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A Common Mistake Size/Communication Trade-Off Specific Tradeoffs

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Designing Efficient Map-Reduce Algorithms

A Common MistakeSize/Communication Trade-OffSpecific Tradeoffs

Jeffrey D. Ullman

Stanford University

- FotoAfrati (NTUA)
- Anish Das Sarma (Google)
- SemihSalihoglu (Stanford)
- U.

Motivating Example

The Drug Interaction ProblemA Failed AttemptLowering the Communication

- 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.

- 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]).

Mapper

for drug 1

Reducer

for {1,2}

Drug 3 data

Drug 2 data

Drug 2 data

Drug 3 data

Drug 1 data

Drug 1 data

{2, 3}

{1, 3}

{2, 3}

{1, 2}

{1, 3}

{1, 2}

Mapper

for drug 2

Reducer

for {1,3}

Mapper

for drug 3

Reducer

for {2,3}

Mapper

for drug 1

Reducer

for {1,2}

Drug 2 data

Drug 3 data

Drug 2 data

Drug 3 data

Drug 1 data

Drug 1 data

{2, 3}

{1, 3}

{2, 3}

{1, 2}

{1, 3}

{1, 2}

Mapper

for drug 2

Reducer

for {1,3}

Mapper

for drug 3

Reducer

for {2,3}

Reducer

for {1,2}

{1, 2}

Drug 1 data

Drug 2 data

Reducer

for {1,3}

Drug 1 data

Drug 3 data

{1, 3}

Reducer

for {2,3}

Drug 2 data

Drug 3 data

{2, 3}

- 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.

- 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.

- 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.

- 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.

- 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 Algorithms

Reducer SizeReplication RateMapping SchemasLower Bounds

- A set of inputs.
- Example: the drug records.

- A set of outputs.
- Example: One output for each pair of drugs.

- 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.

Output 1-2

Drug 1

Output 1-3

Drug 2

Output 1-4

Drug 3

Output 2-3

Drug 4

Output 2-4

Output 3-4

j

j

i

i

=

- 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.

- The average number of key-value pairs created by each mapper is the replication rate.
- Denoted r.

- Represents the communication cost per input.

- 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.

- 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.

- A mapping schema for a problem and a reducer size q is an assignment of inputs to sets of reducers, with two conditions:
- No reducer is assigned more than q inputs.
- For every output, there is some reducer that receives all of the inputs associated with that output.
- Say the reducer covers the output.

- 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.

- 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.

- 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.

- 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.

j = 0

1

2

3

4

i = 0

1

2

3

4

Team 0

j = 0

1

2

3

4

i = 0

1

2

3

4

Team 1

j = 0

1

2

3

4

i = 0

1

2

3

4

Team 2

j = 0

1

2

3

4

i = 0

1

2

3

4

Team 3

j = 0

1

2

3

4

i = 0

1

2

3

4

Team 4

j = 0

1

2

3

4

i = 0

1

2

3

4

Team 5

- 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’.

- 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

- 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.

- 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.

- 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.

- 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.

- 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

Generalized Splitting

One reducer

for each output

b

Splitting

All inputs

to one

reducer

r = replication

rate

2

1

2b/2

2b

21

q = reducer

size

Matrix Multiplication

One-Job MethodTwo-Job MethodComparison

- 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 = jAij Bjk.
- Output Cikdepends on the ith row of A, that is, Aijfor all j, and the kth column of B, that is, Bjkfor all j.

Column k

Row i

=

A

B

C

- 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 Cigand 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.

- 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.

- Total outputs = n2.
- One reducer can cover at most q2/4n2outputs.
- 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.

- 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.

n/g

n/g

=

- 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.

K

K

J

J

I

I

A

B

C

For i in I and k in K, contribution

is j in J Aij × Bjk

- 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.

- 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.

K

K

J

n/2g

J

I

n/g

n/g

I

n/g

n/g

n/2g

A

B

C

2g reducers contribute to

this area, one for each J.

- 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.

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

- 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.

- 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

- 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.

- 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.

- 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.

- Ugly fact: For HD=1, you cannot have a large reducer with all pairs at distance 1; for HD=2, it is possible.

- 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?