10 KiB
| title | chunk | source | category | tags | date_saved | instance |
|---|---|---|---|---|---|---|
| Communication complexity | 2/8 | https://en.wikipedia.org/wiki/Communication_complexity | reference | science, encyclopedia | 2026-05-05T14:40:16.151030+00:00 | kb-cron |
=== Example: EQ === We consider the case where Alice and Bob try to determine whether or not their input strings are equal. Formally, define the Equality function, denoted
E
Q
:
{
0
,
1
}
n
×
{
0
,
1
}
n
→
{
0
,
1
}
{\displaystyle EQ:\{0,1\}^{n}\times \{0,1\}^{n}\rightarrow \{0,1\}}
, by
E
Q
(
x
,
y
)
=
1
{\displaystyle EQ(x,y)=1}
if
x
=
y
{\displaystyle x=y}
. As we demonstrate below, any deterministic communication protocol solving
E
Q
{\displaystyle EQ}
requires
n
{\displaystyle n}
bits of communication in the worst case. As a warm-up example, consider the simple case of
x
,
y
∈
{
0
,
1
}
3
{\displaystyle x,y\in \{0,1\}^{3}}
. The equality function in this case can be represented by the matrix below. The rows represent all the possibilities of
x
{\displaystyle x}
, the columns those of
y
{\displaystyle y}
.
In this table, the function only evaluates to 1 when
x
{\displaystyle x}
equals
y
{\displaystyle y}
(i.e., on the diagonal). It is also fairly easy to see how communicating a single bit divides someone's possibilities in half. When the first bit of
y
{\displaystyle y}
is 1, consider only half of the columns (where
y
{\displaystyle y}
can equal 100, 101, 110, or 111).
=== Theorem: D(EQ) = n === Proof. Assume that
D
(
E
Q
)
≤
n
−
1
{\displaystyle D(EQ)\leq n-1}
. This means that there exists
x
≠
x
′
{\displaystyle x\neq x'}
such that
(
x
,
x
)
{\displaystyle (x,x)}
and
(
x
′
,
x
′
)
{\displaystyle (x',x')}
have the same communication transcript
h
{\displaystyle h}
. Since this transcript defines a rectangle,
f
(
x
,
x
′
)
{\displaystyle f(x,x')}
must also be 1. By definition
x
≠
x
′
{\displaystyle x\neq x'}
and we know that equality is only true for
(
a
,
b
)
{\displaystyle (a,b)}
when
a
=
b
{\displaystyle a=b}
. This yields a contradiction. This technique of proving deterministic communication lower bounds is called the fooling set technique.
== Randomized communication complexity == In the above definition, we are concerned with the number of bits that must be deterministically transmitted between two parties. If both the parties are given access to a random number generator, can they determine the value of
f
{\displaystyle f}
with much less information exchanged? Yao, in his seminal paper answers this question by defining randomized communication complexity. A randomized protocol
R
{\displaystyle R}
for a function
f
{\displaystyle f}
has two-sided error.
Pr
[
R
(
x
,
y
)
=
0
]
>
2
3
,
if
f
(
x
,
y
)
=
0
{\displaystyle \Pr[R(x,y)=0]>{\frac {2}{3}},{\textrm {if}}\,f(x,y)=0}
Pr
[
R
(
x
,
y
)
=
1
]
>
2
3
,
if
f
(
x
,
y
)
=
1
{\displaystyle \Pr[R(x,y)=1]>{\frac {2}{3}},{\textrm {if}}\,f(x,y)=1}
A randomized protocol is a deterministic protocol that uses an extra random string in addition to its normal input. There are two models for this: a public string is a random string that is known by both parties beforehand, while a private string is generated by one party and must be communicated to the other party. A theorem presented below shows that any public string protocol can be simulated by a private string protocol that uses O(log n) additional bits compared to the original. In the probability inequalities above, the outcome of the protocol is understood to depend only on the random string; both strings x and y remain fixed. In other words, if R(x,y) yields g(x,y,r) when using random string r, then g(x,y,r) = f(x,y) for at least 2/3 of all choices for the string r. The randomized complexity is simply defined as the number of bits exchanged in such a protocol. Note that it is also possible to define a randomized protocol with one-sided error, and the complexity is defined similarly.
=== Example: EQ === Returning to the previous example of EQ, if certainty is not required, Alice and Bob can check for equality using only
O
(
log
n
)
{\displaystyle O(\log n)}
messages. Consider the following protocol: Assume that Alice and Bob both have access to the same random string
z
∈
{
0
,
1
}
n
{\displaystyle z\in \{0,1\}^{n}}
. Alice computes
z
⋅
x
{\displaystyle z\cdot x}
and sends this bit (call it b) to Bob. (The
(
⋅
)
{\displaystyle (\cdot )}
is the dot product in GF(2).) Then Bob compares b to
z
⋅
y
{\displaystyle z\cdot y}
. If they are the same, then Bob accepts, saying x equals y. Otherwise, he rejects. Clearly, if
x
=
y
{\displaystyle x=y}
, then
z
⋅
x
=
z
⋅
y
{\displaystyle z\cdot x=z\cdot y}
, so
P
r
o
b
z
[
A
c
c
e
p
t
]
=
1
{\displaystyle Prob_{z}[Accept]=1}
. If x does not equal y, it is still possible that
z
⋅
x
=
z
⋅
y
{\displaystyle z\cdot x=z\cdot y}
, which would give Bob the wrong answer. How does this happen? If x and y are not equal, they must differ in some locations:
{
x
=
c
1
c
2
…
p
…
p
′
…
x
n
y
=
c
1
c
2
…
q
…
q
′
…
y
n
z
=
z
1
z
2
…
z
i
…
z
j
…
z
n
{\displaystyle {\begin{cases}x=c_{1}c_{2}\ldots p\ldots p'\ldots x_{n}\\y=c_{1}c_{2}\ldots q\ldots q'\ldots y_{n}\\z=z_{1}z_{2}\ldots z_{i}\ldots z_{j}\ldots z_{n}\end{cases}}}