What’s Hot and What’s Not:Tracking Most Frequent Items Dynamically

By Graham Cormode

& S. Muthukrishnan

Rutgers University, Piscataway NY

Presented by Tal Sterenzy

Motivation

A basic statistic on database relationship is which items are hot – occur frequently

Dynamically maintaining hot items in the presence of delete and insert transactions.

Examples: DBMS – keep statistics to improve performance Telecommunication networks - network

connections start and end over time

Overview

Definitions Prior work Algorithm description & analysis Experimental results Summery

Formal definition Sequence of n transactions on m items [1…m] - Net occurrence of item i at time t

The number of times it has inserted minus the times it has been deleted

- current frequency of item at time t - most frequent item at time t The k most frequent items at time t are those with

the k largest

in t

1

( ) / ( )m

i i jjf t n t n t

*( ) max ( )i if t f t

( )if t

Finding k hot items

k is a parameter Item i is an hot item if Frequent items that appear a significant

fraction of the entire dataset There can be at most k hot items, and there

can be none Assume basic integrity constraint

( ) 1/(1 )if t k

Our algorithm

highly efficient, randomized algorithm for maintaining hot items in a dynamically changing database

monitors the changes to the data distribution and maintains O(klogklogm)

When queried, we can find all hot items in time O(klogklogm) with probability 1-δ

No need to scan the underlying relation

Small tail assumption Restriction:

are the frequencies of items A set of frequencies has a small tail

if If there are k hot items then small tail

probability holds If small tail probability holds then some top k

items might not be hot We shall analyze our solution in the presence

and absence of this small tail property (STP)

1 .. mf f

( ) 1/(1 )i k if t k

Prior work – why is it not adaptable? All these algorithms hold counters:

incremented when the item is observed decremented or reallocated under certain circumstances

Can’t directly adapt these algorithms for insertions and deletions: the state of the algorithm is different to that reached without

the insertions and deletions of the item.

Work on dynamic data is sparse, and provide no guarantees for the fully dynamic case with deletions

Our algorithm - idea

Do not keep counters of individual items, but rather of subsets of items

Ideas from group testing: Design a number of tests, each of which group

together a number of m items in order to find up to k items which test positive

Here: find k items that are hot Minimize number of tests, where each group

consists of a subset of items

General procedure For each transaction on item i, determine

which subsets it is included in: S(i) Each subset has a counter:

For insertion: increment all S(i) counters For deletion: decrement all S(i) counters

The test will be: does the counter exceed a threshold

Identifying the hot items is done by combining test results from several groups

The challenge is choosing the subsets

Bounding the number of required subsets Finding concise representation of the groups Giving efficiant way to go from results of tests

to the sets of hot items

Lets start with a simple case: k=1 (freq>1/2)

Deterministic algorithm for maintaining majority item

Finding majority item

For insertions only, constant time and space Keep logm+1 counters:

1 counter of items “alive”: The rest are labeled ,one per group Each group represents a bit in the binary

representation of the item Each group consists of half of the items

( ) ( )in t n t1 log... mc c

Finding majority item – cont. bit(i,j) – reports value of jth bit in binary representation of i gt(i, j) – return 1 if i>j, 0 otherwise

Scheme: Insertion of item i: Increment each counter such

that bit(i, j) = 1 in time O(logm). Deletion of i: Decrement each counter such that

bit(i, j) = 1 in time O(logm). Query: If there is a majority, then it is given by

computed in time O(logm).2log

12 ( , / 2)

m jjj

gt c c

jc

jc

Finding majority item – cont. Theorem: The algorithm finds a majority item

if there is one with time O(logm) per operation

The state of the data structure is equivalent if there are I insertion and D deletions, or if there are c = I - D insertions

In case of insertions only: the majority is found

UpdateCounters procedure

int c[0…logm]

UpdateCounters(i,transtype,c[0…logm])c[0]=c[0] + diff

for j=1 to logm do

If (transtype = ins)

c[j] = c[j] + bit(j,i)

Else

c[j] = c[j] - bit(j,i)

FindMajority procedure

FindMajority(c[0 ... log m])

Position = 0, t =1

for j=1 to log m do

if (c[j] > c[0]/2) then

position = position + t

t = 2* t

return position

Randomized constructions for finding hot items Observation: If we select subsets with one hot

item exactly applying the majority algorithm will identify the hot item

Definition: Let [1... ] denote the set of hot items

Set [1... ] is a if | | 1

F m

S m good set S F

How many subsets do we need? Theorem: Picking O(k logk) subsets by drawing m/k

items uniformly from [1…m] means that with constant probability we have included k good subsets S1…Sk such that

Proof: p – pick one item from F

O(k logk) subsets will guarantee with constant probability that we have one of each hot item (coupon’s collector problem)

( )iiF S F

/ 1 /(1 ) (1 )

And for 1 / 2 1/ 4 2 / 3 / 4

m k m km k k m kp

k m m m k mk m p e

Coupon collector problem p is probability that coupon is good X – number of trials required to collect at

least one of each type of coupon Epoch i begins with after i-th success and

ends with (i+1)-th success Xi – number of trials in the i-th epoch Xi distributed geometrically and pi = p(k-i)/k

1 1

0 0 1

1[ ] [ ] ln ( ) ( log )

( )

k k kk

pii i i

k kE X E X k O k O k k

p k i p i

Defining the groups with universal hash functions

The groups are chosen in a pseudo-random way using universal hash functions: Fix prime P > 2k a, b are drawn uniformly from [0…P-1] Then set:

Fact: Over all choices of a and b, for x<>y:

,

, , ,

( ) (( ) mod ) mod 2

{ | ( ) }a b

a b i a b

h x ax b P k

S x h x i

, ,

1Pr( ( ) ( ))

2a b a bh x h yk

Choosing and updating the subsets

We will choose T = logk/δ values of a and b,Which creates 2kT= 2klogk/δ subsets of items

Processing an item i means: To which T sets i belongs? For each one: update logm counters based on bit

representation of i If the set is good, this gives us the hot item

Space requirements

a and b are O(m): O(logk/δ logm) Number of counters: 2k logk/δ (logm + 1) Total space: O(k logk/δ logm)

log(k/δ) choices of a,b

2k subsets

log m + 1 counters

Probability of each hot item being in at least one good subset is at least 1-δ Consider one hot item: For each T repetitions

we put it in one of 2k groups The expected total

frequency of other items: If f<1/(k+1) majority will be found success If f>1/(k+1) majority can’t be found failure Probability of failure < ½ (by Markov inequality) Probability to fail on each T < Probability of any hot items failing at most δ.

i

i j

f 1 1E[f]=( )

2k 2 1 2( 1)

k

k k k

log /1/ 2 /k k

Detecting good subsets Given a subset and it’s associated

counters , it is possible to detect deterministically whether the subset is a good subset

Proof: a subset can fail in two cases: No hot items (assuming STP) : then

More than one hot item: there will be j such that:

a good subset is determined

Sa,b,i

0/( 1) and /( 1)j jc c k c c c k

0 /( 1)c c k

0 log... mc c

ProcessItem procedureInitialize c[0 … 2Tk][0 … log m]Draw a[1 … T], b[1 … T], c=0

ProccessItem(i,transtype,T,k)if (trans = ins) then

c = c + 1 else

c = c – 1 for x = 1 to T do index =2k(x-1)+(i*a[x]+b[x]modP)mod2k UpdateCounters(i,transtype,c[index])

GroupTest procedure

GroupTest(T,k,b)for i=1 to 2Tk do

if c[i][0] > cb position = 0; t =1 for j = 1 to log m do if (c[i][j] > cb and

c[i][0] – c[i][j] > cb) then Skip to next i

if c[i][j] > cb position += t

t = 2 * t output position

Algorithm correctness

With probability at least 1-δ, calling the GroupTest(logk/δ,k,1/k+1) procedure finds all hot items. Time processing item is: O(logk/δ logm) Time to get all hot items is O(k logk/δ logm)

With or without STP, we are still guarenteed to include all hot items with high probability

Without STP, we might output infrequent items

Algorithm correctness – cont.

When will an infrequent item be output? (no STP) A set with 2 hot items or more will be detected A set with one hot item will never fault. Even if

there is a split without the hot item that exceeds the threshold – it will be detected

A set with no hot item, and for all logm splits one half will exceed the threshold and the other not only then the algorithm will fail

Algorithm properties

• The set of counters created with T= log k/ δ can be used to find hot items with parameter k’ for any k’<k with probability of success 1 – δ by calling GroupTest(logk/δ,k,1/(k’+1))

Proof: in the proof of probability for k hot items: 1 ' 1

2 ' 1 2( ' 1)

k

k k k

Experiments GroupTesting algorithm was compared to Loosy

Counting and Frequent algorithms. The authors implemented them so that when an

item is deleted we decrement the corresponding counter if such exist.

The recall is the proportion of the hot items that are found by the method to the total number of hot items.

The precision is the proportion of items identified by the algorithm, which are hot, to number of all output items.

Synthetic data (Recall)

Zipf for hot items: 0 – distributed uniformly , 3 – highly skewed

Synthetic data (Precision)

Zipf for hot items: 0 – distributed uniformly , 3 – highly skewed

Real data (Recall)

Real data was obtained from one of AT&T network for part of a day.

Real Data (Percision)

Real data has no guarantee of having small tail property

Varying frequency at query time

The data structure was build for queries at the 0.5% level, but was then tested with queries ranged from 10% to 0.02%

Conclusions and extensions

New method which can cope with dynamic dataset is proposed.

It’s interesting to try to use the algorithm to compare the differences in frequencies between different datasets.

Can we find combinatorial design that achieve the same properties but in deterministic construction for maintaining hot items?