Lecture 24 Time and Space of
NTM
Time
• For a NDM M and an input x,
TimeM(x) = the minimum # of moves
leading to accepting x
if x ε L(M)
= infinity
if x not in L(M)
Time Bound
A NTM M is said to have a time bound t(n)
if for sufficiently large n and every x ε L(M)
With |x|=n,
TimeM(x) < max {n+1, t(n)} .
Complexity Classes
NTIME(t(n)) = {L(M) | M is a NTM with time
bound t(n)}
c
NP = U c > 0 NTIME(n )
Relationship
• P  NP
• NP ≠ EXP
• NP  EXPOLY
First, note that EXP  DTIME (2 )  DTIME (2 ).
For contradict ion, suppose NP  EXP.
n2 / 2
Consider any L  DTIME (2 ).
n2
Define L'  {x$
| x|2 | x|
| x  L}.
Then L' EXP  NP.
So, L' NTIME (n c ) for some c  0.
Hence, L  NTIME (( n 2 ) c )  NP  EXP.
n2
Theorem
• Speed Up Theorem still holds.
• Hierarchy Theorem may not.
Space
For a NTM M and an input x,
SpaceM(x) = the minimum, over all
computation paths, of maximum
space taken in each work tape
on input x
if x ε L(M)
= infinity
otherwise
Space bound
• A NTM M is said to have a space bound
s(n) if sufficiently large n and every input x
with |x|=n,
SpaceM(x) ≤ max{1, s(n)}
Complexity Classes
• NSPACE(s(n)) = {L(M) | M is a NTM with
space bound s(n)}
• NSPACE = Uc>0 NSPACE(n c)
Relationship
• NP  NSPACE
• PSAPACE = NPSPACE (why?)
Savich’s Theorem
• If s(n) ≥ log n, then
NSPACE (s(n))  DSPACE(s(n) )
The proof will be given in next lecture!
2
Theorems
• Compresion Theorem holds.
• Hierarchy Theorem may not.
Translation Lemma
• Let s1(n), s2(n) and f(n) be fully spaceconstructible functions with s2(n) > n and f(n)
> n. Then
NSPACE(s1(n))  NSPACE(s2(n))
implies
NSPACE(s1(f(n)))  NSPACE(s2(f(n)))
Hierarchy
• NSPACE(n 4 )  DSPACE(n8 )
 DSPACE(n 9 )
≠
 NSPACE(n9 )
• For r > 1 and a > 0,
r+a
r
NSPACE(n ) ≠ NSPACE (n )
Proof of Savitch’s Theorem