Zero-One Frequency Laws

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Vladimir(Vova) Braverman UCLA Joint work with Rafail Ostrovsky Zero-One Frequency Laws

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Zero-One Frequency Laws . Vladimir( Vova ) Braverman UCLA Joint work with Rafail Ostrovsky. Plan:. General m ethod for computing over frequencies with polylog space (Zero-one f requency l aw) Recursive sketching for vectors. Frequencies. Stream. - PowerPoint PPT Presentation

Transcript of Zero-One Frequency Laws

Page 1: Zero-One Frequency Laws

Vladimir(Vova) Braverman

UCLA

Joint work with Rafail Ostrovsky

Zero-One Frequency Laws

Page 2: Zero-One Frequency Laws

• General method for computing over frequencies with polylog space (Zero-one frequency law)

• Recursive sketching for vectors

Plan:

Page 3: Zero-One Frequency Laws

Stream

Frequencies

Frequency Vector

0 0 0 0 0 0 0 011 123 1 2

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Frequency-Based Functions

Frequency Vector0 0 0 1 2 0 0 013

G: N —> R

0 0 G(0)G(1)G(2)G(0)G(0)G(0)G(1)G(3)

G-Sum(V) = ∑ G(mi)

Modified Vector

The objective function

The Data

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D is a a stream p1,…, pm where pj є [n] Frequency mi = |{j: pj = i}|Frequency-based function G-Sum(D) =∑i

G(mi) Fk frequency moment G(mi) = mi

k

A single pass over DSmall (polylog) memory :

(1/ε log(nm))O(1)

The (Basic) Streaming Model

Formal Definition

LimitationsOutput a multiplicative approximation X such that: P(|X- ∑i G(mi) | > ε ∑i G(mi) ) < 2/3

What is needed

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Alon, Matias, Szegedy (STOC 1996, JCSS 1999, Gödel Award 2005)

• Frequency moments G(x) = xk , in particular:

•Polylog-space algorithms for G(x) = x0 and G(x) = x2

•Lower bounds for k>2

•Algorithms for k>2 (large but sublinear memory)

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The open question ofAlon, Matias, Szegedy (1996)

What is the space complexity of estimating other functions G(x)?

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Our Result G(0)=0, G is non-decreasing

Function G : R—> R is in STREAM-POLYLOG classIf there exists an algorithm A such that for any data stream D and for any ε, A makes a single pass over D, uses (1/ε log(nm))O(1)

memory bits and outputs X s.t.P(|X - ∑i G(mi) | > ε ∑i G(mi)) < 2/3.

= min(x, min( |z| : |G(x+z) – G(x)| > εG(x)))G : N —> R is tractable

G is in STREAM-POLYLOG if and only if G is tractable

The Main Result

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Related Work (A subset)Alon, Gibbons, Matias, Szegedy PODS 99Alon, Matias, Szegedy STOC 96Andoni, Krauthgamer, Onak 2010 (arxiv)

Bar-Yossef, Jayram, Kumar, Sivakumar JCSS 2004Bar-Yossef, Jayram, Kumar, Sivakumar, Trevisan RANDOM 2002Beame, Jayram, Rudra STOC 2007Bhuvanagiri, Ganguly, Kesh, Saha SODA 2006Bhuvanagiri, Ganguly ESA 2006Chakrabarti, Do Ba, Muthukrishnan SODA 2007Chakrabarti, Cormode, McGregor STOC 08, SODA 07Chakrabarti, Khot, Sun 2003Chakrabarti, Regev STOC 2011Charikar, Chen, Farach-Colton Th.Comp.Sc. 2004Coppersmith, Kumar SODA 2004Cormode, Datar, Indyk, Muthukrishnan VLDB 2002Comrode, Muthukrishnan J.Alg. 2005Feigenbaum, Kannan, Strauss, Viswanathan FOCS 99Flajolet, Martin JCSS 85

Ganguly 2004, 2011Ganguly, Cormode RANDOM 2007Guha, Indyk, McGregor COLT 2007Guha, McGregor, Venkatasubramanian SODA 06Harvey, Nelson, Onak FOCS 08Indyk FOCS 2000Indyk, Woodruff FOCS 03, STOC 2005Jayram, McGregor, Muthukrishnan, Vee PODS 07Kane, Nelson, Woodruff PODS 2010, SODA 2010Kane, Nelson, Porat, Woodruff STOC 2011Li SODA 2009, KDD 07McGregor, Indyk SODA 2009Monemizadeh, Woodruff SODA 2010Muthukrishnan 2005 Nelson, Woodruff PODS 2011Saks, Sun STOC 2002Woodruff SODA 2004

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Lower Bounds

•Reduction to MultiParty SET-DISJOINTESS problem•The reduction requires monotonicity•Relatively straightforward (see the paper)

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y copies

Lower Bounds (informal)

100

1

010

…0

001

0

….

Assume first that x = k * y

Pick N~ G(x)/G(y)

i

i

i …. i

y copies

j j …. j

The Stream

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Reduction (very informal)If the sets intersect then, by monotonicity, the value of G-Sum is at least NG(y) + G(x) ~ 2G(x)

If do not intersect then the value is at most (N+k)G(y) ~ G(x)

Any constant approximation algorithm for G-Sum MUST recognize the difference

And thus requires N/(k^2) space ([Chakrabarti, Khot, Sun]) which is larger then any polylog

Thus G is not tractable

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• We follow the fundamental idea of Indyk and Woodruff• First we solve a specific case of G-

heavy elements• Then we show that the general case

can be solved by recursive sketching

Upper Bound: Basic Ideas

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Mimic F

G

Certifier H

1 0

IF H=1 RETURN F

ELSE RETURN 0

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G-heavy elementsG(1)

G(1)

G(1)

G(10^10)

G(1)

G(1)

ji

ij yGyG )(100)(

Freq

uenc

y Ve

ctor

of s

ize n

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G(x)=x^2G(x)=x^3/2

Frequencies

Certifier

G3G2G1

If G is “good” then every G-heavy element is

also F2-heavy

111001

1110001

11100001

Mimic F

G

Certifier H1 0

IF H=1 RETURN F

ELSE RETURN 0

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Lemma 0 (very informal)

)1(

)/]([

22

)1(

][

))/(log(

:such that )log(||,

: implies

)()(

then tractableisG IF

O

Snii

O

nii

nmyx

nS [n]S

yGxG

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Proof for L_p (0<p<2)

2/12

/1

i

i

p

i

pi

i

pi

p

yy

yx

x

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Proof (sketch)

wSii

w

w

www

w

w

ii

ySx

SG

xGx

SGyGxG

iw yiS

22225.02||

||)2(

)(2

2

||)2()()(

1ww }22 :{

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Mimic Function

n

1

1

1

1

1 )(||

5.0)1()1(

||

1

1

yGyhG

hPhP

yyh

ii

ii

ii

Mimic F

G

Certifier H1 0

IF H=1 RETURN F

ELSE RETURN 0

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Recursive Sketches

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Lemma 1

Si

iSi

ii vhvX 2

.|)||||(| 2 VVXP

Svvin

jji

}:{1

Let V є Rn be a vector with non-negative entries. Let H є {0,1}n be a random vector with pairwise-independent uniform entries. Let S be s.t.:

Define

Then

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Hadamard product Had(U,V) of two vectors U and V is a vector with entries viui

v1v2

u1u2

v1u1v2u2

vn un vnun

… Had(U,V)

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Lemma 2

),(,

1

0

iii HVHadVVV

i

n

j

ij

il Svvl

}:{1

.|)|||||( 21

tVVXP iii

t

i

ii Sj

ij

Sj

ij

iji vvhX 2

Denote for i=1,2,..,t

Then

tHHH ,...,, 21 are i.i.d. vectors

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Lemma 3

i

jSj

iijii

tt

vhYY

VY

)21(2

||

11

.1.0|)||||(| 0 VVYP

Denote

Then for )( 3

2

t

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The general algorithm (informal)Maintain H1,..,Ht

We can obtain Vi by dropping all stream elements that are not “sampled” For t=O(log(n)), the number of non-zero elements in V t is constant, with constant probabilityThus, given an oracle for “heavy” elements, the sum can be approximated using only log(n) number of calls to “heavy” elements oracle

i

jSj

iijii

tt

vhYY

VY

)21(2

||

11

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The Algorithm for large Frequency moments (informal)The general algorithm works for any “separable” vector, in particular for frequency moments vectorAlso, such oracles for “heavy” elements exist for frequency moments E.g., CountSketch by Charikar, Chen, Farach-Colton, 2004. The final algorithm requires n1-2/k log(n)log(m)log(log…(log(nm))) memory bits

Independently Andoni, Krauthgamer, Onak improved the bound to n1-2/k log(n)log(m) (Precision Sampling: Alex’s talk yesterday)

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NotesWe need to overcome additional technical issues

Heavy elements: from precise values to approximations

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Open problemsCharacterize non-monotonic functions (we made some progress)

Extend the results to sublinear algorithms (o(n) space)

Other models: deletions, sliding windows etc.,

Optimal algorithm for large frequency moments

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Thank you!