CSci 5403 Lecture 8

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CSci 5403
COMPLEXITY THEORY
LECTURE VIII CIRCUIT LOWER BOUNDS
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A CIRCUIT
Is a directed acyclic graph, where
x1
x2
Each node has a label 2 x1,,xn, Æ,Ç,
x0

Each node has in-degree either 0 (inputs), 1 (),
or m (Æ , Ç)
Nodes with out-degree 0 are the outputs of the
circuit.
?
A circuit C with n inputs and m outputs computes
a function C 0,1n ! 0,1m.
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Definition. A circuit family is a sequence of
circuits C1, C2, C3, C4, where for each n, Cn
has n inputs.
We write the family as Cnn 2 N.
The language of a circuit family is the set L,
where x 2 L , Cx(x) 1.
The size of a circuit, C, is the number of
gates.
The size of a circuit family is the function s(n)
Cn
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Definition. A language is in SIZE(t(n)) if it
is decided by a circuit family of size O(t(n)).
P/poly ?c 2 N SIZE(nc)
Theorem. P ½ P/poly.
Corollary. If NP P/Poly, then P ? NP.
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NONUNIFORM ALGORITHMS
A Turing Machine with advice gets an
additional input ?n for each input length n.
M decides L with a(n) advice if there is an
advice sequence ?nn 2 N with ?n ? a(n) such
that 8 n 8 x2 0,1n M(x,?n) 1 , x 2 L.
Definition. L 2 DTIME(t(n))/a(n) if there is a
time O(t(n)) TM that decides L with O(a(n))
advice.
Theorem. P/poly c,d DTIME(nc)/nd
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Theorem. If NP µ P/log then PNP.
Proof.
Suppose ? is the correct advice for all ? 2
0,1n
If M(?,?) says satisfiable, we use ? recursively
to find a satisfying assignment. (padding to
the same length if necessary)
We can try all O(log n) advice strings. If any
produces a satisfying assignment, ? is
satisfiable.
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Theorem (Karp-Lipton). If NP µ P/poly then PH
?2.
Proof.
Suppose NP µ P/poly. We will show that ?2SAT
? 8 X 9 Y. ?(X,Y) is in ?2 thus ?2 ?2 and
PH collapses to the second level.
Since NP µ P/poly, there is a polytime
nonuniform algorithm M and advice strings An
?1?n so that when ? 2 ?2SAT, M(?, X, An) outputs
a Y for any X.
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So if ? 2 ?2SAT then 9 A. 8 X. ?(X,M(?,X,A)) 1.
But if ? ? ?2SAT then even for the right
advice M(?,X,A) cannot satisfy ? for choice of X.
Thus ?2SAT ? 9 A. 8 X. ?(X,M(?,X,A)) 1
Then ?2SAT 2 ?2 and the theorem follows.
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Theorem. Every language has circuits of
size O(n2n).
Theorem. For large enough n, almost all f
0,1n ! 0,1 require circuits of size ?(2n/n)
Proof.
We count the number of circuits of size S.
Gate n has n2 possible inputs both can
be negated or not and have type Æ or Ç.
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Thus the total number of circuits of size S is
at most S2S8S.
The total number of n-bit functions is 22n.
We thus need s2s8s lt 22n/n.
, s(2log s 3) lt 2n - \log n.
( s lt 2n/3n.
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NONUNIFORM HEIRARCHY THEOREM
Theorem. For every t(n) ?(t(n) log t(n)),
SIZE(t(n)) ( SIZE(t(n)2).
Proof.
The last result implies 9 f 0,1l ! 0,1 that
is not computable with 2n/3n gates, but is
with n2n gates.
Define the language L that computes f on
the first log(t(n)) bits of input.
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log(t(n))
f
Can be computed in size t(n) log t(n).
Cannot be computed in size t(n)/3 log t(n).
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