AC class
the "A" in AC stands for "alternating", in reference to alternating turing machines.
the following is an open question:
and size
. if
the claim is true by clm-bool-formula-2 because standard AC circuits of depth 2 are simply DNF or CNF formulas. if
, we apply ac_class.html
times, and get a circuit of depth 2 and size at most
that computes
or
. the size of such a circuit is atleast
, meaning
meaning
and so

(
circuits with constant depth) cannot compute
.
since
cannot compute parity, one can try to empower it by allowing new gates:
is the circuit class in which a circuit is under the restriction of
, but allowed to use parity gates.
however, we can show that this circuit class is still not powerful enough to compute all functions. which is why we try to generalize it again with the following class.
is the circuit class in which a circuit is under the restriction of
, but allowed to use
gates where
.
BROKENand the new task becomes to bound the complexity of functions that
can feasibly compute.
a more formal definition for
is given below:
for
.
[cite:;taken from @complexity_vollmer_1999 definition 4.34]
for
, define
and let
[cite:;taken from @complexity_vollmer_1999 definition 4.5]
a connection to the classes of the NC hierarchy can be stated as follows:
for all
, we have
; hence
.
is there a function in
that every circuit in
with depth 3 that computes it has to be of size atleast
? or even
for some
.
an
circuit is a standard
circuit iff:
, where
is the size of the circuit.
- the circuit is divided into layers, such that edges only connect vertices between subsequent layers.
- in each layer there are only
or
gates (except the input layer).
- in subsequent layers there are only different types of gates (e.g. an AND layer is followed by an OR layer, and vice versa).
every
circuit can be turned into a standard AC circuit such that if the original circuit is of size
and depth
then the new circuit would be of depth
and at most of size
.
note that two consecutive
gates are redundant if you are allowed to have arbitrary fan-in. the same applies to
gates.
let
be a function that can be computed by a standard
circuit of size
and depth
, and let
be a
-random restriction. then with probability of atleast
we can compute
using a very standard circuit of size at most
and depth
(i.e. to every gate in the first non-input layer we have an indegree of at most
).
let
be the circuit that computes
, we prove that with a probability of atleast
all gates in the first layer of
with more than than
ingoing edges, turn into constants as a consequence of the assignment
. the circuit that remains would be a very standard circuit that computes
. let
be a gate with
ingoing edges for
in the first layer. assume that
is an OR gate, then
isnt set to 1 iff
didnt set any of the literals going into
to 1.
is the probability for a literal to not be set to 1, meaning either it gets set to 0 or it stays alive, hence the case where none of the
edges get set to 1 happens with a probability of at most
we get that the probability that a single gate
isnt set to a constant is at most
, and there are at most
such gates, hence the probability that atleast one of them isnt set to a constant is at most
(we know that for every
events
we have
). it follows that with a probability of atleast
all of them are set to constants.
a very standard
circuit of depth
that computes
has to be of size atleast
.
AC circuits by definition have polynomial size, what we mean here is that no AC circuit of polynomial size can compute parity, as we would to extend the class AC to allow for exponential size circuits.
[cite:;variant of @complexity_jukna_2012 theorem 12.3]let
such that
, if there exists a very standard AC circuit of depth
and size
that computes
or
then there exists some very standard AC circuit of depth
and size at most
that computes
or
.
let
, let
be a standard AC circuit that computes
, notice that the gates in the second layer of the circuit compute
-cnf and
-dnf formulas. without loss of generality we assume
-cnf. let
be a random restriction that keeps alive
variables (
). by the switching lemma, after the assignment we can replace every gate
of the second layer with a
-dnf with a probability of atleast
. there is at most
vertices in the sescond layer, therefore the probability that one of them isnt replaceable by a
-dnf after the assignment is at most
hence there exists some assignment
that if we apply we can turn all of the gates in the second layer into
-dnf, and keep alive
variables. after this assignment, the circuit computes
or
, and both the second and third layer compute OR gates. notice that since an OR of OR's is an OR, we can merge those layers and receive a standard layer of depth
.
assume that such a circuit exists of depth BROKEN
[cite:;taken from @complexity_vollmer_1999 definition 4.34]