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The Theory of NP-Completeness

2010/11/30

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Polynomial-time Reductions

We want to solve a problem R; we already have an algorithm for a problem S

We have a transformation function T Correct answer for R on x is “yes”, iff the

correct answer for S on T(x) is “yes” Problem R is polynomially reducible to

(polynomially reduces to) problem S if such a transformation T can be computed in polynomial time (R S)

The point of reducibility: S is at least as hard to solve as R

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Polynomial-time Reductions

We use reductions to prove that a problem is NP-complete

x is an input for R; x’ is an input for S (R S)

T Algorithm for Sx’

Yes or Noanswer

x

Algorithm for R

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NP

PNPC

NP: Non-deterministic Polynomial

P: Polynomial

NPC: Non-deterministic Polynomial Complete

P=NP?

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NP : the class of decision problem which can be solved by a non-deterministic polynomial algorithm.

P: the class of problems which can be solved by a deterministic polynomial algorithm.

NP-hard: the class of problems to which every NP problem reduces. (It is “at least as hard as the hardest problems in NP.”)

NP-complete: the class of problems which are NP-hard and belong to NP.

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Nondeterministic algorithms A nondeterministic algorithm

is an algorithm consisting of two phases: guessing (or choice) and checking.

Furthermore, it is assumed that a nondeterministic algorithm always makes a correct guessing.

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Nondeterministic algorithms

Machines for running nondeterministic algorithms do not exist and they would never exist in reality. (They can only be made by allowing unbounded parallelism in computation.) (Quantum computation may be an approximation.)

Nondeterministic algorithms are useful only because they will help us define a class of problems: NP problems

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NP (Nondeterministic Polynomial) algorithm

If the checking stage of a nondeterministic algorithm is of polynomial time-complexity, then this algorithm is called an NP (nondeterministic polynomial) algorithm.

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NP problem If a decision problem can be solved by a

NP algorithm, this problem is called an NP (nondeterministic polynomial) problem.

NP problems : must be decision problems

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Decision problems The solution is simply “Yes” or “No”. Optimization problem : harder

Decision problem : easier E.g. the traveling salesperson problem

Optimization version:Find the shortest tour (cycle)

Decision version:Is there a tour whose total length is less than or equal to a constant C ?

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Decision version of sorting

Given a1, a2,…, an and c, is there a

permutation of ais ( a1

, a2 , … ,an

)

such that∣a2–a1

∣+∣a3–a2

∣+ … +∣an–

an-1∣＜ C ?

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Decision vs Original Version

We consider decision version problem

D rather than the original problem O

because we are addressing the lower

bound of a problem

and D ∝ O

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To express Nondeterministic Algorithm by three special commands

Choice(S) : arbitrarily chooses one of the elements in set S at one step

Failure : an unsuccessful completion

Success : a successful completion

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Example:Nondeterministic searching Algorithm

input: n elements and a target element xoutput: success if x is found among the n elements; failure, otherwise.

j ← choice(1 : n) /* guessif A(j) = x then success /* check else failure

A nondeterministic algorithm terminates unsuccessfully iff there exist no set of choices leading to a success signal.

The time required for choice(1 : n) is O(1).

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Relationship Between NP and P

It is known PNP. However, it is not known whether P=NP or

whether P is a proper subset of NP It is believed NP is much larger than P

We cannot find a polynomial-time algorithm for many NP problems.

But, no NP problem is proved to have exponential lower bound. (No NP problem has been proved to be not in P.)

So, “does P = NP?” is still an open question! Cook tried to answer the question by

proposing NPC.

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NP-hardA problem X is NP-hard if every NP problem (polynomially) reduces to X.

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NP-complete (NPC)A problem X is NP-complete (NPC) if X∈NP and every NP problem (polynomially) reduces to X.

So, X X NPC NPCif Xif XNP-hard and XNP-hard and XNP NP

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Cook’s Theorem (1971) Prof. Cook Toronto U. Receiving Turing Award (1982) Discussing difficult problems whose

worst case lower bound seems to be in the order of an exponential function

The problems are called NP-complete (NPC) Problems

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Cook’s Theorem (1971) NP = P iff SAT NP = P iff SAT P P That is, NP = P iff the satisfiability That is, NP = P iff the satisfiability

(SAT) problem is a P problem(SAT) problem is a P problem SAT is NP-complete It is the first NP-complete problem Every NP problem reduces to SAT

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SAT is NP-complete Every NP problem can be solved by an

NP algorithm Every NP algorithm can be transformed

in polynomial time to an SAT problem Such that the SAT problem is satisfiable

iff the answer for the original NP problem is “yes”

That is, every NP problem SAT SAT is NP-complete

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Proof of NP-Completeness To show that X is NP-complete

(I) Prove that X is an NP problem (II) Prove that Y NPC, Y X

X NPC Why ?

Transitive property of polynomial-time reduction

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Karp R. Karp showed several NPC

problems, such as 3-STA, node (vertex) cover, and Hamiltonian cycle problems, etc.

Karp received Turing Award in 1985

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NP-Completeness Proof: Reduction

Vertex Cover

Clique 3-SAT

SAT

Chromatic Number

Dominating Set

All NP problems

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NP-Completeness “NP-complete problems”: the

hardest problems in NP Interesting property

If any one NP-complete problem can be solved in polynomial time, then every problem in NP can also be solved in polynomial time (i.e., P=NP)

Many believe P≠NP

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Importance of NP-Completeness

NP-complete problems: considered “intractable”

Important for algorithm designers & engineers Suppose you have a problem to solve

Your colleagues have spent a lot of time to solve it exactly but in vain

See whether you can prove that it is NP-complete If yes, then spend your time developing an

approximation (heuristic) algorithm Many natural problems can be NP-complete

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Some concepts Up to now, none of the NPC problems can be

solved by a deterministic polynomial time algorithm in the worst case.

It does not seem to have any polynomial time algorithm to solve the NPC problems.

The lower bound of any NPC problem seems to be in the order of an exponential function.

The theory of NP-completeness always considers the worst case.

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Caution !

If a problem is NP-complete, its special cases may or may not be of exponential time-complexity.

We consider worst case lower bound in NP-complete.

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Even harder problems:Undecidable Problems

They cannot be solved by guessing and checking.

They are even more difficult than NP problems.

E.G.: Halting problem: Given an arbitrary program with an arbitrary input data (the number of input cases is infinite), will the program terminate or not? It is not NP It is NP-hard (SAT Halting problem)

Some known NPC problems SAT problem 3SAT problem 0/1 knapsack problem Traveling salesperson problem Partition problem Art gallery problem Clique problem Vertex Cover problem Dominating Set problem Chromatic Coloring problem

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The satisfiability (SAT) problem

Def : Given a Boolean formula, determine whether this formula is satisfiable or not.

  A literal (Boolean variable): xi or -xi

A clause (disjunction of variables): ci x1 v x2

v -x3

A formula : conjunctive normal form C1& c2 & … & cm

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• SAT: Give a Boolean expression (formula) in CNF (conjunctive normal form), determine if it is satisfiable.

• 3SAT: Give a Boolean expression in CNF such that each clause has exactly 3 variables (literals), determine if it is satisfiable.

NPC Problems

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The satisfiability (SAT) problem

The satisfiability problem The logical formula :

x1 v x2 v x3

& - x1

& - x2 the assignment :

x1 ← F , x2 ← F , x3 ← Twill make the above formula true

(-x1, -x2 , x3) represents

x1 ← F , x2 ← F , x3 ← T

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The satisfiability problem If there is at least one assignment which

satisfies a formula, then we say that this formula is satisfiablesatisfiable; otherwise, it is unsatisfiableunsatisfiable.

An unsatisfiable formula : x1 v x2

& x1 v -x2

& -x1 v x2

& -x1 v -x2

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Let formula F=c1 : -x1 v -x2 v x3

c2 : x1 v x4

c3 : -x2 v x3 v x4 (resolvent)

Resolution principle

x1 cannot satisfyc1 and c2 at thesame time, so itis deleted to deducea resolvent of c1 andc2. This is called theresolution principle .

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If no new clauses can be deduced satisfiable

-x1 v -x2 v x3 (1)

x1 (2)

x2 (3)

(1) & (2) -x2 v x3 (4)

(4) & (3) x3 (5)

(1) & (3) -x1 v x3 (6)

Satisfiable

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Unsatisfiable If an empty clause is deduced unsatisfiable - x1 v -x2 v x3 (1)

x1 v -x2 (2)

x2 (3)

- x3 (4)  deduce

  (1) & (2) -x2 v x3 (5)

(4) & (5) -x2 (6) (6) & (3) □ (7)

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Nondeterministic SAT Guessing for i = 1 to n do

xi ← choice( true, false )

if E(x1, x2, … ,xn) is true

Checking then success

else failure

Proof of Cook’s Theory Dr. Cook Prove the theory by

showing that all instances of each NP algorithm can reduce to the SAT problem.

Below, we show a simple example that the NP searching algorithm reduces to the SAT problem.

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The NP searching algorithm reduces to the SAT problem

Does there exist a number in { x(1), x(2), …, x(n) }, which is equal to 7?

For example, assume n = 2 The algorithm becomes:

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The NP searching algorithm reduces to SAT

i=1 v i=2& i=1 → i≠2 & i=2 → i≠1 & x(1)=7 & i=1 → SUCCESS & x(2)=7 & i=2 → SUCCESS & x(1)≠7 & i=1 → FAILURE & x(2)≠7 & i=2 → FAILURE & FAILURE → -SUCCESS & SUCCESS

(Guarantees a successful termination when it is possible) & x(1)=7 (Input Data) & x(2)≠7

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The NP searching algorithm reduces to SAT CNF (conjunctive normal form) :

i=1 v i=2 (1) i≠1 v i≠2 (2) x(1)≠7 v i≠1 v SUCCESS (3) x(2)≠7 v i≠2 v SUCCESS (4) x(1)=7 v i≠1 v FAILURE (5) x(2)=7 v i≠2 v FAILURE (6) -FAILURE v -SUCCESS (7) SUCCESS (8) x(1)=7 (9) x(2)≠7 (10)

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The NP searching algorithm reduces to SAT Satisfiable at the following assignment :

i=1 satisfying (1) i≠2 satisfying (2), (4) and (6) SUCCESS satisfying (3), (4) and (8) -FAILURE satisfying (7) x(1)=7 satisfying (5) and (9) x(2)≠7 satisfying (4) and (10)

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The NP searching algorithm reduces to SAT Searching for 7, but x(1)7, x(2)7

CNF :

i=1 v i=2 (1) i1 v i2 (2) x(1)7 v i1 v S U C C E S S ( 3 ) x(2)7 v i2 v S U C C E S S ( 4 ) x(1)=7 v i1 v FAILURE (5 ) x(2)=7 v i2 v FAILURE (6 ) SUCCESS (7) -SUCCESS v -FAILURE (8) x(1) 7 (9) x(2) 7 (10)

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The NP searching algorithm reduces to SAT Apply resolution principle :

(9) & (5) i1 v FAILURE (11) (10) & (6) i2 v FAILURE (12) (7) & (8) -FAILURE (13) (13) & (11) i1 (14) (13) & (12) i2 (15) (14) & (1) i=2 (11) (15) & (16) □ (17)

We get an empty clause unsatisfiable 7 does not exit in x(1) or x(2).

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Searching in CNF with inputs

Searching for 7, where x(1)=7, x(2)=7 CNF : i=1 v i=2 (1) i1 v i2 (2) x(1)7 v i1 v S U C C E S S ( 3 ) x(2)7 v i2 v S U C C E S S ( 4 ) x(1)=7 v i1 v F A I L U R E ( 5 ) x(2)=7 v i2 v F A I L U R E ( 6 ) SUCCESS (7) -SUCCESS v -FAILURE (8) x(1)=7 (9) x(2)=7 (10)

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0/1 Knapsack problem

Given M (weight limit) and P, is there is a solution with a profit larger than P?

This is an NPC problem.

P1 P2 P3 P4 P5 P6 P7 P8

Profit 10 5 1 9 3 4 11 17

Weight

7 3 3 10 1 9 22 15

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Traveling salesperson problem Given: A set of n planar points and

a value LFind: Is there a closed tour which includes all points exactly once such that its total length is less than L?

This is an NPC problem.

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Partition problem

Given: A set of positive integers SFind: Is there a partition of S1 and S2 such that S1S2=, S1S2=S, iS1i=iS2 i(partition S into S1 and S2 such that element sum of S1 is equal to that of S2)

e.g. S={1, 7, 10, 9, 5, 8, 3, 13} S1={1, 10, 9, 8} S2={7, 5, 3, 13}

This problem is NP-complete.

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Art gallery problem: *Given a constant C, is there a guard placement such that the number of guards is less than C and every wall is monitored?*This is an NPC problem.

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NPC Problems CLIQUE(k): Does G=(V,E) contain a

clique of size k?

Definition: A clique in a graph is a set of vertices

such that any pair of vertices are joined by en edge.

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NPC Problems Vertex Cover(k): Given a graph G=(V,

E) and an integer k, does G have a vertex cover with k vertices?

Definition: A vertex cover of G=(V, E) is V’V such

that every edge in E is incident to some vV’.

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Dominating Set(k): Given an graph G=(V, E) and an integer k, does G have a dominating set of size k ?

Definition: A dominating set D of G=(V, E) is

DV such that every vV is either in D or adjacent to at least one vertex of D.

NPC Problems

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• Chromatic Coloring(k): Given a graph G=(V, E) and an integer k, does G have a coloring for k

Definition A coloring of a graph G=(V, E) is a

function f : V { 1, 2, 3,…, k } if (u, v) E, then f(u)f(v).

NPC Problems

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Q&A