Estimating Distinct Elements,
Optimally
David Woodruff
IBM
Based on papers with Piotr Indyk,
Daniel Kane, and Jelani Nelson
Problem Description
• Given a long string of at most n distinct characters, count
the number F0 of distinct characters
• See characters one at a time
• One pass over the string
• Algorithms must use small memory and fast update time
– too expensive to store set of distinct characters
– algorithms must be randomized and settle for an
approximate solution: output F 2 [(1-²)F0, (1+²)F0]
with, say, good constant probability
Algorithm History
• Flajolet and Martin introduced problem, FOCS 1983
– O(log n) space for fixed ε in random oracle model
• Alon, Matias and Szegedy
– O(log n) space/update time for fixed ε with no oracle
• Gibbons and Tirthapura
– O(ε-2 log n) space and O(ε-2) update time
• Bar-Yossef et al
– O(ε-2 log n) space and O(log 1/ε) update time
– O(ε-2 log log n + log n) space and O(ε-2) update time, essentially
– Similar space bound also obtained by Flajolet et al in the random
oracle model
• Kane, Nelson and W
– O(ε-2 + log n) space and O(1) update and reporting time
– All time complexities are in unit-cost RAM model
Lower Bound History
•
Alon, Matias and Szegedy
– Any algorithm requires Ω(log n) bits of space
•
Bar-Yossef
– Any algorithm requires Ω(ε-1) bits of space
•
Indyk and W
– If ε > 1/n1/9, any algorithm needs Ω(ε-2) bits of space
•
W
– If ε > 1/n1/2, any algorithm needs Ω(ε-2) bits of space
•
Jayram, Kumar and Sivakumar
– Simpler proof of Ω(ε-2) bound for any ε > 1/m1/2
•
Brody and Chakrabarti
– Show above lower bounds hold even for multiple passes over the string
Combining upper and lower bounds, the complexity of this problem is:
Θ(ε-2 + log n) space and Θ(1) update and reporting time
Outline for Remainder of Talk
• Proofs of the Upper Bounds
• Proofs of the Lower Bounds
Hash Functions for Throwing Balls
• We consider a random mapping f of B balls into C containers and
count the number of non-empty containers
• The expected number of non-empty containers is C – C(1-1/C)B
• If instead of the mapping f, we use an O(log C/ε)/log log C/ε – wise
independent mapping g, then
– the expected number of non-empty containers under g is the
same as that under f, up to a factor of (1 ± ε)
• Proof based on approximate inclusion-exclusion
– express 1 – (1-1/C)B in terms of a series of binomial coefficients
– truncate the series at an appropriate place
– use limited independence to handle the remaining terms
Fast Hash Functions
•
Use hash functions g that can be evaluated in O(1) time.
•
If g is O(log C/ε)/(log log C/ε)-wise independent, the natural family of
polynomial hash functions doesn’t work
•
We use theorems due to Pagh, Pagh, and Siegel that construct k-wise
independent families for large k, and allow O(1) evaluation time
•
For example, Siegel shows:
– Let U = [u] and V = [v] with u = vc for a constant c > 1, and suppose the machine
word size is Ω(log v)
– Let k = vo(1) be arbitrary
– For any constant d > 0 there is a randomized procedure that constructs a k-wise
independent hash family H from U to V that succeeds with probability 1-1/vd and
requires vd space. Each h 2 H can be evaluated in O(1) time
•
Can show we have sufficiently random hash functions that can be evaluated
in O(1) time and represented with O(ε-2 + log n) bits of space
Algorithm Outline
1.
Set K = 1/ε2
2.
Instantiate a lg n x K bitmatrix A, initializing entries of A to 0
3.
Pick random hash functions f: [n]->[n] and g: [n]->[K]
4.
Obtain a constant factor approximation R to F0 somehow
5.
Update(i): Set A1, g(i) = 1, A2, g(i) = 1, …, Alsb(f(i)), g(i) = 1
6.
Estimator: Let T = |{j in [K]: Alog (16R/K), j = 1}|
Output (32R/K) * ln(1-T/K)/ln(1-1/K)
Space Complexity
•
Naively, A is a lg n x K bitmatrix, so O(ε-2 log n) space
•
Better: for each column j, store the identity of the largest row i(j) for which
Ai, j = 1. Note if Ai,j = 1, then Ai’, j = 1 for all i’ < i
– Takes O(ε-2 log log n) space
•
Better yet: maintain a “base level” I. For each column j, store max(i(j) – I, 0)
– Given an O(1)-approximation R to F0 at each point in the stream, set
I = log R
– Don’t need to remember i(j) if i(j) < I, since j won’t be used in estimator
– For the j for which i(j) ¸ I, about 1/2 such j will have i(j) = I, about one
fourth such j will have i(j) = I+1, etc.
– Total number of bits to store offsets is now only O(K) = O(ε-2) with good
probability at all points in the stream
The Constant Factor Approximation
•
Previous algorithms state that at each point in the stream, with probability 1δ, the output is an O(1)-approximation to F0
– The space of such algorithms is O(log n log 1/δ).
– Union-bounding over a stream of length m gives O(log n log m) total space
•
We achieve O(log n) space, and guarantee the O(1)-approximation R of the
algorithm is non-decreasing
–
–
–
–
–
–
•
Apply the previous scheme on a log n x log n/(log log n) matrix
For each column, maintain the identity of the deepest row with value 1
Output 2i, where i is the largest row containing a constant fraction of 1s
We repeat the procedure O(1) times, and output the median of the estimates
Can show the output is correct with probability 1- O(1/log n)
Then we use the non-decreasing property to union-bound over O(log n) events
We only increase the base level every time R increases by a factor of 2
– Note that the base level never decreases
Running Time
•
Blandford and Blelloch
– Definition: a variable length array (VLA) is a data structure implementing an array
C1, …, Cn supporting the following operations:
• Update(i, x) sets the value of Ci to x
• Read(i) returns Ci
The Ci are allowed to have bit representations of varying lengths len(Ci).
– Theorem: there is a VLA using O(n + sumi len(Ci)) bits of space supporting worst
case O(1) updates and reads, assuming the machine word size is at least log n
•
Store our offsets in a VLA, giving O(1) update time for a fixed base level
•
Occasionally we need to update the base level and decrement offsets by 1
– Show base level only increases after Θ(ε-2) updates, so can spread this work
across these updates, so O(1) worst-case update time
– Copy the data structure, use it for performing this additional work so it doesn’t
interfere with reporting the correct answer
– When base level changes, switch to copy
•
For O(1) reporting time, maintain a count of non-zero containers in a level
Outline for Remainder of Talk
• Proofs of the Upper Bounds
• Proofs of the Lower Bounds
1-Round Communication Complexity
Alice:
What is f(x,y)?
input x
Bob:
input y
• Alice sends a single, randomized message M(x) to Bob
• Bob outputs g(M(x), y) for a randomized function g
• g(M(x), y) should equal f(x,y) with constant probability
• Communication cost CC(f) is |M(x)|, maximized over x and random bits
• Alice creates a string s(x), runs a randomized algorithm A on s(x), and
transmits the state of A(s(x)) to Bob
• Bob creates a string s(y), continues A on s(y), thus computing A(s(x)◦s(y))
• If A(s(x)◦s(y)) can be used to solve f(x,y), then space(A) ¸ CC(f)
The Ω(log n) Bound
•
Consider equality function: f(x,y) = 1 if and only if x = y for x, y 2 {0,1}n/3
•
Well known that CC(f) = Ω(log n) for (n/3)-bit strings x and y
•
Let C: {0,1}n/3 -> {0,1}n be an error-correcting code with all codewords of
Hamming weight n/10
– If x = y, then C(x) = C(y)
– If x != y, then ¢(C(x), C(y)) = Ω(n)
•
Let s(x) be any string on alphabet size n with i-th character appearing in s(x)
if and only if C(x)i = 1. Similarly define s(y)
•
If x = y, then F0(s(x)◦s(y)) = n/10. Else, F0(s(x)◦s(y)) = n/10 + Ω(n)
•
A constant factor approximation to F0 solves f(x,y)
The Ω(ε-2) Bound
•
Let r = 1/ε2. Gap Hamming promise problem for x, y in {0,1}r
– f(x,y) = 1 if ¢(x,y) > 1/(2ε-2)
– f(x,y) = 0 if ¢(x,y) < 1/(2ε-2) - 1/ε
•
Theorem: CC(f) = Ω(ε-2)
– Can prove this from the Indexing function
– Alice has w 2 {0,1}r, Bob has i in {1, 2, …, r}, output g(w, i) = wi
– Well-known that CC(g) = Ω(r)
•
Proof: CC(f) = Ω(r),
– Alice sends the seed r of a pseudorandom generator to Bob, so the parties have
common random strings zi, …, zr 2 {0,1}r
– Alice sets x = coordinate-wise-majority{zi | wj = 1}
– Bob sets y = zi
– Since the zi are random, if xj = 1, then by properties of majority, with good
probability ¢f(x,y) < 1/(2ε-2) - 1/ε, otherwise likely that ¢f(x,y) > 1/(2ε-2)
– Repeat a few times to get concentration
The Ω(ε-2) Bound Continued
•
Need to create strings s(x) and s(y) to have F0(s(x)◦s(y)) decide whether
¢(x,y) > 1/(2ε-2) or ¢(x,y) < 1/(2ε-2) - 1/ε
•
Let s(x) be a string on n characters where character i appears if and only if
xi = 1. Similarly define s(y)
•
F0(s(x)◦s(y)) = (wt(x) + wt(y) + ¢(x,y))/2
– Alice sends wt(x) to Bob
•
A calculation shows a (1+ε)-approximation to F0(s(x)◦s(y)), together with
wt(x) and wt(y), solves the Gap-Hamming problem
•
Total communication is space(A) + log 1/ε = Ω(ε-2)
•
It follows that space(A) = Ω(ε-2)
Conclusion
Combining upper and lower bounds, the streaming complexity of
estimating F0 up to a (1+ε) factor is:
Θ(ε-2 + log n) bits of space and Θ(1) update and reporting time
• Upper bounds based on careful combination of efficient hashing,
sampling and various data structures
• Lower bounds come from 1-way communication complexity