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Efficient Sketches for Earth-Mover Distance, with Applications

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### Efficient Sketches for Earth-Mover Distance, with Applications

David Woodruff

IBM Almaden

Joint work with Alexandr Andoni, Khanh Do Ba, and Piotr Indyk

(Planar) Earth-Mover Distance

- For multisets A, B of points in [∆]2, |A|=|B|=N,

i.e., min cost of perfect matching between A and B

EMD(, ) = 6 + 3√2

Geometric Representation of EMD

- Map A, B to k-dimensional vectors F(A), F(B)
- Image space of F “simple,” e.g., k small
- Can estimate EMD(A,B) from F(A), F(B) via some efficient recovery algorithm E

2 Rk

F

E

≈ EMD(A,B)

Geometric Representation of EMD: Motivation

- Visual search and recognition:
- Approximate nearest neighbor under EMD
- Reduces to approximate NN under simpler distances
- Has been applied to fast image search and recognition in large collections of images [Indyk-Thaper’03, Grauman-Darrell’05, Lazebnik-Schmid-Ponce’06]
- Data streaming computation:
- Estimating the EMD between two point sets given as a stream
- Need mapping F to be linear: adding new point a to A translates to adding F(a) to F(A)
- Important open problem in streaming [“Kanpur List ’06”]

Prior and New Results

Geometric representation of EMD:

Main Theorem

For any ε2(0,1), there exists a distribution over linear mappings F: R∆2!R∆εs.t. for multisets A,Bµ [∆]2 of equal size, we can produce an O(1/ε)-approximation to EMD(A,B) from F(A), F(B) with probability 2/3.

Implications

- Streaming:
- Approximate nearest neighbor:

* N = number of points

* s = number of data points (multisets) to preprocess

α>1 free parameter

Proof Outline

- Old [Agarwal-Varadarajan’04, Indyk’07]:
- Extend EMD to EEMD which:
- Handles sets of unequal size |A| · |B| in a grid of side-length k
- EEMD(A,B) = min|S|=|A| andS µ B EMD(A,S) + k¢|B\S|
- Is induced by a norm ||¢||EEMD, i.e., EEMD(A,B) = ||Â(A) – Â(B)||EEMD, where Â(A)2 R∆2 is the characteristic vector of A
- Decomposition of EEMD into weighted sum of small EEMD’s
- O(1/ε) distortion
- New:
- Linear sketching of “sum-norms”

EMD over [∆]2

EEMD over [∆ε]2

EEMD over [∆ε]2

EEMD over [∆ε]2

+

+ … +

∆O(1) terms

Old Idea [Indyk ’07]

EEMD over [∆ε]2

EEMD over [∆ε]2

EEMD over [∆ε]2

+

+ … +

∆O(1) terms

EMD over [∆]2

EMD over [∆]2

EEMD over [∆1/2]2

EEMD over [∆1/2]2

+ … +

Old Idea [Indyk ’07]

Solve one additional

EEMD problem in [¢1/2]2

2

Should also scale edge

lengths by ¢1/2

Old Idea [Indyk ’07]

- Total cost is the sum of the two phases
- Algorithm outputs a matching, so its cost is at least the EMD cost
- Indyk shows that if we put a random shift of the [¢1/2]2 grid on top of the [¢]2 grid,algorithm’s cost is at most a constant factor times the true EMD cost
- Recursive application gives multiple [¢ε]2 grids on top of each other, and results in O(1/ε)-approximation

Main New Technical Theorem

||M||1, X =

+

+ … +

For normed space X = (Rt, ||¢||X) and M2Xn, denote ||M||1,X = ∑i ||Mi||X.

||M1||X

||M2||X

||Mn||X

Given C > 0 and λ > 0, if C/λ· ||M||1, X· C, there is a distribution over linear mappings

μ: Xn!X(λlog n)O(1)

such that we can produce an O(1)-approximation to ||M||1,X from μ(M) w.h.p.

Proof Outline: Sum of Norms

- First attempt:
- Sample (uniformly) a few Mi’s to compute ||Mi||X
- Problem: sum could be concentrated in 1 block
- Second attempt:
- Sample Mi w/probability proportional to ||Mi||X [Indyk’07]
- Problem: how to do online?
- Techniques from [JW09, MW10]?
- Need to sample/retrieve blocks, not just individual coordinates

…

M2 contains most of mass

…

M1

M2

M3

Mn

Proof Outline: Sum of Norms (cont.)

M = (M1,

M2,

…,

Mn)

M2

S11

- Our approach:
- Split into exponential levels:
- Assume ||M||1, X· C
- Sk = {i2[n] s.t. ||Mi||X2(Tk, 2Tk]}, Tk=C/2k
- Suffices to estimate |Sk| for each level k. How?
- For each level k, subsample from [n] at a rate

such that event Ek (“isolation” of level k)

holds with probability proportional to |Sk|

- Repeat experiment several times, count number of successes

M4, M7

S2

S3

M1, M3, M8, M9

…

Sℓ

M5, M10, Mn

M:

Subsample:

Ek?

Y

N

Proof Outline: Event Ek

- Ek$ “isolation” of level k:
- Exactly one i 2Sk gets subsampled
- Nothing from Sk’ for k’<k
- Verification of trial success/failure
- Hash subsampled elements
- Each cell maintains vector sum of

subsampled Mi’s that hash there

- Ek holds roughly (we “accept”) when:
- 1 cell has X-norm in (0.9Tk, 2.1Tk]
- All other cells have X-norm ≤ 0.9Tk
- Check fails only if:
- Elements from lighter levels contribute a lot to 1 cell
- Elements from heavier levels subsampled and collide
- Both unlikely if hash table big enough
- Under-estimates |Sk|. If |Sk| > 2k/polylog(n), gives O(1)-approximation
- Remark: triangle inequality of norm gives control over impact of collisions

Subsample:

M1

M4

M5

M6

M9

M11

Mn–1

∑

∑

∑

∑

Sketch and Recovery Algorithm

Sketch:

- For every k, the estimator under-estimates |Sk|
- If |Sk| > 2k/polylog n, the estimator is (|Sk|)

- For each level k, create t hash tables
- For each hash table:
- Subsample from [n], including each i2[n] w.p. pk = 2-k
- Each cell maintains sum of Mi’s that hash to it

Recovery algorithm:

- For each level k, count number ck of “accepting” hash tables
- Return ∑kTk · (ck/t) · (1/pk)

{

EMD Wrapup

- We achieve a linear embedding of EMD
- with constant distortion, namely O(1/ε),
- into a space of strongly sublinear dimension, namely ∆ε.
- Open problems:
- Getting (1+ε)-approximation / proving impossibility
- Reducing dimension to logO(1)∆ / proving lower bound

What We Did

- We showed that in a data stream, one can sketch ||M||1,X = ∑i ||Mi||X with space about the space complexity of computing (or sketching) ||¢||X
- This quantity is known as a cascaded norm, written as L1(X)
- Cascaded norms have many applications [CM, JW]
- Can we generalize this? E.g., what about L2(X), i.e., (∑i ||Mi||2X )1/2

Cascaded Norms [JW09]

- No!
- L2(L1), i.e., (∑i ||Mi||21)1/2, requires (n1/2) space, where n is the number of different i, but sketching complexity of L1 is O(log n)
- More generally, for p ¸ 1, Lp(L1), i.e., (∑i ||Mi||p 1)1/p is £(n1-1/p) space
- So, L1(X) is very special

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