∈

∈

∈

∈

= (c+1)+

i

a Α Α' i

a

−

∑

∈

∈

∈

∈

= 1+

i

a Α i

∑

∈

∈

∈

∈

.

(⇐⇐⇐⇐) Suppose a_m+2 +

i

a Α' i

∑

∈

∈∈

∈

= a_m+1 +

i

a Α Α' i

a

−

∑

∈

∈∈

∈

, i.e., B’=A’+{a_m+2}.

Then, (1−−−−c+

i

a A i

∑

∈

∈

∈

∈

)+

i

a Α' i

∑

∈

∈∈

∈

= (c+1)+

i

a Α Α' i

a

−

∑

∈

∈

∈

∈

,

from which

∑

Exercise 5. Read Example 8-14 on page 367 of the textbook.

(1) Give a reduction from Partition to the bin packing problem.

(2) Illustrate the reduction by an example.

(3) Verify the reduction.

Exercise 6. Read Theorem 11.2 on page 518 of Ref. (2).

(1) Give a reduction from Satisfiability to Clique.

(2) Illustrate the reduction by an example.

(3) Verify the reduction.

• •

• • Three Proof Techniques

Restriction

Local Replacement Component Design

• •

• • Restriction

If a problem ΠΠΠΠ contains an NP-hard problem ΠΠΠ’ Π as a special case (i.e., ΠΠΠΠ’ is a restricted subproblem of ΠΠΠΠ), then ΠΠΠΠ is NP-hard.

Ex. Exact Cover

Instance: A finite set S and k subsets S₁, S₂, …, S_k of S.

Question: Is there a subset of {S₁, S₂, …, S_k} that forms a partition of S?

Exact Cover by 3-Sets

Instance: A finite set S with |S|=3p and k 3-

Exact Cover by 3-Sets is a special case of Exact Cover.

3-Dimensional Matching

Instance: A set M⊆⊆⊆⊆W××××X××××Y, where W, X and Y are three disjoint q-element subsets.

Question: Does M contain a matching, i.e., a subset M’⊆⊆⊆⊆M such that |M’|=q and no two elements of M’ agree in any coordinate?

For example, if W={0, 1}, X={a, b}, Y={+++, −+ −−−}, and M={(0, a, ++++), (1, b, +++),+ (1, b, −−−−)}, then the answer is yes (M’={(0, a, +++), (1, b, −+ −−−)}).

3-Dimensional Matching is a special case of Exact Cover by 3-Sets.

For example, the following instance of 3-Dimensional Matching:

W={0, 1}, X={a, b}, Y={++++, −−−}, −

and M={(0, a, ++++), (1, b, +++),+ (1, b, −−−−)}

can be transformed into an instance of Exact Cover by 3-Sets as follows:

S₁ ={0_W, a_X, ++++_Y}, S₂ ={1_W, b_X, ++++_Y}, S₃ ={1_W, b_X, −−−−_Y}, and S=W∪∪∪∪X∪∪∪∪Y={0_W, 1_W, a_X, b_X, ++++_Y, −−−−_Y}.

Therefore,

3-Dimensional Matching is NP-complete.

⇒

⇒⇒

⇒ Exact Cover by 3-Sets is NP-complete.

⇒

⇒⇒

⇒ Exact Cover is NP-complete.

Ex. Hamiltonian Cycle

Instance: An undirected graph G=(V, E).

Question: Does G contain a Hamiltonian Cycle, i.e., an ordering (v₁, v₂, …, v_|V|) of the vertices of G such that (v₁, v_|V|)∈∈∈∈

E and (v_i, v_i+1)∈∈∈∈E for all 1≤≤≤≤i<|V|?

Directed Hamiltonian Cycle

Instance: A directed graph G=(V, A), where A is a set of arcs (i.e., ordered pairs of vertices).

Question: Does G contain a directed Hamiltonian cycle, i.e., an ordering (v₁, v₂, …, v_|V|) of the vertices of G such that (v₁, v_|V|)∈∈∈∈ A and (v_i, v_i+1)∈∈∈∈A for all 1≤≤≤≤i<|V|?

Hamiltonian Cycle is a special case of Directed Hamiltonian Cycle (or Hamiltonian Cycle ∝∝∝∝

Directed Hamiltonian Cycle, where each (u, v)∈∈∈∈E corresponds to two arcs (u, v), (v, u)∈∈∈∈A).

Therefore,

Hamiltonian Cycle is NP-complete.

⇒

⇒⇒

⇒ Directed Hamiltonian Cycle is NP-complete.

Hamiltonian Path between Two Vertices

Instance: An undirected graph G=(V, E) and two distinct vertices u, v∈∈∈∈V.

Question: Does G contain a Hamiltonian path starting at u and ending at v, i.e., an ordering (v₁, v₂, …, v_|V|) of the vertices of G such that u=v₁, v=v_|V|, and (v_i, v_i+1)∈∈∈∈E for all 1≤≤≤≤i<|V|?

Hamiltonian Cycle ∝∝∝∝ Hamiltonian Path between Two Vertices:

if the latter is polynomial time solvable,

then the former is also polynomial time solvable (considering all edges (u, v)∈∈∈∈V for the latter).

⇒⇒⇒⇒ Hamiltonian Path between Two Vertices is

Hamiltonian Path

Instance: An undirected graph G=(V, E).

Question: Does G contain a Hamiltonian path?

Hamiltonian Cycle ∝∝∝∝ Hamiltonian Path:

For each (u, v)∈∈∈∈E, construct an instance of Hamiltonian Path by adding x, y to V and

(x, u), (y, v) to E (thus G’ is induced).

u v u v

x y

G G’

G has a Hamiltonian cycle if and only if G’ has a Hamiltonian x-y path.

Exercise 7. Show the following two problems NP- complete by restriction to Hamiltonian Path and Partition, respectively.

Bounded Degree Spanning Tree

Instance: An undirected graph G=(V, E) and a positive integer k≤≤≤≤|V|−−−−1.

Question: Does G contains a spanning tree in which each node has degree at most k?

0/1 Knapsack

Instance: A finite set U, a “size” s(u)∈∈∈∈Z⁺ and a “value” v(u)∈∈∈∈Z⁺ for each u∈∈∈∈U, a size constraint b∈∈∈∈Z⁺, and a value goal k∈∈∈∈Z⁺.

Question: Is there a subset U’⊆⊆⊆⊆U such that

• •

• • Local Replacement

In order to show ΠΠΠΠ’ ∝∝∝∝ ΠΠΠΠ, local replacement specifies the “basic units” for ΠΠΠΠ’ and replaces them with

others, while constructing a corresponding instance of ΠΠ. ΠΠ

Usually, local replacement has one kind of basic units that each are replaced with the same structure.

Ex. Partition into Triangles

Instance: An undirected graph G=(V, E) with

|V|=3p for some integer p>0.

Question: Is there a partition of V into 3-vertex subsets V₁, V₂, …, V_p, such that each

For example, the answer for the following instance is yes, because V can be partitioned into {1, 2, 3}, {4, 5, 6}, {7, 8, 9} or {1, 4, 7}, {2, 5, 6}, {3, 8, 9}.

1

2 3

4

5 6

7

8 9

Exact Cover by 3-Sets ∝∝∝∝ Partition into Triangles is shown below.

Let a set S, where |S|=3p, and a collection C of 3-element subsets of S denote an arbitrary instance of Exact Cover by 3-Sets.

Construct an instance of Partition into Triangles as follows.

Consider each subset {x_i, y_i, z_i}∈∈∈∈C a basic unit, and replace it with the following structure.

ai[3]

ai[1] ai[2]

ai[4] ai[5]

ai[6]

ai[7] ai[8]

ai[9]

xi yi zi

For example, if S={1, 2, 3, 4, 5, 6} and C= {{1, 4, 6}, {2, 4, 6}, {2, 3, 5}}, then an instance of Partition into Triangles is obtained as follows.

3 5

It is not difficult to check that if Exact Cover by 3-Sets has an answer yes (e.g., {{1, 4, 6}, {2, 3, 5}}

is a partition of S), then Partition into Triangles has an answer yes (the triangles are shown with bold edges).

Also, if Partition into Triangles has an answer yes, then Exact Cover by 3-Sets has an answer yes.

Exercise 8. Read Example 8-9 on page 353 of the textbook.

(1) Give a reduction from Satisfiability to 3-Satisfiability.

(2) Illustrate the reduction by an example.

(3) Verify the reduction.

Sometimes, additional structures are required, while using the technique of local replacement.

Ex. Sequencing within Intervals

Instance: A finite set T of “tasks” and for each t∈∈∈∈T, a “release time”

r(t)∈∈∈∈Z⁺ ∪∪∪∪{0}, a “deadline”

d(t)∈∈∈∈Z⁺, and a “length” l(t)∈∈∈∈Z⁺. Question: Does there exist a feasible schedule for T, i.e., a function f:T→→→→Z⁺ such that for each t∈∈∈∈T, f(t)≥≥≥≥r(t),

f(t)+l(t)≤≤≤≤d(t), and f(t’)+l(t’)≤≤≤≤f(t) or f(t)+l(t)≤≤≤≤f(t’) for each t’∈∈∈∈T−−−−{t}? (It means that the task t, which is “executed”

Partition ∝∝∝∝ Sequencing within Intervals is shown below.

An arbitrary instance of Partition:

a multiset B={b₁, b₂, …, b_n} of positive integers.

Consider each b_i (1≤≤≤≤i≤≤≤≤n) a basic unit, and let

m = _i

i n

b

≤ ≤

∑

1

.

Construct an instance of Sequencing within Intervals as follow:

each b_i corresponds to a task t_i with r(t_i)=0, d(t_i)=m+1, and l(t_i)=b_i.

An additional structure:

a task t% with r t%( )= m/2, d t%( )= (m+1)/2, and l t%( )=1.

⇒

⇒

⇒⇒ m should be even

⇒⇒⇒

⇒ r t%( )= m/2, d t%( )= (m/2)+1

⇒

⇒⇒

⇒ f t%( ) must be m/2.

⇒ Partition has the answer yes if and only if Sequencing within Intervals has the answer yes.

2

m 1

2 m+

0 m+1

t%

2 m

2 m

Time

• •

• • Component Design

While showing ΠΠΠΠ’ ∝∝∝∝ ΠΠΠΠ, component design is similar to local replacement in replacing the structures (i.e., basic units) of ΠΠ’ΠΠ with other structures, in order to obtain an instance of Π

Π ΠΠ.

Usually, component design adopts multiple kinds of basic units, and different basic units are replaced with different structures.

Ex. Vertex Cover

Instance: An undirected graph G=(V, E) and a positive integer k≤≤≤≤|V|.

Question: Does G contain a vertex cover of size at most k, i.e., a subset V’⊆⊆⊆⊆V such that |V’|≤≤≤≤k and for each (u, v)∈∈∈∈E,

For example, {1, 3}, {1, 2, 3} and {1, 2, 4, 5} are three vertex covers of the following graph. If k≥≥≥≥2, the answer is yes. If k=1, the answer is no.

We show below 3-Satisfiability ∝∝∝∝ Vertex Cover.

3-Satisfiability

Instance: A set U of variables and a collection C={c₁, c₂, …, c_m} of clauses over U, where each clause of C contains three literals.

For example, when U={x₁, x₂, x₃} and C= {x₁ ∨∨∨∨x₂ ∨∨∨∨x₃, x

1 ∨∨∨∨x₂ ∨∨∨∨ x

3, x₁ ∨∨∨∨ x

2 ∨∨∨∨x₃}, the answer is yes, becausethe assignment of U: x₁ ←←←←F, x₂ ←←←←F, and x₃ ←←←←T, can satisfy C (i.e., (x₁ ∨∨∨∨x₂ ∨∨∨∨x₃)∧∧∧∧(x

1 ∨∨∨∨x₂ ∨∨∨∨ x

3)∧∧∧∧(x₁ ∨∨∨∨ x

2 ∨∨∨∨x₃) = T).

Let U={u₁, u₂, …, u_n} and C={c₁, c₂, …, c_m} be an arbitrary instance of 3-Satisfiability.

♦ For each u_i ∈∈∈∈U, construct a component T_i = (V_i, E_i), where V_i ={u_i,

ui } and E_i ={(u_i,

ui)}.

♦ For each c_j ∈∈∈∈C, construct a component S_j = (V’_j, E’_j), where V’_j ={a₁[j], a₂[j], a₃[j]} and E’_j ={(a₁[j], a₂[j]), (a₁[j], a₃[j]), (a₂[j], a₃[j])}.

♦ For each c_j ∈∈∈∈C, construct an edge set E’’_j = {(a₁[j], x_j), (a₂[j], y_j), (a₃[j], z_j)}, where x_j, y_j and z_j are the three literals in c_j.

An instance of Vertex Cover can be constructed as G=(V, E) and k=n+2m, where

V = ( ⁿ

i Vi

=1

U ) ∪∪∪∪ ( ^m

j V'j

=1

U ) and

E = ( ⁿ

i Ei

=1

U ) ∪∪∪∪ ( ^m

j E'j

=1

U ) ∪∪∪ (∪ ^m

j E''j

=1

U ).

For example, if U={u₁, u₂, u₃, u₄} and C= {u₁ ∨∨∨∨ u

3 ∨∨∨∨u

4, u

1 ∨∨∨∨u₂ ∨∨∨∨ u

4}, then the following instance of Vertex Cover is constructed, where k=8.

•••

• Each edge in E’’_j represents a satisfying truth assignment for c_j.

For example, (u₁, a₁[1])∈∈∈∈E’’₁ implies that u₁ ←←←←T can satisfy c₁.

•••

• Any vertex cover V’⊆⊆⊆⊆V of G contains at least one from {u_i,

ui } and at least two from {a₁[j], a₂[j], a₃[j]}.

⇒

⇒

⇒⇒ |V’| ≥≥≥≥ n+2m=k

As explained below, C is satisfiable if and only if G has a vertex cover V’⊆⊆⊆⊆V with |V’|≤≤≤≤k.

♣♣♣

♣ C is satisfiable ⇒⇒⇒ V’⇒ ⊆⊆⊆⊆V with |V’|≤≤≤≤k exists

Consider the example above, where u₁ ←←←←T, u₂ ←←←←T, u

3 ←←←←T, and u

4 ←←←←T can satisfy C.

⇒ include u₁, u₂, u

3, u

4 in V’

In order to make V’ a vertex cover, V’ must be augmented with two vertices from each set

{a₁[j], a₂[j], a₃[j]}, while covering all edges in E’’_j.

⇒

⇒⇒

⇒ augment V’ with any two from {a₁[1], a₂[1], a₃[1]} and a₁[2], a₂[2] (or a₁[2], a₃[2]) from {a₁[2], a₂[2], a₃[2]}

(a₁[2] must be included in V’, in order to cover the edge (u , a₁[2]))

♣♣♣

♣ V’⊆⊆⊆⊆V with |V’|≤≤≤≤k exists ⇒⇒⇒ C is satisfiable ⇒

V’ contains exactly k=n+2m vertices: one for each {u_i,

ui } and two for each {a₁[j], a₂[j], a₃[j]}.

Consider the example above, where k=8 and V’={u

1, u₂, u

3, u₄, a₁[1], a₃[1], a₁[2], a₃[2]} is a vertex cover.

⇒⇒⇒

⇒ u

1 ←←←←T, u₂ ←←←←T, u

3 ←←←←T and u₄ ←←←←T can satisfy C (u

1, u₂, u

3, u₄ ∈∈∈ V’) ∈

Since two (e.g., a₁[1] and a₃[1]) from {a₁[j], a₂[j], a₃[j]} are included in V’, the other (e.g., a₂[1]) must be connected to u_i or

ui (e.g., u

3) that is included in V’.

⇒

⇒⇒

⇒ each c_j is satisfiable.

Ex. Minimum Tardiness Sequencing

Instance: A finite set T of “tasks”, where each t∈∈∈∈T has “length” 1 and “deadline”

d(t)∈∈∈∈Z⁺, a partial order p on T, and a non-negative integer r≤≤≤≤|T|.

Question: Is there a “schedule” f: T→→→→{0, 1, …,

|T|−−−−1} such that f(t)≠≠≠≠f(t’) if t≠≠≠≠t’, f(t)<f(t’) if tp t’, and

|{t∈∈∈∈T: f(t)+1>d(t)}| ≤≤≤≤ r?

A task t∈∈∈∈T is tardy, if f(t)+1>d(t).

The schedule f is required not to cause more than r tasks tardy.

Clique

Instance: An undirected graph G=(V, E) and a positive integer k≤≤≤≤|V|.

Question: Does there exist a subset V’⊆⊆⊆⊆V such that |V’|≥≥≥≥k and every two vertices of V’ are adjacent in G?

Let G=(V, E) and k≤≤≤≤|V| be an arbitrary instance of Clique.

An instance of Minimum Tardiness Sequencing can be constructed as follows.

T = V∪∪∪∪E;

r = |E|−−−−k(k−−−−1)/2;

vp e ⇔⇔⇔⇔ v∈∈∈∈V, e∈∈∈∈E, and v is an endpoint of e;

d(v) = |V|+|E| for v∈∈∈∈V, and