Transcript Slide 1

Jeffrey D. Ullman
Stanford University




Foto Afrati (NTUA)
Anish Das Sarma (Google)
Semih Salihoglu (Stanford)
U.
2

Data consists of records for 3000 drugs.
 List of patients taking, dates, diagnoses.
 About 1M of data per drug.

Problem is to find drug interactions.
 Example: two drugs that when taken together
increase the risk of heart attack.

Must examine each pair of drugs and compare
their data.
4

The first attempt used the following plan:
 Key = set of two drugs {i, j}.
 Value = the record for one of these drugs.


Given drug i and its record Ri, the mapper
generates all key-value pairs ({i, j}, Ri), where j is
any other drug besides i.
Each reducer receives its key and a list of the
two records for that pair: ({i, j}, [Ri, Rj]).
5
Mapper
for drug 1
Mapper
for drug 2
Mapper
for drug 3
{1, 2}
Drug 1 data
{1, 3}
Drug 1 data
{1, 2}
Drug 2 data
{2, 3}
Drug 2 data
{1, 3}
Drug 3 data
{2, 3}
Drug 3 data
Reducer
for {1,2}
Reducer
for {1,3}
Reducer
for {2,3}
6
Mapper
for drug 1
Mapper
for drug 2
Mapper
for drug 3
{1, 2}
Drug 1 data
{1, 3}
Drug 1 data
{1, 2}
Drug 2 data
{2, 3}
Drug 2 data
{1, 3}
Drug 3 data
{2, 3}
Drug 3 data
Reducer
for {1,2}
Reducer
for {1,3}
Reducer
for {2,3}
7
{1, 2}
Drug 1 data
Drug 2 data
Reducer
for {1,2}
{1, 3}
Drug 1 data
Drug 3 data
Reducer
for {1,3}
{2, 3}
Drug 2 data
Drug 3 data
Reducer
for {2,3}
8





3000 drugs
times 2999 key-value pairs per drug
times 1,000,000 bytes per key-value pair
= 9 terabytes communicated over a 1Gb
Ethernet
= 90,000 seconds of network use.
9

They grouped the drugs into 30 groups of 100
drugs each.
 Say G1 = drugs 1-100, G2 = drugs 101-200,…, G30 =
drugs 2901-3000.
 Let g(i) = the number of the group into which drug i
goes.
10


A key is a set of two group numbers.
The mapper for drug i produces 29 key-value
pairs.
 Each key is the set containing g(i) and one of the
other group numbers.
 The value is a pair consisting of the drug number i
and the megabyte-long record for drug i.
11


The reducer for pair of groups {m, n} gets that
key and a list of 200 drug records – the drugs
belonging to groups m and n.
Its job is to compare each record from group m
with each record from group n.
 Special case: also compare records in group n with
each other, if m = n+1 or if n = 30 and m = 1.

Notice each pair of records is compared at
exactly one reducer, so the total computation is
not increased.
12


The big difference is in the communication
requirement.
Now, each of 3000 drugs’ 1MB records is
replicated 29 times.
 Communication cost = 87GB, vs. 9TB.
13
1.
A set of inputs.
 Example: the drug records.
2.
A set of outputs.
 Example: One output for each pair of drugs.
3.
A many-many relationship between each
output and the inputs needed to compute it.
 Example: The output for the pair of drugs {i, j} is
related to inputs i and j.
15
Output 1-2
Drug 1
Drug 2
Drug 3
Drug 4
Output 1-3
Output 1-4
Output 2-3
Output 2-4
Output 3-4
16
j
j
i

i
=
17

Reducer size, denoted q, is the maximum
number of inputs that a given reducer can have.
 I.e., the length of the value list.


Limit might be based on how many inputs can
be handled in main memory.
Or: make q low to force lots of parallelism.
18

The average number of key-value pairs created
by each mapper is the replication rate.
 Denoted r.

Represents the communication cost per input.
19




Suppose we use g groups and d drugs.
A reducer needs two groups, so q = 2d/g.
Each of the d inputs is sent to g-1 reducers, or
approximately r = g.
Replace g by r in q = 2d/g to get r = 2d/q.
Tradeoff!
The bigger the reducers,
the less communication.
20


What we did gives an upper bound on r as a
function of q.
A solid investigation of map-reduce algorithms
for a problem includes lower bounds.
 Proofs that you cannot have lower r for a given q.
21

A mapping schema for a problem and a reducer
size q is an assignment of inputs to sets of
reducers, with two conditions:
1. No reducer is assigned more than q inputs.
2. For every output, there is some reducer that
receives all of the inputs associated with that
output.

Say the reducer covers the output.
22


Every map-reduce algorithm has a mapping
schema.
The requirement that there be a mapping
schema is what distinguishes map-reduce
algorithms from general parallel algorithms.
23






d drugs, reducer size q.
Each drug has to meet each of the d-1 other
drugs at some reducer.
If a drug is sent to a reducer, then at most q-1
other drugs are there.
Thus, each drug is sent to at least (d-1)/(q-1)
reducers, and r > (d-1)/(q-1).
Half the r from the algorithm we described.
Better algorithm gives r = d/q + 1, so lower
bound is actually tight.
24

The problem with the algorithm dividing inputs
into g groups is that members of a group
appear together at many reducers.
 Thus, each reducer can only productively compare
about half the elements it gets.

Better: use smaller groups, with each reducer
getting many little groups.
 Eliminates almost all the redundancy.
25







Assume d inputs.
Let p be a prime, where p2 divides d.
Divide inputs into p2 groups of d/p2 inputs each.
Name the groups (i, j), where 0 < i, j < p.
Use p(p+1) reducers, organized into p+1 teams
of p reducers each.
For 0 < k < p, group (i, j) is sent to the reducer
i+kj (mod p) in group k.
In the last team (p), group (i, j) is sent to
reducer j.
26
j=0
1
2
3
4
i=0
1
2
3
4
Team 0
27
j=0
1
2
3
4
i=0
1
2
3
4
Team 1
28
j=0
1
2
3
4
i=0
1
2
3
4
Team 2
29
j=0
1
2
3
4
i=0
1
2
3
4
Team 3
30
j=0
1
2
3
4
i=0
1
2
3
4
Team 4
31
j=0
1
2
3
4
i=0
1
2
3
4
Team 5
32







Let two inputs be in groups (i, j) and (i’, j’).
If the same group, these inputs obviously share
a reducer.
If j = j’, then they share a reducer in team p.
If j  j’, then they share a reducer in team k
provided i + kj = i’ + kj’ (all arithmetic modulo p).
Equivalently, (i-i’) = k(j-j’).
But since j  j’, (j-j’) has an inverse modulo p.
Thus, team k = (i-i’)(j-j’)-1 has a reducer for
which i + kj = i’ + kj’.
33




The replication rate r is p+1, since every input is
sent to one reducer in each team.
The reducer size q = p(d/p2) = d/p, since each
reducer gets p groups of size d/p2.
Thus, r = d/q + 1.
(d/q + 1) - (d-1)/(q-1) < 1 provided q < d.
 But if q > d, we can do everything in one reducer,
and r = 1.

The upper bound r < d/q + 1 and the lower
bound r > (d-1)/(q-1) differ by less than 1, and
are integers, so they are equal.
34



Given a set of bit strings of length b, find all
those that differ in exactly one bit.
Example: For b=2, the inputs are 00, 01, 10, 11,
and the outputs are (00,01), (00,10), (01,11),
(10,11).
Theorem: r > b/log2q.
 (Part of) the proof later.
36




We can use one reducer for every output.
Each input is sent to b reducers (so r = b).
Each reducer outputs its pair if both its inputs
are present, otherwise, nothing.
Subtle point: if neither input for a reducer is
present, then the reducer doesn’t really exist.
37



Alternatively, we can send all inputs to one
reducer.
No replication (i.e., r = 1).
The lone reducer looks at all pairs of inputs that
it receives.
38


Assume b is even.
Two reducers for each string of length b/2.
 Call them the left and right reducers for that string.



String w = xy, where |x| = |y| = b/2, goes to the
left reducer for x and the right reducer for y.
If w and z differ in exactly one bit, then they will
both be sent to the same left reducer (if they
disagree in the right half) or to the same right
reducer (if they disagree in the left half).
Thus, r = 2; q = 2b/2.
39

Lemma: A reducer of size q cannot cover more
than (q/2)log2q outputs.
 Induction on b; proof omitted.




(b/2)2b outputs must be covered.
There are at least p = (b/2)2b/((q/2)log2q) =
(b/q)2b/log2q reducers.
Sum of inputs over all reducers > pq =
b2b/log2q.
Replication rate r = pq/2b = b/log2q.
 Omits possibility that smaller reducers help.
40
Algorithms Matching Lower Bound
Generalized Splitting
One reducer
for each output
b
Splitting
All inputs
to one
reducer
r = replication
rate
2
1
21
2b/2
2b
q = reducer
size
41


Assume n  n matrices AB = C.
Aij is the element in row i and column j of matrix
A.
 Similarly for B and C.


Cik = j Aij  Bjk.
Output Cik depends on the ith row of A, that is,
Aij for all j, and the kth column of B, that is, Bjk for
all j.
43
Column k
Row i
=
A
B
C
44




Important fact: If a reducer covers outputs Cik
and Cfg, then it also covers Cig and Cfk.
Why? This reducer has all of rows i and f of A as
inputs and also has all of columns k and g of B
as inputs.
Thus, it has all the inputs it needs to cover Cig
and Cfk.
Generalizing: Each reducer covers all the
outputs in the “rectangle” defined by a set of
rows and a set of columns of matrix C.
45
46


If a reducer gets q inputs, it gets q/n rows or
columns.
Maximize the number of outputs covered by
making the input “square.”
 I.e., #rows = #columns.

q/2n rows and q/2n columns yield q2/4n2
outputs covered.
47






Total outputs = n2.
One reducer can cover at most q2/4n2 outputs.
Therefore, 4n4/q2 reducers.
4n4/q total inputs to all the reducers, divided by
2n2 total inputs = 2n2/q replication rate.
Example: If q = 2n2, one reducer suffices and
the replication rate is r = 1.
Example: If q = 2n (minimum possible), then
r = n.
48




Divide rows of the first matrix into g groups of
n/g rows each.
Also divide the columns of the second matrix
into g groups of n/g columns each.
g2 reducers, each with q = 2n2/g inputs
consisting of a group of rows and a group of
columns.
r = g = 2n2/q.
49
n/g
n/g
=
50


A better way: use two map-reduce jobs.
Job 1: Divide both input matrices into
rectangles.
 Reducer takes two rectangles and produces partial
sums of certain outputs.

Job 2: Sum the partial sums.
51
K
J
K
J
I
I
A
B
C
For i in I and k in K, contribution
is j in J Aij × Bjk
52





Divide the rows of the first matrix A into g
groups of n/g rows each.
Divide the columns of A into 2g groups of n/2g.
Divide the rows of the second matrix B into 2g
groups of n/2g rows each.
Divide the columns of B into g groups of n/g.
Important point: the groups of columns for A
and rows for B must have indices that match.
53

Reducers correspond to an n/g by n/2g
rectangle in A (with row indices I, column
indices J) and an n/2g by n/g rectangle in B
(with row indices J and column indices K).
 Call this reducer (I,J,K).
 Important point: there is one set of indices J that
plays two roles.
 Needed so only rectangles that need to be multiplied are
given a reducer.
54
K
J
J
I
K
n/2g
n/g
n/g
I
n/g
n/g
n/2g
A
B
C
2g reducers contribute to
this area, one for each J.
55


Convention: i, j, k are individual rows and/or
column numbers, which are members of groups
I, J, and K, respectively.
Mappers Job 1:
 Aij -> key = (I,J,K) for any group K; value = (A,i,j,Aij).
 Bjk -> key = (I,J,K) for any group I; value = (B,j,k,Bjk).

Reducers Job 1: For key (I,J,K) produce
xiJk =  j in J Aij  Bjk for all i in I and k in K.
56

Mappers Job 2: xiJk -> key = (i,k), value = xiJk.

Reducers Job 2: For key (i,k), produce output Cik
= J xiJk.
57

The two methods (one or two map-reduce jobs)
essentially do the same computation.
 Every Aij is multiplied once with every Bjk.
 All terms in the sum for Cik are added together
somewhere, only once.

2 jobs requires some extra overhead of task
management.
58


One-job method: r = 2n2/q; there are 2n2
inputs, so total communication = 4n4/q.
Two-job method with parameter g:
Job 2: Communication = (2g)(n2/g2)(g2) = 2n2g.
Number of reducers
contributing to
each output
Number of output squares
Area of each square
59

Job 1 communication:





2n2 input elements.
Each generates g key-value pairs.
So another 2n2g.
Total communication = 4n2g.
Reducer size q = (2)(n2/2g2) = n2/g2.
 So g = n/q.
 Total communication = 4n3/q.
 Compares favorably with 4n4/q for the one-job approach.
60






Represent problems by mapping schemas
Get upper bounds on number of covered
outputs as a function of reducer size.
Turn these into lower bounds on replication
rate as a function of reducer size.
For HD = 1 and all-pairs problems: exact match
between upper and lower bounds.
1-job matrix multiplication analyzed exactly.
But 2-job MM yields better total
communication.
61

Get matching upper and lower bounds for the
Hamming-distance problem for distances
greater than 1.
 Ugly fact: For HD=1, you cannot have a large
reducer with all pairs at distance 1; for HD=2, it is
possible.

Consider all inputs of weight 1 and length b.
62
1.
2.
3.
Give an algorithm that takes an input-output
mapping and a reducer size q, and gives a
mapping schema with the smallest replication
rate.
Is the problem even tractable?
A recent extension by Afrati, Dolev, Korach,
Sharma, and U. lets inputs have weights, and
the reducer size limits the sum of the weights
of the inputs received.
 What can be extended to this model?
63