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NP-completeness. Class of hard problems. Outline. Introduction Problems and Languages Turing machines and algorithms P and NP Satisfiability problem Reduction Other NP-complete problems. Problems. A problem can be defined as a mapping from the instance to the answer. Examples:

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### NP-completeness

### Reduction

### Ground Reduction

Class of hard problems

Outline

- Introduction
- Problems and Languages
- Turing machines and algorithms
- P and NP

- Satisfiability problem
- Reduction
- Other NP-complete problems

NP-complete Problems

Problems

- A problem can be defined as a mapping from the instance to the answer.
- Examples:
- Shortest path problem
{<graph G, node u, node v>} {a shortest path P}

- 0/1 Knapsack problem
{<set S of objects, capacity C>} {the maximum value}

- Closest-pair-of-points problem
{<set P of points>} {<point q, point r>, which is the closest pair of points in P}.

- Satisfiable problem
{<boolean formula>} {T, F}

- Shortest path problem

NP-complete Problems

Decision Problems

- A decision problem is a problem whose answer is either “yes” or “no”.
- Example: Satisfiable problem
{<boolean formula>} {T, F}

- Any problem can be described in term of a decision problem.

NP-complete Problems

Reformulate Problems as Decision Problems

- Shortest path problem
- {<graph G, node u, node v, int k>} {T, F}
- The answer is T iff there is no path between u and v, which is shorter than k.

- 0/1 Knapsack problem
- {<set S of objects, capacity C, value V>} {T, F}
- The answer is T iff there is no selection of objects fitted in the sack which yields higher value than V.

NP-complete Problems

Encoding of Decision Problems

- An encoding of a problem is a mapping from the set of instances of the problem to a set of strings.
- Example: shortest path
- The encoding of <graph G, node u, node v, path p> in the Shortest path problem is a string which is used to describe G, u, v, and p.
- G can be described by a string of its adjacency matrix.
- u and v can be described by two strings, indicating the node labels.
- p can be described by a sequence of node in the path.

- The encoding of <graph G, node u, node v, path p> in the Shortest path problem is a string which is used to describe G, u, v, and p.

NP-complete Problems

Abstract/ Concrete problems

A set of instances

Mapping

(encoding)

Abstract

problem

Mapping

(encoding)

A set of solutions

Concrete

problem

NP-complete Problems

Encoding of Problems

- Is X a prime ?
f:{X |X>0 and is a binary number}{Y,N}

f(x) = Y if X is prime

f(x) = N otherwise

- Is a 3-CNF expression X satisfiable?
f:{X | X is a 3-CNF expression} {Y, N}

f(x) = Y if X is satisfiable

f(x) = N otherwise

NP-complete Problems

Why encoding is important?

- Encoding determines the instance size.
- Example: an integer n
- n encoded as one number
=> instance size = 1

- n encoded as binary number
=> instance size = log2 n

- n encoded as unary number
=> instance size = n

- n encoded as one number

NP-complete Problems

Languages and Decision Problems

- Let be an alphabet.
- A language over is a set of strings created from symbols in .
- B={0,1}* is the set of all binary strings.
- { is a palindrome} is a language over B.

- A yes-instance (or no-instance) is an instance whose answer is “yes” (or “no”).
- An instance must be either a yes- or a no-instance.

NP-complete Problems

Languages and Decision Problems

- A decision problem can be encoded in a language {| is an encoding of a yes- instance}.
- The shortest path problem can be encoded as {e(G, u, v, p>)| p is the shortest path between u and v in the graph G}.

NP-complete Problems

Turing machines (TM)

- A Turing machine is a machine which takes a string on its input tape, and decides whether to accept or to reject the string.
- A Turing machine contains
- A finite-state control with
- A start state
- A halt state
- Transition function: state x symbol state x symbol x {R,L}

- A tape with a Read/Write headwhich can move left and right.

- A finite-state control with

NP-complete Problems

Deterministic/ Nondeterministic Turing machines

- Difference
- A transition is deterministic or nondeterministic.

- What is nondeterminism?
- Given a state of a machine/program, there is more than one possible next step.

- Examples
- Deterministic: functions.
- Nondeterministic: relations.

NP-complete Problems

Example of a DTM

s

/@,R

/,L

1/,R

p1

q1

/,R

0/,R

@/,R

0/0,L

1/1,L

0/0,R

1/1,R

p2

p4

q2

0/,L

/,L

/1,L

p3

h

NP-complete Problems

How a DTM works

s

/,L

/@,R

1/,R

p1

q1

@

1

0

0

0

1

0

0

0

/,R

0/,R

@/,R

0/0,L

1/1,L

0/0,R

1/1,R

p2

p4

q2

/1,L

/,L

0/,L

On the input 0001000,

the TM halts.

h

p3

NP-complete Problems

How a DTM works

s

/,L

/@,R

1/,R

p1

q1

@

0

0

0

0

0

0

0

/,R

0/,R

@/,R

0/0,L

1/1,L

0/0,R

1/1,R

p2

p4

q2

/1,L

On the input 0000000,

the TM hangs.

/,L

0/,L

h

p3

NP-complete Problems

Example of NTM

- Let L={ww| w{0,1}*}

r0

@/,R

0/@,R

0/0,R

1/1,R

0/0,L

1/1,L

/,L

s

q0

t0

/@,R

0/0,L

1/1,L

0/,L

0/,L

/,R

/@,L

@/,L

v

p

u

h

/,R

/,R

1/,L

1/,L

0/0,R

1/1,R

q1

t1

0/0,L

1/1,L

/,L

@/,R

1/@,R

0/0,R

1/1,R

r1

NP-complete Problems

Accept/Decide

- Let T be a Turing machine.
- For a string over ,
- T accepts iff T halts on with output “1”.
- T rejects iff T halts on with output “0”.
- T hangs on iff T dose not halts on .

- For a language L over ,
- T accepts L iff, for any string in L, T accepts .
- T decides iff
- for any string in L, T accepts , and
- for any string inL, T rejects .

NP-complete Problems

Time complexity

- A language L is accepted/decided in O(f(n)) by a Turing machine T if, for any length-n string in L, is accepted/decided in k·f(n) by a Turing machine T, for some constant k.
- A language L is accepted/decided in polynomial time by a Turing machine T if, L is accepted/decided in O(nk) by a Turing machine T, for some constant k.

NP-complete Problems

Church-Turing Thesis

- If there is a Turing machine deciding a problem in O(f), then there is an algorithm solving in O(f).
- Turing machine algorithm

NP-complete Problems

Class of P

- If is a problem in P, given any instance I of ,
- there is an algorithm which solves I in O(nk), or
- there is a Turing machine which accepts e(I) in O(nk) steps,
where n is the size of e(I) and k is a constant.

- What does this mean?
- If is in P, is easy because it can be solved in polynomial time.

NP-complete Problems

Class of NP

- If is a problem in P, given any instance I and any certificate C of ,
- there is an algorithm which verifies, in O(nk), that C is the answer of I, for any C and I.
- there is a Turing machine which verifies, in O(nk), that C is the answer of I, for any C and I.
- there is a nondeterministic Turing machine which accepts e(I), in O(nk), for any I.
where n is the size of e(I) and k is a constant.

What does this mean?

- If is in NP, is not necessarily easy because it can only be verified in polynomial time.

NP-complete Problems

Can be solved in polynomial time

Prove by showing the algorithm.

Cannot be solved in polynomial time

Prove by showing that there is no polynomial-time algorithm.

Can verify the answer in polynomial time.

Can usually be solved by an exponential-time algorithm.

Brute-force algorithm: try every possible answers.

Easy problems / Hard problemsNP-complete Problems

Is X more difficult than Y?

Reduction

- Let L1 and L2 be languages over 1 and 2, respectively. L1 is (polynomial-time) reducible to L2, denoted by L1L2, if there is a TM M computing a function f: 1*2* such that wL1 f(w)L2 in polynomial time.
- Let P1 and P2 be problems. P1 is (polynomial-time) reducible to P2, denoted by P1P2, if there is an algorithm computing a function f: 1*2* such that w is a yes-instance of P1 f(w) is a yes-instance of P2 in polynomial time.

NP-complete Problems

Meaning of Reduction

P1 is reducible to P2

if DTM computing, in polynomial time, a function f: 1*2* such that w is a yes-instance of P1 f(w) is a yes-instance of P2.

- If you can map yes-instances of problem A to yes-instances of problem B, then
- we can solve A in polynomial time if we can solve B in polynomial time.
- it doesn’t mean we can solve B in polynomial time if we can solve A in polynomial time.

NP-complete Problems

Satisfiability problem (SAT)

- Given a Boolean expression , is satisfiable?
- (A~BC)(~A~C)B~C is not satisfiable.

NP-complete Problems

3CNF-SAT Problem

- Given an expression in 3CNF, is satisfiable?
- Given an expression in 3CNF, is there a truth assignment of propositions in which makes true?
- Example:
- =(PQR)(PQR)(PSR)(PSR)
- is satisfiable when
- P=t, Q=f, R=f, S=t or
- P=f, Q=f, R=f, S=t

NP-complete Problems

Clique or Complete Subgraph

- A graph G=(V,E) has a clique if there is W V such that for all nodes a and b in W, there is an edge between a and b.

C

B

F

A

E

D

NP-complete Problems

Clique Problem

- Given a graphG=(V,E) and an integer k, is there a clique of size k in G?
- Example: Is there a clique of size 3, 4, or 5, in the graph below?

C

B

F

A

E

D

NP-complete Problems

3CNF-SAT Clique

Proof:

To prove that 3CNF-SAT Clique,

- Find a function f such that
- a 3CNF expression is satisfiable
a graph G has a clique of size k, where <G,k>=f() and

- there is a TM computing f in polynomial time.
f can be defined as follows:

Let = (111213)(212223)…(n1n2n3)

f() = <G,n>, where G=(V, E),

- V={ij|0<i<n+1, 0<j<4}, and
- E={(ij, km)| i≠k and ij≠~km}

- a 3CNF expression is satisfiable

NP-complete Problems

Building a graph from a 3CNF exp.

(PQR)(PSR)(PSR)

~Q

P

R

~P

P

~S

S

~R

~R

NP-complete Problems

3CNF-SAT Clique (cont’d)

- Show is satisfiableG has clique of size n
- Let G has a clique of size n.
- From the definition of G, there is no edge between vertices
- in the same disjunctive clause, and
- representing a literal and its negation.

- A link between two nodes means that the two literals can be true at the same time, and they are in different disjunctive clauses.

NP-complete Problems

Satisfiability and Clique

(PQR)(PQR)(PQR)(PQR)

Q

P

R

~P

P

Q

Q

R

~R

~P

~R

Q

NP-complete Problems

Satisfiability and Clique

(PQR)(PQR)(PQR)(PQR)

Q

P

R

~P

P

Q

Q

R

~R

~P

~R

Q

NP-complete Problems

Reflexivity of reduction

Theorem:

Let L be a language over . L L.

Proof:

Let L be a language over .

Let f be an identity function from **.

Then, there is a TM computing f.

Because f is an identity function, wL f(w)=wL.

By the definition, L L.

NP-complete Problems

Property of reduction

Theorem:

Let L1 and L2 be languages over .

If L1L2, thenL1L2.

Proof:

Let L1 and L2 be languages over .

Because L1L2, there are a function f such that wL1 f(w)L2, and a TM T computing f in polynomial time.

wL1 f(w)L2.

By the definition,L1L2.

NP-complete Problems

Transitivity of reduction

Theorem: Let L1, L2 and L3 be languages over .

If L1 L2 and L2 L3, then L1 L3.

Proof: Let L1, L2 and L3 be languages over .

There is a function f such that wL1 f(w)L2, and a TM T1 computing f in polynomial time because L1 L2.

There is a function g such that wL2 g(w)L3, and a TM T2 computing g in polynomial time because L2 L3.

wL1 f(w)L2 g(f(w))L3, and T1T2 computes g(f(w)) in polynomial time .

By the definition, L1 L3.

NP-complete Problems

Using reduction to prove P/NP

Theorem: If L2 is in P/NP, and L1L2, then L1 is also P/NP.

Proof:

Let L1 and L2 be languages over , L1L2, and L2 be in P/NP.

Because L2 is in P/NP, there is a DTM/NTM T2 accepting L2 in polynomial time.

Because L1L2, there is a DTM T1 computing a function f such that wL1 f(w)L2 in polynomial time.

Construct a DTM/NTM T=T1T2. We show that T accepts L1in polynomial time.

- If wL1, T1 in T computes f(w)L2 and T2 in T accepts f(w), in polynomial time.
- If wL1, T1 in T computes f(w) L2 and T2 in T does not accept f(w) in polynomial time.
Thus, L1 is also in P/NP.

NP-complete Problems

Using reduction to prove co-NP.

Theorem:

If L2 is in co-NP, and L1L2, then L1 is also in co-NP.

Proof:

Let L1 and L2 be languages over , L1L2, and L2 be in co-NP.

Because L2 is in co-NP,L2 is in NP.

Because L1L2,L1L2. Then,L1 is in NP.

Thus, L1 is co-NP.

NP-complete Problems

NP-completeness

- A language (or problem) L is NP-hard if, for any language L' in NP, L' ≤ L.
- A language (or problem) L is NP-complete if it is in NP and is NP-hard.

NP-complete Problems

Why interested in NP-complete problem

- Implication on the problem “is P=NP ?”
- If an NP-complete problem is in P, then P=NP.
- Why?
- If L is NP-complete, any language in NP is reducible to L.
- Since L is in P, any language in NP is reducible to a language in P.
- Then, any language in NP is in P.

NP-complete Problems

Lemma

If L is a language such that L’ ≤ L for some L’ ∈ NPC, then L is NP-hard.

Moreover, if L ∈ NP, then L ∈ NPC.

Proof

Since L’ is NP-complete, for all L’’ ∈ NP, we have L’’ ≤ L’ .

BecauseL’ ≤ L, L’’ ≤ L by transitivity.

Then,L is NP-hard.

If L ∈ NP, we also have L ∈ NPC.

NP-complete Problems

SAT is NP-complete

SAT is NP-complete

- To prove that SAT is NP-complete:
- Prove SAT is in NP
- Prove any language in NP is reducible to SAT.

- SAT is in NP: Informal proof
- Given a certificate which is a truth assignment of propositions which makes the expression true, there is an algorithm that can verify the answer in polynomial time.

NP-complete Problems

SAT is NP-complete (cont’d)

Prove any language in NP is reducible to SAT.

Proof:

Let L be any language in NP.

Then, there is an NTM T which accepts L in polynomial time p(n), where n is the input size.

To show that L ≤ SAT, we find a function f such that x L f(x) SAT.

That is, to find a function f such that x L f(x) is satisfiable.

NP-complete Problems

SAT is NP-complete (cont’d)

Proof (cont’d):

That is, to find a function f such that T accepts x f (x) is satisfiable.

Let f (x)=g1(x)g2(x)g3(x)g4(x)g5(x)g6(x)g7(x).

f (x) describes the sequence of configuration changes which leads to the halt state of T.

f(x) contains propositions Qi,j, Hi,k, and Si,k,j .

Qi,j: T is in state qj after i moves.

Hi,k: T’s tape head is on square k after i moves.

Si,k,j: Symbol jis on square k after i moves.

NP-complete Problems

SAT is NP-complete (cont’d)

Proof (cont’d):

g1(x): the initial configuration of T at the beginning.

g2(x): accepting configuration of T after N steps.

g3(x): possible transitions of T.

g4(x): hanging configuration.

g5(x): describe that T is in exactly one state at a time.

g6(x): describe that each square of the tape contains exactly one symbol.

g7(x): describe that change can only made on the tape square on which the tape head is located.

NP-complete Problems

SAT is NP-complete (cont’d)

Proof (cont’d):

g1(x): describe the initial configuration of T at the beginning.

n N

Q0,0 H0,0 S0,0,0 S0,k, ik S0,k, 0

k=1k=n+1

∆ 0 1 1 0 1 ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆

q0

NP-complete Problems

SAT is NP-complete (cont’d)

Proof (cont’d):

g2(x): describe accepting configuration of T after N steps.

QN,a

ha

NP-complete Problems

SAT is NP-complete (cont’d)

Proof (cont’d):

g3(x): describe possible transitions of T.

For each transition

δ(qj, l)= (qt1, z1, D1), δ(qj, l)= (qt2, z2, D2)

(Qi,j Hi,k Si,k,l)((Qi+1,t1 Hi+1,d1(k) Si+1,k,z1)

V (Qi+1,t2 Hi+1,d2(k) Si+1,k,z2))

NP-complete Problems

SAT is NP-complete (cont’d)

Proof (cont’d):

g4(x): describe hanging configuration.

For each transitionδ(qj, l)= ø

(Qi,j Hi,k Si,k,l) (Qi+1,j Hi+1,k Si+1,k,l)

NP-complete Problems

SAT is NP-complete (cont’d)

Proof (cont’d):

g5(x): T is in exactly one state at a time.

For each time i and states qj0, qj1, qj2, …, qj0≠ qj1 ≠ qj2 ≠ … :

(Qi,j0 v Qi,j1 v Qi,j2 v …)

- At least one state
(~Qi,j0v~Qi,j1)(~Qi,j1v~Qi,j2)(~Qi,j0v~Qi,j2)…

- Never two state at the same time

NP-complete Problems

SAT is NP-complete (cont’d)

Proof (cont’d):

g6(x): state that each square of the tape contains exactly one symbol.

For each time i,square k,and symbols 0, 1, 2, …, 0≠ 1 ≠ 2 ≠ … :

(Si,k,0 v Si,k,1 v Si,k,2 v …)

- At least one symbol in a cell
(~Si,k,0v~Si,k,1) (~Si,k,1v~Si,k,2) (~Si,k,0v~Si,k,2) …

- Never two symbols in a cell at the same time

NP-complete Problems

SAT is NP-complete (cont’d)

Proof (cont’d):

g7(x): state that change can only made on the tape square on which the tape head is located.

For each i, k, and l :

(~Hi,kSi,k,l) (Si+1,k,l)

Hi,k v ~Si,k,l v Si+1,k,l

NP-complete Problems

SAT is NP-complete (cont’d)

Proof (cont’d):

From the definition of f, f(x) is true iff T accepts x.

f(x) can be computed in polynomial time.

Thus, SAT is NP-complete.

NP-complete Problems

3CNF-SAT is NP-complete

First, show 3CNF-SAT is in NP.

Let be a 3CNF expression,and

truth assignments C be a certificate for .

There is an algorithm to verify that is true under the assignments C, by substituting the value of each literal and evaluating the expression.

This algorithm takes polynomial time of the length of .

NP-complete Problems

3CNF-SAT is NP-complete

Second, show SAT ≤ 3CNF-SAT.

Let be a Boolean expression.

We will construct a 3CNF expression f() such that f() is satisfiable iff is satisfiable.

To createf():

(A) Create an abstract syntax tree of so that internal nodes are logical operators ,v, ~, ,or , and leaf nodes are literals (x or ~x).

Then, create , from the abstract syntax tree which is satisfiable iff is satisfiable.

NP-complete Problems

3CNF-SAT is NP-complete

:y1

:y2

v:y3

:y4

p

q

~s

v:y5

:y6

~s

r

q

~r

= ((p q) (((~s v r) (q ~r)) v ~s)

= y1 (y1 (y2 y3))

(y2 (p q ))

(y3 (y4 v~s))

(y4 (y5 y6))

(y5 (~s vr ))

(y6 (q ~r))

is satisfiable iff is satisfiable.

NP-complete Problems

3CNF-SAT is NP-complete

(B) Convert into CNF.

P Q is transformed to (P Q) (Q P).

P Q is transformed to ~P v Q.

~(P v Q) is transformed to ~P ~Q.

~(P Q) is transformed to ~P v ~Q.

= (y1 (y2 y3)) =(y1 (y2 y3)) ((y2 y3) y1)

=(~y1 v (y2 y3)) (~y2 v ~y3 v y1)

= (~y1 v y2) (~y1 v y3) (y1 v ~y2 v ~y3)

= (~y1 v y2 v y3) (~y1 v y2 v ~y3) (~y1 v y2 v y3)

(~y1 v ~y2 v y3) (y1 v ~y2 v ~y3)

NP-complete Problems

3CNF-SAT is NP-complete

f() = , and f() is satisfiable iff is satisfiable.

Next, we need to show that f is computable in polynomial time.

In step (A):

the abstract syntax tree can be created in polynomial time of the length of ,

the expression can be created also in polynomial time of the number of node in the tree and the length of , which is also a polynomial time of the length of .

In step (B):

can be created, using Boolean logic identities, in polynomial time of the length of , which is also a polynomial time of the length of .

That is, SAT ≤ 3CNF-SAT.

NP-complete Problems

CLIQUE is NP-complete

Proof:

3CNF-SAT is NP-complete.

3CNF-SAT ≤ CLIQUE.

CLIQUEis NP-complete.

NP-complete Problems

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