[PDF] - ranges from 1 to the n um b er of i clauses

SLIDE 1 Indep enden t Set Problem Input: a graph G and a lo w er b

und

k . Output: \y es" i there are at least k indep endent no des

f

G; i.e., no des with no edges in terconnecting. Reduction from: 3SA T.

Clearly

, this problem is in N P ; just guess k no des and c hec k that they ha v e no edges among them. The Reduction T ak e a 3-SA T instance suc h as (x + y + z )(

x

+

z

+ w ).

Create

no de [i; j ] for the j th literal in the ith clause.

✦

i ranges from 1 to the n um b er

f

clauses | certainly O (n), where n = the input length.

✦

j = 1; 2,

r

3.

Edges

among the three no des with a common i prev en t more than

ne
f

them b eing c hosen in an indep enden t set.

Edges

b et w een no des for an y literal and its complemen t.

✦

In

ur

little example: [1; 1] and [2; 1] are connected (x and

x);

[1; 3] and [2; 2] are also connected (z and

z

).

Pic

k k = n um b er

f

clauses. Pro

f

the Reduction is Correct

First,

supp

se

w e ha v e a satisfying truth assignmen t for the v ariables.

✦

Pic k

ne

true literal from eac h clause (there could b e more, but not few er).

✦

The no des corresp

nding

to these literals form an indep enden t set

f

size k .

✦

Wh y? The

nly

edges among them w

uld

connect no des for dieren t clauses, and these w

uld

ha v e to go b et w een a literal and its complemen t, b

th
f

whic h could not ha v e b een selected. 1

SLIDE 2

No

w, supp

se

w e ha v e an indep enden t set

f

size k .

✦

This set cannot ha v e more than

ne

no de from an y

ne

clause.

✦

This set cannot c ho

se

no des corresp

nding

to a literal and its complemen t.

✦

Th us, it tells us a truth assignmen t for enough

f

the v ariables that ev ery clause is made true. Coping With Complexit y When faced with an NP-complete problem, there are three things w e can do: 1. Appr

ximate.

F

r

example, do w e need an absolutely maxim um

size

indep enden t set?

✦

P erhaps a greedy heuristic (grab an y no de w e see as long as it has no edges connected it to those w e'v e selected already) will get an indep enden t set that is big enough? 2. R estrict. Do w e really need to solv e the problem in all its generalit y? Or could a sp ecial case that has a p

lynomial

algorithm serv e

ur

needs?

✦

Example, while 3SA T is NP-complete, the 2SA T problem (clauses

f

2 literals

nly)

has a subtle, linear-time algorithm. 3. T

ugh

It Out. Sometimes w e are

nly

in terested in problem instances that are small enough that the exp

nen

tial gro wth do esn't

v

erwhelm

ur

resources.

✦

Query

ptimization

algorithms are lik e that: ev erything is NP-complete, but database queries tend to b e v ery small.

✦

T ra v eling Salesman is an un usual NP- complete problem b ecause it is in fact v ery easy to solv e ev en 1000-cit y problems. Th us, it is used b y man y snak e-oil salesmen to demonstrate that their fa v

rite

algorithmic metho dology \b eats" NP-completeness (e.g., Hopgo

d

with neural nets, Adelman with DNA algorithms). Out Bey

nd

N P There is no end to the n um b er

f

complexit y classes that can b e in v en ted b y mathematically 2

SLIDE 3 inclined academics desirous

f

gaining ten ure. Some

f

these are actually in teresting. Co-NP A language/problem is in Co-NP if its complemen t is in N P .

If

P = N P , then Co-NP = N P .

✦

Wh y? b ecause the complemen t

f

a problem in P is surely in P , since w e can just complemen t the answ er in

ne

more step.

Ho

w ev er, if P 6= N P , as w e assume, then Co- NP 6= N P is lik ely , although not certain.

Apparen

t example: The complemen t

f

SA T (i.e., all Bo

lean

expressions that are not satisable, plus the \garbage" that is not a w ell-formed expression) app ears not to b e in N P .

✦

While w e can guess a satisfying truth assignmen t and c hec k that w e guessed righ t in p

lynomial

time, there is no w a y to \guess wh y there is no suc h assignmen t."

✦

Note that the nonsatisable expressions are the negations

f

the tautolo gies (expressions that are alw a ys true), so tautology testing is another example

f

a Co-NP problem that app ears not to b e in N P . PSP A CE A TM that uses no more than p(n) space

n

input

f

length n, for some p

lynomial

p, is said to b e in PSP A CE.

Y
u

migh t think that it matters whether the TM is deterministic

r

nondeterministic, but it do esn't! See b elo w.

A

PSP A CE TM can tak e exp

nen

tial time b efore accepting.

Ho

w ev er, if it tak es more than k p(n) mo v es, where k = sum

f

the n um b er

f

states and tap e sym b

ls,

then it has rep eated an ID and so has a shorter sequence

f

mo v es leading to acceptance if it accepts at all. Example The tautology problem is in PSP A CE. 3

SLIDE 4

Use

linear space to en umerate all p

ssible

truth assignmen ts,

ne

at a time (i.e., run a coun ter in binary).

Chec

k eac h assignmen t, sa y \no" if y

u

nd

ne

that do esn't mak e the expression true, and sa y \y es" if y

u

reac h the end. PSP A CE-complete Problems While P

N

P

PSP

A CE is

b

vious (remem b er that PSP A CE includes nondeterministic TM's), it is not ev en kno wn whether P = PSP A CE.

Sa

y a problem L is PSP A CE-c

mplete

if ev ery problem in PSP A CE p

lynomial-t

ime reduces to L.

✦

Th us, if L is in P , then P = PSP A CE; if L is in N P , then N P = PSP A CE. Example QBF (Quantie d Bo

le

an F

rmulas)

is a PSP A CE- complete problem.

Example
f

a QBF: (8x)(9y )(x

y

+

xy

).

✦

This instance

f

QBF has answ er \y es" (true), b ecause w e can pic k y to b e the complemen t

f

x. Sa vitc h's Theorem: Equiv alence

f

Determini sti c and Nondetermini sti c PSP A CE Key ideas: 1. If a PSP A CE NTM accepts, it do es so within k p(n) steps. 2. A sim ulating DTM uses a recursiv e algorithm to answ er questions

f

the form: \do es ID

`

* ID

in

at most 2 i steps?"

Basis

: i = 0. Chec k if

=
r
`
.
Induction

: F

r

eac h p

ssible
[ID
f

length at most p(n)], recursiv ely c hec k if

`

*

in

at most 2 i1 mo v es and

`

*

in

at most 2 i1 mo v es.

✦

Return \y es" if an y suc h

found;

return \no" if not.

✦

Y

u

need

nly
ne

\stac k frame"

f

length p(n) to generate and store eac h p

ssible
(use

a coun ter in base k ). 4

SLIDE 5

Clinc

her: W e can limit the stac k to p(n) log 2 k recursiv e calls, taking a total

f

p 2 (n) log 2 k space, a p

lynomial

if p(n) is.

✦

Wh y? That is enough to answ er the question \do es

`

*

in

at most 2 p(n) log 2 k = k p(n) mo v es?"

✦

Let

b

e the initial ID, and (using a coun ter)

b

e an y

f

the p

ssible

accepting ID's

f

length p(n).

✦

Remem b er, if acceptance

ccurs,

k p(n) mo v es is enough. 5