Download Crash Course on Data Stream Algorithms

Crash Course on Data Stream Algorithms
Part I: Basic Definitions and Numerical Streams
Andrew McGregor
University of Massachusetts Amherst
1/24
Goals of the Crash Course
I
Goal: Give a flavor for the theoretical results and techniques from
the 100’s of papers on the design and analysis of stream algorithms.
2/24
Goals of the Crash Course
I
Goal: Give a flavor for the theoretical results and techniques from
the 100’s of papers on the design and analysis of stream algorithms.
“When we abstract away the application-specific details, what are
the basic algorithmic ideas and challenges in stream processing?
What is and isn’t possible?”
2/24
Goals of the Crash Course
I
Goal: Give a flavor for the theoretical results and techniques from
the 100’s of papers on the design and analysis of stream algorithms.
“When we abstract away the application-specific details, what are
the basic algorithmic ideas and challenges in stream processing?
What is and isn’t possible?”
I
Disclaimer: Talks will be theoretical/mathematical but shouldn’t
require much in the way of prerequisites.
2/24
Goals of the Crash Course
I
Goal: Give a flavor for the theoretical results and techniques from
the 100’s of papers on the design and analysis of stream algorithms.
“When we abstract away the application-specific details, what are
the basic algorithmic ideas and challenges in stream processing?
What is and isn’t possible?”
I
I
Disclaimer: Talks will be theoretical/mathematical but shouldn’t
require much in the way of prerequisites.
Request:
2/24
Goals of the Crash Course
I
Goal: Give a flavor for the theoretical results and techniques from
the 100’s of papers on the design and analysis of stream algorithms.
“When we abstract away the application-specific details, what are
the basic algorithmic ideas and challenges in stream processing?
What is and isn’t possible?”
I
I
Disclaimer: Talks will be theoretical/mathematical but shouldn’t
require much in the way of prerequisites.
Request:
I
If you get bored, ask questions. . .
2/24
Goals of the Crash Course
I
Goal: Give a flavor for the theoretical results and techniques from
the 100’s of papers on the design and analysis of stream algorithms.
“When we abstract away the application-specific details, what are
the basic algorithmic ideas and challenges in stream processing?
What is and isn’t possible?”
I
I
Disclaimer: Talks will be theoretical/mathematical but shouldn’t
require much in the way of prerequisites.
Request:
I
I
If you get bored, ask questions. . .
If you get lost, ask questions. . .
2/24
Goals of the Crash Course
I
Goal: Give a flavor for the theoretical results and techniques from
the 100’s of papers on the design and analysis of stream algorithms.
“When we abstract away the application-specific details, what are
the basic algorithmic ideas and challenges in stream processing?
What is and isn’t possible?”
I
I
Disclaimer: Talks will be theoretical/mathematical but shouldn’t
require much in the way of prerequisites.
Request:
I
I
I
If you get bored, ask questions. . .
If you get lost, ask questions. . .
If you’d like to ask questions, ask questions. . .
2/24
Outline
Basic Definitions
Sampling
Sketching
Counting Distinct Items
Summary of Some Other Results
3/24
Outline
Basic Definitions
Sampling
Sketching
Counting Distinct Items
Summary of Some Other Results
4/24
Data Stream Model
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Stream: m elements from universe of size n, e.g.,
hx1 , x2 , . . . , xm i =
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3, 5, 3, 7, 5, 4, . . .
Data Stream Model
I
Stream: m elements from universe of size n, e.g.,
hx1 , x2 , . . . , xm i =
I
3, 5, 3, 7, 5, 4, . . .
Goal: Compute a function of stream, e.g., median, number of
distinct elements, longest increasing sequence.
5/24
Data Stream Model
I
Stream: m elements from universe of size n, e.g.,
hx1 , x2 , . . . , xm i =
I
I
3, 5, 3, 7, 5, 4, . . .
Goal: Compute a function of stream, e.g., median, number of
distinct elements, longest increasing sequence.
Catch:
1. Limited working memory, sublinear in n and m
5/24
Data Stream Model
I
Stream: m elements from universe of size n, e.g.,
hx1 , x2 , . . . , xm i =
I
I
3, 5, 3, 7, 5, 4, . . .
Goal: Compute a function of stream, e.g., median, number of
distinct elements, longest increasing sequence.
Catch:
1. Limited working memory, sublinear in n and m
2. Access data sequentially
5/24
Data Stream Model
I
Stream: m elements from universe of size n, e.g.,
hx1 , x2 , . . . , xm i =
I
I
3, 5, 3, 7, 5, 4, . . .
Goal: Compute a function of stream, e.g., median, number of
distinct elements, longest increasing sequence.
Catch:
1. Limited working memory, sublinear in n and m
2. Access data sequentially
3. Process each element quickly
5/24
Data Stream Model
I
Stream: m elements from universe of size n, e.g.,
hx1 , x2 , . . . , xm i =
I
I
3, 5, 3, 7, 5, 4, . . .
Goal: Compute a function of stream, e.g., median, number of
distinct elements, longest increasing sequence.
Catch:
1. Limited working memory, sublinear in n and m
2. Access data sequentially
3. Process each element quickly
I
Origins in 70s but has become popular in last ten years because of
growing theory and very applicable.
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Why’s it become popular?
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Practical Appeal:
I
I
Faster networks, cheaper data storage, ubiquitous data-logging
results in massive amount of data to be processed.
Applications to network monitoring, query planning, I/O efficiency
for massive data, sensor networks aggregation. . .
6/24
Why’s it become popular?
I
Practical Appeal:
I
I
I
Faster networks, cheaper data storage, ubiquitous data-logging
results in massive amount of data to be processed.
Applications to network monitoring, query planning, I/O efficiency
for massive data, sensor networks aggregation. . .
Theoretical Appeal:
I
I
Easy to state problems but hard to solve.
Links to communication complexity, compressed sensing,
embeddings, pseudo-random generators, approximation. . .
6/24
Outline
Basic Definitions
Sampling
Sketching
Counting Distinct Items
Summary of Some Other Results
7/24
Sampling and Statistics
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Sampling is a general technique for tackling massive amounts of data
8/24
Sampling and Statistics
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Sampling is a general technique for tackling massive amounts of data
I
Example: To compute the median packet size of some IP packets,
we could just sample some and use the median of the sample as an
estimate for the true median. Statistical arguments relate the size of
the sample to the accuracy of the estimate.
8/24
Sampling and Statistics
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Sampling is a general technique for tackling massive amounts of data
I
Example: To compute the median packet size of some IP packets,
we could just sample some and use the median of the sample as an
estimate for the true median. Statistical arguments relate the size of
the sample to the accuracy of the estimate.
I
Challenge: But how do you take a sample from a stream of unknown
length or from a “sliding window”?
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Reservoir Sampling
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Problem: Find uniform sample s from a stream of unknown length
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Reservoir Sampling
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Problem: Find uniform sample s from a stream of unknown length
Algorithm:
I
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Initially s = x1
On seeing the t-th element, s ← xt with probability 1/t
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Reservoir Sampling
I
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Problem: Find uniform sample s from a stream of unknown length
Algorithm:
I
I
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Initially s = x1
On seeing the t-th element, s ← xt with probability 1/t
Analysis:
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What’s the probability that s = xi at some time t ≥ i?
9/24
Reservoir Sampling
I
I
Problem: Find uniform sample s from a stream of unknown length
Algorithm:
I
I
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Initially s = x1
On seeing the t-th element, s ← xt with probability 1/t
Analysis:
I
What’s the probability that s = xi at some time t ≥ i?
„
«
„
«
1
1
1
1
P [s = xi ] = × 1 −
× ... × 1 −
=
i
i +1
t
t
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Reservoir Sampling
I
I
Problem: Find uniform sample s from a stream of unknown length
Algorithm:
I
I
I
Initially s = x1
On seeing the t-th element, s ← xt with probability 1/t
Analysis:
I
What’s the probability that s = xi at some time t ≥ i?
„
«
„
«
1
1
1
1
P [s = xi ] = × 1 −
× ... × 1 −
=
i
i +1
t
t
I
To get k samples we use O(k log n) bits of space.
9/24
Priority Sampling for Sliding Windows
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Problem: Maintain a uniform sample from the last w items
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Priority Sampling for Sliding Windows
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Problem: Maintain a uniform sample from the last w items
Algorithm:
1. For each xi we pick a random value vi ∈ (0, 1)
10/24
Priority Sampling for Sliding Windows
I
I
Problem: Maintain a uniform sample from the last w items
Algorithm:
1. For each xi we pick a random value vi ∈ (0, 1)
2. In a window hxj−w +1 , . . . , xj i return value xi with smallest vi
10/24
Priority Sampling for Sliding Windows
I
I
Problem: Maintain a uniform sample from the last w items
Algorithm:
1. For each xi we pick a random value vi ∈ (0, 1)
2. In a window hxj−w +1 , . . . , xj i return value xi with smallest vi
3. To do this, maintain set of all elements in sliding window whose v
value is minimal among subsequent values
10/24
Priority Sampling for Sliding Windows
I
I
Problem: Maintain a uniform sample from the last w items
Algorithm:
1. For each xi we pick a random value vi ∈ (0, 1)
2. In a window hxj−w +1 , . . . , xj i return value xi with smallest vi
3. To do this, maintain set of all elements in sliding window whose v
value is minimal among subsequent values
I
Analysis:
10/24
Priority Sampling for Sliding Windows
I
I
Problem: Maintain a uniform sample from the last w items
Algorithm:
1. For each xi we pick a random value vi ∈ (0, 1)
2. In a window hxj−w +1 , . . . , xj i return value xi with smallest vi
3. To do this, maintain set of all elements in sliding window whose v
value is minimal among subsequent values
I
Analysis:
I
The probability that j-th oldest element is in S is 1/j so the
expected number of items in S is
1/w + 1/(w − 1) + . . . + 1/1 = O(log w )
10/24
Priority Sampling for Sliding Windows
I
I
Problem: Maintain a uniform sample from the last w items
Algorithm:
1. For each xi we pick a random value vi ∈ (0, 1)
2. In a window hxj−w +1 , . . . , xj i return value xi with smallest vi
3. To do this, maintain set of all elements in sliding window whose v
value is minimal among subsequent values
I
Analysis:
I
The probability that j-th oldest element is in S is 1/j so the
expected number of items in S is
1/w + 1/(w − 1) + . . . + 1/1 = O(log w )
I
Hence, algorithm only uses O(log w log n) bits of memory.
10/24
Other Types of Sampling
I
Universe sampling: For a random i ∈R [n], compute
fi = |{j : xj = i}|
11/24
Other Types of Sampling
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Universe sampling: For a random i ∈R [n], compute
fi = |{j : xj = i}|
I
Minwise hashing: Sample i ∈R {i : there exists j such that xj = i}
11/24
Other Types of Sampling
I
Universe sampling: For a random i ∈R [n], compute
fi = |{j : xj = i}|
I
Minwise hashing: Sample i ∈R {i : there exists j such that xj = i}
I
AMS sampling: Sample xj for j ∈R [m] and compute
r = |{j 0 ≥ j : xj 0 = xj }|
11/24
Other Types of Sampling
I
Universe sampling: For a random i ∈R [n], compute
fi = |{j : xj = i}|
I
Minwise hashing: Sample i ∈R {i : there exists j such that xj = i}
I
AMS sampling: Sample xj for j ∈R [m] and compute
r = |{j 0 ≥ j : xj 0 = xj }|
P
Handy when estimating quantities like i g (fi ) because
X
E [m(g (r ) − g (r − 1))] =
g (fi )
i
11/24
Outline
Basic Definitions
Sampling
Sketching
Counting Distinct Items
Summary of Some Other Results
12/24
Sketching
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Sketching is another general technique for processing streams
13/24
Sketching
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Sketching is another general technique for processing streams
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Basic idea: Apply a linear projection “on the fly” that takes
high-dimensional data to a smaller dimensional space. Post-process
lower dimensional image to estimate the quantities of interest.
13/24
Estimating the difference between two streams
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Input: Stream from two sources hx1 , x2 , . . . , xm i ∈ ([n] ∪ [n])m
14/24
Estimating the difference between two streams
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Input: Stream from two sources hx1 , x2 , . . . , xm i ∈ ([n] ∪ [n])m
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Goal: Estimate difference between distribution of red values and blue
values, e.g.,
X
|fi − gi |
i∈[n]
where fi = |{k : xk = i}| and gi = |{k : xk = i}|
14/24
p-Stable Distributions and Algorithm
I
Defn: A p-stable distribution µ has the following property:
for X , Y , Z ∼ µ and a, b ∈ R :
aX + bY ∼ (|a|p + |b|p )1/p Z
e.g., Gaussian is 2-stable and Cauchy distribution is 1-stable
15/24
p-Stable Distributions and Algorithm
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Defn: A p-stable distribution µ has the following property:
for X , Y , Z ∼ µ and a, b ∈ R :
I
aX + bY ∼ (|a|p + |b|p )1/p Z
e.g., Gaussian is 2-stable and Cauchy distribution is 1-stable
Algorithm:
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Generate random matrix A ∈ Rk×n where Aij ∼ Cauchy, k = O(−2 ).
15/24
p-Stable Distributions and Algorithm
I
Defn: A p-stable distribution µ has the following property:
for X , Y , Z ∼ µ and a, b ∈ R :
I
aX + bY ∼ (|a|p + |b|p )1/p Z
e.g., Gaussian is 2-stable and Cauchy distribution is 1-stable
Algorithm:
I
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Generate random matrix A ∈ Rk×n where Aij ∼ Cauchy, k = O(−2 ).
Compute sketches Af and Ag incrementally
15/24
p-Stable Distributions and Algorithm
I
Defn: A p-stable distribution µ has the following property:
for X , Y , Z ∼ µ and a, b ∈ R :
I
aX + bY ∼ (|a|p + |b|p )1/p Z
e.g., Gaussian is 2-stable and Cauchy distribution is 1-stable
Algorithm:
I
I
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Generate random matrix A ∈ Rk×n where Aij ∼ Cauchy, k = O(−2 ).
Compute sketches Af and Ag incrementally
Return median(|t1 |, . . . , |tk |) where t = Af − Ag
15/24
p-Stable Distributions and Algorithm
I
Defn: A p-stable distribution µ has the following property:
for X , Y , Z ∼ µ and a, b ∈ R :
I
e.g., Gaussian is 2-stable and Cauchy distribution is 1-stable
Algorithm:
I
I
I
I
aX + bY ∼ (|a|p + |b|p )1/p Z
Generate random matrix A ∈ Rk×n where Aij ∼ Cauchy, k = O(−2 ).
Compute sketches Af and Ag incrementally
Return median(|t1 |, . . . , |tk |) where t = Af − Ag
Analysis:
I
By the 1-stability property for Zi ∼ Cauchy
X
X
|ti | = |
Ai,j (fj − gj )| ∼ |Zi |
|fj − gj |
j
j
15/24
p-Stable Distributions and Algorithm
I
Defn: A p-stable distribution µ has the following property:
for X , Y , Z ∼ µ and a, b ∈ R :
I
e.g., Gaussian is 2-stable and Cauchy distribution is 1-stable
Algorithm:
I
I
I
I
aX + bY ∼ (|a|p + |b|p )1/p Z
Generate random matrix A ∈ Rk×n where Aij ∼ Cauchy, k = O(−2 ).
Compute sketches Af and Ag incrementally
Return median(|t1 |, . . . , |tk |) where t = Af − Ag
Analysis:
I
By the 1-stability property for Zi ∼ Cauchy
X
X
|ti | = |
Ai,j (fj − gj )| ∼ |Zi |
|fj − gj |
j
I
j
For k = O(−2 ), since median(|Zi |) = 1, with high probability,
X
X
(1 − )
|fj − gj | ≤ median(|t1 |, . . . , |tk |) ≤ (1 + )
|fj − gj |
j
j
15/24
A Useful Multi-Purpose Sketch: Count-Min Sketch
16/24
A Useful Multi-Purpose Sketch: Count-Min Sketch
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Heavy Hitters: Find all i such that fi ≥ φm
16/24
A Useful Multi-Purpose Sketch: Count-Min Sketch
I
I
Heavy Hitters: Find all i such that fi ≥ φm
P
Range Sums: Estimate i≤k≤j fk when i, j aren’t known in advance
16/24
A Useful Multi-Purpose Sketch: Count-Min Sketch
I
I
I
Heavy Hitters: Find all i such that fi ≥ φm
P
Range Sums: Estimate i≤k≤j fk when i, j aren’t known in advance
Find k-Quantiles: Find values q0 , . . . , qk such that
q0 = 0,
qk = n,
and
X
i≤qj −1
16/24
fi <
jm X
≤
fi
k
i≤qj
A Useful Multi-Purpose Sketch: Count-Min Sketch
I
I
I
Heavy Hitters: Find all i such that fi ≥ φm
P
Range Sums: Estimate i≤k≤j fk when i, j aren’t known in advance
Find k-Quantiles: Find values q0 , . . . , qk such that
q0 = 0,
qk = n,
and
X
i≤qj −1
I
fi <
jm X
≤
fi
k
i≤qj
Algorithm: Count-Min Sketch
I
Maintain an array of counters ci,j for i ∈ [d] and j ∈ [w ]
16/24
A Useful Multi-Purpose Sketch: Count-Min Sketch
I
I
I
Heavy Hitters: Find all i such that fi ≥ φm
P
Range Sums: Estimate i≤k≤j fk when i, j aren’t known in advance
Find k-Quantiles: Find values q0 , . . . , qk such that
q0 = 0,
qk = n,
and
X
i≤qj −1
I
fi <
jm X
≤
fi
k
i≤qj
Algorithm: Count-Min Sketch
I
I
Maintain an array of counters ci,j for i ∈ [d] and j ∈ [w ]
Construct d random hash functions h1 , h2 , . . . hd : [n] → [w ]
16/24
A Useful Multi-Purpose Sketch: Count-Min Sketch
I
I
I
Heavy Hitters: Find all i such that fi ≥ φm
P
Range Sums: Estimate i≤k≤j fk when i, j aren’t known in advance
Find k-Quantiles: Find values q0 , . . . , qk such that
q0 = 0,
qk = n,
and
X
i≤qj −1
I
fi <
jm X
≤
fi
k
i≤qj
Algorithm: Count-Min Sketch
I
I
I
Maintain an array of counters ci,j for i ∈ [d] and j ∈ [w ]
Construct d random hash functions h1 , h2 , . . . hd : [n] → [w ]
Update counters: On seeing value v , increment ci,hi (v ) for i ∈ [d]
16/24
A Useful Multi-Purpose Sketch: Count-Min Sketch
I
I
I
Heavy Hitters: Find all i such that fi ≥ φm
P
Range Sums: Estimate i≤k≤j fk when i, j aren’t known in advance
Find k-Quantiles: Find values q0 , . . . , qk such that
q0 = 0,
qk = n,
X
and
fi <
i≤qj −1
I
jm X
≤
fi
k
i≤qj
Algorithm: Count-Min Sketch
I
I
I
I
Maintain an array of counters ci,j for i ∈ [d] and j ∈ [w ]
Construct d random hash functions h1 , h2 , . . . hd : [n] → [w ]
Update counters: On seeing value v , increment ci,hi (v ) for i ∈ [d]
To get an estimate of fk , return
f˜k = min ci,hi (k)
i
16/24
A Useful Multi-Purpose Sketch: Count-Min Sketch
I
I
I
Heavy Hitters: Find all i such that fi ≥ φm
P
Range Sums: Estimate i≤k≤j fk when i, j aren’t known in advance
Find k-Quantiles: Find values q0 , . . . , qk such that
q0 = 0,
qk = n,
X
and
fi <
i≤qj −1
I
jm X
≤
fi
k
Algorithm: Count-Min Sketch
I
I
I
I
Maintain an array of counters ci,j for i ∈ [d] and j ∈ [w ]
Construct d random hash functions h1 , h2 , . . . hd : [n] → [w ]
Update counters: On seeing value v , increment ci,hi (v ) for i ∈ [d]
To get an estimate of fk , return
f˜k = min ci,hi (k)
i
I
i≤qj
Analysis: For d = O(log 1/δ) and w = O(1/2 )
h
i
P fk − m ≤ f˜k ≤ fk ≥ 1 − δ
16/24
Outline
Basic Definitions
Sampling
Sketching
Counting Distinct Items
Summary of Some Other Results
17/24
Counting Distinct Elements
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Input: Stream hx1 , x2 , . . . , xm i ∈ [n]m
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Goal: Estimate the number of distinct values in the stream up to a
multiplicative factor (1 + ) with high probability.
18/24
Algorithm
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Algorithm:
1. Apply random hash function h : [n] → [0, 1] to each element
19/24
Algorithm
I
Algorithm:
1. Apply random hash function h : [n] → [0, 1] to each element
2. Compute φ, the t-th smallest value of the hash seen where t = 21/2
19/24
Algorithm
I
Algorithm:
1. Apply random hash function h : [n] → [0, 1] to each element
2. Compute φ, the t-th smallest value of the hash seen where t = 21/2
3. Return r̃ = t/φ as estimate for r , the number of distinct items.
19/24
Algorithm
I
Algorithm:
1. Apply random hash function h : [n] → [0, 1] to each element
2. Compute φ, the t-th smallest value of the hash seen where t = 21/2
3. Return r̃ = t/φ as estimate for r , the number of distinct items.
I
Analysis:
1. Algorithm uses O(−2 log n) bits of space.
19/24
Algorithm
I
Algorithm:
1. Apply random hash function h : [n] → [0, 1] to each element
2. Compute φ, the t-th smallest value of the hash seen where t = 21/2
3. Return r̃ = t/φ as estimate for r , the number of distinct items.
I
Analysis:
1. Algorithm uses O(−2 log n) bits of space.
2. We’ll show estimate has good accuracy with reasonable probability
P [|r̃ − r | ≤ r ] ≤ 9/10
19/24
Accuracy Analysis
1. Suppose the distinct items are a1 , . . . , ar
20/24
Accuracy Analysis
1. Suppose the distinct items are a1 , . . . , ar
2. Over Estimation:
P [r̃ ≥ (1 + )r ] = P [t/φ ≥ (1 + )r ] = P φ ≤
20/24
t
r (1 + )
Accuracy Analysis
1. Suppose the distinct items are a1 , . . . , ar
2. Over Estimation:
P [r̃ ≥ (1 + )r ] = P [t/φ ≥ (1 + )r ] = P φ ≤
3. Let Xi = 1[h(ai ) ≤
P φ≤
t
r (1+) ]
and X =
P
t
r (1 + )
Xi
t
= P [X > t] = P [X > (1 + )E [X ]]
r (1 + )
20/24
Accuracy Analysis
1. Suppose the distinct items are a1 , . . . , ar
2. Over Estimation:
P [r̃ ≥ (1 + )r ] = P [t/φ ≥ (1 + )r ] = P φ ≤
3. Let Xi = 1[h(ai ) ≤
P φ≤
t
r (1+) ]
and X =
P
t
r (1 + )
Xi
t
= P [X > t] = P [X > (1 + )E [X ]]
r (1 + )
4. By a Chebyshev analysis,
P [X > (1 + )E [X ]] ≤
20/24
1
≤ 1/20
2 E [X ]
Accuracy Analysis
1. Suppose the distinct items are a1 , . . . , ar
2. Over Estimation:
P [r̃ ≥ (1 + )r ] = P [t/φ ≥ (1 + )r ] = P φ ≤
3. Let Xi = 1[h(ai ) ≤
P φ≤
t
r (1+) ]
and X =
P
t
r (1 + )
Xi
t
= P [X > t] = P [X > (1 + )E [X ]]
r (1 + )
4. By a Chebyshev analysis,
P [X > (1 + )E [X ]] ≤
1
≤ 1/20
2 E [X ]
5. Under Estimation: A similar analysis shows P [r̃ ≤ (1 − )r ] ≤ 1/20
20/24
Outline
Basic Definitions
Sampling
Sketching
Counting Distinct Items
Summary of Some Other Results
21/24
Some Other Results
Correlations:
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Input: h(x1 , y1 ), (x2 , y2 ), . . . , (xm , ym )i
I
Goal: Estimate strength of correlation between x and y via the
distance between joint distribution and product of the marginals.
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Result: (1 + ) approx in Õ(−O(1) ) space.
22/24
Some Other Results
Correlations:
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Input: h(x1 , y1 ), (x2 , y2 ), . . . , (xm , ym )i
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Goal: Estimate strength of correlation between x and y via the
distance between joint distribution and product of the marginals.
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Result: (1 + ) approx in Õ(−O(1) ) space.
Linear Regression:
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Input: Stream defines a matrix A ∈ Rn×d and b ∈ Rd×1
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Goal: Find x such that kAx − bk2 is minimized.
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Result: (1 + ) estimation in Õ(d 2 −1 ) space.
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Some More Other Results
Histograms:
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Input: hx1 , x2 , . . . , xm i ∈ [n]m
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Goal: Determine B bucket histogram H : [m] → R minimizing
X
(xi − H(i))2
i∈[m]
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Result: (1 + ) estimation in Õ(B 2 −1 ) space
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Some More Other Results
Histograms:
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Input: hx1 , x2 , . . . , xm i ∈ [n]m
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Goal: Determine B bucket histogram H : [m] → R minimizing
X
(xi − H(i))2
i∈[m]
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Result: (1 + ) estimation in Õ(B 2 −1 ) space
Transpositions and Increasing Subsequences:
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Input: hx1 , x2 , . . . , xm i ∈ [n]m
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Goal: Estimate number of transpositions |{i < j : xi > xj }|
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Goal: Estimate length of longest increasing subsequence
√
Results: (1 + ) approx in Õ(−1 ) and Õ(−1 n) space respectively
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Thanks!
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Blog: http://polylogblog.wordpress.com
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Lectures: Piotr Indyk, MIT
http://stellar.mit.edu/S/course/6/fa07/6.895/
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Books:
“Data Streams: Algorithms and Applications”
S. Muthukrishnan (2005)
“Algorithms and Complexity of Stream Processing”
A. McGregor, S. Muthukrishnan (forthcoming)
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