Computational Models — Lecture...

Computational Models — Lecture 12

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Roded Sharan.

Tel Aviv University.

June, 2018

Roded Sharan (TAU) Computational Models, Lecture 12 June, 2018 1 / 49

Talk Outline

I NP-completeness

I Satisfiability

I Cook-Levin theorem

I CSAT ∈ NPCI 3SAT ∈ NPCI Clique is NPC

I Hamiltonian Path and Hamiltonian Cycle are NP-Complete

Sipser’s book, 7.4–7.5


Section 1

NP Completeness


NP Completeness

P

NP

NP-complete

The class of NP-complete languages are

I “hardest” languages in NPI If any NP-complete L ∈ P, then NP = P.

Question 1Are there NP-complete languages?


Polynomial-time reducibility

Definition 2 (poly-time computable functions)

A function f : Σ∗ 7→ Σ∗ is polynomial-time computable, if there is a poly-timedeterministic TM thatI starts with input w , and

I halts with f (w) on tape.

Definition 3 (poly-time reduction)

A polynomial-time computable f : Σ∗ 7→ Σ∗ is a poly-time reduction fromlanguage A to B, if x ∈ A⇐⇒ f (x) ∈ B for every x ∈ Σ∗.Is such a reduction from A to B exists, we say that A is poly- time mappingreducible to B, denoted A ≤P B.

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


Example: CLIQUE ≤P IND-SET

Proof:

Definition 4The complement of a graph G = (V ,E) is a graph Gc = (V ,Ec), whereEc = {(v1, v2) : v1, v2 ∈ V and (v1, v2)6∈E}.

The reduction f (G, k) from CLIQUE to IND-SET simply computes thecomplement of the graph and outputs (Gc , k).

f satisfies:

I U is a clique in G ⇐⇒ U is a independent set in Gc .

I computable in polynomial time!

♣

Remark 5Same reduction shows that IND-SET ≤P CLIQUE


Reductions to P

Theorem 6

If A ≤P B and B ∈ P then A ∈ P.

Proof:

I 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.

I Let MB be the poly-time decider for B.On input y , the TM MB makes at most cB · |y |aB steps.

Algorithm 7 (Decider MA for A)

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

I MA decides A

I Since |f (x)| ≤ cf |x |af , running time of MB(x), is at mostcB · (cf · |x |af )aB = (cB · caB

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

♣ Roded Sharan (TAU) Computational Models, Lecture 12 June, 2018 7 / 49

What A ≤P B tells us about B?

Question 8Assume that {0n1n : n ≥ 0} ≤P L. Does it yield that L ∈ P?

Answer: No. (Reduction in the wrong direction!)

Let L = HTM,ε and define f (x) =

{Mstop, x ∈ {0n1n : n ∈ N}Mno−stop, otherwise. .

A ≤P B does tell us that B is “at least as hard" as A.


NP completeness, formal definition

Definition 9 (NP-complete)

A language B is NP-complete, if

I B ∈ NP, and

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

Let NPC denote the class of all NP-complete languages.

Compare to

Definition 10 (RE-complete)

A language B is RE-complete, if

I B ∈ RE , and

I Every A ∈ RE is mapping reducible to B.


Why NP completeness?

Theorem 11

If B ∈ NPC and B ∈ P, then P = NP.

Proof: Immediately follows by Thm 6. ♣

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

Question 12Is 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 13ANP ∈ NPC

Proof:

I Clearly ANP ∈ NP.

I Let L ∈ NP, let V be a verifier for L and let p ∈ poly be a bound on therunning time of V (i.e., V(x , ·) halts within p(|x |) steps, for every x ∈ Σ∗).

I Define f (x) = 〈V, x ,1p(|x|)〉.I f is poly-time computable

I x ∈ L⇐⇒ f (x) ∈ ANP.

♣


Finding additional NP-complete languages

Theorem 14Assume 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.


Section 2

Satisfiability


Boolean variables

I A Boolean variable assumes values

I TRUE (written 1), and FALSE (written 0).

I Boolean operations:

I and: ∧I or: ∨I not: ¬

I Examples:

0 ∧ 1 = 00 ∨ 1 = 1

0 = 1


Boolean formulas and SAT

A Boolean formula is an expression involving Boolean variables andoperations.

φ = (x ∧ y) ∨ (x ∧ z)

Definition 15 (satisfiable formula)

A formula is satisfiable, if some Boolean assignment to its variables, makesthe formula evaluate to 1.

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

x = 0y = 1z = 0

The language of satisfied formulas:

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


SAT ∈ NPC

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

Theorem 16 (Cook-Levin (early 70s))SAT ∈ NPC.

I The “most important" NP-complete language.

I It is easy to see that SAT ∈ NP


Section 3

Proving SAT ∈ NPC


The proof, high level

I Let L ∈ NP and let N = (Q,Σ, Γ, δ,q0,qa,qr ) be an t-time NTM thataccepts L, for some t ∈ poly.

I Given the string w ∈ {0,1}∗, construct in time O(t(|w |)2) a formula φN,wsuch that:

φN,w ∈ SAT⇐⇒ N accepts w

I Hence, the mapping w 7→ φN,w is a poly-time reduction from L to SAT,establishing L ≤P SAT.

I In the following fix L, N and w ∈ {0,1}n and let t = t(n).

I We assume for simplicity that M(w) halts after exactly t − 1 steps.


The configuration-history Tableau

Consider the t-by-t Tableau that describes a possible accepting computationhistory of N on input w .

cell[1,1]

cell[1,t(n)]

q 0 0 10 0 . . .

1 2 3 . . . t(n)

0 0

0

1

I First row represents initial configuration of N on input w .I i ’th row represents the i-th configuration in a possible computation of N

on input w .


The formula φN,w

I Let S = Q ∪ Γ (the alphabet of the configuration history).

I φN,w uses Boolean variables {xi,j,s}i,j∈[t],s∈S.

φN,w = φCell(N) ∧ φStart(w) ∧ φMove(N) ∧ φAccept(N)

I Given an assignment z for φN,w , let T (z) be the t × t Tableau, defined bysetting the j-th cell in i ’th configuration to s, if xi,j,s = 1 in z.

(T (z) is undefined, if xi,j,s′ = xi,j,s = 1 for some s 6= s′ ∈ S, or xi,j,s = 0for all s ∈ S).

I T (z) will represents a (possible) accepting execution of N(w) iff z is ana satisfying assignment for φN,w .


The formula φCell(N)

φCell(N) guarantees that the variables encode legal configurations:

I Each cell (i , j) has at least one letter:∨

s∈S xi,j,s.

I No cell (i , j) has two or more letters∧

s 6=s′∈S xi,j,s ∧ xi,j,s′ .

Together:

φCell(N) =∧i,j

(∨s∈S

xi,j,s

)∧

∧s 6=s′∈S

xi,j,s ∧ xi,j,s′

Claim 17If an assignment z satisfies φCell(N), then T (z) is defined.


The formula φStart(w)

φStart(w) guarantees that the first row encodes the initial configuration (i.e.,q0w).

φ start(w) =x1,1,q0 ∧ x1,2,w1 ∧ x1,3,w2 ∧ . . . ∧ x1,n+1,wn

∧ x1,n+2, ∧ . . . ∧ x1,t,

Claim 18If z satisfies φCell(N) ∧ φStart(w), then the first line of T (z) is q0w t . . .t︸︷︷︸

t−n−1

.


The formula φMove(N)

φMove(N) is the “heart” of φN,w . To construct it, we employ locality ofcomputations.

Observation: Configuration C, with head location h, yields configuration C′

(with respect to δ), if the following holds.

I C′i =Ci for any i /∈ {h − 1,h,h + 1}I C′h−1,h,h+1 is consistent (with respect to δ) with Ch−1,h,h+1.

We check that each configuration in T (z) yields the next one, by inducinglocal “checks” on z.


φMove(N) – Rectangles

I A rectangle is a 2× 3 configuration sub-table.

I Assume δ(q1,a) = {(q1,b,R)} and δ(q1,b) = {(q2, c, L), (q2,a,R)}.I (some) Legal 2× 3 rectangles:

a q1 b a q1 b a a q1q2 a c a a q2 a a b

a b a b b b a a ba b q2 c b b a a b

I (some) Illegal 2× 3 rectangles:

a b a a q1 b b q1 ba a a q1 a a q2 b q2

I There is a constant number of legal rectangles (determined by δ).

I Denote this set by C = C(δ)

(we will be somewhat liberal.allowing some illegal if notation simpler)


φMove(N) – Characterizing legal rectangles

The formula “verifies" that all 2× 3 rectangles in the Tableau are in the list C:

1. a b c∗ b ∗

2.a q bq’ a b’

a q ba b’ q’

(L,q′,b′) ∈ δ(q,b) (R,q′,b′) ∈ δ(q,b)

3. q ∗ ∗∗ ∗ ∗

∗ ∗ q∗ ∗ ∗

• Some rectangles in C are clearly illegal.

• For rectangles on the left-most and right-most side of Tableau, we useslightly different first type rectangles.


φMove(N)– formal definition

I For each entry (i , j) ∈ [t ]× [t ] and c ∈ C, let φMove,i,j,c be the formulataking the value 1 iff the 2× 3 table of cells in the Tableau whoseupper-left corner is (i , j) is c.

For instance, for entry (1,1) and c =a q1 bq2 a d ,

let φMove,1,1,c = x1,1,a ∧ x1,2,q1 ∧ x1,3,b ∧ x2,1,q2 ∧ x2,2,a ∧ x2,3,d

I Finally, let φMove(N) =∧

(i,j)∨

c∈C φMove,i,j,c .

Claim 19If z satisfies φCell(N) ∧ φStart(w) ∧ φMove(N), then T (z) is a possible configurationhistory of N(w).

Proof: By induction on the row index. Base case: z satisfies φCell(N) ∧φStart(w).Assume configuration defined in rows 1, . . . , i is possible and head is in cell j .The configuration of rows 1, . . . , i + 1 is also possible: Cells of indices not in{j − 1, j , j + 1}, by first type of rectangles in C. Other cells, by second typerectangles in C. Q: Why do we need the third type of cells?


The formula φAccept(N)

φAccept(N) guarantees that some row encodes an accepting configuration ( i.e.,uqav ):

φAccept(N) =∨i,j

xi,j,qa

Claim 20If z satisfies φN,w = φCell(N) ∧ φStart(w) ∧ φMove(N) ∧ φAccept(N), then T (z) is anaccepting configuration history of N(w).


Correctness of reduction

I The transformation w 7→ φN,w is computable in time O(n2c).

I An assignment satisfying φN,w , corresponds to an acceptingconfiguration history of N(w).

I An accepting configuration history of N(w) corresponds to anassignment satisfying φN,w . (?)

Therefore, N accepts w iff φN,w ∈ SAT. ♣.

I For complete details, consult Sipser chapter 7.4.


Section 4

CSAT ∈ NPC


CSAT

It is useful to consider a special version of SAT.

I A literal is a variable or negated variable: x or x .

I A clause is several literals joined by ∨s: (x1 ∨ x2 ∨ x3)

I A Boolean formula is in conjunctive normal form (CNF) if it consists ofclauses, connected with ∧s.

I For example: (x1 ∨ x2 ∨ x3 ∨ x4) ∧ (x3 ∨ x5 ∨ x6) ∧ (x3 ∨ x6)

I The language of satisfied CNF formulas:

CSAT = {〈φ〉 : φ is satisfiable CNF formula}

I Clearly, CSAT ≤P SAT (no need Cook-Levin for that).

I The proof of Cook-Levin can be modified to show that CSAT is NPC(take home exercise..)


Section 5

3SAT ∈ NPC


3SAT

Definition 21A Boolean formula is in k-CNF form, if it is a CNF formula, and all clauseshave 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}

I Clearly 3SAT ≤P SAT.

I and 3SAT ∈ NP.

I We show next that CSAT ≤P 3SAT, yielding that 3SAT ∈ NPC.I Interestingly, 2SAT ∈ P


CSAT ≤P 3SAT

I The reduction maps CNF formulae to 3CNF ones “clause by clause”.

I A clause with d ≤ 3 literals is mapped to equivalent clause with 3 literals.

(x1 ∨ x2) 7−→ (x1 ∨ x2 ∨ x2)

I A clause with d > 3 literals is mapped to d − 2 clauses, built on theoriginal literals together with (d − 3) new variables.

φ = (x1 ∨ x2 ∨ x3 ∨ x4 ∨ x8) 7−→φ3 = (x1 ∨ x2 ∨ y1) ∧ (y1 ∨ x3 ∨ y2) ∧ (y2 ∨ x4 ∨ x8)


The reduction works!

Claim 22φ has a satisfying assignment iff φ3 does.

Proof’s idea: (for case d > 3)⇐ Given a satisfying assignment for φ3, use the settings of the xi ’s to getsatisfying assignment for φ. Why does it work? For each set of clauses thatcame from a clause in φ, note that each yi sets exactly one clause to true.There are k − 2 clauses but only k − 3 yi ’s. So at least one clause needs tobe satisfied by one of the xj or xj literals. Set this literal to true in φ.

⇒ An assignment satisfying φ, makes at least one literal per clause happy. Inthe “φ3 clause” of this literal the new variables are under no constraints. Usethe d − 2 new variables to set the remaining d − 2 clauses to “true”.


CSAT ≤P 3SAT, cont.

φ = (x1 ∨ x2 ∨ x3 ∨ x4 ∨ x8) 7−→φ3 = (x1 ∨ x2 ∨ y1) ∧ (y1 ∨ x3 ∨ y2) ∧ (y2 ∨ x4 ∨ x8)

I Doing the above mapping to each clause of a CNF formula, we get a3CNF that is satisfied iff the original one is.

I Since this mapping is polynomial time (?), we get CSAT ≤P 3SAT. ♣.


Section 6

Clique is NPC


3SAT ≤P CLIQUE

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

Claim 233SAT ≤P CLIQUE.

Since CLIQUE ∈ NP, it follows that CLIQUE ∈ NPC.Since IND-SET ∈ NP, and CLIQUE ≤P IND-SET, it follows thatIND-SET ∈ NPC.

Proof’s idea: We’ll construct a poly-time reduction f that maps 3CNF formulaeφ to pairs 〈G, k〉 of graphs and numbers.

The function f will have the property that φ is satisfiable, iff G has a clique ofsize k .


Proving 3SAT ≤P CLIQUE

On input φ, a 3CNF formula, the mapping reduction is defined as follows:

let k be the number of clauses in φ.

We construct a graph G = G(φ), see below, and output (G, k).

The graph G is defined as follows:

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

I Each triple corresponds to a clause of φ

I Each node in a triple corresponds to a literal in corresponding clause.

Ongoing example:

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


Nodes of G

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

Add a node per literal


Edges of G

Add edges between all vertex pairs, exceptI within same tripleI between contradictory literals (e.g., x3 and x3)

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


φ ∈ 3SAT =⇒ 〈G, k〉 ∈ CLIQUE

Proof: Suppose φ is satisfiable by an assignment ψ.With respect to ψ:I At least one literal is assigned to 1 in every clause (?)I Select a 1-literal in every tuple;I These literals can be joined by edges (?)

Yielding a k -clique in G.

♣Roded Sharan (TAU) Computational Models, Lecture 12 June, 2018 41 / 49

〈G, k〉 ∈ CLIQUE =⇒ φ ∈ 3SAT

Proof: Suppose G has a k -clique.

I No two of the cliques nodes are in the same triple. (?)I G has 3k vertexes and k clauses, so each triple has exactly one clique

node.I Assign 1 to each node in cliqueI Assignment has no contradictions (?)

Yielding a satisfying assignment to φ.

♣ Roded Sharan (TAU) Computational Models, Lecture 12 June, 2018 42 / 49

Recap

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

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

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

=⇒ 3SAT ≤P CLIQUE. ♣


Section 7

Hamiltonian Path and Hamiltonian Cycle areNPC


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 24HAMPATH ∈ NPC.

Soon


Hamiltonian Cycle

A Hamiltonian cycle in a graph is an Hamiltonian path that ends up where itstarts (i.e., s = t).

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

Clearly HAMCYCLE ∈ NP. We will show that HAMPATH ≤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 t to s?


HAMPATH ≤P HAMCYCLE.

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


CSAT ≤P HAMPATH

For any CNF formula φ with clauses c1, . . . , c` and variables x1, . . . , xk , weconstruct a directed graph G with vertices s and t , such that

φ ∈ CSAT⇐⇒ 〈G, s, t〉 ∈ HAMPATH

thus establishing

Theorem 25HAMPATH,HAMCYCLE ∈ NPC

Turn to a separate pdf presentation:tau-cm2017a.wdfiles.com/local--files/course-schedule/ham-reduction-new.pdf


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