Non-Uniform ACC Circuit Lower Bounds

Non-Uniform
ACC Circuit Lower Bounds
Ryan Williams     
IBM Almaden
TexPoint fonts used in EMF.
Read the TexPoint manual before you delete this box.: 
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The Circuit Class 
ACC
 
An ACC circuit family 
{
C
n
} 
has the following properties:
  Each 
C
n
 takes n bits of input and outputs a bit
  The gates of 
C
n
  can be 
AND, OR, NOT, MODm
     
 
MODm(x
1
,…,
x
t
) = 1  iff  
i
 
x
i
 is divisible by m
  
C
n
 has constant depth (independent of n)
 
Remarks
1.
The default size of n
th
 circuit:  
polynomial in n
2.
This is a 
non-uniform
 model of computation
(Can compute some undecidable languages)
3.
ACC circuits can be efficiently simulated by
       
constant-depth threshold circuits
Brief History of Circuit Lower Bounds
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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What We Prove: Theorem 1
 
EXP
NP 
 
=  Exponential Time with an NP oracle
 
Theorem 1
 
  
There is a problem 
Q
 in 
EXP
NP
 
such that for
every 
d
, m there is an 
ε
 > 0 such that 
Q
 fails to have 
ACC
circuits with 
MODm gates, depth 
d,
 and size 
2
n
ε
 
Corollary
 
 
Any problem that is 
EXP
NP
-complete under ACC
reductions cannot be solved with circuits of the above type
 
Remark   
It has been known for 30+ years that
 
EXP
NP
NP
         
cannot have 
unrestricted 
subexponential-size circuits
What We Prove: Theorem 2
 
Theorem 2
 
 
There is a problem 
Q 
in 
NEXP
 such that
 Q
 fails
to have
 
n
poly(log n)
 size 
ACC circuits of any constant depth
Corollary
 
 
Any problem that is 
NEXP
-complete under ACC
reductions cannot be solved with polynomial size ACC
 
Remark   
It has been known for 30+ years that 
NEXP
NP
 
 
  
   cannot have 
unrestricted  
polynomial-size
 
circuits
 
[KW’95] 
ZPEXP
NP
 doesn’t have polysize circuits
 
[BFT’98] 
MAEXP
 doesn’t have polysize circuits
 
NEXP 
= Nondeterministic Exponential Time
How We Prove It: Intuition
 
ACC circuits should be much weaker than 
NEXP
.
Find some 
nice property
 of ACC that 
NEXP
 doesn’t have.
Turn this into a proof!
 
Nondeterministic Time Hierarchy Theorem
 [SFM ’78]
For functions 
t
, 
T
 such that 
t(n+1) 
o(T(n))
,
NTIME[t(n)] 
(
 NTIME[T(n)]
 
Corollary
 
  
There are 
NTIME[2
n
]
 problems that can’t be
 
      
nondeterministically solved in O(
2
n
/n)
 time.
 
Can problems recognized by ACC circuits
“be nondeterminstically solved” in O
(2
n
/n
) time?
 
What could this mean??  
ACC is a non-uniform class…
Circuit Satisfiability
 
Let C be a class of Boolean circuits
 
C = {formulas}, C = {ANDs of ORs}, C = {ACC circuits}
 
 
 
 
 
  
C-SAT 
is NP-complete, for essentially all interesting 
C
 
  
C-SAT 
is solvable in 
O(2
n
 
¢
|K|) 
time
The C-SAT Problem:
Given a circuit  
K(x
1
,…,x
n
)
 
2
 
C
, is there an assignment
(a
1
, …, a
n
) 
2
 {0,1}
n
  
such that  
K(a
1
,…,a
n
) =1 
?
Turning Positives Into Negatives
“A slightly faster algorithm for C-SAT implies
lower bounds against C-Circuits solving NEXP”
 
O
(
2
n
 
/
n
1
0
)
 
0
1
0
1
1
0
1
0
1
0
0
 
1
Turning Positives Into Negatives
A faster C-SAT algorithm uncovers a “weakness”
in representing computations with C-circuits
O
(
2
n
 
/
n
1
0
)
0
1
0
1
1
0
1
0
1
0
0
1
A faster C-SAT algorithm uncovers a “strength”
in less-than-exponential algorithms!
O
(
2
n
 
/
n
1
0
)
0
1
0
1
1
0
1
0
1
0
0
1
Turning Positives Into Negatives
Outline of Proofs
 
1.
Show that faster ACC-SAT algorithms imply
ACC Circuit Lower Bounds
2.
Design faster ACC-SAT algorithms!
 
Theorem
 [STOC’10]
  
Let a(n) 
¸
 n
!
(1)
Circuit SAT with n inputs and a(n) size in 
O(2
n
/a(n
)) time
)
  
NEXP
 
*
 
P/poly
 
Proof Idea:
  
Show that if both
Circuit SAT in 
O(2
n
/a(n
)) time   and  
NEXP
 
µ
 
P/poly
then 
 
NTIME[2
n
] 
µ
 
NTIME[o(2
n
)]
 
 
  
(contradiction)
Outline of Proofs
 
1.
Show that faster ACC-SAT algorithms imply
ACC Circuit Lower Bounds
 
Theorem
  (Example)
If ACC Circuit SAT with n inputs and 
2
n
ε
 size is in 
O(2
n
/n
10
)
time, then 
EXP
NP
 doesn’t have 
2
O(n
ε
)
 
size ACC circuits.
 
2.
Design faster ACC-SAT algorithms
 
Theorem
  
For all d, m there’s an 
ε
 > 0 such that
ACC Circuit SAT with n inputs, depth d, MODm gates, and
2
n
ε
 size can be solved in 
2
n - 

n
ε
)  
time
Outline of Talk
  Faster ACC SAT Algorithms
    
)
 
2
n
ε
 size
 ACC Lower Bounds for 
EXP
NP
  Faster ACC SAT Algorithms
  Lower Bounds for NEXP
  Future Work
Fast ACC SAT 
) 
ACC Lower Bounds
 
Theorem
  
If ACC SAT with n inputs and 
2
n
ε
 size is in 
O(2
n
/n
10
)
time, then 
EXP
NP
 doesn’t have 
2
O(n
ε
)
 
size ACC circuits.
 
Proof Idea
   
Show that if both:
ACC Circuit SAT with n inputs and 
2
n
ε
 size is in 
O(2
n
/n
10
)
EXP
NP
 has 
2
O(n
ε
)
 
size ACC circuits
then 
 
NTIME[2
n
] 
µ
 
NTIME[o(2
n
)]
    
(contradiction!)
For every L 
2
 
NTIME[2
n
], reduce L to Succinct3SAT with a
tight reduction, then use ACC Circuits + ACC-SAT to solve
Succinct3SAT in 
faster 
than 
2
n
 time.
 
Theorem
  
If ACC SAT with n inputs and 
2
n
ε
 size is in 
O(2
n
/n
10
)
time, then 
EXP
NP
 doesn’t have 
2
O(n
ε
)
 
size ACC circuits.
 
Work with a “compressed” version of the 3SAT problem
 
Succinct 3SAT
:
 
Given a circuit C, let F be the string obtained by
evaluating C on all possible assignments in lexicographical
order. Does F encode a satisfiable 3CNF formula?
 
Theorem
 [PY]
  
Succinct 3SAT is 
NEXP
-complete.
 
We say 
Succinct 3SAT has ACC satisfying assignments
 
if
for 
every 
C that encodes a satisfiable 3CNF F,
there is an ACC circuit D of 
2
|C|
ε
 size such that
 
D encodes an assignment A which satisfies F.
Evaluating D on all of its possible assignments generates A
“Compressible formulas have compressible sat assignments”
 
Theorem
  
If ACC SAT with n inputs and 
2
n
ε
 size is in 
O(2
n
/n
10
)
time, then 
EXP
NP
 doesn’t have 
2
O(n
ε
)
 
size ACC circuits.
 
For every L 
2
 
NTIME[2
n
], reduce L to Succinct3SAT with a
tight reduction, then use ACC Circuits + ACC-SAT to solve
Succinct3SAT in 
faster 
than 
2
n
 time.
 
Lemma 1
  If 
EXP
NP
 has 
2
n
ε
 size ACC circuits then
Succinct 3SAT has ACC satisfying assignments
.
 
Proof
  
The following can be computed in 
EXP
NP
:
 
On input (C, i), find the lexicographically first satisfying
assignment to F encoded by C, then output its 
i-
th bit.
Hence there are 
2
|C|
ε
 size ACC circuits that take (C, i) as input,
and output the i-th bit of a satisfying assignment to F
encoded by C.
 
Theorem
  
If ACC SAT with n inputs and 
2
n
ε
 size is in 
O(2
n
/n
10
)
time, then 
EXP
NP
 doesn’t have 
2
O(n
ε
)
 
size ACC circuits.
 
For every L 
2
 
NTIME[2
n
], reduce L to Succinct3SAT with a
tight reduction, then use ACC Circuits + ACC-SAT to solve
Succinct3SAT in 
faster 
than 
2
n
 time.
 
Lemma 1
  If 
EXP
NP
 
has 
2
n
ε
 size ACC circuits then
Succinct 3SAT has ACC satisfying assignments.
 
Lemma 2
 [FLvMV ’05] 
For all L 
2
 
NTIME[2
n
]
, there is a
polytime reduction 
R
L
 from L to Succinct 3SAT such that:
 
-  
x 
2
 L  
R
L
(x) = 
C
x
 encodes a satisfiable 3CNF formula
 
-  
C
x
 has size at most poly(n)
 
-  
C
x
 has at most n + 4 log n inputs
 
Theorem
  
If ACC SAT with n inputs and 
2
n
ε
 size is in 
O(2
n
/n
10
)
time, then 
EXP
NP
 doesn’t have 
2
O(n
ε
)
 
size ACC circuits.
 
First Try at the Proof
   
Let L 
2
 
NTIME[2
n
] be arbitrary.
We want to produce a faster NTIME algorithm for L.
- By Lemma 2, there is a reduction 
R
L 
from L to Succinct 3SAT.
Given x, let 
C
x
 be the circuit generated by 
R
L
(x
), and
let  
φ
x
 be the exponentially long formula encoded by
 
C
x
 
- By Lemma 1, 
Succinct3SAT
 has ACC satisfying assignments:
there is an ACC circuit 
Y
 of subexponential size such that
Y(i
) 
outputs the 
i-
th bit of a satisfying assignment for 
φ
x
.
 
Now we will give an 
NTIME[o(2
n
)] algorithm for L
 
 
 
 
 
 
Given input x of length n,
Guess
 an ACC circuit Y of 
2
n
ε
 size
Y
(
j
)
 
i
s
 
s
u
p
p
o
s
e
d
 
t
o
 
o
u
t
p
u
t
 
t
h
e
 
j
-
t
h
 
b
i
t
 
o
f
 
a
 
s
a
t
i
s
f
y
i
n
g
 
a
s
s
i
g
n
m
e
n
t
 
f
o
r
 
φ
x
Construct the following circuit 
D
 which has O(
2
n
ε
) size:
o(2
n
) algorithm for L 
2
 
NTIME[2
n
]?
 
Using ACC SAT algorithm: determine satisfiability of 
D
 in 
o(2
n
) time??
 
             
i 
            
n + 4 log n input bits
 
C
x
 
D outputs 1 iff the assignment encoded by Y
does not
 satisfy the 
i
-th clause of 
φ
x
 
Y
 
s
 
Y
 
Y
a
b
 
t
 
u
c
 
Constant size circuit
 
O
u
t
p
u
t
s
 
t
h
e
 
i
t
h
 
c
l
a
u
s
e
 
o
f
 
3
C
N
F
 
φ
x
 
O
u
t
p
u
t
s
 
a
s
s
i
g
n
m
e
n
t
s
 
t
o
 
t
h
e
 
 
 
 
v
a
r
i
a
b
l
e
s
 
a
,
 
b
,
 
c
 
o
f
 
φ
x
C
x
 may not
be an ACC
circuit!
How to Get an ACC Circuit out of C
x
 
Lemma 2
 [FLvMV ’05] 
For all L 
2
 
NTIME[2
n
], there is a
polytime
 reduction 
R
L
 from L to Succinct 3SAT such that:
 
-  x 
2
 L  
R
L
(x) = 
C
x
 encodes a satisfiable 3CNF formula
 
-  
C
x
 has size at most poly(n)
 
-  
C
x
 has at most n + 4 log n inputs
 
The circuits 
C
x
 are constructible in polynomial time!
We already are assuming 
EXP
NP
 
has 
2
n
ε
 size ACC circuits
Therefore there 
are
 
2
n
ε
 size ACC circuits 
C’
x
 equivalent to 
C
x
We can guess 
C’
x
 … but how do we verify 
C’
x
(i) = 
C
x
(i) for all i?
Use ACC SAT algorithm?  
This looks ridiculous.
  
C
x 
 
is an 
arbitrary
circuit, how could we use an ACC SAT algorithm on it?
 
Lemma 3
   
If
  
P
 has ACC circuits of size s(n)
and
  
ACC SAT is in 
o(2
n
) time
, 
then
 for all L in 
NTIME[2
n
], there
is an ntime 
o(2
n
) algorithm that, on every input x, prints an
ACC circuit 
C’
x
 of size s(O(n)) that is 
equivalent
 to the circuit 
C
x
 
Proof Sketch:
 Exploit the P-uniformity of 
C
x
Guess two ACC circuits 
G
 and 
H
1.
G
 encodes 
gate information of 
C
x 
:
G(x, j) = (
j
1
, 
j
2
, g)
 where the two inputs to gate 
j
 in 
C
x
 are
from gates 
j
1
 and 
j
2
, and the gate type of gate 
j
 is 
g
2.
H
 encodes 
gate values of 
C
x
:
H(x, i, j) = the output of gate j when 
C
x
 is evaluated on i
Given a correct 
H
,  
C
x
(i) = H(x, i, j*)
  
where j* = output gate
Must verify G(x,
¢
) is correct, and H(x,
¢
,
¢
) is correct.
How to Get an ACC Circuit out of C
x
Outline of Talk
  Faster ACC SAT Algorithms
    
)
 
2
n
ε
 size
 ACC Lower Bounds for 
EXP
NP
  Faster ACC SAT Algorithms
  Lower Bounds for NEXP
  Future Work
New Algorithm for ACC-SAT
 
Ingredients:
1.
A known representation of ACC
[Yao ’90, Beigel-Tarui’94]
  Every ACC function
f : {
0,1}
n
 ! 
{0,1} 
can be put in the form
f
(
x
1
,
.
.
.
,
x
n
)
 
=
 
g
(
h
(
x
1
,
.
.
.
,
x
n
)
)
-
 
 
h
 
i
s
 
a
 
s
p
a
r
s
e
"
 
p
o
l
y
n
o
m
i
a
l
,
 
 
h
(
x
1
,
.
.
.
,
x
n
)
 
2
 
{
0
,
,
K
}
-
  K
 is not “too large” 
(quasipolynomial in circuit size)
-
  
g : {0,...,K}  
!
 {0,1}
 can be an arbitrary function
 
2.
“Fast Fourier Transform” for multilinear polynomials:
 
Given a multilinear polynomial h in its coefficient
representation, the value h(x) can be computed over
all points x 
2
 
{
0,1}
n
 
in 
2
n
 poly(n) 
time.
The ACC Satisfiability Algorithm
 
Theorem
  
For all d, m there’s an 
ε
 > 0 such that 
ACC[m]  SAT
with depth d, n inputs, 
2
n
ε
 size can be solved in 
2
n - 

n
ε
)  
time
 
Proof:
 
 
      
n inputs
 K = 2
n
O(
ε
)
size
 
            
n-n
ε
 
 inputs
C
Size
2
n
ε
g
 
 
Beigel and Tarui
 
T
a
k
e
 
a
n
 
O
R
 
o
f
 
a
l
l
 
a
s
s
i
g
n
m
e
n
t
s
t
o
 
t
h
e
 
f
i
r
s
t
 
n
ε
 
i
n
p
u
t
s
 
o
f
 
C
C
2
n
ε
C
2
n
ε
C
2
n
ε
C
2
n
ε
C
2
n
ε
C
2
n
ε
 
 
Fast Fourier Transform
 
For small 
ε
 > 0, e
valuate h
on all 
2
n - n
ε
 
assignments in
2
n -
n
ε
 
poly(n)  time
 
n
-
n
ε
 
i
n
p
u
t
s
 
2
2n
ε
 
size
 
h
Outline of Talk
  Faster ACC SAT Algorithms
    
)
 
2
n
ε
 size
 ACC Lower Bounds for 
EXP
NP
  Faster ACC SAT Algorithms
  Lower Bounds for NEXP
  Future Work
 
Theorem
  
If ACC SAT with n inputs, 
n
O(1
)
 size is in 
O(2
n
/n
10
)
time, then 
NEXP
 doesn’t have 
n
O(1
) 
size ACC circuits.
 
Proceed just as with 
EXP
NP
, but use the following lemma:
Lemma
 [IKW’02]
 
 If 
NEXP 
µ
 ACC
 then Succinct 3SAT has
poly-size circuits encoding its satisfying assignments.
 
The proof applies work on “hardness versus randomness”
1. If 
EXP 
µ
 P/poly
 then EXP = MA 
[BFNW93]
2. If Succinct 3SAT does 
not
 have sat assignment circuits then
in 
NTIME[2
n
] we can 
guess a hard function and verify it
  Can derandomize MA infinitely often with n bits of advice:
EXP = MA 
µ
 i.o.-
NTIME[2
n
]/n 
µ
 i.o.-
SIZE(n
k
)
(this is a contradiction)
 
Theorem
  
If ACC SAT with n inputs, 
n
O(1
)
 size is in 
O(2
n
/n
10
)
time, then 
NEXP
 doesn’t have 
n
O(1
) 
size ACC circuits.
 
Proceed just as with 
EXP
NP
, but use the following lemma:
Lemma
 [IKW’02]
 
 If 
NEXP 
µ
 ACC
 then Succinct 3SAT has
poly-size circuits encoding sat assignments.
 
Lemma
  
If 
P 
µ
 ACC  
then all poly-size 
unrestricted
 circuit
families have equivalent poly-size ACC circuit families.
Corollary
  
If 
NEXP
 µ
 ACC
 then Succinct 3SAT has poly-size
ACC circuits encoding its satisfying assignments.
 
This is all we need for the previous proof to go through.
Also works for quasipolynomial size circuits.
 
Outline of Talk
  Faster ACC SAT Algorithms
    
)
 
2
n
ε
 size
 ACC Lower Bounds for 
EXP
NP
  Faster ACC SAT Algorithms
  Lower Bounds for NEXP
  Future Work
Future Work
 
Can NEXP be replaced with EXP?
Appears we need the 
deterministic
 time
hierarchy for this. Couldn’t 
guess
 circuits then…
 
Can ACC be replaced with TC0?
We only need new TC0 SAT algorithms!
 
Is Uniform ACC 
 NP?
 
Is it possible that 
circuit lower bounds
 could help
provide 
faster algorithms for NP problems
?
 
 
Thank you
for waking up for this talk!
(I am not sure that I did.)
TexPoint fonts used in EMF.
Read the TexPoint manual before you delete this box.: 
A
A
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Theorem 2
 Let a(n) 
¸
 n
!
(1)
.  Suppose we are given a circuit C
and are promised that it is either 
unsatisfiable,
 or at least
½ of its assignments are satisfying. Determine which.
If this problem is in 
O(2
n
/a(n
)) time then NEXP 
*
 P/poly.
Proof Idea: 
Same as Theorem 1, but 
replace 
the succinct
reduction 
R
L 
from L to 3SAT with a 
succinct PCP reduction
 
Lemma 3
 
[BGHSV’05]
  
For all L 
2
 
NTIME(2
n
),
there is a reduction 
S
L
 from L to 
MAX CSP
 such that:
 
  x 
2
 L  
)
  All constraints of 
S
L
(x
) are satisfiable
  x 
 L  
)
  At most ½ of the constraints are satisfiable
1. |
S
L
(x
)| = 
2
n
 poly(n)
2. The i-th constraint of 
S
L
(x
) is computable in poly(n) time.
 
 
 
 
Weak Derandomization Suffices
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Non-Uniform ACC circuits play a crucial role in circuit lower bound research. Explore the history, theorems, and implications behind these circuits in the context of complexity theory.

  • Circuit theory
  • Complexity
  • Non-Uniform ACC
  • Lower bounds

Uploaded on Feb 19, 2025 | 0 Views


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  1. Non-Uniform ACC Circuit Lower Bounds MOD6 MOD6 Ryan Williams IBM Almaden

  2. The Circuit Class ACC An ACC circuit family {Cn} has the following properties: Each Cn takes n bits of input and outputs a bit The gates of Cn can be AND, OR, NOT, MODm MODm(x1, ,xt) = 1 iff i xi is divisible by m Cn has constant depth (independent of n) Remarks 1. The default size of nth circuit: polynomial in n 2. This is a non-uniform model of computation (Can compute some undecidable languages) 3. ACC circuits can be efficiently simulated by constant-depth threshold circuits

  3. Brief History of Circuit Lower Bounds Prove P NP by proving NP does not have polynomial size Boolean circuits. The simplicity and combinatorial nature of circuits should make it easier to prove impossibility results. Ajtai, Furst-Saxe-Sipser (early 80 s) PARITY of n bits AC0 (i.e., poly(n) size ACC without MODm) Razborov, Smolensky (late 80 s) MOD3 AC0 with MOD2 (PARITY) gates MODp AC0 with MODq gates, p q prime Barrington (late 80 s) Suggested ACC as the next step Conjecture Majority ACC Conjecture (early 90 s) NP ACC Conjecture (late 90 s) NEXP ACC

  4. What We Prove: Theorem 1 EXPNP = Exponential Time with an NP oracle Theorem 1 There is a problem Q in EXPNP such that for every d, m there is an > 0 such that Q fails to have ACC circuits with MODm gates, depth d, and size 2n Corollary Any problem that is EXPNP-complete under ACC reductions cannot be solved with circuits of the above type Remark It has been known for 30+ years that EXPNPNP cannot have unrestricted subexponential-size circuits

  5. What We Prove: Theorem 2 NEXP = Nondeterministic Exponential Time Theorem 2 There is a problem Q in NEXP such that Q fails to have npoly(log n) size ACC circuits of any constant depth Corollary Any problem that is NEXP-complete under ACC reductions cannot be solved with polynomial size ACC Remark It has been known for 30+ years that NEXPNP cannot have unrestricted polynomial-sizecircuits [KW 95] ZPEXPNPdoesn t have polysize circuits [BFT 98] MAEXPdoesn t have polysize circuits

  6. How We Prove It: Intuition ACC circuits should be much weaker than NEXP. Find some nice property of ACC that NEXPdoesn t have. Turn this into a proof! Nondeterministic Time Hierarchy Theorem [SFM 78] For functions t, T such that t(n+1) o(T(n)), NTIME[t(n)] ( ( NTIME[T(n)] Corollary There are NTIME[2n]problems that can t be nondeterministically solved in O(2n/n) time. Can problems recognized by ACC circuits be nondeterminstically solved in O(2n/n) time? What could this mean?? ACC is a non-uniform class

  7. Circuit Satisfiability Let C be a class of Boolean circuits C = {formulas}, C = {ANDs of ORs}, C = {ACC circuits} The C-SAT Problem: Given a circuit K(x1, ,xn)2C, is there an assignment (a1, , an) 2 {0,1}n such that K(a1, ,an) =1 ? C-SAT is NP-complete, for essentially all interesting C C-SAT is solvable in O(2n |K|) time

  8. Turning Positives Into Negatives A slightly faster algorithm for C-SAT implies lower bounds against C-Circuits solving NEXP Size =nc O(2n /n10) Size =nc NEXP x1 0 1 0 1 1 0 1 0 1 0 0 x1 1 xn xn

  9. Turning Positives Into Negatives A faster C-SAT algorithm uncovers a weakness in representing computations with C-circuits Size =nc O(2n /n10) x1 0 1 0 1 1 0 1 0 1 0 0 NEXP 1 xn

  10. Turning Positives Into Negatives A faster C-SAT algorithm uncovers a strength in less-than-exponential algorithms! Size =nc O(2n /n10) x1 0 1 0 1 1 0 1 0 1 0 0 NEXP 1 xn

  11. Outline of Proofs 1. Show that faster ACC-SAT algorithms imply ACC Circuit Lower Bounds 2. Design faster ACC-SAT algorithms! Theorem [STOC 10]Let a(n) n!(1) Circuit SAT with n inputs and a(n) size in O(2n/a(n)) time )NEXP*P/poly Proof Idea: Show that if both Circuit SAT in O(2n/a(n)) time and NEXP P/poly then NTIME[2n] NTIME[o(2n)] (contradiction)

  12. Outline of Proofs 1. Show that faster ACC-SAT algorithms imply ACC Circuit Lower Bounds Theorem (Example) If ACC Circuit SAT with n inputs and 2n size is in O(2n/n10) time, then EXPNPdoesn t have 2O(n ) size ACC circuits. 2. Design faster ACC-SAT algorithms TheoremFor all d, m there s an > 0 such that ACC Circuit SAT with n inputs, depth d, MODm gates, and 2n size can be solved in 2n - (n ) time

  13. Outline of Talk Faster ACC SAT Algorithms ) 2n size ACC Lower Bounds for EXPNP Faster ACC SAT Algorithms Lower Bounds for NEXP Future Work

  14. Fast ACC SAT ) ) ACC Lower Bounds Theorem If ACC SAT with n inputs and 2n size is in O(2n/n10) time, then EXPNPdoesn t have 2O(n ) size ACC circuits. Proof Idea Show that if both: ACC Circuit SAT with n inputs and 2n size is in O(2n/n10) EXPNP has 2O(n ) size ACC circuits then NTIME[2n] NTIME[o(2n)](contradiction!) For every L 2 2 NTIME[2n], reduce L to Succinct3SAT with a tight reduction, then use ACC Circuits + ACC-SAT to solve Succinct3SAT in faster than 2n time.

  15. Theorem If ACC SAT with n inputs and 2n size is in O(2n/n10) time, then EXPNPdoesn t have 2O(n ) size ACC circuits. Work with a compressed version of the 3SAT problem Succinct 3SAT: Given a circuit C, let F be the string obtained by evaluating C on all possible assignments in lexicographical order. Does F encode a satisfiable 3CNF formula? Theorem [PY]Succinct 3SAT is NEXP-complete. We say Succinct 3SAT has ACC satisfying assignments if for every C that encodes a satisfiable 3CNF F, there is an ACC circuit D of 2|C| size such that D encodes an assignment A which satisfies F. Evaluating D on all of its possible assignments generates A Compressible formulas have compressible sat assignments

  16. Theorem If ACC SAT with n inputs and 2n size is in O(2n/n10) time, then EXPNPdoesn t have 2O(n ) size ACC circuits. For every L 2 2 NTIME[2n], reduce L to Succinct3SAT with a tight reduction, then use ACC Circuits + ACC-SAT to solve Succinct3SAT in faster than 2n time. Lemma 1 If EXPNP has 2n size ACC circuits then Succinct 3SAT has ACC satisfying assignments. Proof The following can be computed in EXPNP: On input (C, i), find the lexicographically first satisfying assignment to F encoded by C, then output its i-th bit. Hence there are 2|C| size ACC circuits that take (C, i) as input, and output the i-th bit of a satisfying assignment to F encoded by C.

  17. Theorem If ACC SAT with n inputs and 2n size is in O(2n/n10) time, then EXPNPdoesn t have 2O(n ) size ACC circuits. For every L 2 2 NTIME[2n], reduce L to Succinct3SAT with a tight reduction, then use ACC Circuits + ACC-SAT to solve Succinct3SAT in faster than 2n time. Lemma 1 If EXPNPhas 2n size ACC circuits then Succinct 3SAT has ACC satisfying assignments. Lemma 2 [FLvMV 05] For all L 2NTIME[2n], there is a polytime reduction RL from L to Succinct 3SAT such that: - x 2 L RL(x) = Cx encodes a satisfiable 3CNF formula - Cx has size at most poly(n) - Cx has at most n + 4 log n inputs

  18. Theorem If ACC SAT with n inputs and 2n size is in O(2n/n10) time, then EXPNPdoesn t have 2O(n ) size ACC circuits. First Try at the Proof Let L 2 NTIME[2n] be arbitrary. We want to produce a faster NTIME algorithm for L. - By Lemma 2, there is a reduction RL from L to Succinct 3SAT. Given x, let Cx be the circuit generated by RL(x), and let x be the exponentially long formula encoded by Cx - By Lemma 1, Succinct3SAT has ACC satisfying assignments: there is an ACC circuit Y of subexponential size such that Y(i) outputs the i-th bit of a satisfying assignment for x. Now we will give an NTIME[o(2n)] algorithm for L

  19. o(2n) algorithm for L 2 NTIME[2n]? Given input x of length n, Guess an ACC circuit Y of 2n size Y(j) is supposed to output the j-th bit of a satisfying assignment for x Construct the following circuit D which has O(2n ) size: D outputs 1 iff the assignment encoded by Y does not satisfy the i-th clause of x Constant size circuit Outputs assignments to the variables a, b, c of x Y Y Y s a t b u c Cx Outputs the ith clause of 3CNF x i n + 4 log n input bits Using ACC SAT algorithm: determine satisfiability of D in o(2n) time??

  20. How to Get an ACC Circuit out of Cx Lemma 2 [FLvMV 05] For all L 2 NTIME[2n], there is a polytime reduction RL from L to Succinct 3SAT such that: - x 2 L RL(x) = Cx encodes a satisfiable 3CNF formula - Cx has size at most poly(n) - Cx has at most n + 4 log n inputs The circuits Cx are constructible in polynomial time! We already are assuming EXPNP has 2n size ACC circuits Therefore there are 2n size ACC circuits C x equivalent to Cx We can guess C x but how do we verify C x(i) = Cx(i) for all i? Use ACC SAT algorithm? This looks ridiculous. Cx is an arbitrary circuit, how could we use an ACC SAT algorithm on it?

  21. How to Get an ACC Circuit out of Cx Lemma 3IfP has ACC circuits of size s(n) and ACC SAT is in o(2n) time, then for all L in NTIME[2n], there is an ntime o(2n) algorithm that, on every input x, prints an ACC circuit C x of size s(O(n)) that is equivalent to the circuit Cx Proof Sketch: Exploit the P-uniformity of Cx Guess two ACC circuits G and H 1. G encodes gate information of Cx : G(x, j) = (j1, j2, g) where the two inputs to gate j in Cx are from gates j1 and j2, and the gate type of gate j is g 2. H encodes gate values of Cx: H(x, i, j) = the output of gate j when Cx is evaluated on i Given a correct H, Cx(i) = H(x, i, j*) where j* = output gate Must verify G(x, ) is correct, and H(x, , ) is correct.

  22. Outline of Talk Faster ACC SAT Algorithms ) 2n size ACC Lower Bounds for EXPNP Faster ACC SAT Algorithms Lower Bounds for NEXP Future Work

  23. New Algorithm for ACC-SAT Ingredients: 1. A known representation of ACC [Yao 90, Beigel-Tarui 94] Every ACC function f : {0,1}n ! ! {0,1} can be put in the form f(x1,...,xn) = g(h(x1,...,xn)) - his a sparse" polynomial, h(x1,...,xn) 2 2{0, ,K} - Kis not too large (quasipolynomial in circuit size) - g : {0,...,K} ! ! {0,1} can be an arbitrary function 2. Fast Fourier Transform for multilinear polynomials: Given a multilinear polynomial h in its coefficient representation, the value h(x) can be computed over all points x 2{0,1}nin 2n poly(n) time.

  24. The ACC Satisfiability Algorithm TheoremFor all d, m there s an > 0 such that ACC[m] SAT with depth d, n inputs, 2n size can be solved in 2n - (n ) time Proof: 22n size Take an OR of all assignments to the first n inputs of C C Size 2n C C C C C C n inputs K = 2nO( ) size n-n inputs 2n 2n 2n 2n 2n 2n g n-n inputs For small > 0, evaluate h on all 2n - n assignments in 2n -n poly(n) time Fast Fourier Transform h

  25. Outline of Talk Faster ACC SAT Algorithms ) 2n size ACC Lower Bounds for EXPNP Faster ACC SAT Algorithms Lower Bounds for NEXP Future Work

  26. Theorem If ACC SAT with n inputs, nO(1) size is in O(2n/n10) time, then NEXP doesn t have nO(1) size ACC circuits. Proceed just as with EXPNP, but use the following lemma: Lemma [IKW 02] If NEXP ACC then Succinct 3SAT has poly-size circuits encoding its satisfying assignments. The proof applies work on hardness versus randomness 1. If EXP P/poly then EXP = MA [BFNW93] 2. If Succinct 3SAT does not have sat assignment circuits then in NTIME[2n] we can guess a hard function and verify it Can derandomize MA infinitely often with n bits of advice: EXP = MA i.o.-NTIME[2n]/n i.o.-SIZE(nk) (this is a contradiction)

  27. Theorem If ACC SAT with n inputs, nO(1) size is in O(2n/n10) time, then NEXP doesn t have nO(1) size ACC circuits. Proceed just as with EXPNP, but use the following lemma: Lemma [IKW 02] If NEXP ACC then Succinct 3SAT has poly-size circuits encoding sat assignments. Lemma If P ACC then all poly-size unrestricted circuit families have equivalent poly-size ACC circuit families. Corollary If NEXP ACC then Succinct 3SAT has poly-size ACC circuits encoding its satisfying assignments. This is all we need for the previous proof to go through. Also works for quasipolynomial size circuits.

  28. Outline of Talk Faster ACC SAT Algorithms ) 2n size ACC Lower Bounds for EXPNP Faster ACC SAT Algorithms Lower Bounds for NEXP Future Work

  29. Future Work Can NEXP be replaced with EXP? Appears we need the deterministic time hierarchy for this. Couldn t guesscircuits then Can ACC be replaced with TC0? We only need new TC0 SAT algorithms! Is Uniform ACC NP? Is it possible that circuit lower bounds could help provide faster algorithms for NP problems?

  30. Thank you for waking up for this talk! (I am not sure that I did.)

  31. Weak Derandomization Suffices Theorem 2 Let a(n) n!(1). Suppose we are given a circuit C and are promised that it is either unsatisfiable, or at least of its assignments are satisfying. Determine which. If this problem is in O(2n/a(n)) time then NEXP * * P/poly. Proof Idea: Same as Theorem 1, but replace the succinct reduction RL from L to 3SAT with a succinct PCP reduction Lemma 3[BGHSV 05] For all L 2 NTIME(2n), there is a reduction SL from L to MAX CSP such that: x 2 2 L ) ) All constraints of SL(x) are satisfiable x L ) ) At most of the constraints are satisfiable 1. |SL(x)| = 2n poly(n) 2. The i-th constraint of SL(x) is computable in poly(n) time.

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