• 沒有找到結果。

# Algorithm Design and Analysis NP Completeness (2)

N/A
N/A
Protected

Share "Algorithm Design and Analysis NP Completeness (2)"

Copied!
77
0
0

(1)

### Algorithm Design and AnalysisNP Completeness (2)

Yun-Nung (Vivian) Chen

(2)

### • Proving NP-Completeness

• 3-CNF-SAT

• Clique

• Vertex Cover

• Independent Set

• Traveling Salesman Problem

(3)

Complexity

(4)

(3) (2)

(1)

(4)

initial

configuration

(5)

### Non-Deterministic Polynomial

polynomial

“solved” in non-deterministic polynomial time

= “verified” in polynomial time

(6)

## Polynomial-Time Reduction

Textbook Chapter 34.3 – NP-completeness and reducibility

(7)

### First NP-Complete Problem –SAT (Satisfiability)

• Input: a Boolean formula with variables

• Output: whether there is a truth assignment for the variables that satisfies the input Boolean formula

Stephan A. Cook [FOCS 1971] proved that

SAT can be solved in non-deterministic polynomial time → SAT ∈ NP

If SAT can be solved in deterministic polynomial time, then so can any NP problems → SAT ∈ NP-hard

(8)

### Reduction

• Problem A can be reduced (in polynomial time) to Problem B

= Problem B can be reduced (in polynomial time) from Problem A

• We can find an algorithm that solves Problem B to help solve Problem A

• If problem B has a polynomial-time algorithm, then so does problem A

• Practice: design a MULTIPLY() function by ADD(), DIVIDE(), and SQUARE()

What is the complexity of Algorithm for A?

Algorithm for B

? Instance 𝛽 of B Answer for 𝛽 ? Answer for 𝛼 Instance 𝛼 of A

Algorithm for A

(9)

### Reduction

• A reduction is an algorithm for transforming a problem instance into another

• Definition

• Reduction from A to B implies A is not harder than B

• A ≤p B if A can be reduced to B in polynomial time

• Applications

• Designing algorithms: given algorithm for B, we can also solve A

• Classifying problems: establish relative difficulty between A and B

• Proving limits: if A is hard, then so is B

Algorithm to decide B Reduction

Algorithm

Instance 𝛽 of B Yes

Instance 𝛼 of A

Algorithm to decide A No

This is why we need it for proving NP-completeness!

(10)

### Questions

• If A is an NP-hard problem and B can be reduced from A, then B is an NP- hard problem?

• If A is an NP-complete problem and B can be reduced from A, then B is an NP-complete problem?

• If A is an NP-complete problem and B can be reduced from A, then B is an NP-hard problem?

(11)

### Problem Difficulty

• Q: Which one is harder?

• A: They have equal difficulty.

• Proof:

• PARTITION ≤p KNAPSACK

• KNAPSACK ≤p PARTITION

KNAPSACK: Given a set 𝑎1, … , 𝑎𝑛 of non-negative integers, and an integer 𝐾, decide if there is a subset 𝑃 ⊆

1, 𝑛 such that σ𝑖∈𝑃 𝑎𝑖 = 𝐾.

PARTITION: Given a set of 𝑛 non-negative integers

𝑎1, … , 𝑎𝑛 , decide if there is a subset 𝑃 ⊆ 1, 𝑛 such that σ𝑖∈𝑃 𝑎𝑖 = σ𝑖∉𝑃𝑎𝑖.

Polynomial-time reducible?

Polynomial-time reducible?

(12)

### Polynomial Time Reduction

• PARTITION ≤p KNAPSACK

• If we can solve KNAPSACK, how can we use that to solve PARTITION?

• KNAPSACK ≤p PARTITION

• If we can solve PARTITION, how can we use that to solve KNAPSACK?

KNAPSACK: Given a set 𝑎1, … , 𝑎𝑛 of non-negative integers, and an integer 𝐾, decide if there is a subset 𝑃 ⊆

1, 𝑛 such that σ𝑖∈𝑃 𝑎𝑖 = 𝐾.

PARTITION: Given a set of 𝑛 non-negative integers

𝑎1, … , 𝑎𝑛 , decide if there is a subset 𝑃 ⊆ 1, 𝑛 such that σ𝑖∈𝑃 𝑎𝑖 = σ𝑖∉𝑃𝑎𝑖.

Polynomial-time reducible?

Polynomial-time reducible?

(13)

### PARTITION ≤pKNAPSACK

• If we can solve KNAPSACK, how can we use that to solve PARTITION?

• Polynomial-time reduction

• Set

KNAPSACK: Given a set 𝑎1, … , 𝑎𝑛 of non-negative integers, and an integer 𝐾, decide if there is a subset 𝑃 ⊆ 1, 𝑛 such that σ𝑖∈𝑃 𝑎𝑖 = 𝐾.

PARTITION: Given a set of 𝑛 non-negative integers 𝑎1, … , 𝑎𝑛 , decide if there is a

subset 𝑃 ⊆ 1, 𝑛 such that σ𝑖∈𝑃 𝑎𝑖 = σ𝑖∉𝑃𝑎𝑖.

5 6 7 8 5 6 7 8

p-time reduction

PARTITION instance KNAPSACK instance with

(14)

### PARTITION ≤pKNAPSACK

• If we can solve KNAPSACK, how can we use that to solve PARTITION?

• Polynomial-time reduction

• Set

• Correctness proof: KNAPSACK returns yes if and only if an equal-size partition

5 6 7 8 5 6 7 8

p-time reduction

PARTITION instance KNAPSACK instance with

Algorithm to decide KNAPSACK

P-time Reduction

Instance 𝛽 of KNAPSACK Instance 𝛼 of Yes

PARTITION

Algorithm to decide PARTITION No

(15)

### KNAPSACK ≤pPARTITION

• If we can solve PARTITION, how can we use that to solve KNAPSACK?

• Polynomial-time reduction

• Set

5 6 7 8 5 6 7 8

p-time reduction

PARTITION instance 48 52 KNAPSACK instance with

KNAPSACK: Given a set 𝑎1, … , 𝑎𝑛 of non-negative integers, and an integer 𝐾, decide if there is a subset 𝑃 ⊆ 1, 𝑛 such that σ𝑖∈𝑃 𝑎𝑖 = 𝐾.

PARTITION: Given a set of 𝑛 non-negative integers 𝑎1, … , 𝑎𝑛 , decide if there is a

subset 𝑃 ⊆ 1, 𝑛 such that σ𝑖∈𝑃 𝑎𝑖 = σ𝑖∉𝑃𝑎𝑖.

(16)

### KNAPSACK ≤pPARTITION

• If we can solve PARTITION, how can we use that to solve KNAPSACK?

• Polynomial-time reduction

• Set

• Correctness proof: PARTITION returns yes if and only if there is a subset

Algorithm to decide PARTITION

P-time Reduction

Instance 𝛽 of PARTITION Instance 𝛼 of Yes

KNAPSACK

Algorithm to decide KNAPSACK No

5 6 7 8 5 6 7 8

p-time reduction

PARTITION instance 48 52 KNAPSACK instance with

(17)

### KNAPSACK ≤pPARTITION

• Polynomial-time reduction

• Set

• Correctness proof: PARTITION returns yes if and only if there is a subset 𝑃 ⊆ 1, 𝑛 such that σ𝑖∈𝑃 𝑎𝑖 = 𝐾

• “if” direction

𝑎1 𝑎2 𝑎3 𝑎4 𝑎1 𝑎4 𝑎5 𝑎2 𝑎3 𝑎6

PARTITION returns yes!

(18)

### KNAPSACK ≤pPARTITION

• Polynomial-time reduction

• Set

• Correctness proof: PARTITION returns yes if and only if there is a subset 𝑃 ⊆ 1, 𝑛 such that σ𝑖∈𝑃 𝑎𝑖 = 𝐾

• “only if” direction

• Because , if PARTITION returns yes, each set has 4𝐻 + 𝐾

• 𝑎1, … , 𝑎𝑛 must be divided into 2𝐻 − 𝐾 and 𝐾

𝑎6

𝑎2 𝑎3 𝑎5 𝑎1 𝑎4 𝑎1 𝑎4 𝑎2 𝑎3

(19)

### Reduction for Proving Limits

• Definition

• Reduction from A to B implies A is not harder than B

• A ≤p B if A can be reduced to B in polynomial time

• NP-completeness proofs

• Goal: prove that B is NP-hard

• Known: A is NP-complete/NP-hard

• Approach: construct a polynomial-time reduction algorithm to convert 𝛼 to 𝛽

• Correctness: if we can solve B, then A can be solved → A ≤p B

• B is no easier than A → A is NP-hard, so B is NP-hard

If the reduction is not p-time, does this argument hold?

Algorithm to decide B Reduction

Algorithm

Instance 𝛽 of B Yes

Instance 𝛼 of A

Algorithm to decide A No

(20)

## Proving NP-Completeness

(21)

### Formal Language Framework

• Focus on decision problems

• A language L over σ is any set of strings made up of symbols from σ

• Every language L over σ is a subset of σ

• An algorithm A accepts a string if

• The language accepted by an algorithm A is the set of strings

• An algorithm A rejects a string x if

The formal-language framework allows us to express concisely the relation between decision problems and algorithms that solve them.

(22)

### Proving NP-Completeness

• NP-Complete (NPC): class of decision problems in both NP and NP-hard

• In other words, a decision problem L is NP-complete if 1.L ∈ NP

2.L ∈ NP-hard (that is, L’ ≤p L for every L’ ∈ NP)

L1 L2 L3 :

L

p

all NP problems How to prove L is NP-hard ?

L1 L2 L3 :

p

all NP problems

L known

NPC problem

p

held by definition

Goal: prove polynomial- time reduction

(23)

### Polynomial-Time Reducible

• If are languages s.t. , then L2 ∈ P implies L1 ∈ P.

A2 Transform

function f 𝑥 𝑓 𝑥

A1

(24)

(25)

### Circuit Satisfiability Problem

• Given a Boolean combinational circuit composed of AND, OR, and NOT gates, is it satisfiable?

• Satisfiable: there exists an assignment s.t. outputs = 1

Satisfiable Unsatisfiable

(26)

### CIRCUIT-SAT

• CIRCUIT-SAT can be solved in non-deterministic polynomial time

→ ∈ NP

• If CIRCUIT-SAT can be solved in deterministic polynomial time, then so can any NP problems

→ ∈ NP-hard

• (proof in textbook 34.3)

• CIRCUIT-SAT is NP-complete

CIRCUIT-SAT = {<C>: C is a satisfiable Boolean combinational circuit}

(27)

### Karp’s NP-Complete Problems

1. CNF-SAT

2. 0-1 INTEGER PROGRAMMING 3. CLIQUE

4. SET PACKING 5. VERTEX COVER 6. SET COVERING

7. FEEDBACK ARC SET 8. FEEDBACK NODE SET

9. DIRECTED HAMILTONIAN CIRCUIT

10.UNDIRECTED HAMILTONIAN CIRCUIT 11.3-SAT

12.CHROMATIC NUMBER 13.CLIQUE COVER

14.EXACT COVER

15.3-dimensional MATCHING 16.STEINER TREE

17.HITTING SET 18.KNAPSACK

19.JOB SEQUENCING 20.PARTITION

21.MAX-CUT

(28)

(29)

### Formula Satisfiability Problem (SAT)

• Given a Boolean formula Φ with variables, is there a variable assignment satisfying Φ

• ∧ (AND), ∨ (OR), ¬ (NOT), → (implication), ↔ (if and only if)

• Satisfiable: Φ is evaluated to 1

(30)

### SAT

• Is SAT ∈ NP-Complete?

• To prove that SAT is NP-Complete, we show that

• SAT ∈ NP

• SAT ∈ NP-hard (CIRCUIT-SAT ≤p SAT)

1) CIRCUIT-SAT is a known NPC problem

2) Construct a reduction f transforming every CIRCUIT-SAT instance to an SAT instance 3) Prove that x ∈ CIRCUIT-SAT iff f(x) ∈ SAT

4) Prove that f is a polynomial time transformation

SAT = {Φ | Φ is a Boolean formula with a satisfying assignment }

(31)

### SAT ∈ NP

• Polynomial-time verification: replaces each variable in the formula with the corresponding value in the certificate and then evaluates the expression

initial configuration

polynomial

(32)

### SAT ∈ NP-Hard

1)CIRCUIT-SAT is a known NPC problem

2)Construct a reduction f transforming every CIRCUIT-SAT instance to an SAT instance

• Assign a variable to each wire in circuit C

• Represent the operation of each gate using a formula, e.g.

• Φ = AND the output variable and the operations of all gates

(33)

### SAT ∈ NP-Hard

• Prove that x ∈ CIRCUIT-SAT ↔ f(x) ∈ SAT

• x ∈ CIRCUIT-SAT → f(x) ∈ SAT

• f(x) ∈ SAT → x ∈ CIRCUIT-SAT

• f is a polynomial time transformation CIRCUIT-SAT ≤p SAT → SAT ∈ NP-hard

(34)

## Polynomial-Time Verification

Chapter 34.1 – Polynomial-time

Chapter 34.2 – Polynomial-time verification

(35)

### Abstract Problems

• Example of a decision problem, PATH

• I: a set of problem instances

• S: a set of problem solutions

• Q: abstract problem, defined as a binary relation on I and S

G1, s1, t1, k1 G2, s2, t2, k2

Q: PATH

Is there a path with the length ≤ k?

I: <G, src, dest, k>

All graphs with arbitrary src, dest, and the path length k

1 (Yes) 0 (No) S: {0, 1}

(36)

### Problem Instance Encoding

• Convert an abstract problem instance into a binary string fed to a computer program

• A concrete problem is polynomial-time solvable if there exists an algorithm that solves any concrete instance of length n in time for some constant k

• Solvable = can produce a solution

Encoder

Abstract Problem Instance

Binary Strings

(Concrete Problem Instance)

(37)

### Decision Problem Representation

• I: a set of problem instances

• Q: a decision problem

= a language L over s.t.

str1 str2 str3

Q: decision problem I: a set of problem

instances

1 (Yes) 0 (No) S: {0, 1}

some strings represent no meaningful instances

(38)

### P in Formal Language Framework

• An algorithm A accepts a string if

• An algorithm A rejects a string if

• An algorithm A accepts a language L if A accepts every string

• If the string is in L, A outputs yes.

• If the string is not in L, A may output no or loop forever.

• An algorithm A decides a language L if A accepts L and A rejects every string

• For every string, A can output the correct answer.

A decision problem Q can be defined as a language L over s.t.

(39)

### P in Formal Language Framework

• Class P: a class of decision problems solvable in polynomial time

• Given an instance x of a decision problem Q, its solution Q(x) (i.e., YES or NO) can be found in polynomial time

• An alternative definition of P:

• P is the class of language that can be accepted in polynomial time

P = {L ⊆ {0,1}* | there exists an algorithm that decides L in polynomial time}

P = {L | L is accepted by a polynomial algorithm}

(40)

### Hamiltonian-Cycle Problem

• Problem: find a cycle that visits each vertex exactly once

• Formal language:

• Is this language decidable?

• Is this language decidable in polynomial time?

• Given a certificate – the vertices in order that form a Hamiltonian cycle in G, how much time does it take to verify that G indeed contains a Hamiltonian cycle?

HAM-CYCLE = {<G> | G has a Hamiltonian cycle}

Yes

Probably not

(41)

### Verification Algorithm

• Verification algorithms verify memberships in language

HAM-CYCLE = {<G> | G has a Hamiltonian cycle}

Verification Algorithm Is y a Hamiltonian cycle in the graph (encoded in x)?

Certificate y

YES

x is in HAM-CYCLE Input x

There exists a certificate for each YES instance

(42)

### Verification Algorithm

• Verification algorithms verify memberships in language

Certificate y

NO

No conclusion Input x

??

HAM-CYCLE = {<G> | G has a Hamiltonian cycle}

Verification Algorithm Is y a Hamiltonian cycle in the graph (encoded in x)?

(43)

### Non-Deterministic Polynomial

polynomial

“solved” in non-deterministic polynomial time

= “verified” in polynomial time

(44)

### Polynomial-Time Reducible

• If are languages s.t. , then L2 ∈ P implies L1 ∈ P.

A2 Transform

function f 𝑥 𝑓 𝑥

A

(45)

### Proving NP-Completeness

• NP-Complete (NPC): class of decision problems in both NP and NP-hard

• In other words, a decision problem L is NP-complete if 1.L ∈ NP

2.L ∈ NP-hard (that is, L’ ≤p L for every L’ ∈ NP)

L1 L2 L3 :

L

p

all NP problems How to prove L is NP-hard ?

L1 L2 L3 :

p

all NP problems

L known

NPC problem

p

held by definition

Goal: prove polynomial- time reduction

(46)

### Proving NP-Completeness

• L ∈ NPC iff L ∈ NP and L ∈ NP-hard

• Proof of L in NPC:

• Prove L ∈ NP

• Prove L ∈ NP-hard

1) Select a known NPC problem C

2) Construct a reduction f transforming every instance of C to an instance of L 3) Prove that

4) Prove that f is a polynomial time transformation

(47)

### • “Computers and Intractability” by Garey and Johnson includes more than 300 NP-complete problems

• All except SAT are proved by Karp’s polynomial-time reduction

(48)

## Proving NP-Completeness

Chapter 34.5 – NP-complete problems

(49)

• A → B: A ≤p B

CIRCUIT-SAT SAT

3-CNF-SAT

CLIQUE SUBSET-SUM

VERTEX-COVER HAM-CYCLE

TSP

(50)

### 3-CNF-SAT Problem

• 3-CNF-SAT: Satisfiability of Boolean formulas in 3-conjunctive normal form (3- CNF)

• 3-CNF = AND of clauses, each of which is the OR of exactly 3 distinct literals

• A literal is an occurrence of a variable or its negation, e.g., x1 or ¬x1

→ satisfiable

(51)

### 3-CNF-SAT

• Is 3-CNF-SAT ∈ NP-Complete?

• To prove that 3-CNF-SAT is NP-Complete, we show that

• 3-CNF-SAT ∈ NP

• 3-CNF-SAT ∈ NP-hard (SAT ≤p 3-CNF-SAT)

1) SAT is a known NPC problem

2) Construct a reduction f transforming every SAT instance to an 3-CNF-SAT instance 3) Prove that x ∈ SAT iff f(x) ∈ 3-CNF-SAT

4) Prove that f is a polynomial time transformation

3-CNF-SAT = {Φ | Φ is a Boolean formula in 3-conjunctive normal form (3-CNF) with a satisfying assignment }

We focus on the reduction construction from now on, but remember that a full proof requires showing that all other conditions are true as well

(52)

### SAT ≤p3-CNF-SAT

a) Construct a binary parser tree for an input formula Φ and introduce a variable yi for the output of each internal node

(53)

### SAT ≤p3-CNF-SAT

b) Rewrite Φ as the AND of the root variable and clauses describing the operation of each node

(54)

### SAT ≤p3-CNF-SAT

c) Convert each clause Φi’ to CNF

• Construct a truth table for each clause Φi

• Construct the disjunctive normal form for ¬Φi

• Apply DeMorgan’s Law to get the CNF formula Φi’’

𝒚𝟏 𝒚𝟐 𝒚𝟐 Φ1 ¬Φ1

1 1 1 0 1

1 1 0 1 0

1 0 1 0 1

1 0 0 0 1

0 1 1 1 0

0 1 0 0 1

0 0 1 1 0

𝒚𝟏 𝒚𝟐 𝒚𝟐 Φ1

1 1 1 0

1 1 0 1

1 0 1 0

1 0 0 0

0 1 1 1

0 1 0 0

0 0 1 1

(55)

### SAT ≤p3-CNF-SAT

d) Construct Φ’’’ in which each clause Ci exactly 3 distinct literals

• 3 distinct literals:

• 2 distinct literals:

• 1 literal only:

• Φ’’’ is satisfiable iff Φ is satisfiable

• All transformation can be done in polynomial time

• → 3-CNF-SAT is NP-Complete

(56)

### Clique Problem

• A clique in G = (V, E) is a complete subgraph of G

• Each pair of vertices in a clique is connected by an edge in E

• Size of a clique = # of vertices it contains

• Optimization problem: find a max clique in G

• Decision problem: is there a clique with size larger than k

Does G contain a clique of size 4?

Does G contain a clique of size 5?

Does G contain a clique of size 6?

Yes Yes No

(57)

### CLIQUE ∈ NP-Complete

• Is CLIQUE ∈ NP-Complete?

• Construct a reduction f transforming every 3-CNF-SAT instance to a CLIQUE instance

• a graph G s.t. Φ with k clauses is satisfiable  G has a clique of size k

CLIQUE = {<G, k>: G is a graph containing a clique of size k}

3-CNF-SAT ≤p CLIQUE

p

(58)

### CLIQUE ∈ NP-Complete

• Polynomial-time reduction:

• Let be a Boolean formula in 3-CNF with k clauses, and each 𝐶𝑟 has exactly 3 distinct literals 𝑙1𝑟, 𝑙2𝑟, 𝑙3𝑟

• For each , introduce a triple of vertices 𝑣1𝑟, 𝑣2𝑟, 𝑣3𝑟 in V

• Build an edge between 𝑣𝑖𝑟, 𝑣𝑗𝑠 if both of the following hold:

• 𝑣𝑖𝑟 and 𝑣𝑗𝑠 are in different triples

• 𝑙𝑖𝑟 is not the negation of 𝑙𝑗𝑠

(59)

### 3-CNF-SAT ≤p CLIQUE

• Correctness proof: Φ is satisfiable → G has a clique of size k

• If Φ is satisfiable

• → Each Cr contains at least one 𝑙𝑖𝑟 = 1 and such literal corresponds to 𝑣𝑖𝑟

• → Pick a TRUE literal from each Cr forms a set of V’ of k vertices

• → For any two vertices 𝑣𝑖𝑟, 𝑣𝑗𝑠 ∈ 𝑉(𝑟 ≠ 𝑠), edge (𝑣𝑖𝑟, 𝑣𝑗𝑠) ∈ 𝐸, because 𝑙𝑖𝑟 = 𝑙𝑗𝑠 = 1 and they cannot be complements

(60)

### 3-CNF-SAT ≤p CLIQUE

• Correctness proof: G has a clique of size k → Φ is satisfiable

• G has a clique V’ of size k

• → V’ contains exactly one vertex per triple since no edges connect vertices in the same triple

• → Assign 1 to each 𝑙𝑖𝑟 where 𝑣𝑖𝑟 ∈ 𝑉′ s.t. each Cr is satisfiable, and so is Φ

(61)

### Vertex Cover Problem

• A vertex cover of G = (V, E) is a subset V’ ⊆ V s.t. if (w, v) ∈ E, then w ∈ V’ or v

∈ V’

• A vertex cover “covers” every edge in G

• Optimization problem: find a minimum size vertex cover in G

• Decision problem: is there a vertex cover with size smaller than k

Does G have a vertex cover of size 11?

Does G have a vertex cover of size 7?

Does G have a vertex cover of size 6?

Yes Yes No

(62)

### VERTEX-COVER ∈ NP-Complete

• Is VERTEX-COVER ∈ NP-Complete?

• Construct a reduction f transforming every CLIQUE instance to a VERTEX- COVER instance (polynomial-time reduction)

• Compute the complement of G

• Given G = <V, E>, Gcis defined as <V, Ec> s.t. Ec= {(u,v) | (u,v) ∉ E}

• a graph G has a clique of size k  Gc has a vertex cover of size |V| - k

VERTEX-COVER = {<G, k>: G is a graph containing a vertex cover of size k}

CLIQUE ≤p VERTEX-COVER

p

(63)

### CLIQUE ≤p VERTEX-COVER

• Correctness proof:

a graph G has a clique of size k → Gc has a vertex cover of size |V| - k

• If G has a clique V’ ⊆ V with |V’| = k

• → for all 𝑤, 𝑣 ∈ 𝐸𝑐, at least one of 𝑤 or 𝑣 ∉ 𝑉′

• → 𝑤 ∈ 𝑉 − 𝑉′ or 𝑣 ∈ 𝑉 − 𝑉′ (or both)

• → edge 𝑤, 𝑣 is covered by 𝑉 − 𝑉′

• → 𝑉 − 𝑉′ forms a vertex cover of Gc, and |V - V'| = |V| - k

(64)

### CLIQUE ≤p VERTEX-COVER

• Correctness proof:

Gc has a vertex cover of size |V| - k → a graph G has a clique of size k

• If Gc has a vertex cover V’ ⊆ V with |V’| = |V| - k

• → for all 𝑤, 𝑣 ∈ 𝑉, if 𝑤, 𝑣 ∈ 𝐸𝑐, then 𝑤 ∈ 𝑉′ or 𝑣 ∈ 𝑉′ or both

• → for all 𝑤, 𝑣 ∈ 𝑉, if 𝑤 ∉ 𝑉′ and 𝑣 ∉ 𝑉′, 𝑤, 𝑣 ∈ 𝐸

• → 𝑉 − 𝑉′ is a clique where |V - V'| = k

(65)

### Independent-Set Problem

• An independent set of G = (V, E) is a subset V’ ⊆ V such that G has no edge between any pair of vertices in V’

• A vertex cover “covers” every edge in G

• Optimization problem: find a maximum size independent set

• Decision problem: is there an independent set with size larger than k

Does G have an independent set of size 1?

Does G have an independent set of size 4?

Does G have an independent set of size 5?

(66)

### IND-SET ∈ NP-Complete

• Is IND-SET ∈ NP-Complete?

• Practice by yourself (textbook problem 34-1)

IND-SET = {<G, k>: G is a graph containing an independent set of size k}

(67)

### CLIQUE, VERTEX-COVER, IND-SET

• The following are equivalent for G = (V, E) and a subset V’ of V:

1) V' is a clique of G

2) V-V' is a vertex cover of Gc 3) V' is an independent set of Gc

(68)

### Traveling Salesman Problem (TSP)

• Optimization problem: Given a set of cities and their pairwise distances, find a tour of lowest cost that visits each city exactly once.

• Decision problem: is there a traveling salesman tour with cost at most k

(69)

### TSP ∈ NP-Complete

• Is TSP ∈ NP-Complete?

• Construct a reduction f transforming every HAM-CYCLE instance to a TSP instance (polynomial-time reduction)

• G contains a Hamiltonian cycle h = <v1, v2, …, vn, v1>  <v1, v2, …, vn, v1> is a traveling-salesman tour with cost 0

TSP = {<G, c, k>: G = (V,E) is a complete graph, c is a cost function for edges, G has a traveling-salesman tour with cost at most k}

HAM-CYCLE ≤p TSP

u

v y

w z

u

v y

w z

p

0

1

(70)

### HAM-CYCLE ≤p TSP

• Correctness proof: x ∈ HAM-CYCLE → f(x) ∈ TSP

• If Hamiltonian cycle is h = <v1, v2, …, vn, v1>

• → h is also a tour in the transformed TSP instance

• → The distance of the tour h is 0 since there are n consecutive edges in E, and so has distance 0 in f(x)

• → f(x) ∈ TSP (f(x) has a TSP tour with cost ≤ 0)

u

v y

u

v y

p

0

1

(71)

### HAM-CYCLE ≤p TSP

• Correctness proof: f(x) ∈ TSP → x ∈ HAM-CYCLE

• After reduction, if a TSP tour with cost ≤ 0 as <v1, v2, …, vn, v1>

• → The tour contains only edges in E

• → Thus, <v1, v2, …, vn, v1> is a Hamiltonian cycle

u

v y

w z

u

v y

w z

p

0

1

(72)

### TSP Challenges

• Mona Lisa TSP: \$1,000 Prize for 100,000-city

(73)

### Strategies for NP-Complete/NP-Hard Problems

• NP-complete/NP-hard problems are unlikely to have polynomial-time

solutions (unless P = NP), we must sacrifice either optimality, efficiency, or generality

• Approximation algorithms: guarantee to be a fixed percentage away from the optimum

• Local search: simulated annealing (hill climbing), genetic algorithms, etc

• Heuristics: no formal guarantee of performance

• Randomized algorithms: use a randomizer (random number generator) for operation

• Pseudo-polynomial time algorithms: e.g., DP for 0-1 knapsack

• Exponential algorithms/Branch and Bound/Exhaustive search: feasible only when the problem size is small

• Restriction: work on some special cases of the original problem. e.g., the maximum independent set problem in circle graphs

(74)

74

(75)

### Concluding Remarks

• Proving NP-Completeness: L ∈ NPC iff L ∈ NP and L ∈ NP-hard

• Polynomial-time verification

• Step-by-step approach for proving L in NPC:

• Prove L ∈ NP

• Prove L ∈ NP-hard

• Select a known NPC problem C

• Construct a reduction f transforming every instance of C to an instance of L

• Prove that

• Prove that f is a polynomial time transformation L ∈ NP

• Strategies for NP-complete/NP-hard problems

P ≠ NP

P = NP

CIRCUIT -SAT

SAT

3-CNF- SAT

CLIQUE SUBSET- SUM VERTEX

-COVER HAM- CYCLE

TSP

(76)

### Concluding Remarks

• Proving NP-Completeness: L ∈ NPC iff L ∈ NP and L ∈ NP-hard

• Polynomial-time verification

• Step-by-step approach for proving L in NPC:

• Prove L ∈ NP

• Prove L ∈ NP-hard

1) Select a known NPC problem C

2) Construct a reduction f transforming every instance of C to an instance of L 3) Prove that

4) Prove that f is a polynomial time transformation

• Strategies for NP-complete/NP-hard problems

CIRCUIT-SAT

SAT

3-CNF-SAT

CLIQUE SUBSET-

SUM

VERTEX- COVER HAM-CYCLE

TSP

(77)

## Question?

Important announcement will be sent to

@ntu.edu.tw mailbox & post to the course website

• Algorithmic design methods to solve problems efficiently (polynomial time).. • Divide

• The randomized bipartite perfect matching algorithm is called a Monte Carlo algorithm in the sense that. – If the algorithm ﬁnds that a matching exists, it is always correct (no

• An algorithm for such a problem whose running time is a polynomial of the input length and the value (not length) of the largest integer parameter is a..

• The randomized bipartite perfect matching algorithm is called a Monte Carlo algorithm in the sense that.. – If the algorithm ﬁnds that a matching exists, it is always correct

• Consider an algorithm that runs C for time kT (n) and rejects the input if C does not stop within the time bound.. • By Markov’s inequality, this new algorithm runs in time kT (n)

• By definition, a pseudo-polynomial-time algorithm becomes polynomial-time if each integer parameter is limited to having a value polynomial in the input length.. • Corollary 42

• Consider an algorithm that runs C for time kT (n) and rejects the input if C does not stop within the time bound.. • By Markov’s inequality, this new algorithm runs in time kT (n)

• The randomized bipartite perfect matching algorithm is called a Monte Carlo algorithm in the sense that.. – If the algorithm ﬁnds that a matching exists, it is always correct

• Consider an algorithm that runs C for time kT (n) and rejects the input if C does not stop within the time bound.. • By Markov’s inequality, this new algorithm runs in time kT (n)

Here, a deterministic linear time and linear space algorithm is presented for the undirected single source shortest paths problem with positive integer weights.. The algorithm

We are not aware of any existing methods for identifying constant parameters or covariates in the parametric component of a semiparametric model, although there exists an

In particular, we present a linear-time algorithm for the k-tuple total domination problem for graphs in which each block is a clique, a cycle or a complete bipartite graph,

If P6=NP, then for any constant ρ ≥ 1, there is no polynomial-time approximation algorithm with approximation ratio ρ for the general traveling-salesman problem...

The set of concrete decision problems that are polynomial-

• An algorithm is any well-defined computational procedure that takes some value, or set of values, as input and produces some value, or set of values, as output.. • An algorithm is

 If SAT can be solved in deterministic polynomial time, then so can any NP problems  SAT ∈ NP-hard..  If A is an NP-hard problem and B can be reduced from A, then B is an

The well-known halting problem (a decision problem), which is to determine whether or not an algorithm will terminate with a given input, is NP-hard, but

O(log 2 k/ log log k)-Approximation Algorithm for Directed Steiner Tree: A Tight

Given a graph and a set of p sources, the problem of finding the minimum routing cost spanning tree (MRCT) is NP-hard for any constant p &gt; 1 [9].. When p = 1, i.e., there is only

• Given a (singly) linked list of unknown length, design an algorithm to find the n-th node from the tail of the linked list. Your algorithm is allowed to traverse the linked

Hwang, “An O (nk)-time algorithm for computing the reliability of a circular consecutive-k-out-of-n:F system,” IEEE Trans on Reliability, Vol.. Shantikumar, “Recursive algorithm

This study used ETABS and DRAWRC to design a concrete-steel building containing fluid viscous damper, and conducted nonlinear time-history analysis to obtain

A parallel route building algorithm for the vehicle routing and scheduling problem with time windows, European Journal of Operational Research, vol. A tabu search