The Average Case Complexity of Counting Distinct Elements

1
The Average Case Complexity of Counting Distinct
Elements

• David Woodruff
• IBM Almaden

2
Problem Description

• Given a data stream of n insertions of records,
  count the number F0 of distinct records
• One pass over the data stream
• Algorithms must use small amount of memory and
  have fast update time
• too expensive to store set of distinct records
• implies algorithms must be randomized and must
  settle for an approximate solution output
• F 2 (1-²)F0, (1²)F0 with constant
  probability

3
The Data Stream

• How is the data in the stream organized?
• Usually, assume worst-case ordered
• In this case, (1/²2) bits are necessary and
  sufficient to estimate F0
• As the quality of approximation improves to say,
  ² 1, the quadratic dependence is a major
  drawback
• Sometimes a random-ordering can be assumed
• Suppose we are mining salaries, and they are
  ordered alphabetically by surname. If there is no
  correlation between salaries and surname, then
  the stream of salaries is ordered randomly
• The backing sample architecture assumes data
  randomly ordered by design (Gibbons, Matias,
  Poosala)
• This model is referred to as the Random-Order
  Model (Guha, McGregor)
• Unfortunately, even in this model, we still need
  ?(1/²2) bits to estimate F0
• (Chakrabarti, Cormode, McGregor).
  Intuitively this is because the data is still
  worst-case.

4
Random Data Model

• In an attempt to bypass the ?(1/²2) bound, we
  propose to study the case when the data comes
  from an underlying distribution.
• Problem 3 of Muthukrishnans book Provide
  inproved estimates for Lp sums including distinct
  element estimation if input stream has
  statistical properties such as being Zipfian.
• There is a distribution defined by probabilities
  pi, 1 i m, ?i pi 1.
• The next item in the stream is chosen
  independently of previous
• items, and is i with probability pi.
• We call this the Random-Data Model.
• The Random-Data Model is contained in the
  Random-Order Model.

5
Random Data Model

• This model for F0 was implicitly studied before
• by Motwani and Vassilvitskii when the
  distribution is Zipfian. This distribution is
  useful for estimating WebGraph statistics
• sampling-based algorithms used in practice impose
  distributional assumptions, without which they
  have poor performance (Charikar et al)
• Generalized Inverse Gaussian Poisson (GIGP) model
  studies sampling-based estimators when
  distribution is uniform and Zipfian
• by Guha and McGregor for estimating the density
  function of an unknown distribution, which is
  useful in learning theory

6
Further Restriction

• We focus on the case when each probability pi
  1/d for an unknown value of d (so uniform over a
  subset of m)
• Captures the setting of sampling with replacement
  a set of unknown cardinality
• For a certain range of d, we show that one can
  beat the space lower bound that holds for
  adversarial data and randomly-ordered data
• For another choice of d, we show the lower bound
  for adversarial and randomly-ordered data also
  applies in this setting
• Distribution fairly robust in the sense that
  other distributions with a few heavy items and
  remaining items that are approximately uniform
  have the same properties above

7
Our Upper Bound

• 1-pass algorithm with an expected O(d(log
  1/²)/(n²2) log m) bits of space, whenever d
  1/²2 and d n. The per-item processing time is
  constant.
• Recall the distribution is uniform over a
  d-element subset of m, and we see n samples
  from it, so this is a typical setting of
  parameters.
• Notice for n even slightly larger than d, the
  algorithm does much better than the ?(1/²2) lower
  bound in other data stream models.
• One can show for every combination of known
  algorithms with different space/time tradeoffs
  for F0 in the adversarial model, our algorithm is
  either better in space or in time.

8
Our Lower Bound

• Our main technical result is that if n and d are
  (1/²2), then even estimating F0 in the random
  data model requires ?(1/²2) space
• Lower bound subsumes previous lower bounds,
  showing that even for a natural (random) choice
  of data, the problem is hard
• Our choice of distribution for showing the lower
  bound was used in subsequent work by Chakrabarti
  and Brody, which turned out to be useful for
  establishing an ?(1/²2) lower bound for constant
  pass algorithms for estimating F0

9
Techniques Upper Bound

• Very simple observation
• Since d n, each item should have frequency n/d
  in the stream.
• If n/d is at least ?(1/²2), we can just compute n
  and the frequency of the first item in the stream
  to get a (1²)-approximation to d
• Using a balls-and-bins occupancy bound of Kamath
  et al, a good estimation of d implies a good
  estimation to F0
• If n/d is less than 1/²2, we could instead store
  the first O(1/²2) items (hashed appropriately for
  small space), treat them as a set S, and count
  the number of items in the remaining part of the
  stream that land in S
• Correct, but unnecessary if d is much less than n.

10
Techniques Upper Bound

• Instead record the first item x in the stream,
  and find the second position i of x in the
  stream.
• Position i should occur at roughly the d-th
  position in the stream, so i provides a constant
  factor approximation to d
• Since n ?(d), position i should be in the first
  half of the stream with large constant
  probability.
• Now store the first i/(n²2) distinct stream
  elements in the second half of stream, treat them
  as a set S, and count the remaining items in the
  stream that occur in S.
• Good enough for (1²)-approximation
• Space is O(log n (i log m)/(n²2)) ¼ O(log n
  (d log m)/(n²2))

11
Techniques Upper Bound

• Space is O(log n (d log m)/(n²2)), but we can
  do better.
• For each j in m, sample j independently with
  probability 1/(i²2). In expectation new
  distribution is now uniform over d/(i²2) items.
  If j is sampled, say j survives.
• Go back to the previous step store the first
  i/(n²2) distinct surviving stream elements in the
  second half of stream, treat them as a set S, and
  count the remaining items in the stream that
  occur in S.
• Since only (1/²2) items survive, can store S
  with only (i log 1/²)/(n²2) bits by hashing item
  IDs down to a range of size, say, 1/²5
• We estimate the distributions support size in
  the sub-sampled stream, which is roughly d/(i²2).
  We can get a (1²)-approximation to this quantity
  provided it is at least ?(1/²2), which it is with
  high probability. Then scale by i²2 to estimate
  d, and thus F0 by previous reasoning.
• Constant update time

12
Techniques Lower Bound

• Consider the uniform distribution ¹ on d 1,
  2, , d. d and n are (1/²2)
• Consider a stream of n samples from ¹ where we
  choose n so that
• 1-(1-1/d)n/2
  in 1/3, 2/3
• Let X be the characteristic vector of the first
  n/2 stream samples on the universe d. So Xi 1
  if and only if item i occurs in these samples.
• Let Y be the characteristic vector of the second
  n/2 stream samples on the universe d. So Yi 1
  if and only if item i occurs in these samples.
• Let wt(X), wt(Y) be the number of ones in vectors
  X, Y.
• We consider a communication game. Alice is given
  X and wt(Y), while Bob is given Y and wt(X), and
  they want to know if
• Delta(X,Y) wt(X) wt(Y)-2wt(X)wt(Y)/d
• Alice and Bob should solve this problem with
  large probability, where the probability is over
  the choice of X and Y, and their coin tosses
  (which can be fixed).

13
Techniques Lower Bound

• Strategy
• (1) show the space complexity S of the streaming
  algorithm is at least the one-way communication
  complexity CC of the game
• (2) lower bound CC
• Theorem S CC
• Proof Alice runs the streaming algorithm on a
  random stream aX generated by her characteristic
  vector X. Alice transmits the state to Bob
• Bob continues the computation of the
  streaming algorithm on a random stream aY with
  characteristic vector Y
• At the end the algorithm estimates F0 of a
  stream whose elements are in the support of X or
  Y (or both)
• Notice that the two halves of the stream are
  independent in the random data model, so the
  stream generated has the right distribution

14
Techniques Lower Bound

• We show the estimate of F0 can solve the
  communication game
• Remember, the communication game is to decide
  whether
• Delta(X,Y) wt(X) wt(Y)-2wt(X)wt(Y)/d
• Note the quantity on the right is the expected
  value of Delta(X,Y)
• Some Lemmas
• (1) Prd/4 wt(X), wt(Y) 3d/4 1-o(1)
• (2) Consider the variable X, distributed as
  X, but conditioned on wt(X) k. The variable Y
  is distributed as Y, but conditioned on wt(Y)
  r.
• X and Y are uniform over k and r bit strings,
  respectively
• Choose k, r to be integers in d/4, 3d/4
• Then for any constant d gt 0, there is a constant
  gt 0, so that
• Pr(X,Y) E(X, Y) d1/2 1-
  d
• Follows from standard deviation of
  hypergeometric distribution

15
Techniques Lower Bound

• (X,Y) 2F0(aX?aY)-wt(X)-wt(Y)
• Note that Bob has wt(X) and wt(Y), so he can
  compute
• wt(X) wt(Y) wt(X)wt(Y)/d
• If F is the output of the streaming algorithm,
  Bob simply computes
• 2F-wt(X)-wt(Y) and checks if it is
  greater than
• F 2 (1-²)F0, (1²)F0 with large constant
  probability
• 2F-wt(X)-wt(Y) 2 2F0-wt(X)-wt(Y)-2²F0,,
  2F0-wt(X)-wt(Y)2²F0
• (X,Y)-2²F0, (X,Y)
  2²F0
• 
  (X,Y)-(²d), (X,Y) (²d)
• (X,Y) - (d1/2), (X,Y) (d1/2)

16
Techniques Lower Bound

• By our lemmas, and using that E(X, Y) ,
  with large constant probability either (X,Y) gt
  d1/2 or (X,Y) lt - d1/2
• By previous slide, Bobs value 2F-wt(X)-wt(Y) is
  in the range
• (X,Y) - (d1/2), (X,Y) (d1/2)
• If (X,Y) gt d1/2, then 2F-wt(X)-wt(Y) gt
  d1/2 -(d1/2)
• If (X,Y) lt - d1/2, then 2F-wt(X)-wt(Y) lt -
  d1/2 (d1/2)
• So Bob can use the output of the streaming
  algorithm to solve the communication problem, so
  S CC

17
Lower Bounding CC

• Communication game Alice is given X and wt(Y),
  while Bob is given Y and wt(X), and they want to
  know if
• Delta(X,Y) wt(X) wt(Y)-2wt(X)wt(Y)/d
• Here X,Y have the distribution of being
  independent characteristic vectors of a stream on
  n/2 samples, where
• 1-(1-1/d)n/2 2 1/3, 2/3

With large probability, Prd/4 wt(X), wt(Y)
3d/4 1-o(1) By averaging, a correct protocol
is also correct with large probability for fixed
weights i and j in d/4, 3d/4, so we can assume
X is a random string of Hamming weight i, and Y a
random string of Hamming weight j.
18
Lower Bounding CC
Y such that wt(Y) j
0 1 0 0 0 1 1 1
1 1 1 1 0 0 0 0

1 0 0 0 1 0 1 1
0 0 1 1 0 0 1 1
1 0 1 0 1 1 0 0
X such that wt(X) i
Since i, j 2 d/4, 3d/4, one can show that the
fraction of 1s in each row is in
1/2-o(1),1/2o(1) Recall that an entry is 1
if and only if (X,Y) i j 2ij/d
With large probability, the message M that Alice
sends has the property that many different X
cause Alice to send M. Say such an M is large.
We show that for any large M, the fraction of 1s
in most of the columns is in, say, 1/10, 9/10.
Then Bob doesnt know what to output.
Since each row is roughly balanced, the expected
fraction of 1s in each column is in 1/2-o(1),
1/2o(1).
19
Lower Bounding CC
Since each row is roughly balanced, the expected
fraction of 1s in each column is in 1/2-o(1),
1/2o(1). But the variance could be huge,
e.g., the matrix may look like this, in which
case Bob can easily output the answer.
1 1 1 1 0 0 0 0
1 1 1 1 0 0 0 0

1 0 0 0 1 0 1 1
0 0 1 1 0 0 1 1
1 0 1 0 1 1 0 0
20
Lower Bounding CC

• Can show this doesnt happen by the second-moment
  method.
• Let Vy be the fraction of 1s in the column
  indexed by y.
• Let S be the set of x which cause Alice to send a
  large message M.
• Consider random Y. Let Cu 1 if and only if
  (u,Y) gt . So V (1/S) sumu in S Cu
• VarV (1/S2) sumu,v in S (ECu Cv
  ECuECv).
• Notice that ECuECv 2 1/4-o(1), 1/4o(1)
• while ECu Cv Pr(u,Y) gt (v,Y) gt
  ½.
• Since S is a large set, most pairs u, v 2 S have
  Hamming distance in d/5, 4d/5.
• A technical lemma shows that Pr(u,Y) gt
  (v,Y) gt is a constant strictly less than 1.
• Hence, ECu Cv is a constant strictly less than
  1/2, and since this happens for most pairs u, v,
  we get that VarV is a constant strictly less
  than 1/4.

21
Lower Bounding CC

• Fraction of 1s in a random column is just V
  (1/S) sumu in S Cu
• Let be a small positive constant. By
  Chebyshevs inequality,
• PrV-1/2 gt 1/2- lt PrV-EV gt ½--o(1)
  VarV/(1/2-)2 o(1)
• But we showed VarV is a constant strictly less
  than ¼, so this probability is a constant
  strictly less than 1 for small enough .
• Hence, for a random column Y, the fraction of 1s
  is in , 1- with constant probability.
• It follows that with some constant probability
  Bob outputs the wrong answer.
• Hence, most messages of Alice must be small, so
  one can show that there must be 2?(d) of them, so
  that communication is ?(d) ?(1/²2).

22
Conclusions

• Introduced random data model, and studied
  F0-estimation under distributions uniform over a
  subset of d items
• For a certain range of d, we show that one can
  beat the space lower bound that holds for
  adversarial data and randomly-ordered data
• For another choice of d, we show the lower bound
  for adversarial and randomly-ordered data also
  applies in this setting
• Are there other natural distributions that admit
  more space-efficient algorithms in this model?
  Are there other useful ways of bypassing the
  ?(1/²2) lower bound?