Computational Models - Lecture...

Computational Models - Lecture 11 1

Handout Mode

Iftach Haitner and Yishay Mansour.

Tel Aviv University.

May 7/9, 2012

1Based on frames by Benny Chor, Tel Aviv University, modifying frames byMaurice Herlihy, Brown University.

Iftach Haitner and Yishay Mansour (TAU) Computational Models Lecture 11 May 7/9, 2012 1 / 64

Talk Outline

Reminder – deterministic and nondeterministic time classes

Additional NP languages

The class co-NP

P Verses NP

NP-completeness

Satisfiability

Reductions

Sipser’s book, 7.4–7.5

Reminder – Deterministic Time

Definition 1 (deterministic Time)

A deterministic TM M runs in time t , where t : N 7→ N, if for every inputw , the number of steps that M(w) uses is at most t(|w |).

Definition 2 ( DTIME)

For t : N 7→ N, letDTIME(t) = {L ⊆ Σ∗ : L is decided by an O(t)-time single tape TM}

The bound on the running time is required to hold also for input not inthe language (i.e., L).

Definition 3 ( P)

P =⋃

c≥0 DTIME(nc)

Reminder – NonDeterministic Time

Definition 4 (nondeterministic Time)

A non-deterministic TM N runs in time t , where t : N 7→ N, if for everyinput w , the maximum number of steps that N(w) uses on any branchof its computation tree, is at most t(|w |).

Notice that also non-accepting branches must reject within therequired time.

Definition 5 ( NTIME)

For t 7→ N 7→ N let NTIME(t) ={L ⊆ Σ∗ : L is decided by an O(t)-time single tape NTM}

Definition 6 ( NP)

NP =⋃

c≥0 NTIME(nc)

Reminder – NonDeterministic Time, cont.

Definition 7 (verifier)

A verifier for a language L, is an algorithm V withL = {w : V accepts 〈w , c〉 for some string c}.

Theorem 8A language is in NP iff it has a polynomial time verifier.

Section 1

Additional NP Languages

CLIQUE

A clique in a graph is a subgraph where every two nodes areconnected by an edge.A k-clique is a clique of size k .

Question 9What is the largest k-clique in the figure?

CLIQUE cont.

CLIQUE = {〈G, k〉 : G is an undirected graph with a k-clique}

Theorem 10CLIQUE ∈ NP

Proof’s idea: The clique is the certificate.

Algorithm 11 ( V)

On input (〈G, k〉, c)Accept if c is a k-clique subgraph of G;Otherwise Reject.

SUBSET-SUM

A collection of non-negative integers x1, . . . , xk

A target number t

Question: does some subcollection add up to t?

Example 12

{4,11,16,21,27},25) ∈ SUBSET-SUM because 4 + 21 = 25.

{4,11,16,21,27},26) /∈ SUBSET-SUM (why?)

SUBSET-SUM cont.

SUBSET-SUM = {〈S = {x1, . . . , xk}, t〉 : ∃{y1, . . . , y`} ⊆S :

∑`j=1 yj = t}

Collections are sets: repetitions not allowed.

Theorem 13SUBSET-SUM ∈ NP

Proof’s idea: The subset is the certificate.

Algorithm 14 ( V)

On input (〈S = {x1, . . . , xk}, t〉, c):Accept if the following holds (otherwise Reject):

1 c is a collection of numbers summing to t .2 c is a subset of S

Theorem 15There is a pseudo-polynomial algorithm for SUBSET-SUM

Graph characterization of SUBSET-SUM

Definition 16For 〈S = {x1, . . . , xk}, t〉, let G(S, t) = (V ,E)

V = V0 ·⋃

V1 . . . ·⋃

Vk , where Vi = {vi ,0, . . . , vi ,t}

E = E1 ·⋃. . . ·

⋃Ek , where

Ei = {(vi−1,j , vi ,j), (vi−1,j , vi ,j+xi) : j ∈ {0, . . . , t}}

Claim 17vi ,j is reachable from v0,0 in G(S, t), iff〈{x1, . . . , xi}, j〉 ∈ SUBSET-SUM.

Proof?

Pseudo-polynomial algorithm for SUBSET-SUM

Algorithm 18

Input 〈S, t〉

Return TRUE iff vk ,t is reachable from v0,0 in G(S, t).

Correctness

Size of G(S, t) is O(|S| · t), and therefore the resulting algorithmfor SUBSET-SUM runs in time poly(|S| · t).

Not poly(log2 t) but poly(t) (i.e., pseudo-polynomial)

Independent Set

An independent set in a graph is a set of vertexes, no two of which arelinked by an edge.

A k-IS is an independent set of size k .

Question 19What is the largest k-IS in the figure?

IND-SET cont.

IND-SET = {〈G, k〉 : G contains an independent set of size k}

Theorem 20IND-SET ∈ NP

Proof’s idea: The independent set is the certificate.

Algorithm 21 ( V)

On input (〈G, k〉, c)Accept if c is a k-IS of G (no edges between nodes in c, and |c| = k);Otherwise Reject.

KNAPSACK

A collection of non-negative integers x1, . . . , xk (size)

A collection of non-negative integers y1, . . . , yk (benefit)

A capacity B

A target number t

Question: does some S ⊂ [1, k ] has∑

i∈S xi ≤ B and∑

i∈S yi ≥ t?

Example 22

({4,11,16,21,27}, {7,3,9,5,35},B = 20, t = 16) ∈ KNAPSACKbecause S = {1,3}.

KNAPSACK cont.

Theorem 23KNAPSACK ∈ NP

Proof’s idea: The subset is the certificate.

Algorithm 24 ( V)

On input (〈{x1, . . . , xk}, {y1, . . . , yk},B, t〉, c):Accept if the following holds (otherwise Reject):

i∈c xi ≤ B2

∑i∈c yi ≥ t .

3 c is a subset of [1, . . . , k ]

Pseudo-polynomial algorithm for KNAPSACK

Theorem 25There is a pseudo-polynomial algorithm for KNAPSACK

Running time O(k · t)

Not poly(log2 t) but poly(t) (i.e., pseudo-polynomial)

Algorithm 26 (KNAPSACK)

A(0,0) = 0 and A(0,p) = ∞ for p 6= 0.For i = 1 to k DOFor p = 0 to t DOA(i ,p) = min{A(i − 1,p), xi + A(i − 1,p − yi)}ENDTest t ≤ max{p|A(k ,p) ≤ B}.

Section 2

The Class coNP

The Class co-NP

CLIQUE and SUBSET-SUM seem not to be members of NP .

It is harder to efficiently verify that something does not exist than toefficiently verify that something does exist..

Definition 27 ( co-NP)

co-NP = {L : L ∈ NP}.

So far, no one knows if co-NP is distinct from NP.

Claim 28P ⊆ co-NP .

Proof?

L ∈ P =⇒ L ∈ P =⇒ L ∈ NP =⇒ L ∈ co-NP

Section 3

P vs. NP

P Vs. NP

P P=NP

The question P?= NP is one of the great unsolved mysteries in

contemporary mathematics.

Most computer scientists believe the two classes are not equal

Most bogus proofs show them equal (why?)

Observations

If P differs from NP , then the distinction between P and NP \ P ismeaningful and important.

languages in P are tractable

languages in NP \ P are intractable

Until we can prove that P 6= NP, there is no hope of proving that aspecific language lies in NP \ P .

Nevertheless, we can prove statements of the form

If A ∈ NP \ P then B ∈ NP \ P.

If co-NP 6= NP then P 6= NP .

Section 4

NP Completeness

NP-complete

The class of NP-complete languages are“hardest” languages in NP

“least likely” to be in P

If any NP-complete L ∈ P, then NP = P .Such languages, “carry on their backs” the burden of all of NP.

Question 29Are there NP-complete languages?

Polynomial-Time Computable Functions

Definition 30 (poly-time computable functions)

A function f : Σ∗ 7→ Σ∗ is polynomial-time computable, if there is apoly-time deterministic TM that

starts with input w , and

halts with f (w) on tape.

Polynomial-Time Reducibility

Definition 31A language A is polynomial time mapping reducible to B, denotedA ≤P B, if exists poly-time computable f such that

w ∈ A ⇐⇒ f (w) ∈ B.

for every w ∈ Σ∗.The function f is called a polynomial-time reduction from A to B.

The mapping f efficiently converts questions about membership in A tomembership in B.

Reductions to P

Theorem 32

If A ≤P B and B ∈ P then A ∈ P.

Proof:

Let f the reduction from A to B, computed by TM Mf .On input x , the TM Mf makes at most cf · |x |af steps.Let MB be the poly-time decider for B.On input y , the TM MB makes at most cB · |y |aB steps.

Algorithm 33 (Decider MA for A)

On input x , return MB(f (x))

MA decides ARunning time of MB(x), is at mostcB · (cf · |x |af )aB = (cB · caB

f ) · |x |af ·aB ∈ poly(|x |)Hence, A ∈ P

♣Iftach Haitner and Yishay Mansour (TAU) Computational Models Lecture 11 May 7/9, 2012 27 / 64

Remarks

Question 34Assume that {0n1n : n ≥ 0} ≤P L. Does it yield that L ∈ P?

L = HTM

If x = 0n1n then f (x) = Mstop

Otherwise f (x) = Mrun.

NP Completeness, Formal Definition

Definition 35 ( NP-complete)

A language B is NP-complete, if

B ∈ NP , and

Every A ∈ NP is poly-time reducible to B (i.e., A ≤P B)

We let NPC denote the class of all NP-complete languages.

Compare to

Definition 36 (RE-Complete)

A language B is RE-complete, if

B ∈ RE , and

Every A ∈ RE is mapping reducible to B.

Why NP Completeness?

Theorem 37

If B ∈ NPC and B ∈ P, then P = NP .

Proof: Immediately follows by Thm 32. ♣

To show P = NP (and make an instant fortune, seewww.claymath.org/millennium/P_vs_NP/), suffices to find apolynomial-time algorithm for any NP-complete problem.

Question 38Is NPC empty?

NPC Is Not Empty

ANP = {〈M, x ,1n〉 : M is a TM ∧ ∃c ∈Σ∗ s.t. M(x , c) accepts within n steps}

Theorem 39ANP ∈ NPC

Proof:

Clearly ANP ∈ NP

Let L ∈ NP , let V be a verifier for L and let p ∈ poly be a bound onthe running time of V (i.e., V(w , ·) halts within p(|w |) steps, forevery w ∈ Σ∗)).

Define f (x) = 〈V, x ,1p(|x|)〉.

Clearly f is poly-time computable and x ∈ L ⇐⇒ f (x) ∈ ANP.

Finding Additional NP-complete Languages

Theorem 40Assume that

1 B ∈ NP

2 A ∈ NPC and A ≤P B

then B ∈ NPC.

Proof: Home exercise . . .♣

We would like to find L ∈ NPC that is “natural" and “easy" to reduce to.

The most useful such language is the language of satisfied formulas.

Section 5

Satisfiability

Boolean Variables

A Boolean variable assumes valuesI TRUE (written 1), and FALSE (written 0).

Boolean operations:I and: ∧I or: ∨I not: ¬

Examples:

0 ∧ 1 = 0

0 ∨ 1 = 1

Boolean Formulas and SATA Boolean formula is an expression involving Boolean variables andoperations.

φ = (x ∧ y) ∨ (x ∧ z)

Definition 41 (satisfiable formula)

A formula is satisfiable, if some assignment of 0s and 1s to thevariables makes the formula evaluate to 1.

The formula φ = (x ∧ y) ∨ (x ∧ z) is satisfiable by the assignment

The language of satisfied formulas:

SAT = {〈φ〉 : φ is a satisfiable Boolean formula}

SAT ∈ NPC

SAT = {〈φ〉 : φ is satisfiable Boolean formula}

Theorem 42 (Cook-Levin (early 70s))

SAT ∈ NPC.

The “most important" NP-complete language.

It is easy to see that SAT ∈ NP

For the proof of other part wait for next week . . .

The Language 3SATIt is useful to consider a special version of SAT

A literal is a variable or negated variable: x or x .A clause is several literals joined by ∨s: (x1 ∨ x2 ∨ x3)

A Boolean formula is in conjunctive normal form (CNF) if itconsists of clauses, connected with ∧s.For example: (x1 ∨ x2 ∨ x3 ∨ x4) ∧ (x3 ∨ x5 ∨ x6) ∧ (x3 ∨ x6)

Definition 43A Boolean formula is in k-CNF form, if it is a CNF formula, and allclauses have k literals.

Example of 3CNF: (x1 ∨ x2 ∨ x3) ∧ (x3 ∨ x5 ∨ x6) ∧ (x3 ∨ x6 ∨ x4)

The language of satisfied 3CNF formulas:

3SAT = {〈φ〉 : φ is satisfiable 3CNF formula}

3SAT ∈ NPC

It is trivial that 3SAT ≤P SAT

Clearly 3SAT ∈ NP

We later show that SAT ≤P 3SAT, yielding that 3SAT ∈ NPC.

Since 3SAT has more structure than SAT, it is simpler to reduce itto other languages

In the following we prove that several NP languages are NPC, byshowing a reduction from 3SAT to them

1SAT,2SAT ∈ P

Question 44Is 1SAT ∈ P ?

Question 45Is 2SAT ∈ P ?

A Graph characterization for 2CNF

Definition 46For 2CNF formula φ with variables x1, . . . , xk , define G(φ) = (V ,E) as

V = {x1, x1, . . . , xk , xk}

E = {(`, `′), (`′, `) : (` ∨ `′) ∈ φ} (taking x = x).

Claim 47

G(φ) contains path from ` to `′ =⇒ in any satisfying assignment of φwith ` = 1, it holds that `′ = 1.

Proof?

Claim 48

G(φ) contains path from ` to `′ =⇒ G(φ) contains path from `′ to `

Proof?

A Graph characterization for 2CNF , cont

Definition 49A variable x in a a 2CNF φ is bad, if there are paths in G(φ) from x tox and from x to x .

Claim 50A 2CNF formula is satisfiable iff it contains no bad variables.

φ has bad =⇒ φ is unsatisfiable. Proof? by Claim 47φ has no bad variables =⇒ φ is satisfiable. Proof?

Algorithm 511 Pick unassigned literal ` ∈ V that has no path from ` to `2 Assign 1 to all literals reachable from `, and 0 to their negations.3 Repeat.

Can there be conflicts?Same iteration? No since φ has no bad variablesDifferent iterations? No by Claim 48.

Poly-time algorithm for 2SAT

Algorithm 52 ( TwoSatSolver)

Input: 2CNF φReturn TRUE if there exist no bad variables in φ:

Efficiency?

Correctness?

Section 6

Clique

3SAT ≤P CLIQUE

CLIQUE = {〈G, k〉 : G is an undirected graph with a k-clique}

Claim 533SAT ≤P CLIQUE

Since CLIQUE ∈ NP, it follows that CLIQUE ∈ NPCProof’s idea: We’ll construct a poly-time reduction f that maps 3CNFformulae φ to graphs and numbers 〈G, k〉.

The function f will have the property that φ is satisfiable, iff G has aclique of size k .

Proving 3SAT ≤P CLIQUE

Let φ be a 3CNF formula with k clauses.On going example:

(x1 ∨ x2 ∨ x3) ∧ (x3 ∨ x5 ∨ x6) ∧ (x3 ∨ x6 ∨ x4)

We define a graph G = G(φ) as follows:

Nodes in G are organized into triples t1, . . . , tk .

Each triple corresponds to a clause of φ

Each node in a triple corresponds to a literal in correspondingclause.

Nodes of G

φ = (x1 ∨ x2 ∨ x3) ∧ (x3 ∨ x5 ∨ x6) ∧ (x3 ∨ x6 ∨ x4)

Add a node per literal

Edges of G

φ = (x1 ∨ x2 ∨ x3) ∧ (x3 ∨ x5 ∨ x6) ∧ (x3 ∨ x6 ∨ x4)

Add edges between all vertex pairs, exceptwithin same triplebetween contradictory literals (e.g., x3 and x3)

φ ∈ 3SAT =⇒ 〈G, k〉 ∈ CLIQUE

Proof: Suppose φ is satisfiable by an assignment ψ.With respect to ψ:

At least one literal is assigned to 1 in every clause (why?)

Select a 1-literal in every tuple;These literals can be joined by edges (why?)Yielding a k-clique in G. ♣

〈G, k〉 ∈ CLIQUE =⇒ φ ∈ 3SAT

Proof: Suppose G has a k-clique.

No two of the cliques nodes are in the same triple. (why?)G has k vertexes and k clauses, so each triple has exactly oneclique node.Assign 1 to each node in clique;Assignment has no contradictions (why?)Yielding a satisfying assignment to φ.

We’ve constructed a poly-time computable function f .

We saw that f has the property that φ ∈ 3SAT iff f (φ) ∈ CLIQUE.

Therefore, f is a poly-time reduction from 3SAT to CLIQUE

=⇒ 3SAT ≤P CLIQUE. ♣

Section 7

Independent Set

Definition 54An independent set in a graph is a set of vertexes, no two of which arelinked by an edge.

The language of independent sets:

IND-SET = {〈G, k〉 : G contains an independent set of size k}

Clearly IND-SET ∈ NP. We show that CLIQUE ≤P IND-SET anddeduce that IND-SET ∈ NPC

CLIQUE ≤P IND-SET

Proof:

Definition 55The complement of a graph G = (V ,E) is a graph Gc = (V ,Ec), where

Ec = {(v1, v2) : v1, v2 ∈ V and (v1, v2)6∈E}.

Claim 56If U is an independent set in G, then U is a clique in Gc.

The reduction from CLIQUE to IND-SET simply compute thecomplement of the graph. ♣

Remark 57Same reduction shows that IND-SET ≤P CLIQUE

Independent Set 7→ Clique

Section 8

Hamiltonian Paths and Cycles

Reminder – Hamiltonian Path

A Hamiltonian path in a directed G visits each node exactly once.

HAMPATH = {〈G, s, t〉 : G has Hamiltonian path from s to t}

Theorem 58HAMPATH ∈ NPC.

Not today . . .

Hamiltonian Cycle

A Hamiltonian Cycle in a graph is an Hamiltonian path that ends upwhere it starts (i.e., s = t).

HAMCYCLE = {〈G〉 : G has Hamiltonian cycle}

Clearly HAMCYCLE ∈ NP . We will show thatHAMPATH ≤P HAMCYCLE, and deduce that HAMCYCLE ∈ NPC

HAMPATH ≤P HAMCYCLE.

Proof’s idea:

Hey, is the new vertex really needed? Why not just add an edge from tto s?

HAMPATH ≤P HAMCYCLE.

Why the new vertex really needed? Why not just add an edge from t tos?

HAMCYCLE ≤P HAMPATH.

Claim 59HAMCYCLE ≤P HAMPATH.

Left as an easy (recommended) exercise.

Section 9

Traveling Salesman

Input: set of cities C and set of inter-city distances DGoal: find a an hamiltonian cycle (TS tour) of total distance atmost k

TRAVELING-SALESMAN ={〈C,D, k〉 : (C,D) has a TS tourof total distance ≤ k}

HAMCYCLE ≤P TRAVELING-SALESMAN

Proof: Given a directed graph G = (V ,E), construct travelingsalesman instance.

The cities are identical to the nodes of the original graph, C = V .The distance of going from v1 to v2 is 1 if (v1, v2) ∈ E , and 2otherwise.The bound on the total distance of a tour is k = |V |.

correctness:=⇒ Suppose G has a Hamiltonian cycle.The distance assigned by the reduction to all edges in this cycle is1. Thus in (C,D) there is a traveling salesman tour of total distance|V | = k , namely (C,D, k) ∈ TRAVELING-SALESMAN.⇐= Suppose (C,D) has a traveling salesman tour of total distance|V | = k .Tour cannot contain any edge of distance 2. Therefore it gives aHamiltonian cycle in G.

Efficiency: reduction done in quadratic time (filling up distancesfor all edges of the complete graph).

Chains of Reductions: NPC ProblemsSAT

IntegerProg 3SAT

Clique

IndepSet

VertexCover

SetCover

3ExactCover

Knapsack

Scheduling

3Color HamPath

HamCircuit

TRAVELING-SALESMAN

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