# sketching - Technical University of ... â€¢ Sketching â€¢ CountMin sketch Today Sketching...

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Streaming: Sketching Inge Li Gørtz

• Sketching

• CountMin sketch

Today

Sketching

• Sketching. create compact sketch/summary of data.

• Example. Durand and Flajolet 2003.

• Condensed the whole Shakespeares’ work

• Estimated number of distinct words: 30897 (correct answer is 28239, ie. relative error of 9.4%).

• Composable.

• Data streams and with sketches and

• There exists an efficiently computable function such that

S1 S2 sk(S1) sk(S2)

f

sk(S1 ∪ S2) = f(sk(S1), sk(S2))

Sketching

CountMin Sketch

• Frequency estimation. Construct a sketch such that can estimate the frequency of any element

fi i ∈ [n] .

Frequency Estimation

• Fixed array of counters of width w and depth d. Counters all initialized to be zero.

• Pariwise independent hash function for each row .

• When item arrives increment counter of in all rows.

hi : [n] → [w]

x hi(x)

CountMin Sketch

d

w

h1

h4

h2 h3

CountMin Sketch

d

w

h1

h4

h2 h3

h1(x) h2(x)

h3(x)

h4(x)

• Fixed array of counters of width w and depth d. Counters all initialized to be zero.

• Pariwise independent hash function for each row .

• When item arrives increment counter of in all rows.

hi : [n] → [w]

x hi(x)

CountMin Sketch

d

w

h1

h4

h2 h3

h2(y)

h4(y)

h3(y)

h1(y)

• Fixed array of counters of width w and depth d. Counters all initialized to be zero.

• Pariwise independent hash function for each row .

• When item arrives increment counter of in all rows.

• Estimate frequency of y: return minimum of all entries y hash to.

hi : [n] → [w]

x hi(x)

CountMin Sketch

d

w

h1

h4

h2 h3

h2(y)

h4(y)

h3(y)

h1(y)

• Fixed array of counters of width w and depth d. Counters all initialized to be zero.

• Pariwise independent hash function for each row .

• When item arrives increment counter of in all rows.

• Estimate frequency of y: return minimum of all entries y hash to.

hi : [n] → [w]

x hi(x)

• The estimator has the following property:

•

• with probability at least

̂fi

̂fi ≥ fi ̂fi ≤ fi + 2m /w

1 − (1/2)d

CountMin Sketch

d

w

h1

h4

h2 h3

h2(y)

h4(y)

h3(y)

h1(y)

Algorithm 1: CountMin

Initialize d independent hash functions hj : [n] ! [w]. Set counter Cj(b) = 0 for all j 2 [d] and b 2 [w]. while Stream S not empty do

if Insert(x) then for j = 1 . . . n do

Cj(hj(x)) = +1 end

else if Frequency(i) then return f̂i = minj2[d] Cj(hj(i)).

end end

1

• Claim. with probability at least

• Consider a fixed element .

• Let . Then

• Expected value of

• Want to bound

̂fi ≤ fi + 2m /w 1 − (1/2)d

i Zj = C(hj(i))

b = hj(i)

Zj = ∑ s:hj(s)=b

fs

Zj

E[Zj] = E ∑ s:hj(s)=b

fs

P[Zj ≥ fi + 2m /w] = P[Zj − fi ≥ 2m /w]

= ≤ E[Zj − fi]

2m /w

CountMin Sketch: Analysis

≤ fi + m w

= fi + 1 w ∑

s:s≠i

fs

= E[Zj] − fi

2m /w ≤

( fi + m /w) − fi 2m /w

= 1 2

• Claim. with probability at least

• Consider a fixed element . We have .

• What is the probability that ?

•

• Thus

̂fi ≤ fi + 2m /w 1 − (1/2)d

i P[Zj − fi ≥ 2m /w] ≤ 1/2

̂fi ≥ fi + 2m /w

P[ ̂fi ≥ fi + 2m /w] = P[Z1 ≥ fi + 2m /w ∩ Z2 ≥ fi + 2m /w ∩ ⋯ ∩ Zd ≥ fi + 2m /w]

= P[Z1 ≥ fi + 2m /w] ⋅ P[Z2 ≥ fi + 2m /w] ⋅ ⋯ ⋅ P[Zd ≥ fi + 2m /w]

≤ 1 2

⋅ 1 2

⋅ ⋯ ⋅ 1 2

= ( 12 ) d

P[ ̂fi ≤ fi + 2m /w] ≥ 1 − ( 12 ) d

CountMin Sketch: Analysis

• Use and .

• The estimator has the following property:

•

• with probability at least

• Space. words.

• Query and processing time.

w = 2/ε d = lg(1/δ) ̂fi

̂fi ≥ fi ̂fi ≤ fi + εm 1 − δ

O(dw) = O(2 lg(1/δ)/ε) = O(lg(1/δ)/ε)

O(d) = O(lg(1/δ))

CountMin Sketch: Analysis

d

w

h1

h4

h2 h3

h2(y)

h4(y)

h3(y)

h1(y)

• We can use the CountMin Sketch to solve e.g.:

• Heavy hitters: List all heavy hitters (elements with frequency at least m/k).

• Range(a,b): Return (an estimate of) the number of elements in the stream with value between a and b.

• Exercise.

• How can we solve heavy hitters with a single CountMin sketch?

• What is the space and query time?

Applications of CountMin Sketch

• Dyadic intervals. Set of intervals:

{[ j n 2i

+ 1,…, ( j + 1) n 2i

] | 0 ≤ i ≤ lg n, 0 ≤ j ≤ 2i−1}

Dyadic Intervals

[1,2] [3,4] [5,6] [7,8]

1 2 3 4 5 6 7 8

[1,4] [5,8]

[1,8]

9 10 11 12 13 14 15 16

[9,10] [11,12] [13,14] [15,16]

[9,12] [13,16]

[9,16]

[1,16]

• Heavy Hitters. Store a CountMin Sketch for each level in the tree of dyadic intervals

• On a level: Treat all elements in same interval as the same element.

Heavy Hitters

[1,2] [3,4] [5,6] [7,8]

1 2 3 4 5 6 7 8

[1,4] [5,8]

[1,8]

9 10 11 12 13 14 15 16

[9,10] [11,12] [13,14] [15,16]

[9,12] [13,16]

[9,16]

[1,16]

• Heavy Hitters. Store a CountMin Sketch for each level in the tree of dyadic intervals

• On a level: Treat all elements in same interval as the same element.

Heavy Hitters

[1,2] [3,4] [5,6] [7,8]

1 2 3 4 5 6 7 8

[1,4] [5,8]

[1,8]

9 10 11 12 13 14 15 16

[9,10] [11,12] [13,14] [15,16]

[9,12] [13,16]

[9,16]

[1,16]

• Heavy Hitters. Store a CountMin Sketch for each level in the tree of dyadic intervals

• On a level: Treat all elements in same interval as the same element.

• To find heavy hitters:

• traverse tree from root.

• only visit children with frequency ≥ m/k.

Heavy Hitters

[1,2] [3,4] [5,6] [7,8]

1 2 3 4 5 6 7 8

[1,4] [5,8]

[1,8]

9 10 11 12 13 14 15 16

[9,10] [11,12] [13,14] [15,16]

[9,12] [13,16]

[9,16]

[1,16]

• Heavy Hitters. Store a CountMin Sketch for each level in the tree of dyadic intervals

• On a level: Treat all elements in same interval as the same element.

• To find heavy hitters:

• traverse tree from root.

• only visit children with frequency ≥ m/k.

Heavy Hitters

[1,2] [3,4] [5,6] [7,8]

1 2 3 4 5 6 7 8

[1,4] [5,8]

[1,8]

9 10 11 12 13 14 15 16

[9,10] [11,12] [13,14] [15,16]

[9,12] [13,16]

[9,16]

[1,16]

• Heavy Hitters. Store a CountMin Sketch for each level in the tree of dyadic intervals

• On a level: Treat all elements in same interval as the same element.

• To find heavy hitters:

• traverse tree from root.

• only visit children with frequency ≥ m/k.

Heavy Hitters

[1,2] [3,4] [5,6] [7,8]

1 2 3 4 5 6 7 8

[1,4] [5,8]

[1,8]

9 10 11 12 13 14 15 16

[9,10] [11,12] [13,14] [15,16]

[9,12] [13,16]

[9,16]

[1,16]

• Heavy Hitters. Store a CountMin Sketch for each level in the tree of dyadic intervals

• On a level: Treat all elements in same interval as the same element.

• To find heavy hitters:

• traverse tree from root.

• only visit children with frequency ≥ m/k.

Heavy Hitters

[1,2] [3,4] [5,6] [7,8]

1 2 3 4 5 6 7 8

[1,4] [5,8]

[1,8]

9 10 11 12 13 14 15 16

[9,10] [11,12] [13,14] [15,16]

[9,12] [13,16]

[9,16]

[1,16]

• Heavy Hitters. Store a CountMin Sketch for each level in the tree of dyadic intervals

• On a level: Treat all elements in same interval as the same element.

• To find heavy hitters:

• traverse tree from root.

• only visit children with frequency ≥ m/k.

Heavy Hitters

[1,2] [3,4] [5,6] [7,8]

1 2 3 4 5 6 7 8

[1,4] [5,8]

[1,8]

9 10 11 12 13 14 15 16

[9,10] [11,12] [13,14] [15,16]

[9,12] [13,16]

[9,16]

[1,16]

• Heavy Hitters. Store a CountMin Sketch for each level in the tree of dyadic intervals

• On a level: Treat all elements in same inte

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