8.7 KiB
| title | chunk | source | category | tags | date_saved | instance |
|---|---|---|---|---|---|---|
| Communication complexity | 3/8 | https://en.wikipedia.org/wiki/Communication_complexity | reference | science, encyclopedia | 2026-05-05T14:40:16.151030+00:00 | kb-cron |
Where x and y agree,
z
i
∗
x
i
=
z
i
∗
c
i
=
z
i
∗
y
i
{\displaystyle z_{i}*x_{i}=z_{i}*c_{i}=z_{i}*y_{i}}
so those terms affect the dot products equally. We can safely ignore those terms and look only at where x and y differ. Furthermore, we can swap the bits
x
i
{\displaystyle x_{i}}
and
y
i
{\displaystyle y_{i}}
without changing whether or not the dot products are equal. This means we can swap bits so that x contains only zeros and y contains only ones:
{
x
′
=
00
…
0
y
′
=
11
…
1
z
′
=
z
1
z
2
…
z
n
′
{\displaystyle {\begin{cases}x'=00\ldots 0\\y'=11\ldots 1\\z'=z_{1}z_{2}\ldots z_{n'}\end{cases}}}
Note that
z
′
⋅
x
′
=
0
{\displaystyle z'\cdot x'=0}
and
z
′
⋅
y
′
=
Σ
i
z
i
′
{\displaystyle z'\cdot y'=\Sigma _{i}z'_{i}}
. Now, the question becomes: for some random string
z
′
{\displaystyle z'}
, what is the probability that
Σ
i
z
i
′
=
0
{\displaystyle \Sigma _{i}z'_{i}=0}
? Since each
z
i
′
{\displaystyle z'_{i}}
is equally likely to be 0 or 1, this probability is just
1
/
2
{\displaystyle 1/2}
. Thus, when x does not equal y,
P
r
o
b
z
[
A
c
c
e
p
t
]
=
1
/
2
{\displaystyle Prob_{z}[Accept]=1/2}
. The algorithm can be repeated many times to increase its accuracy. This fits the requirements for a randomized communication algorithm. This shows that if Alice and Bob share a random string of length n, they can send one bit to each other to compute
E
Q
(
x
,
y
)
{\displaystyle EQ(x,y)}
. In the next section, it is shown that Alice and Bob can exchange only
O
(
log
n
)
{\displaystyle O(\log n)}
bits that are as good as sharing a random string of length n. Once that is shown, it follows that EQ can be computed in
O
(
log
n
)
{\displaystyle O(\log n)}
messages.
=== Example: GH === For yet another example of randomized communication complexity, we turn to an example known as the gap-Hamming problem (abbreviated GH). Formally, Alice and Bob both maintain binary messages,
x
,
y
∈
{
−
1
,
+
1
}
n
{\displaystyle x,y\in \{-1,+1\}^{n}}
and would like to determine if the strings are very similar or if they are not very similar. In particular, they would like to find a communication protocol requiring the transmission of as few bits as possible to compute the following partial Boolean function,
GH
n
(
x
,
y
)
:=
{
−
1
⟨
x
,
y
⟩
≤
n
+
1
⟨
x
,
y
⟩
≥
n
.
{\displaystyle {\text{GH}}_{n}(x,y):={\begin{cases}-1&\langle x,y\rangle \leq {\sqrt {n}}\\+1&\langle x,y\rangle \geq {\sqrt {n}}.\end{cases}}}
Clearly, they must communicate all their bits if the protocol is to be deterministic (this is because, if there is a deterministic, strict subset of indices that Alice and Bob relay to one another, then imagine having a pair of strings that on that set disagree in
n
−
1
{\displaystyle {\sqrt {n}}-1}
positions. If another disagreement occurs in any position that is not relayed, then this affects the result of
GH
n
(
x
,
y
)
{\displaystyle {\text{GH}}_{n}(x,y)}
, and hence would result in an incorrect procedure. A natural question one then asks is, if we're permitted to err
1
/
3
{\displaystyle 1/3}
of the time (over random instances
x
,
y
{\displaystyle x,y}
drawn uniformly at random from
{
−
1
,
+
1
}
n
{\displaystyle \{-1,+1\}^{n}}
), then can we get away with a protocol with fewer bits? It turns out that the answer somewhat surprisingly is no, due to a result of Chakrabarti and Regev in 2012: they show that for random instances, any procedure that is correct at least
2
/
3
{\displaystyle 2/3}
of the time must send
Ω
(
n
)
{\displaystyle \Omega (n)}
bits worth of communication, which is to say essentially all of them.