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COMPLEXITY THEORY
CSci 5403
LECTURE VII: DIAGONALIZATION
Theorem (Cantor). Let ƒ : S
!
P
(S) map elements
of the set S to subsets of S.
Then ƒ is not onto.
In particular, let D
ƒ
= { y  y
∉
ƒ(y) }.
There is no
s
2
S such that ƒ(s) = D
ƒ
:
Suppose ƒ(s) = D
ƒ
.
Then s
2
D
ƒ
,
s
∉
ƒ(s)
,
s
∉
D
ƒ
.
Example: Let S = {0,1,2} and let ƒ(0) = Ø, ƒ(1) = {1,2},
ƒ(2) = {0,1}.
y
0
2
ƒ(y)? 1
2
ƒ(y)? 2
2
ƒ(y)?
0
N
N
N
1
N
Y
Y
2
Y
Y
N
D
ƒ
Y
N
Y
Theorem (Turing). The diagonal language
D
TM
= {
h
M
i
 M does not accept on input
h
M
i
}
is undecidable.
Proof.
Let ƒ
TM
: {0,1}*
!
P
({0,1}*) map a TM’s code
to the language it decides.
If D
TM
is decidable, there must be a TM H such that
ƒ
TM
(
h
H
i
) = D
TM
.
But D
TM
is the diagonal set for ƒ
TM
.
Cantor, Gödel, and Turing used it… must be good!
(Insert cheesy 1970s theme music)
TIME HEIRARCHY THEOREM
Theorem. Let f and g be time constructible,
and
let g(n) =
ω
(f(n)
2
). Then DTIME(f(n))
(
DTIME(g(n)).
Corollary. P
≠
EXP.
Proof.
Let D
f
= {
h
M
i
 M does not accept
h
M
i
in
f(
h
M
i
) steps. }.
D
f
∉
DTIME(f(n)), since it is
the diagonal set for L
f
mapping a TM to the set it
decides in time f(n).
But recall that we can simulate a TM that runs in
time t in time O(t
2
).
Thus D
f
2
DTIME(f(n)
2
).
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SPACE HEIRARCHY THEOREM
Theorem. Let f and g be space constructible,
and
let g(n) =
ω
(f(n)). Then SPACE(f(n))
(
SPACE(g(n)).
Corollary. L
≠
PSPACE.
Proof.
Analogous.
We can simulate a machine
using space s(n) in space O(s(n)).
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 Spring '08
 Sturtivant,C
 Computational complexity theory, SATH, NTIME HEIRARCHY THEOREM

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