Hierarchy Theorems

Hierarchy Theorems

2

: ( ) log1

( )( ( ))

log log

:

n

Definitiof N N f n n

f nO f n

n n

n

n n

A function , where is calledspace constructible if the function that maps tothe binary representation of is computablein space

eg. , , : space constructible

Space Hierarchy Theorem

:( ( ))

( )

:

( )

For any space constructible function , thereexists a language that is dicidable in spac

(Space Hierarchy

e but not in space .

Theorem)f N N

A O

Theor

o

e

f nf n

m

( ( ))(

:

( ))Construct an space algorithm that decidesa language space

O f n BA o f

roof

n

p

"

( )

10*

On input 1.2. Mark off tape space,

3. If is not of the form for som

if later stages overattempt

e , REJECT.

4. Simulate

to use mor

on w

e, then

hile counting the numbero

REJECT.

f step

B wn w

f n

w M M

M w

( )2s used in the simulation. If the count

exceeds , REJ5. If accepts, REJECT;

else if rejects, A

ECT

CCE "P

.

T.

f n

MM

'

may have an arbitrary tape alphabet and has a fixed tape alphabet, so we represent each cell of with several cells on tape.

M BM

B s

( ) ( )Thus, if runs in space, then uses spaceto simulate , for some constant that depends on

.

M g n B dg nM d

M

( ( ))Let be the lang

is a decider becauuage that decide

se it halts in s. is

decida

a

b

limited t

le

ime.

in A B

O f n

BA

0

0 0

0

( )lim 0 . . , ( ) ( )

( )

( ) ( ( ))

( )10

By contradiction, assume is decidable in by a TM .

Thus, can simulate using space . For largeenough , is run on input .

wi

n

n

g nn s t n n dg

Ag n

B

o f n M

B M dg n

n f nf n

n B M

( ) ( ( ))

( ( ))Hence, does not

ll halt and

decide

give opposite answerof on the same input.

Therefore, is not decidable in spac.

e.M A

g n o f nM

A o f n

■

1 2 1

2 2

1 2

2 , : ( )( ( ))

(

:

( )) ( ( )).

For any functions where is and is space constructible.

f f N N f no f n

Corollar

fSPACE f n SPACE f

y

n

Ö

1 21 2( ) ),0

:(SPACE n SPACE n

Corollary Ös

: ( ) lg1

( ) ( ( )

:

:

)

A function , where is called timeconstructible if the function that maps to the binaryrepresentation of is computable in time .

n

Corollary

Defin

NL SPACE

t N N t n n n

t n O t

e

n

Ö

::

( ( ))( )

( )log ( )

(Time hierarchy Theorem)For any time constructible function , thereexists a language that is decidable in time

but not in time .

t NTheorem

NA O t nt n

ot n

"

( )log ( )

3,4,0

5

: On input

1.

2. Store the value in a binary

counter. Decrease this counter before each stepused to carry out stages If the counters , then REJECT.

3. If is not of the

prooB w

n w

w

f

t nt n

10* form for some TM , REJECT.

4. Simulate on .5. If accepts, then REJECT;

else if rejects, then ACCEPT "

M M

M wM

M

1 2 1

22

2

1 2

2 , : ( )( ( )

:

)

log ( )( ( )) ( ( ))

For any functions , where is

and is time constructible

t t N N t no t n

tt n

TI

Corollary

ME t n TIME t n

Ö

1 21 21 ( ) ( )

:, TIME

Corollan M n

ryTI E Ö

:PCorol

EXlary

PTIMEÖ

Relativization :

An oracle is a language , An oracle TM is anordinary TM with an extra tape called the oracle tape.

can query a string on the oracle tape to see if thestring is a member of in a singl

ADefinitio

A

nA M

Me step.

.

,

.

.

: the class of languages decidable with a poly timeoracle TM that uses oracle

Similary, for

S T SAT

A

Aeg NP P co -

PA

NP P

ANP

[1.

:

2.

A A

B B

TheoreA P NPB P NP

m

T. Baker, J . Gill, R. Solovay, SIAM J . Comp. 1975There exists an oracle such that There exists an oracle such that

]

{ | ( )}

1.

2.

{01,10,1} ?

:

Goal: Construct oracle , such that

Let be

It is c

For any oracle

lear that

If , th , Let

e

I

nf

TQBF TQ

B

BF

TQBF TQBF

TQB

B

F TQBF

B

B

A TQBFNP NPSPACE PSPACE P

P NP

proof

B L w x B x w

P NP

B

B N

L

P P

. then has all the strings of length why?B

B

B

L NPx B L x

Limits of the Diagonalization Method

1 2, , .

:

..

For simplicity, assume runs in

Let be a list of all polynomial tim(OTM)

The construction of

eoracle Turing M

proceeds in stages, where stage constru

achi

cts a par

ne

t o

s

f

ii

BB

M n

Claim L

Bi

M

P

B

M

, which makes sure that doesn't decide B

i BM L

,

11

2choose greater than the length of anystring so far in is large enou

Initially, Let stage :

Extend information about so that M accept whenever

gh so

(i.e. whenever there is no

B ni

n B

n i

Bi

B

nB n n

L

string of length in )n B

1

If queries whose status has already beendetermined, we

Run on respond to its oracle queries as follows(unt

respond consistentlyIf 's status is undetermined

il halt

, we respond NOand d

s)

n

i

i

i

M

M y

y

M◈

◈

(1) 1

(2

(2)

)

1

If a string of length in , then will accept If knows all string of length are not in , then reject , since has no string ofle

eclare

No way to know will happen,ngth

ni

ni B

n B MMi n B

Mn

y B

L

2 since cannot

check all strings stops within stepsn ii

MiM n

11

1

If accept , we delcare all the string of length and so

If reject , we find a string of length that hasn't queried and delc

n

i

ni

nB

ni i

Must exist there 2 strings

and M can not query them all

Mn B L

M n M

1

;

After all stages, declare any string whose statusremains undetermined by all stages

are that stri

Thus, no

ng to be in so

repeat next stag

poly time deci

e

des with oracle B

nB

OTM L B

B

B

L

i

■

Circuit Complexity :

Boolean Circuit is a collection of gates and inputsconnected by wires

Definition

x1 x2 x3

output

1 2( , ,...){0,1

:

}( ) 1

A circuit family is an infinite list of circuits , where has input variables

We say that decides a language over if for every string , where is t

n

n

DefinitiC

C C C nC A

w w A f

o

f Cn

n

i w

Size complexity, dep

he le

th co

ng

mp

th

le

f

y

o

xit

w

:

:

22 2n

n n fThe

n

n

orem

proof

[Shannon 1949]

For any there is an -ary Boolean function s.t. no Boolean circuit with

By contradiction.

of fewer gates cancompute it.

Suppose all -variable Boolean function can

1 2

2

2

2

, ,..., ,0,1, ,2

n

n

n

mn

xn

x x

There are -variable Boolean functi

be

computed by circuits with or fewer ga

o

tes.

n.¬∨ ,∧

2

2

2

22 2 2

22

54

2

log [( 5) ] [log ( 5) 2log ][log( 5) 2 2log

( 5

2 ]log

2 [1 ] 22

[( 5) ] .

2

)

n

n

m

m

nn

n nn

n m m n mn n n

n

n m

n m m

choices for each gete:

Thus, at most

There a

possible circuits of siz

re boolean functions.T

e

here 2 22

22

2 ..

n n

nn

n're fewer than circuits of size at most

Thus some circuits have size

5 possibilitiescan be any ofthe gates

n

m

2

0

: ( )( ( )) ( ( )).

( , , , ,

:

):

, , ( )accept rejet

t N N t n nA TIME t n A

Theorem

proofO t n

M Q q q q A t nw n M

Let be a function, where If , then has circuit complexity

Let decide in time and let be an input of length to .

⊔0⊔⊔0 1q00Cell[1,1] start configuration

Cell[t(n),1]Accept position

t(n)-thconfiguration

.

( )

PQ P

t n

Each cell contains a tape symbol in or a combinationof a state and a tape symbol in

Once accept, move the head to the left most cell onthe -th configuration.

( 1, 1) ( 1, ) ( 1, 1

)

)

( ,

i j i j i

j

j

i

[ , ][ -1, -1], [ -1, ], [ -1, 1]

with M's transition functions depends on

c i jc i j c i j c i j

[ , , ]

( )

[ -1, -1]

[ , ]

k Q k

cell i j

light i j

ce

s cell i j s

There are several gates for each cell.

If is on, contains the symbol , Only one light would be on per cel

Let , Create lights for each cell.

Suppose ,

l.

[ -1, ] [ -1, 1][ , ]

.

ll i j cell i ja b c cell i j s

and contain , and respectively; conatins according to

[ -1, -1, ] [ -1, , ][ -1, 1, ] [ , , ].

light i j a light i j blight i j c light i j S

Wire the circuit so that, if , and are on, then so is

[ -1, -1][ -1, ] [ -1, 1]

[ , ] .

cell i jcell i j cell i j

cell i j s

There may be several differentsettings of ,

, maycause to contain

0 0 1

0

0 10q0… 0ql

1q0… 1ql

0 1 … 0 1 …

0 1 …

…

0 1

2

1

[1,1, ] [1, ,1] 2,...,,...,

[1,2,0],..., [1, 1,0],..., [1, , ]

1

n

n

light q w light i i nw w

light light n NOTw w light j

n j t

The is on, Then are connected to inputs and

are connected by gates to inputs . is on forò

2

3

[ ( ),1, ].

( )

( ))

.

(acceptlight t n q

O

n

t nk

We designate the output gate to be the one attachedto Size of the tableau is . Each cell has at most gate ~ a constant.

■

:

:

{

-

-

}

-

- |

is -complete

It is obvious that is in Next we need to show any language in isr

is a satisfiable Boolean circui

educible to

t

Nee

CIRCUIT SAT NP

CIRCU

Theorem

proofIT

CIR

SAT NPA NP

CIRCUIT

CUIT

SA

SA C

T

T C

( )d to give a poly time recudtion :

such hatBoolean circuit is satisfiable

ff w Cw A C

,, it has a poly time verifier whose input has

the form , where may be the certificateshowi

To construct , we obtain the circuit simulating using the method

ng tha

in the

t

previous Theor

A NP Vx c c

f V

x A

m.

The only remaining inu

Fill in the inputs to

pts to the circuit cor

the circuit that correspond to with symbols

respondto the certificate . Call this cir

of

cuit

xw

c C

If is satisfiable, a certificate exists, so Conversely, if is in , a certificate existsso i

The construction of the circuit can be done intime(poly in ).The

s satis

running time of

fi

the

able

C w A

n

w AC

2

2

( )

( )

verifier is for some so the size of the circuit constucted is

The structure of the circuit is quite simpleso the running time of the reduction is

k

k

k

n kO n

O n

■

Monotone Circuits

,

,

1(

:

)0

Circuits that do not use ( only)Monotone circuits can only computemonotone functions

if has a -

o/w

- is -

-gate

is

complete

ve

n k

k CLI

NOT

DefiQUE NP

G k CLIQUECLIQUE

e

G n

G

n

rtex

1,2 1,3 2,3 1

3

1 3,

,3 1,2 1

3 1

,3 2,3

1 0 1( , ,

.

2

)

(

?

) ?

For an -vertex graph , we can use boolean

variable to describe

,

An -verte

, , represe

x gra

it

p

nt

.

nn G

eg G CLIQUE G

x x x GCLIQU

n

E x x x

2

h can have possible

is exponen

-

tial!n

n k CLIQUEk

n

S

1 2

3

1

8

4,

:

2

n k

cn

c nCLIQUE k n

Theorem

(Razborov's 1985)-- 1990 There is a constant such that for large enough allmonotone circuits for wit

Rolf Ne

h have

siz

vanlinna P

e

rize

1{ ,..., }:

A sunflower is a family of sets calledpetals, each of cardinality at most , such that all pairsof sets in the family have the same intersection

p

Definitip S S

l

on

( -1) !: (Erd s-Rado lemma)

Let be a family of more than nonemptysets, each of size Then must contain asunf

or less.er

low

lLemma

pZ

Z M ll

ö

:

1

1

By induction on For , different singletons from a sunflower.

Suppose :Consider a maximal subset of , where sets in aredisjoint, i.e. every set in intersects some set in

ll p

lD Z

pr

D

oo

Z

f

D D

Richard Rado 1906 --1989 Paul Erdös (3/26, 1913 – 9/20,1996)

( -1)

( 1)

!l

D p

S

D p S p

ZZ p l

l

S

D

If contains sets then it contains a sunflowerwith empty coreElse, let be the union of al

Note that intersects every set in has more t

l sets

,

han

in

sets

1

1

-

' {

( -

' ( -1) ( -

1) !

'-{ }: ' '}

( 1) (

1)

1)!(

'!

-1)

Let th

Thus, there is an element of that intersects more

than sets in

By ind. has a sunfl has more th

is element be

o

Let a

an

d

n

l

ll

dZ S d S

Sp l

p l Zp l

Z d

Zp l

S

Z

wer of petalsThus, has a -petal sunflower

pZ p

■

Plucking a sunflower replacing the sets in the sunflower by its core

x1 x2 x1 x3 x2 x3

x1x2 x1x3 x1x3 x2x3

{x1x2, x1x3}, {x1x3, x2x3}

{x1x2x3, x1x3}

1(

({ ,

( 1,...

,..., )

, )

})

, where 's are subsets of with nodes each; and there are 's ( )

Crude Circui

means that is computes the OR

t:

can be seen as a crude circu

f

it

o

m i

n

ij

k

i

CC X X X Vl M X m M

x CC

CC Sk

i j

S n

subcircuits, each indicating whether thecorresonding set in the list is a clique

1 18 8

4

log ( -1) !

2 12

Recall in Razborov's Thm Define , ,

It is clear that

l

k n

l n p n n M p ll k

( ) ( ) CC CCX Y

( ( )) pluckCC X Y

11 2

1 2

( ({ : , , }

(1) ( ) ( )( ( ))

(2) ( ) ( )

{ , ,..., }

:

{ ,

)

,

)

of 2 crude circuits and is definedto be

of 2 crude circuits and :

If

i j i

l

j i jCC pluck X Y X Y a

OR CC CCCC pluck

AND CC CC

X X XY Y

Approxima

nd X Y

t

l

ion

X Y

X YX Y

X Y

XY

2

1 1 2

2

1 1 1

2

1

{ ,..., ,

,...,

., }

}

..At most 's

i j

l

l l l

YlM X YX Y X Y

X Y X Y

Positive examples for -clique

There are smallest graphs with -clique

kn kk

( -1

(

)

-1)

Neg

partition the vertices into groupsThere is no edge in each gr

ative examples: (largest

oupThere are edges for ve

g

r

raph w

ti

ithout -cliqu

ces in different grou s

e)

p

n

k

k

k. .....

. .....

. .....

. .....12

3

K-1

…

22 ( 1)

:Each approximation step introduces at most

false positivep n

Lem

M k

ma

1

:

( .., 2

)

, )p i

OR

Z

pro

Z

i

Z ZZ

of

approximation :

creates an error if every has verticesof the same color, but has distinct colors

1 2

... ...

... p M

OR

Z Z Z ZZ

1 2

1

p p

Z Z

Z Z

1,...,Replace

with , by the plucking procedure

pZ ZZ

1

1

1 1

Pr ( ) ... ( ) ( )

Pr ( ) ... ( ) | ( )

Pr ( )

( )

r )| ( ) P

2

(

p

p

i

p

iii

i

p

R Z R Z R Z

R Z R Z R Z

R Z R Z

R XX

R Z

Let stand for the event that

Consider verti

there are repe

ces in Z . Th

atedco

e prob

lors in se

. tha

t

t they1.

-1?

k

have the

same color is

11 2

|

1 1 1| |Pr ( )22 1 1 2

Pr ( ) ... (

|( )2

(

(

)

)

-1

)

i

ii

p

p

i

i

n

Z

Z lR Zk k

R Z R Z R Z

R Z

k

means at least one of the pairs of vertices in Z have the same color

Since there are at mo

Thus,

st d

iffe

2 2

2

( 1)

2 ( 1

1

)p n

p n

M

pOR

k

M k

k

There are at most pluckin

rent maximalnon

gs

Approximation

- -clique, each plucking introduces at most e

introd

rr

uces at

ors.

er most

rors

2

2

1. ( ({ : , }))2

2 ( -1)

2 ( - )

)

1

approximation :

Throw away with size>No error introduced

3 Execute pluckings at most errors are introduced in each pluckingat most errors ar

p n

p n

iAND

CC pluck X Y X YX Y l

M k

M k

i

X Y

e introduced ineach approximation

■

( ) ( ) CC CCX Y

2 -1- -1

:

-Each approximation step introduces at most

false negative

Lem

l

m

k

a

n lM

:Plucking doesn't introduce false negative, sincereplacing a set in crude circuit by a sub

Thus appro

set c

ximat

an onlyinc

ion

rease the acc

doesn't

epted gra

introduc

p

e falsenegativ

s

e

h

sOR

proof

({ :

- | |- | |

, })( ) ( )

ANDCC pluck X Y X Y

CC CCZ X

Z

Y l

nk Z

Consider approximation :1. accepts a

positive example. accepted by and 2. Deleting sets of size> may introduce

several false negat

There ar

ives

e

X YX Y

2

2

- -1 - | || |- -1 - | |

- -1- -1

n l n ZZ l Mk l k Z

n lMk l

There are at most

such sets. Thus, at most false negative

3. Plucking d

posi

oesn'

tive examples containin

t create false negati

g Z

ve.■

:Every crude circuit either is identically oroutputs on at least half of the neg

FalseTr atu ive example e

Lemma

:

| |1

If the crude circuit is not identically False, then itaccepts at least those graphs that have a

of vertices, with But in the proof of

clique

. we know at leas

onsome s

t halet

f ofth

X lLem

proof

Xma

Pr ( ) 12

e colorings assign different colors to the vertices in ( )R X

X

, at least half of the negative examples have aclique at x and are accepThus

ted■

l k l

18

4,

2

: (Razborov 85)There is a constant s.t. for large enough n allmonotone circuits for with has

sizen k

cn

cCLI

Theor

QU

e

E k n

m

11 1 11 88 8 34

2

log ( -1) !

- -1- -

:

1

Recall: , , , large enough

1. If the crude circuit is identically falseSince each approximation step introduces at most

false negatives

nlp n n l n k n M p l nn

n lMk l

proof

18

,

112

2

2

,- -

2.

1- -1

2 ( -1)( 1)1

2If the cr

Thus, the monotone circuit for is

ude circuit have false positives andeach

at least

creates false positive

Thus,

st

t

ep

he

n

n k

ck

p n

CLIQUEnk

n cn lMk l

Mkk

181 2

2

13

( -1) 22

2 ( -1)circuit size

with

n pp ck

n

km n

M kc

■