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datovky
ds2notes
Commits
defbcae6
Commit
defbcae6
authored
Apr 30, 2021
by
Parth Mittal
Browse files
wrote countmin and the AMS estimator for distinct
parent
9744cd1b
Changes
1
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streaming/streaming.tex
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defbcae6
...
...
@@ 43,7 +43,7 @@ us with a small set $C$ containing $F_k$, and hence lets us solve the frequent
elements problem efficiently.
\algo
{
FrequencyEstimate
}
\algalias
{
Misra/Gries Algorithm
}
\algin
the data stream
$
\alpha
$
, the target for the estimator
$
k
$
\algin
the data stream
$
\alpha
$
, the target for the estimator
$
k
$
.
\:\em
{
Init
}
:
$
A
\=
\emptyset
$
.
\cmt
{
an empty map
}
\:\em
{
Process
}
(
$
x
$
):
\:
:If
$
x
\in
$
keys(
$
A
$
),
$
A
[
x
]
\=
A
[
x
]
+
1
$
.
...
...
@@ 93,12 +93,138 @@ $\vert C \vert = \vert$keys($A$)$\vert \leq k  1$, and a keyvalue pair can
be stored in
$
\O
(
\log
n
+
\log
m
)
$
bits.
\qed
\subsection
{
The CountMin sketch
}
\subsection
{
The CountMin Sketch
}
We will now look at a randomized streaming algorithm that solves the
frequency estimation problem. While this algorithm can fail with some
probability, it has the advantage that the output on two different streams
can be easily combined.
\algo
{
FrequencyEstimate
}
\algalias
{
CountMin Sketch
}
\algin
the data stream
$
\alpha
$
, the accuracy
$
\varepsilon
$
,
the error parameter
$
\delta
$
.
\:\em
{
Init
}
:
$
C
[
1
\ldots
t
][
1
\ldots
k
]
\=
0
$
, where
$
k
\=
\lceil
2
/
\varepsilon
\rceil
$
and
$
t
\=
\lceil
\log
(
1
/
\delta
)
\rceil
$
.
\:
: Choose
$
t
$
independent hash functions
$
h
_
1
,
\ldots
h
_
t :
[
n
]
\to
[
k
]
$
, each
from a 2independent family.
\:\em
{
Process
}
(
$
x
$
):
\:
:For
$
i
\in
[
t
]
$
:
$
C
[
i
][
h
_
i
(
x
)]
\=
C
[
i
][
h
_
i
(
x
)]
+
1
$
.
\algout
Report
$
\hat
{
f
}_
a
=
\min
_{
i
\in
t
}
C
[
i
][
h
_
i
(
a
)]
$
.
\endalgo
We will now look at a randomized streaming algorithm that performs the same task
Note that the algorithm needs
$
\O
(
tk
\log
m
)
$
bits to store the table
$
C
$
, and
$
\O
(
t
\log
n
)
$
bits to store the hash functions
$
h
_
1
,
\ldots
h
_
t
$
, and hence
uses
$
\O
(
1
/
\varepsilon
\cdot
\log
(
1
/
\delta
)
\cdot
\log
m
+
\log
(
1
/
\delta
)
\cdot
\log
n
)
$
bits. It remains to show that it computes
a good estimate.
\endchapter
\lemma
{
$
f
_
a
\leq
\hat
{
f
}_
a
\leq
f
_
a
+
\varepsilon
m
$
with probability
$
\delta
$
.
}
\proof
Clearly
$
\hat
{
f
}_
a
\geq
f
_
a
$
for all
$
a
\in
[
n
]
$
; we will show that
$
\hat
{
f
}_
a
\leq
f
_
a
+
\varepsilon
m
$
with probability at least
$
\delta
$
.
For a fixed element
$
a
$
, define the random variable
$$
X
_
i :
=
C
[
i
][
h
_
i
(
a
)]

f
_
a
$$
For
$
j
\in
[
n
]
\setminus
\{
a
\}
$
, define the
indicator variable
$
Y
_{
i, j
}
:
=
[
h
_
i
(
j
)
=
h
_
i
(
a
)
]
$
. Then we can see that
$$
X
_
i
=
\sum
_{
j
\neq
a
}
f
_
j
\cdot
Y
_{
i, j
}$$
Note that
$
\E
[
Y
_{
i, j
}
]
=
1
/
k
$
since each
$
h
_
i
$
is from a 2independent family,
and hence by linearity of expectation:
$$
\E
[
X
_
i
]
=
{
\vert\vert
f
\vert\vert
_
1

f
_
a
\over
k
}
=
{
\vert\vert
f
_{

a
}
\vert\vert
_
1
\over
k
}$$
And by applying Markov's inequality we obtain a bound on the error of a single
counter:
$$
\Pr
[
X
_
i >
\varepsilon
\cdot
m
]
\geq
\Pr
[
X
_
i >
\varepsilon
\cdot
\vert\vert
f
_{

a
}
\vert\vert
_
1
]
\leq
{
1
\over
k
\varepsilon
}
\leq
1
/
2
$$
Finally, since we have
$
t
$
independent counters, the probability that they
are all wrong is:
$$
\Pr\left
[
\bigcap
_
i X
_
i >
\varepsilon
\cdot
m
\right
]
\leq
1
/
2
^
t
\leq
\delta
$$
\qed
\section
{
Counting Distinct Elements
}
We continue working with a stream
$
\alpha
[
1
\ldots
m
]
$
of integers from
$
[
n
]
$
,
and define
$
f
_
a
$
(the frequency of
$
a
$
) as before. Let
$
d
=
\vert
\{
j : f
_
j >
0
\}
\vert
$
. Then the distinct elements problem is
to estimate
$
d
$
.
\subsection
{
The AMS Algorithm
}
Define
${
\tt
tz
}
(
x
)
:
=
\max\{
i
\mid
2
^
i
$
~divides~
$
x
\}
$
(i.e. the number of trailing zeroes in the base2 representation of
$
x
$
).
\algo
{
DistinctElements
}
\algalias
{
AMS
}
\algin
the data stream
$
\alpha
$
, the accuracy
$
\varepsilon
$
,
the error parameter
$
\delta
$
.
\:\em
{
Init
}
: Choose a random hash function
$
h :
[
n
]
\to
[
n
]
$
from a 2independent
family.
\:
:
$
z
\=
0
$
.
\:\em
{
Process
}
(
$
x
$
):
\:
:If
${
\tt
tz
}
(
h
(
x
))
> z
$
:
$
z
\=
{
\tt
tz
}
(
h
(
x
))
$
.
\algout
$
\hat
{
d
}
\=
2
^{
z
+
1
/
2
}$
\endalgo
\lemma
{
The AMS algorithm is a
$
(
3
,
\delta
)
$
estimator for some constant
$
\delta
$
.
}
\proof
For
$
j
\in
[
n
]
$
,
$
r
\geq
0
$
, let
$
X
_{
r, j
}
:
=
[
{
\tt
tz
}
(
h
(
j
))
\geq
r
]
$
, the
indicator that is true if
$
h
(
j
)
$
has at least
$
r
$
trailing
$
0
$
s.
Now define
$$
Y
_
r
=
\sum
_{
j : f
_
j >
0
}
X
_{
r, j
}
$$
How is our estimate related to
$
Y
_
r
$
? If the algorithm outputs
$
\hat
{
d
}
\geq
2
^{
a
+
1
/
2
}$
, then we know that
$
Y
_
a >
0
$
. Similarly, if the
output is smaller than
$
2
^{
a
+
1
/
2
}$
, then we know that
$
Y
_
a
=
0
$
. We will now
bound the probabilities of these events.
For any
$
j
\in
[
n
]
$
,
$
h
(
j
)
$
is uniformly distributed over
$
[
n
]
$
(since
$
h
$
is
$
2
$
independent). Hence
$
\E
[
X
_{
r, j
}
]
=
1
/
2
^
r
$
. By linearity of
expectation,
$
\E
[
Y
_{
r
}
]
=
d
/
2
^
r
$
.
We will also use the variance of these variables  note that
$${
\rm
Var
}
[
X
_{
r, j
}
]
\leq
\E
[
X
_{
r, j
}^
2
]
=
\E
[
X
_{
r, j
}
]
=
1
/
2
^
r
$$
And because
$
h
$
is
$
2
$
independent, the variables
$
X
_{
r, j
}$
and
$
X
_{
r, j'
}$
are independent for
$
j
\neq
j'
$
, and hence:
$${{
\rm
Var
}}
[
Y
_{
r
}
]
=
\sum
_{
j : f
_
j >
0
}
{
\rm
Var
}
[
X
_{
r, j
}
]
\leq
d
/
2
^
r
$$
Now, let
$
a
$
be the smallest integer such that
$
2
^{
a
+
1
/
2
}
\geq
3
d
$
. Then we
have:
$$
\Pr
[
\hat
{
d
}
\geq
3
d
]
=
\Pr
[
Y
_
a >
0
]
=
\Pr
[
Y
_
a
\geq
1
]
$$
Using Markov's inequality we get:
$$
\Pr
[
\hat
{
d
}
\geq
3
d
]
\leq
\E
[
Y
_
a
]
=
{
d
\over
2
^
a
}
\leq
{
\sqrt
{
2
}
\over
3
}
$$
For the other side, let
$
b
$
be the smallest integer so that
$
2
^{
b
+
1
/
2
}
\leq
d
/
3
$
. Then we have:
$$
\Pr
[
\hat
{
d
}
\leq
d
/
3
]
=
\Pr
[
Y
_{
b
+
1
}
=
0
]
\leq
\Pr
[
\vert
Y
_{
b
+
1
}

\E
[
Y
_{
b
+
1
}
]
\vert
\geq
d
/
2
^{
b
+
1
}
]
$$
Using Chebyshev's inequality, we get:
$$
\Pr
[
\hat
{
d
}
< d
/
3
]
\leq
{{
\rm
Var
}
[
Y
_
b
]
\over
(
d
/
2
^{
b
+
1
}
)
^
2
}
\leq
{
2
^{
b
+
1
}
\over
d
}
\leq
{
\sqrt
{
2
}
\over
3
}$$
\qed
The previous algorithm is not particularly satisfying  by our analysis it
can make an error around
$
94
\%
$
of the time (taking the union of the two bad
events). However we can improve the success probability easily; we run
$
t
$
independent estimators simultaneously, and print the median of their outputs.
By a standard use of Chernoff Bounds one can show that the probability that
the median is more than
$
3
d
$
is at most
$
2
^{

\Theta
(
t
)
}$
(and similarly also
the probability that it is less than
$
d
/
3
$
).
Hence it is enough to run
$
\O
(
\log
(
1
/
\delta
))
$
copies of the AMS estimator
to get a
$
(
3
,
\delta
)
$
estimator for any
$
\delta
>
0
$
. Finally, we note that
the space used by a single estimator is
$
\O
(
\log
n
)
$
since we can store
$
h
$
in
$
\O
(
\log
n
)
$
bits, and
$
z
$
in
$
\O
(
\log
\log
n
)
$
bits, and hence a
$
(
3
,
\delta
)
$
estimator uses
$
\O
(
\log
(
1
/
\delta
)
\cdot
\log
n
)
$
bits.
\endchapter
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