Hierarchy Theorems
description
Transcript of Hierarchy Theorems
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
■