Recall:
Theorem (Tape Compression)
Lecture 4:
Let L be recognized by an s(n)-space-bounded TM M.
Then for all c > 0 there exists a
(c·s(n) + 1)-space-bounded TM M 0 that accepts L.
Finish Tape Compression Theorem
Tape Reduction
Linear Speedup
Coarse Space/Time Hierarchy Theorems
Preparation for Space/Time Hierarchy Theorem
Corollary
∀c > 0 : (D/N)SPACE(s(n)) = (D/N)SPACE(c·s(n)+1)
Proof: exercise.
Lecture 4 : Overview
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Lecture 4 : Tape Compression/Reduction
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Linear Speedup of TMs
Theorem (Tape Reduction)
Lemma
If L is accepted by a s(n)-space-bounded k-tape TM,
then it also can be accepted by a s(n)-space-bounded
TM with one work tape.
Let M be a det. TM and r > 0. Then there exists an
equivalent TM M 0 s.t.
M stops on w after t steps ⇒
M 0 stops on w after n + d nr e + 6d rt e steps (n = |w|)
Proof Sketch
Construction similar to proof of tape compression theorem.
Proof
Method: Transition from Γ to Γ k−1 × Z, where Z contains
information about the head positions on the work tapes of M.
Idea: one symbol in ΓM 0 corresponds to r symbols in ΓM
M 0 has k + 1 tapes if M has k
E.g. Z = {0, ..., 2k−1 − 1} ≈ string of k − 1 bits
1. M 0 encodes input over ΓM 0 on the additional tape
bit l = 1 ⇔ head of tape l at this position
2. Then moves back the head to the beginning
Details: exercise
Lecture 4 : Tape Compression/Reduction
2
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3. M 0 simulates M
Lecture 4 : Speedup
n steps
d nr e steps
6d rt e steps
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Proof (continued)
Simulation of r M-steps by 6 M 0 -steps at a time:
r
M
Theorem (Linear Speedup)
M’
1.
2.
Let L be accepted by a t(n)-time-bounded det. TM such
that
3.
4.
Step 1..4:
lim
M 0 stores the M head position within the current Γ r block
n→∞
M 0 gathers information about neighbor cells in its state by
performing a 4-step dance
n
= 0,
t(n)
then forall c > 0 there exists a (c · t(n))-time-bounded
det. TM that accepts L.
Step 5,6:
M 0 adjusts in 2 steps at most 2 cells and the M 0 head position
within the reached block (result of r M-steps stored in ∆ 0 )
2
total: n + d nr e + 6d rt e steps
Lecture 4 : Speedup
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Proof
Choose r > 12/c, r > 2 and consider M 0 of the previous
lemma. M 0 simulates M in time
t(n)
n
e
n+d e+6·d
r
r
n>0,r>2
6
2n + 6 + 6 ·
t(n) r
Lecture 4 : Speedup
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Remark
The proofs of the lemma and theorem do not depend on
determinism!
Corollary
n
limn→∞ t(n)
=0 ⇒
∀c > 0 : (D/N)TIME(t(n)) = (D/N)TIME(c · t(n))
For all but finitely many n (n ∈ o(t(n))!) this is
c
6
6 t(n) + t(n) < c · t(n)
2
r
n
If limn→∞ t(n)
=C>0
(e.g. t(n) = D · n)
speedup to time (1 + ε) · n is possible for all ε > 0.
For the finite number of exceptions: read input and
discriminate cases in ∆ 0
Lecture 4 : Speedup
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Proof: exercise.
Lecture 4 : Coarse Space/Time Hierarchies
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Proof (continued)
Theorem (Coarse Space/Time Hierarchies)
Then define:
For every recursive function f : N → N there exists a
recursive language L with L 6∈ DTIME(f(n))
resp. L 6∈ DSPACE(f(n)).
L := {xi ∈ Σ∗ | Mi does not accept xi in 6 f(|xi |) steps}
L is recursive: simple simulation!
Proof
L 6∈ DTIME(f(n)): otherwise:
We present a recursive language and show it is not in
DTIME(f(n)) by diagonalization (DSPACE part:
analogous).
∃f(n)-time-bounded det. TM Mj which accepts L (*)
Def.
xj ∈ L ⇔ Mj does not accept xj in 6 f(|xj |) steps
Consider Σ = {0, 1}. Let xi be the i-th word in the
canonical order of Σ∗ and Mi the TM with code xi .
(*)
⇔ xj 6∈ L(Mj ) = L
Contradiction!
Lecture 4 : Coarse Space/Time Hierarchies
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Preparation for Space/Time Hierarchy Theorems
Lecture 4 : Coarse Space/Time Hierarchies
2
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Preparation for Space/Time Hierarchy Theorems
Definition (Time & Space Constructible Functs.)
A function t : N → N is time constructible ⇔
∃ det. TM that on all inputs w halts exactly in
t(|w|) steps.
A function s : N → N is space constructible ⇔
∃ s(n)-space-bounded det. TM which on every
input w halts in a configuration in which on one
work tape exactly s(n) cells are non-blank.
Definition (Gödel Word)
A Gödel word over {0, 1} of a TM M has the form:
1m < M >, where m > 0 and < M > is a 0-1-encoding
of M starting with 0.
Examples
n, n3 , 2n are space and time constructible.
Lecture 4 : Space/Time Constructability
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Lecture 4 : Space/Time Constructability
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Theorem (Space Hierarchy)
Let s1 , s2 be space-constructible with
s1 (n), s2 (n) > log2 n for all n. Then
Lemma
L accepted by an s(n)-space-bounded TM M,
s(n) space-constructible, and
s(n) > log2 n for all n
s1 (n)
= 0 ⇒ DSPACE(s1 (n)) ( DSPACE(s2 (n))
n→∞ s2 (n)
lim
In other words:
⇒
f, g space-constructible, ∀n : f(n), g(n) > log2 (n), and
f ∈ o(g)
∃ s(n)-space-bounded TM M 0 that accepts L and stops
on all inputs
⇒ ∃L ∈ DSPACE(g) − DSPACE(f)
For the proof we need a lemma:
Lecture 4 : Space Hierarchy
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Proof
Observation (Exercise):
∃c > 1 : ∀ but finitely many n: #reachable configs. 6 cs(n)
Construction of M 0 :
TM M 0 simulates M with an additional counter with base c
on an extra tape of length s(n) (space-constructible!)
M 0 stops whenever M stops
M 0 increments the counter for each M-step
M 0 stops and rejects just before the counter exceeds s(n)
symbols
M 0 accepts L, is s(n)-space bounded, and stops on all inputs
Lecture 4 : Space Hierarchy
2
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Lecture 4 : Space Hierarchy
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