Algorithm Design and Analysis NP Completeness (1)

Algorithm Design and Analysis NP Completeness (1)

Yun-Nung (Vivian) Chen

http://ada.miulab.tw

Outline

• Decision Problems v.s. Optimization Problems

• Complexity Classes

• P v.s. NP

• NP, NP-Complete, NP-Hard

• Polynomial-Time Reduction

Algorithm Design & Analysis

• Design Strategy

• Divide-and-Conquer

• Dynamic Programming

• Greedy Algorithms

• Graph Algorithms

• Analysis

• Amortized Analysis

• NP-Completeness

3

Polynomial Time Algorithms

• For an input with size n, the worst-case running time is for some constant k

• Problems that are solvable by polynomial-time algorithms as being tractable, easy, or efficient

• Problems that require superpolynomial time as being intractable, or hard, or inefficient

Four Color Problem

• Use total four colors s.t. the neighboring parts have different colors

5

Four Color Problem (after 100 yrs)

• Finally proven (with the help of computers) by Kenneth Appel and Wolfgang Haken in 1976

• Their algorithm runs in O(n²) time

• First major theorem proved by a computer

• Open problems remain...

• Linear time algorithms to find a solution

• Concise, human-checkable, mathematical proofs

Planar k-Colorability

• Given a planar graph G (e.g., a map), can we color the vertices with k colors such that no adjacent vertices have the same color?

• k = 1?

• k = 2?

• k = 3?

• k ≥ 4?

7

How hard is it when k = 3?

Can we know its level of difficulty before solving it?

Planar k-Colorability

Decision Problems v.s.

Optimization Problems

9

Decision Problems

• Definition: the answer is simply “yes” or “no” (or “1” or “0”)

• MST: Given a graph 𝐺 = 𝑉, 𝐸 and a bound 𝐾, is there a spanning tree with a cost at most 𝐾?

• KNAPSACK: Given a knapsack of capacity 𝐶, a set of objects with weights and values, and a target value 𝑉, is there a way to fill the knapsack with at least 𝑉 value?

Optimization Problems

• Definition: each feasible solution has an associated value, and we wish to find a feasible solution with the best value (maximum or minimum)

• MST-OPT: Given a graph 𝐺 = 𝑉, 𝐸 , find the minimum spanning tree of 𝐺

• KNAPSACK-OPT: Given a knapsack of capacity 𝐶 and a set of objects with weights and values, fill the knapsack so as to maximize the total value

11

Which is Easier? Why?

How to convert an optimization problem to a related decision problem?

Imposing a (lower or upper) bound on the value to be optimized

Difficulty Levels

• Every optimization problem has a decision version that is no harder than the optimization problem.

• Using A_opt to solve A_dec

• check if the optimal value ≤ k, constant overhead

• Using A_dec to solve A_opt

• apply binary search on the value range, logarithmic overhead

13

A_opt: given a graph, find the length of the shortest path

A_dec: given a graph, determine whether there is a path ≤ k

P v.s. NP

Textbook Chapter 34 – NP-Completeness

Algorithm Design

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

• Divide and conquer

• Dynamic programming

• Greedy

• “Hard” problems without known efficient algorithms

• Hamilton, knapsack, etc.

15

Complexity Classes

• Can we decide whether a problem is “too hard to solve” before investing our time in solving it?

• Idea: decide which complexity classes the problem belongs to via reduction

• 已知問題A很難。若能證明問題B至少跟A一樣難，那麼問題B也很難。

To Solve v.s. Not to Solve

• Algorithm design

• Design algorithms to solve computational problems

• Mostly concerned with upper bounds on resources

17

• Complexity theory

• Classify problems based on their difficulty and identify relationships between classes

• Mostly concerned with lower bounds on resources

upper bound

Problem

Problem B

Problem A

Problem B is no easier than A

lower bound

Complexity Classes

• A complexity class is “a set of problems of related resource-based complexity”

• Resource = time, memory, communication, ...

• Focus: decision problems and the resource of time

P

• The class P consists of all the problems that can be solved in polynomial time.

• Sorting

• Exact string matching

• Primes

• …

• Polynomial time algorithm

• For inputs of size n, their worst-case running time is for some constant k

19

NP

• NP consists of the problems that can be solved in non-deterministically polynomial time.

• NP consists of the problems that can be “verified” in polynomial time.

• P consists of the problems that can be solved in (deterministically) polynomial time.

Non-Deterministic

Deterministic Algorithm

21

initial

configuration

Non-Deterministic Algorithm

initial

configuration

Non-Deterministic Bubble Sort

23

This is not a randomized algorithm.

Non-Deterministic-Bubble-Sort(n) for i = 1 to n

for j = 1 to n – 1

if A[j] < A[i+1] then

Either exchange A[j] and A[i+1] or do nothing

Vertex Cover Problem (路燈問題)

• Input: a graph G

• Output: a smallest vertex subset of G that covers all edges of G.

• Known to be NP-complete

Illustration

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Vertex Cover (Decision Version)

• Input: a Graph G and an integer k.

• Output: Does G contain a vertex cover of size no more than k?

• Original problem → optimization problem

• 原先的路燈問題是要算出放路燈的方法

• Yes/No → decision problem

• 問k盞路燈夠不夠照亮整個公園

Non-Deterministic Algorithm

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Non-Deterministic-Vertex-Cover(G, k) set S = {}

for each vertex x of G

non-deterministically insert x to S if |S| > k

output no

if S is not a vertex cover output no

output yes

Algorithm Correctness

• If the correct answer is yes, then there is a computation path of the algorithm that leads to yes.

• 至少有一條路是對的

• If the correct answer is no, then all computation paths of the algorithm lead to no.

Non-Deterministic-Vertex-Cover(G, k) set S = {}

for each vertex x of G

non-deterministically insert x to S if |S| > k

output no

if S is not a vertex cover output no

output yes

Non-Deterministic Problem Solving

29

initial

configuration

correct answer

Non-Deterministic Polynomial

polynomial

“solved” in non-deterministic polynomial time

P ⊆ NP or NP ⊆ P?

• P ⊆ NP

• A problem solvable in polynomial time is verifiable in polynomial time as well

• Any NP problem can be solved in (deterministically) exponential time?

• Yes

• Any NP problem can be solved in (deterministically) polynomial time?

• Open problem

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Why?

US$1,000,000 Per Problem

• http://www.claymath.org/millennium-problems

Millennium Problems

• Yang–Mills and Mass Gap

• Riemann Hypothesis

• P vs NP Problem

• Navier–Stokes Equation

• Hodge Conjecture

• Poincaré Conjecture (solved by Grigori Perelman)

• Birch and Swinnerton-Dyer Conjecture

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Grigori Perelman

Fields Medal (2006), declined Millennium Prize (2010), declined

Vinay Deolalikar

• Aug 2010 claimed a proof of P is not equal to NP.

If P = NP

• problems that are verifiable → solvable

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• public-key cryptography will be broken

Widespread belief in P ≠ NP

“If P = NP, then the world would be a profoundly different place than we usually assume it to be. There would be no special value in “creative leaps,” no fundamental gap between solving a problem and recognizing the solution once it's found. Everyone who could appreciate a symphony would be Mozart; everyone who could follow a step-by-step argument would be Gauss...” – Scott Aaronson, MIT

Travelling Salesman (2012)

NP, NP-Complete, NP-Hard

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NP-Hardness

• A problem is NP-hard if it is as least as hard as all NP problems.

• In other words, a problem X is NP-hard if the following condition holds:

• If X can be solved in (deterministic) polynomial time, then all NP problems can be solved in (deterministic) polynomial time.

NP-Completeness (NPC)

• A problem is NP-complete if

• it is NP-hard and

• it is in NP.

• In other words, an NP-complete problem is one of the “hardest” problems in the class NP.

• In other words, an NP-complete problem is a hardness representative problem of the class NP.

• Hardest in NP → solving one NPC can solve all NP problems (“complete”)

• It is wildly believed that NPC problems have no polynomial-time solution

→ good reference point to judge whether a problem is in P

• We can decide if a problem is “too hard to solve” by showing it is as hard as an NPC problem

• We then focus on designing approximate algorithms or solving special cases

39

Complexity Classes

• Class P: class of problems that can be solved in

• Class NP: class of problems that can be verified in

• Class NP-hard: class of problems that are “at least as hard as all NP problems”

• Class NP-complete: class of problems in both NP and NP-hard

More Complexity Classes

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undecidable: no algorithm; e.g.

halting problem

https://www.youtube.com/watch?v=wGLQiHXHWNk

An Undecidable Problem – Halting Problem

• Halting problem is to determine whether a program 𝑝 halts on input 𝑥

• Proof for undecidable via a counterexample

• Suppose ℎ can determine whether a program 𝑝 halts on input 𝑥

• ℎ(𝑝, 𝑥) = return (p halts on input x)

• Define g(p) = if h(p,p) is 0 then return 0 else HANG

• → g(g) = if h(g,g) is 0 then return 0 else HANG

• Both cases contradict the assumption:

1.g halts on g: then h(g,g)=1, which would make g(g) hang

2.g does not halt on g: then h(g,g)=0, which would make g(g) halt

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Complexity Classes

• Which one is in P?

Shortest Simple Path Longest Simple Path

Euler Tour Hamitonian Cycle

LCS with 2 Input Sequences LCS with Arbitrary Input Sequences Degree-Constrained Spanning Tree Minimal Spanning Tree

Candy Crush is NP-Hard

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Sudoku is NPC

46

Minesweeper Consistency is NPC

• Minesweeper Consistency: Given a state of what purports to be a Minesweeper games, is it logically consistent?

47

Polynomial-Time Reduction

Textbook Chapter 34.3 – NP-completeness and reducibility

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

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• 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

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 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

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

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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!

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?

Problem Difficulty

• Q: Which one is harder?

• A: They have equal difficulty.

• Proof:

• PARTITION ≤_p KNAPSACK

• KNAPSACK ≤_p PARTITION

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

1, 𝑛 such that σ_𝑖∈𝑃 𝑎_𝑖 = 𝐾.

PARTITION: Given a set of 𝑛 non-negative integers

𝑎₁, … , 𝑎_𝑛 , decide if there is a subset 𝑃 ⊆ 1, 𝑛 such that σ_𝑖∈𝑃 𝑎_𝑖 = σ_𝑖∉𝑃𝑎_𝑖.

Polynomial-time reducible?

Polynomial-time reducible?

Polynomial Time Reduction

• PARTITION ≤_p KNAPSACK

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

• KNAPSACK ≤_p PARTITION

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

1, 𝑛 such that σ_𝑖∈𝑃 𝑎_𝑖 = 𝐾.

PARTITION: Given a set of 𝑛 non-negative integers

𝑎₁, … , 𝑎_𝑛 , decide if there is a subset 𝑃 ⊆ 1, 𝑛 such that σ_𝑖∈𝑃 𝑎_𝑖 = σ_𝑖∉𝑃𝑎_𝑖.

Polynomial-time reducible?

Polynomial-time reducible?

PARTITION ≤ _p KNAPSACK

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

• Polynomial-time reduction

• Set

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

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

subset 𝑃 ⊆ 1, 𝑛 such that σ_𝑖∈𝑃 𝑎_𝑖 = σ_𝑖∉𝑃𝑎_𝑖.

5 6 7 8 5 6 7 8

p-time reduction

PARTITION instance KNAPSACK instance with

PARTITION ≤ _p KNAPSACK

• 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

KNAPSACK ≤ _p PARTITION

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

• Polynomial-time reduction

• Set

• Add ,

57

5 6 7 8 5 6 7 8

p-time reduction

PARTITION instance 48 52 KNAPSACK instance with

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

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

subset 𝑃 ⊆ 1, 𝑛 such that σ_𝑖∈𝑃 𝑎_𝑖 = σ_𝑖∉𝑃𝑎_𝑖.

KNAPSACK ≤ _p PARTITION

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

• Polynomial-time reduction

• Set

• Add ,

• 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

KNAPSACK ≤ _p PARTITION

• Polynomial-time reduction

• Set

• Add ,

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

• “if” direction

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𝑎₁ 𝑎₂ 𝑎₃ 𝑎₄ 𝑎₁ 𝑎₄ 𝑎₅ 𝑎₂ 𝑎₃ 𝑎₆

PARTITION returns yes!

KNAPSACK ≤ _p PARTITION

• Polynomial-time reduction

• Set

• Add ,

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

• “only if” direction

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

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

𝑎₆

𝑎₂ 𝑎₃ 𝑎₅ 𝑎₁ 𝑎₄ 𝑎₁ 𝑎₄ 𝑎₂ 𝑎₃

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

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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

Proving NP-Completeness

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

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

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 How to prove L is NP-hard ?

L₁ L₂ L₃ :

≤_p

L known

NPC problem

≤_p

held by definition

Goal: prove polynomial- time reduction

Polynomial-Time Reducible

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

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A₂ Transform

function f 𝑥 𝑓 𝑥

A₁

P v.s. NP

• If one proves that SAT can be solved by a polynomial-time algorithm, then NP = P.

• If somebody proves that SAT cannot be solved by any polynomial-

time algorithm, then NP ≠ P.

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

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Satisfiable Unsatisfiable

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}

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

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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

Karp’s NP-Complete Problems

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

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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 }

SAT ∈ NP

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

73 initial

configuration

polynomial

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

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

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CIRCUIT-SAT ≤_p SAT → SAT ∈ NP-hard

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

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

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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

SAT ≤ _p 3-CNF-SAT

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

SAT ≤ _p 3-CNF-SAT

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

79

SAT ≤ _p 3-CNF-SAT

c)Convert each clause Φ_i’ to CNF

• Construct a truth table for each clause Φ_i’

• Construct the disjunctive normal form for ¬Φ_i’

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

𝒚_𝟏 𝒚_𝟐 𝒚_𝟐 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

𝒚_𝟏 𝒚_𝟐 𝒚_𝟐 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

SAT ≤ _p 3-CNF-SAT

d)Construct Φ’’’ in which each clause C_i 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

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Concluding Remarks

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

• 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

P ≠ NP

P = NP

Question?

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