Distinct items: • Given a stream , where , count the number of distinct items (so we are in the cash

register model)

• Example: 3 5 7 4 3 4 3 4 7 5 9• 5 distinct elements: 3 4 5 7 9 (we only want the count of distinct elements, and not

the set of distinct elements)

• In terms of frequency moments estimation, this is the problem of estimating

• The easy deterministic solutions with space and ( number of distinct elements)

• Deterministic exact solution requires space in the worst case• How about deterministic approximate solutions? And exact randomized?

• Can we do better with randomization and approximation?

Counting distinct elements (Flajolet—Martin 1985)

• Let be a random hash function: For each , value is uniformly distributed in

• What is the relation between the minimum of and the number of distinct elements

(We will do two proofs on the board, one algebraic and one pictorial)

• Moreover, the variance can also be bounded via (Fun problem: I only know an algebraic proof for this, but there could be a pictorial one too given the suggestive-looking rhs)

Counting distinct elementsFirst algorithm• Pick random hash function • Find the minimum of • Output

• Estimator has high variance. Improving the estimator by averaging:

Second algorithm• Run parallel independent copies of the first algorithm• Set ( is the estimate given by the th copy)• Return

Counting distinct elements• Space complexity of the first algorithm: To compute the minimum we just need to keep one real number in the memory. But need to limit precision

• So the space requirement

• Not quite: also need to account for the memory requirements for a random hash function

• What property of random hash function did we really use?

Counting distinct elements• Pick from a 2-wise independent hash function family mapping for a prime

( is chosen large to reduce round off errors)• set of distinct elements

• New estimator: • No longer clear that , but does provide useful informationLemma (probability is over the random choice of )Proof (1) First, prove :

Union bound

Counting distinct elements(2) Prove :

• Define indicator if (this is the good event)

otherwise• and so • We now upper bound by using the pairwise independence of the and Chebyshev’s inequality (proof on the board; also in the book page 297)

Boosting the success probability• Take the median of the means estimator • But doesn’t seem to give a -factor approximation approximation only within factors and

• A related estimator [BJKST 2004]:

• pairwise independent hash function family of functions of type • , so we can take , and have bits decription• So the probability that a random is injective is

• Maintain the smallest hash values the th smallest hash value at the end of the stream The new estimator (BJKST estimator) is

Analyzing the BJKST estimator• Requirements to maintain the BJKST estimator:

– Space – Update time

• We assume (satisfied if true for )

• Recall that the set of distinct elements in the stream• We separately upper bound and using the Chebyshev inequality

Analyzing the BJKST estimator

• I.e., contains at least elements less than (using )

• For , define if and otherwise

• For

• , ,

•

Chebyshev

Analyzing the BJKST estimator• Similarly,

• Thus,

• And now we can apply the median trick: Run parallel independent copies of the algorithm to compute and output their median

Theorem The output of the above algorithm is an -approximation of . It uses space and update time per streaming element

Very powerful: A variant needs 128 bytes for all works of Shakespeare, ≈1/10 [Durand--Flajolet 2003]

• What streaming model does the above algorithm require?

Counting distinct elements (strict turnstile model)

• What about the strict turnstile model? • with integers• Frequency vector nonnegative • The previous algorithm requires cash register model

• A different but closely related algorithm that works in the strict turnstile model

• We will only give the basic idea and not the full details of the proof

Counting distinct elements (strict turnstile model)

• set of distinct elements• First reduce the problem to its decision version: • Input: stream , parameters, and an additional parameter • Output:

– YES if – NO if – Arbitrary otherwise

• Solution of the decision version gives a solution of the general problem with a slight blow up in the space:

• Run parallel versions of the decision problem with • A total of copies

Algorithm for the decision version of counting distinct elements

Basic algorithm

• Choose a random set by picking each element independently with probability :

for all

• Maintain

• Output YES if else output NO

Decision version of counting distinct elements (analysis idea)

Lemma For and if if

Proof

Full algorithm• Run independent parallel copies of the basic algorithm for sufficiently large

constant : Sample independently, and maintain for each

• if the ’th instance of the basic algorithm gives otherwise

• Output YES (i.e. declare ) if • Output NO otherwise

• An application of the Chernoff bound using the independence of the shows that this provides an -approximation

• Space requirement? • Use 2-wise independent sampling to choose • Total space requirement is

Counting distinct elements• Why didn’t we just maintain whether or not ?

• is a linear sketch• Allows for negative • So works in the (strict) turnstile model

• The problem of computing is by now very well understood: space complexity with update time This is optimal up to constant factors [Kane et al. 2010]