Wh y? Because while T ( n ) steps on a 3 computer ma y - - PDF document

SLIDE 1 The Class

• f

Languages P

• If

a (deterministic) TM M has some p

• lynomial

p(n) suc h that M nev er mak es more than p(n) mo v es when presen ted with input

• f

length n, then M is said to b e a p

• lynomial-time

TM.

• P

is the set

• f

languages that are accepted b y p

• lynomial-

tim e TM's.

• Equiv

alen tly , P is the set

• f

problems that can b e solv ed b y a real computer b y a p

• lynomial-

time algorithm.

✦

Wh y? Because while T (n) steps

• n

a computer ma y b ecome T 3 (n) steps

• n

a TM, T (n) cannot b e a p

• lynomial

unless T 3 (n) is.

✦

Man y famili ar problems are in P : graph reac habilit y (transitiv e closure), matrix m ultiplication (is this matrix the pro duct

• f

these

• ther

t w

• matrices?),

etc. The Class

• f

Languages N P

• A

nondeterministic TM that nev er mak es more than p(n) mo v es in an y sequence

• f

c hoices for some p

• lynomial

p) is said to b e a p

• lynomial-time

NTM.

• N

P is the set

• f

languages that are accepted b y p

• lynomial
• tim

e NTM's.

• Man

y problems are in N P but app ear not to b e in P : TSP (is there a tour

• f

all the no des in a graph with total edge w eigh t

• k

?), SA T (do es this Bo

• lean

expression ha v e a satisfying assignmen t

• f

its v ariables?), CLIQUE (do es this graph ha v e a set

• f

k no des with edges b et w een ev ery pair?).

• One
• f

the great mathematical questions

• f
• ur

age: Is there an ything in N P that is not in P ? NP-Complete Problems If w e can't resolv e the \P = N P question, w e can at least demonstrate that certain problems in N P are \hardest," in the sense that if an y

• ne
• f

them w ere in P , then P = N P .

• Called

NP-c

• mplete

problems.

• In

tellectual lev erage: eac h NP-complete problem's apparen t dicult y reenforces the b elief that they are all hard. 1

SLIDE 2 Metho d for Pro ving NP-complete Problems

• P
• lynomial-

tim e reductions (PTR): tak e time that is some p

• lynomial

in the input size to con v ert instances

• f
• ne

problem to instances

• f

another.

✦

Of course, the same algorithm con v erts non-instances

• f
• ne

to non-instances

• f

the

• ther.
• If

P 1 PTR to P 2 , and P 2 is in P , then so is P 1 .

✦

Wh y? Com bine the PTR and P 2 test to get a p

• lynomial
• tim

e algorithm for P 1 .

• Start

b y sho wing every problem in N P has a PTR to SA T (= satisabilit y

• f

a Bo

• lean

form ula).

✦

Th us, if SA T is in P , ev erything in N P is in P ; i.e., P = N P !

• Then,

more problems can b e pro v en NP- complete b y sho wing that SA T PTRs to them, directly ,

• r

indirectly .

✦

Key p

• in

t: the comp

• sition
• f

an y nite n um b er

• f

PTR's is a PTR.

• Don't

forget that y

• u

also need to sho w the problem is in N P (usually easy , but necessary). Reduction

• f

An y L in N P to SA T Assume L = L(M ) for some NTM M that is time- b

• unded

b y p

• lynomial

p(n).

• Key

idea: if w ,

• f

length n, is in L, then there is a sequence

• f

p(n) + 1 ID's, eac h

• f

length p(n) + 1, that demonstrates acceptance

• f

w .

✦

W ell not exactly: the accepting sequence migh t b e shorter. If so, extend ` to allo w

• `
• if
• is

an ID with an accepting state.

✦

Still not exactly: some ID's will b e shorter than p(n) + 1 sym b

• ls.

P ad those

• ut

with blanks.

• No

w, w e can imagine a square arra y

• f

sym b

• ls

X ij , for i and j ranging from to p(n), where X ij is the sym b

• l

in p

• sition

j

• f

the ith ID. 2

SLIDE 3

• Giv

en string w , construct a Bo

• lean

expression that sa ys \these X ij 's represen t an accepting computation

• f

w .

✦

V ery Imp

• rtan

t: The construction m ust b e carried

• ut

in time p

• lynomial

in n = jw j. In fact, w e need

• nly

O (1) w

• rk

p er X ij [or O

• p

2 (n)

• total]

✦

Another imp

• rtan

t principle: The

• utput

cannot b e longer than the amoun t

• f

time tak en to generate it, so w e are sa ying that the Bo

• lean

expression will ha v e O (1) \stu " p er X ij .

• The

prop

• sitional

v ariables in the desired expression are named y i;j;Y , whic h w e should in terpret as an assertion that the sym b

• l

X ij is Y .

• The

desired expression E (w ) is S ^ M ^ F = starts, mo v es, and nishes righ t. Starts Righ t

• S

is the AND

• f

eac h

• f

the prop er v ariables: y 0;0;q ^ y 0;1;a 1 ^

• ^

y 0;n;a n ^ y 0;n+1;B ^

• ^

y 0;p(n);B

✦

Here, q is the start state, w = a 1 ; : : : ; a n , and B is the blank. Mo v es Righ t Key idea: the v alue

• f

X i;j dep ends

• nly
• n

the three sym b

• ls

ab

• v

e it, to the northeast, and the north w est.

• Ho

w ev er, since the comp

• nen

ts

• f

a mo v e (next state, new sym b

• l,

and head direction) m ust come from the same NTM c hoice, w e need rules that sa y: when X i1;j is the state, then all three

• f

X i;j 1 , X ij , and X i;j +1 are determined from X i1;j 1 , X i1;j , and X i1;j +1 b y

• ne

c hoice

• f

mo v e.

• W

e also ha v e rules that sa y when the head is not near X ij , then X ij = X i1;j .

• Details

in the reader. The essen tial p

• in

t is that w e can write an expression for eac h X ij in O (1) time.

✦

Therefore, this expression is O (1) long, indep enden t

• f

n. 3

SLIDE 4 Finishes Righ t Key idea: w e dened the TM to rep eat its ID

• nce

it accepts, so w e can b e sure the last ID has an accepting state if the TM accepts.

• F

is therefore the OR

• f

all v ariables y p(n);j;q where q is an accepting state. Conjunctiv e Normal F

• rm

W e no w kno w SA T is NP-complete. Ho w ev er, when reducing to

• ther

problems, it is con v enien t to use a restricted v ersion

• f

SA T, called 3SA T, where the Bo

• lean

expression is the AND

• f

clauses, and eac h clause consists

• f

exactly 3 liter als.

• A

literal is a v ariable

• r

a negated v ariable.

✦

E.g., x

• r

:y . W e shall sometimes use the common con v en tion where

• x

represen ts the negation

• f

x.

• A

clause is the OR

• f

literals.

✦

E.g., (x _

• y

_ z ).

✦

W e shall

• ften

follo w common con v en tion and use + for _ in clauses, e.g., (x +

• y

+ z ), and also use juxtap

• sition

(lik e a m ultipli cation) for ^.

• An

expression that is the AND

• f

clauses is in c

• njunctive

normal form (CNF). CSA T Satisabilit y for CNF expressions is NP complete.

• The

pro

• f

(reduction from SA T) is simple, b ecause the expression S ^ M ^ F w e deriv ed is already the pro duct (AND)

• f

expressions whose size is O (1), i.e., not dep enden t

• n

n = jw j.

• First,

w e can push all the :'s do wn the expression un til they apply

• nly

to v ariables; i.e., an y expression is con v erted to an AND- OR expression

• f

literals.

✦

Use DeMor gan 's laws: :(E ^ F ) = (:E ) _ (:F ) and :(E _ F ) = (:E ) ^ (:F ).

✦

Changes the size

• f

the expression b y

• nly

a constan t factor (b ecause extra :'s and p

• ssibly

paren theses are in tro duced). 4

SLIDE 5

• Next,

distribute the OR's

• v

er the AND's, to get a CNF expression.

✦

This pro cess can exp

• nen

tiate the size

• f

an expression.

✦

Ho w ev er, since this pro cess

• nly

needs to b e applied to expressions

• f

size O (1), the result ma y b e h uge expressions, but expressions whose lengths are indep enden t

• f

n and therefore still O (1)! 3-CNF and 3SA T

• A

Bo

• lean

expression is in 3-CNF if it is the pro duct (AND)

• f

clauses, and eac h clause consists

• f

exactly 3 literals.

✦

Example: (x +

• y

+ z )(

• x

+ w +

• z

).

• The

problem 3SA T is satisabilit y for 3-CNF expressions. Reducing CSA T to 3SA T It w

• uld

b e nice if there w ere a w a y to turn an y CNF expression in to an equiv alen t 3-CNF expression, but there isn't.

• T

ric k: w e don't ha v e to turn a CNF expression E in to an equiv alen t 3-CNF expression F , w e just need to kno w that F is satisable if and

• nly

if E is satisable.

• W

e turn E in to 3-CNF F in p

• lynomial

time b y in tro ducing new v ariables.

✦

If the clause is to

• long,

in tro duce extra v ariables. Example: (u + v + w + x + y + z ) b ecomes (u + v + a)(

• a

+ w + b)(

• b

+ x + c)(

• c

+ y + z ).

✦

A clause

• f
• nly

t w

• ,

lik e (x + y ) can b ecome (x + y + a)(x + y +

• a).

✦

A clause

• f
• ne,

lik e (x) can b ecome (x + a + b)(x + a +

• b)(x

+

• a

+ b)(x +

• a

+

• b).

✦

See the reader for explanations

• f

wh y these transformations preserv e satisabilit y , and can b e carried

• ut

in p

• lynomial

time.

• Th

us 3SA T is NP-complete. This problem pla ys a role similar to PCP for pro ving NP- completeness

• f

problems. 5