Algorithms for massive data sets
Lecture 2 (Mar 14, 2004)
Yossi Matias & Ely Porat
(partially based on various presentations & notes)
Negative Result for Sampling
[Charikar, Chaudhuri, Motwani, Narasayya 2000]
Theorem: Let E be estimator for D(X) examining r<n
values in X, possibly in an adaptive and randomized
order. Then, for any   e  r, E must have relative error
nr 1
ln
2r

with probability at least .
• Example
– Say, r = n/5
– Error 20% with probability 1/2
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Scenario Analysis
Scenario A:
– all values in X are identical (say V)
– D(X) = 1
Scenario B:
– distinct values in X are {V, W1, …, Wk},
– V appears n-k times
– each Wi appears once
– Wi’s are randomly distributed
– D(X) = k+1
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Proof
• Little Birdie – one of Scenarios A or B only
• Suppose
– E examines elements X(1), X(2), …, X(r) in that order
– choice of X(i) could be randomized and depend arbitrarily on
values of X(1), …, X(i-1)
• Lemma
n  i  k 1
P[ X(i)=V | X(1)=X(2)=…=X(i-1)=V ] 
n  i 1
• Why?
– No information on whether Scenario A or B
– Wi values are randomly distributed
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Proof (continued)
• Define EV – event {X(1)=X(2)=…=X(r)=V}
PΕV  

r
 PX (i)  V | X (1)  X (2)  ...  X (i  1)  V 
i 1
r

i 1
n  i  k 1  n  r  k 


n  i 1
 nr 
r
r
k 

  2kr 
 1 

exp



n

r
n

r




• Last inequality because
1  Z  exp( 2Z ), for 0  Z  1 / 2
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Proof (conclusion)
nr 1
ln
• Choose k 
to obtain
2r

• Thus:
PEV   
 
PEV   
– Scenario A  P EV  1
– Scenario B 
• Suppose
– E returns estimate Z when EV happens
– Scenario A  D(X)=1
– Scenario B  D(X)=k+1
– Z must have worst-case error > k
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Randomized Approximation
(based on [Flajolet-Martin 1983, Alon-Matias-Szegedy 1996])
Theorem: For every c > 2 there exists an algorithm that, given a
sequence A of n members of U={1,2,…,u}, computes a
number d’ using O(log u) memory bits, such that the
probability that max(d’/d,d/d’) > c is at most 2/c.
A bit vector BV will represent the set
Let b be smallest integer s.t. 2^b > u.
Let F = GF(2^b). Let r,s be random from F.
Pr(h(a)=k)
k
For a in A, let h(a) = r ·a + s = 101****10….0
k
Set k’th bit.
Bit vector : 0000101010001001111
Estimate is 2^{max bit set}.
b
01
k
u-1
Randomized Approximation (2)
(based on [Indyk-Motwani 1998])
• Algorithm SM – For fixed t, is D(X) >> t?
– Choose hash function h: U[1..t]
– Initialize answer to NO
– For each x i , if h( x i) = t, set answer to YES
• Theorem:
– If D(X) < t, P[SM outputs NO] > 0.25
– If D(X) > 2t, P[SM outputs NO] < 0.136 = 1/e^2
Analysis
• Let – Y be set of distinct elements of X
• SM(X) = NO  no element of Y hashes to t
• P[element hashes to t] = 1/t
• Thus – P[SM(X) = NO] = (1  1 / t )
|Y |
• Since |Y| = D(X),
– If D(X) < t, P[SM(X) = NO] > (1  1 / t)t > 0.25
– If D(X) > 2t, P[SM(X) = NO] < (1  1 / t ) 2t < 1/e^2
• Observe – need 1 bit memory only!
Boosting Accuracy
• With 1 bit 
can probabilistically distinguish D(X) < t from D(X) > 2t
• Running O(log 1/δ) instances in parallel 
reduces error probability to any δ>0
• Running O(log n) in parallel for t = 1, 2, 4, 8 …, n 
can estimate D(X) within factor 2
• Choice of factor 2 is arbitrary 
can use factor (1+ε) to reduce error to ε
• EXERCISE – Verify that we can estimate D(X) within
factor (1±ε) with probability (1-δ) using space
O(log
n

2
 log
1

)
Sampling: Basics
• Idea: A small random sample S of the data often well-represents all the data
– For a fast approx answer, apply the query to S & “scale” the result
– E.g., R.a is {0,1}, S is a 20% sample
select count(*) from R where R.a = 0
select 5 * count(*) from S where S.a = 0
1101
1111000
01111101
1101011
0110
R.a
Red =
in S
Est. count = 5*2 = 10, Exact count = 10
• Leverage extensive literature on confidence intervals for sampling
Actual answer is within the interval [a,b] with a given probability
E.g., 54,000 ± 600 with prob  90%
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Sampling versus Counting
• Observe
– Count merely abstraction – need subsequent analytics
– Data tuples – X merely one of many attributes
– Databases – selection predicate, join results, …
– Networking – need to combine distributed streams
• Single-pass Approaches
– Good accuracy
– But gives only a count -- cannot handle extensions
• Sampling-based Approaches
– Keeps actual data – can address extensions
– Strong negative result
Distinct Sampling for Streams
[Gibbons 2001]
• Best of both worlds
– Good accuracy
– Maintains “distinct sample” over stream
– Handles distributed setting
• Basic idea
– Hash – random “priority” for domain values
– Tracks O 2 log  1  highest priority values seen
– Random sample of tuples for each such value
– Relative error  with probability 1  
Hash Function
• Domain U = [0..m-1]
• Hashing
– Random A, B from U, with A>0
– g(x) = Ax + B (mod m)
– h(x) – # leading 0s in binary representation of g(x)
• Clearly – 0  h( x)  log m
• Fact
Ph(x)  l   2 (l 1)
Overall Idea
• Hash  random “level” for each domain value
• Compute level for stream elements
• Invariant
– Current Level – cur_lev
– Sample S – all distinct values scanned so far of
level at least cur_lev
• Observe
– Random hash  random sample of distinct values
– For each value  can keep sample of their tuples
Algorithm DS (Distinct Sample)

• Parameters – memory size M  O  2 log  1
• Initialize – cur_lev0; Sempty
• For each input x
– L  h(x)
– If L>cur_lev then add x to S
– If |S| > M
• delete from S all values of level cur_lev
• cur_lev  cur_lev +1
• Return 2
cur _ lev
| S |

Analysis
• Invariant – S contains all values x such that
h( x)  cur _ lev
• By construction
Ph( x)  cur _ lev   2cur _ lev
• Thus
E| S |  2
 cur _ lev
 D( X )
• EXERCISE – verify deviation bound
Hot list queries
• Why is it interesting:
– Top ten – best seller list
– Load balancing
– Caching policies
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Hot list queries
• Let use sampling
edoejddkaklsadkjdkdkpryekfvcuszldfoasd
djkkdkvza
k3d2jvza
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Hot list queries
• The question is:
– How to sample if we don’t know our sample size?
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Gibbons & Matias’ algorithm
Hotlist:
Produced values:
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0
1
2
5
0
1
3
0
1
0
3
a
b
c
d
p = 1.0
a b a c a a b d b a d d
21
Gibbons & Matias’ algorithm
Need to replace
one value
Hotlist:
Produced values:
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0
1
2
5
0
1
3
0
1
0
3
a
b
c
d
p = 1.0
a b a c a a b d b a d d
e
22
Gibbons & Matias’ algorithm
Throw biased
coins with
probability f
Hotlist:
Replace counts
by number
of seen heads
Produced values:
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Multiply p with some
amount f
0
1
2
5
4
0
1
3
1
0
0
3
2
a
b
c
d
p = 0.75 (f = 0.75)
a b a c a a b d b a d d
e
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Gibbons & Matias’ algorithm
Replace a value which
has zero count
Hotlist:
Produced values:
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0
1
2
5
4
0
1
3
0
1
0
3
2
a
b
e
d
p = 0.75
Count/p is an estimate of number of
a b a c a a b d b a d d e
times a value has been seen. E.g., the
value ‘a’ has been seen 4/p = 5.33 times
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Counters
• How many bits need to count?
– Prefix code
– Approximated counters
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Rarity
• Paul goes fishing.
• There are many different fish species U={1,..,u}
• Paul catch one fish at a time atU
• Ct[j]=|{ai| ai=j,i≤t}| number of time catches the species j
• Species j is rare at time t if it appears only once
• [t]=|{j| Ct[j]=1}|/u
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Rarity
• Why is it interesting?
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Again lets use sampling
U={1,2,3,4,5,6,7,8,9,10,11,12…u}
U’={4,9,13,18,24}
Xt[i]=|{t|aj=U’[i],j≤t}|
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Again lets use sampling
Xi[t]=|{t|aj=Xi,j≤t}|
[t]=|{Ct[i]| Ct[i]=1}|/u :‫תזכורת‬
’[t]=|{Xt[i]| Xt[i]=1}|/k
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Rarity
• But [t] need to be at least 1/k to get a
good estimator.
| { j | ct [ j ]  1} |
 [t ] 
| { j | Ct [ j ]  0} |
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Min-wise independent hash functions
• Family of hash functions H[n]->[n]
call Min-wise independent
• If for any X [n] and xX
1
PrhH [h( x)  min h( X )] 
|X|
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