csci 4325 / 6339 theory of computation zhixiang chen
TRANSCRIPT
CSCI 4325 / 6339Theory of Computation
Zhixiang Chen
Chapter 8Space Complexity
Topics
Savitch’s Theorem PSPACE Class, PSPACE-Completeness,
The TQBE Problem L and NL NL-Completeness, the PATH Problem NL=co-NL
Space Complexity
Definition. Let M be a DTM that halts on all inputs. The space
complexity of M is the function where f(n) is the maximum number of tape cells that M scans on any input of length n. We say M runs in space f(n).
If M is a NTM wherein all branches halt on all inputs, we define its space complexity f(n) to be the maximum number of tape cells that M scans on any branch of its computation for any input of length n.
,: NNf
Space Complexity Classes
Definition
Let be a function. We define space complexity classes
L is a language decided by an O(f(n)) space DTM}
L is a language decided by an O(f(n)) space NTM}
RNf :
|{))(( LnfSPACE
|{))(( LnfNSPACE
Examples
).(nSPACESAT
).(nNSPACEALLNFA
Savitch’s Theorem Theorem. For any function
where we have
,: RNf
,)( nnf
)).(())(( 2 nfSPACEnfNSPACE
Proof.
For space O(f(n)), the total number of config’s is
Use “binary-search” to decide whether C-start yields c-accept in steps. Use a stack to maintain the recursive calls.
.2 ))(( nfO
Initial Config C-start
Final Config c-accept
mid Config c-middle
))((2 nfO
PSPACE vs. NPSPACE
Definition
By Savitch’s Theorem,
).(
).(
k
k
k
k
nNSPACENPSPACE
nSPACEPSPACE
.NPSPACEPSPACE
A New World
P
NP
NP-complete
PSPACEEXPTIME
PSPACE-Completeness
Definition
A language B is PSPACE-complete if 1. B is in PSPACE
2. Every language in PSPACE is ploy-time reducible to B.
If M merely satisfies condition 2, we say B is PSPACE-hard.
The TQBF Problem
Quantified Boolean functions or
and is a Boolean CNF formula.
is true if exists some value of is true. is true if for all values of is true.
Example
],[2211 nnxxx , i i
][1 x
][1 x
,1x,1x
)].()[( yxyxyx
TQBF
TQBF is the set of all true fully quantified Boolean formulas.
The TQBF problem is to determine whether a fully quantified Boolean formula is true or false.
TQBF is PSPACE-complete.
Proof. We can view the evaluation process of a given fully
quantified formula as a tree, performing depth-first search of the tree gives a poly-space algorithm. So, TQBF is in PSPACE.
To prove TQBF is PSPACE-hard, we need to construct a fully quantified Boolean formula to simulate the computation of a given space TM M so that )( knO
. accepts only and if trueis )( , wMww
steps 2in accept - start - accepts M
s.config' 2 are there space )()(
)(
k
k
nO
nOk
ccw
nO
)]2/,,()}[,(),,{(),(),,(
Then,
steps. most at in yields iff ),,(
Define
43211143121
2121
tcccmmcccmtcc
tcctcc
formula.Boolean quantifiedfully size |) a is )(
.2logafter stop Will. expand toContinue
)]2/2,,()}[,(),,{(),(
)2,accept-,start-()(
iff
432111431
poly(|ww
stepsdn
cccmmcccm
ccw
kdn
dn
dn
k
k
k
steps 2in accept - start - accepts M
.constant somefor 22Let )(
)(
k
kk
nO
dnnO
ccw
d
The Classes L and NL
Sublinear space TM: A TM M with one read-only input tape and read/write work tape. The read head remains in the portion of the input tape containing the input. The space used on the work tape contributes to the space complexity.
Definition
Example
n) SPACE(logNL
n) SPACE(log
L
n) SPACE(log}0 |10{ k kA k
The PATH problem
.graph given ain node a to node a frompath
directed a is here whether tdetermine tois problem PATH The
} to frompath
directed a has graph that directed a is |,,{
Gts
ts
GtsGPATH
PATH is in P Easy
PATH is in NL Starting a node s, the machine records only the
current node and nondeterministically guesses the next node among those adjacent to the current node. Repeat until t is found, or all m guesses have been done, where m is the number of nodes. This is done in NL.
New Configurations
If M is a TM that has a separate read-only input tape and w is an input, a configuration of M on w is the setting the state, the work tape, and the positions of the two tape heads. The input w is not part of the configuration of M on w.
q.||
,2 is on M of ionsconfigurat of
number then space, )(in runs M If))((
wn
nw
nfnfO
NL-Completeness
Log space transducer: A TM with a r-only input tape, and r/w work tape and a w-only output tape. A log space transducer computes a function using log space of work tape, the result of the function is written on the output tape.
Log space reduction:
B. A to reducing
function computable space log a is thereif ,BA L
A language B is NL-complete if
However, L=?NL is also open.
B. toreducible space log is NLin A Every (2)
and NL, B )1(
NL.L then L,in is language complete-NLany If
. then , and if
LALBBA L
PATH is NL-Complete
Proof. PATH is in NL is known Need to show every A in NL is log space reducible to
PATH. Let the log space NTM M computing A. Simulate the computation of M on w, find all possible configurations of M on w to express nodes for a directed graph. Let s be and t be the start and accept configurations of M on w. Have an edge from one node to another node is the first yields the second. M accepts w iff there is a directed path from the start node to
the accept node.
. since , PPATHPNLL
NL=Co-NL
Proof. Since PATH is in NL, it suffices to show
.-
}|{-
NLcoNL
NLAANLco
.NLPATH
Suppose we know that the number of nodes in G that are reachable from s is c.
Nondeterministically select exactly c nodes reachable from s, not including t, and prove that each is reachable from s by guessing the path. If this is true, then the remaining nodes, including t, is not reachable from s, so accept. Otherwise, reject.
How to compute the number of nodes reachable from s?
.in not isit then ,in shown been not has and found, If rejects.
branch n computatio thisso found,been not has all ,not is in be to
verifiednodes ofnumber total theif loop,inner theof completion At the
counted. is in nodes ofnumber thely,Additional .in is then
yes, If edge.an is her test whet,in be to verified nodeseach For
. from most at distance ofpath a guessingby verifiedis guess positive
Each .in iseach whether guess andG in nodes theall through go to
loopinner an use ,in is vnode a whether decide To : find To )3(
.1},{ )2(
.||Let . distancein
from reachabke nodes ofset thebe let ,0 )1(
11
1
11
00
ii
iii
ii
i
i
ii
ii
i
AAv
AcA
AAv
(u,v)Au
si
A
AA
csA
Aci
sAmi