a common mistake size/communication trade-off specific tradeoffs

Designing Efficient Map-Reduce Algorithms

A Common MistakeSize/Communication Trade-OffSpecific Tradeoffs

Jeffrey D. UllmanStanford University

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Research Is Joint Work of Foto Afrati (NTUA) Anish Das Sarma (Google) Semih Salihoglu (Stanford) U.

Motivating Example

The Drug Interaction ProblemA Failed AttemptLowering the Communication

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The Drug-Interaction Problem 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.

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Initial Map-Reduce Algorithm 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]).

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Example: Three Drugs

Mapperfor drug

2

Mapperfor drug

1

Mapperfor drug

3

Drug 1 data{1, 2} Reducer

for {1,2}

Reducerfor

{2,3}

Reducerfor

{1,3}

Drug 1 data{1, 3}

Drug 2 data{1, 2}

Drug 2 data{2, 3}

Drug 3 data{1, 3}

Drug 3 data{2, 3}

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Example: Three Drugs

Mapperfor drug

2

Mapperfor drug

1

Mapperfor drug

3

Drug 1 data{1, 2} Reducer

for {1,2}

Reducerfor

{2,3}

Reducerfor

{1,3}

Drug 1 data{1, 3}

Drug 2 data{1, 2}

Drug 2 data{2, 3}

Drug 3 data{1, 3}

Drug 3 data{2, 3}

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Example: Three DrugsDrug 1 data{1, 2}

Reducerfor

{1,2}

Reducerfor

{2,3}

Reducerfor

{1,3}Drug 1 data

Drug 2 data

Drug 2 data{2, 3}

Drug 3 data{1, 3}

Drug 3 data

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What Went Wrong? 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.

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The Improved Algorithm 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.

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The Map Function 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.

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The Reduce Function 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.

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The New Communication Cost The big difference is in the communication

requirement. Now, each of 3000 drugs’ 1MB records is

replicated 29 times. Communication cost = 87GB, vs. 9TB.

Theory of Map-Reduce AlgorithmsReducer SizeReplication RateMapping SchemasLower Bounds

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A Model for Map-Reduce Algorithms1. 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.

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Example: Drug Inputs/Outputs

Drug 1

Drug 2

Drug 3

Drug 4

Output 1-2

Output 1-3

Output 2-4

Output 1-4

Output 2-3

Output 3-4

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Example: Matrix Multiplication

=i

jj

i

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Reducer Size 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.

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Replication Rate The average number of key-value pairs created

by each mapper is the replication rate. Denoted r.

Represents the communication cost per input.

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Example: Drug Interaction 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.

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Upper and Lower Bounds on r 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.

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Proofs Need Mapping Schemas 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.

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Mapping Schemas – (2) 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.

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Example: Drug Interactions 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.

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The Better Algorithm 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.

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Optimal Algorithm for All-Pairs 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.

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Example: Teams for p = 5

i = 0

1

2

1

3

432

4

j = 0

Team 0

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Example: Teams for p = 5

i = 0

1

2

1

3

432

4

j = 0

Team 1

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Example: Teams for p = 5

i = 0

1

2

1

3

432

4

j = 0

Team 2

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Example: Teams for p = 5

i = 0

1

2

1

3

432

4

j = 0

Team 3

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Example: Teams for p = 5

i = 0

1

2

1

3

432

4

j = 0

Team 4

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Example: Teams for p = 5

i = 0

1

2

1

3

432

4

j = 0

Team 5

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Why It Works 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’.

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Why It Is Optimal 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.

The Hamming-Distance = 1 Problem

The Exact Lower BoundMatching Algorithms

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Definition of HD1 Problem 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.

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Algorithm With q=2 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.

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Algorithm with q = 2b

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.

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Splitting Algorithm 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.

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Proof That r > b/log2q 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.

Algorithms Matching Lower Bound

q = reducersize

b

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21 2b/2 2b

All inputsto onereducer

One reducerfor each output Splitting

Generalized Splitting

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r = replicationrate

Matrix Multiplication

One-Job MethodTwo-Job MethodComparison

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Matrix Multiplication 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.

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Computing One Output Value

=Row i

Column k

A B C

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Reducers Cover Rectangles 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.

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The Responsibility of One Reducer

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Upper Bound on Output Size 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.

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Lower Bound on Replication Rate 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.

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Matching Algorithm 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.

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Picture of One Reducer

=n/g

n/g

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Two-Job Map-Reduce Algorithm

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.

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Picture of First Job

I

J

J

K

I

K

A CB

For i in I and k in K, contributionis j in J Aij × Bjk

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First Job – Details 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.

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Reducers for First Job 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.

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The Reducer (I,J,K)

I

J

J

K

I

K

A CB

n/g n/gn/g

n/gn/2g

n/2g

2g reducers contribute tothis area, one for each J.

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Job 1: Details 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.

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Job 2: Details

Mappers Job 2: xiJk -> key = (i,k), value = xiJk.

Reducers Job 2: For key (i,k), produce output Cik = J xiJk.

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Comparison: Computation Cost 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.

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Comparison: Communication Cost

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 output squares

Area of each square

Number of reducerscontributing toeach output

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Communication Cost – Continued

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.

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Summary 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.

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Research Questions 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.

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Research Questions – (2)1. Give an algorithm that takes an input-output

mapping and a reducer size q, and gives a mapping schema with the smallest replication rate.

2. Is the problem even tractable?3. 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?