Theta join (M-bucket-I algorithm explained)

Processing Theta Joins using MapReduce

by Minsub Yim

Processing pipeline at a reducer

Goal: We want to minimize job completion time. Since it’s a function of both input and output, we need a way to model both inputs and outputs to a reducer.

Reducer Join OutputMapper Output

time = f(input size) time = f(output size)

Receive Mapper Output

Sort input by key

Read input

Run join algorithm

Send join output

Theta Join Model

S_id Value

1 5

2 6

3 6

4 8

5 8

6 10

Dataset S Dataset TT_id Value

1 5

2 5

3 6

4 8

5 8

6 10

Assuming join condition: S.value = T.value

Theta Join Model

S_id Value

1 5

2 6

3 6

4 8

5 8

6 10

Dataset S Dataset TT_id Value

1 5

2 5

3 6

4 8

5 8

6 10

Assuming join condition: S.value = T.value

5 5 6 8 8 105668810

[ Join Matrix M ]

: tuple satisfying the join condition

ST

Theta Join Model (Examples)

5 5 6 8 8 1056688

10

Join condition: S.value <= T.value

ST 5 5 6 8 8 10

5668810

Join condition: abs (S.value - T.value) < 2

ST 5 5 6 8 8 10

5668810

Join condition: S.value = T.value

ST

Goal Revisited

• We want to minimize job completion time

• We need to assign every true cell to exactly one reducer. (find a mapping from M to R)

Goal Revisited

• We want to minimize job completion time

• We need to assign every true cell to exactly one reducer. (find a mapping from M to R)

• Goal: Find a mapping from the join matrix M to reducers that minimizes job completion time

Mappings from join matrix to reducers

5 5 6 8 8 1056688

10


ST

(1)

(2)

(3)

(4)

[R1] Input: S1, T1, T2 Output: 2 tuples ![R2] Input: S2, S3, T3 Output: 2 tuples ![R3] Input: S4, S5, T4, T5 Output: 4 tuples ![R4] Input: S6, T6 Output: 1 tuple !Max-Reducer-Input: 4 Max-Reducer-Output: 4

5 5 6 8 8 105668810


ST

(1)(2)

(3)(4)

[R1] Input: S1, S4, S5, T1, T4, T5 Output: 3 tuples ![R2] Input: S2, S4, T3,T5 Output: 2 tuples ![R3] Input: S1, S5, T2, T4 Output: 2 tuples ![R4] Input: S3, S6, T3, T6 Output: 2 tuples !MRI: 6 MRO: 3

(1)

(1)

(2)

(3)

(4)

Stndard equi-join algorithm Random


5 5 6 8 8 1056688

10


ST

(1)(2)

(3)

[R1] Input: S1, S2, T1, T2 Output: 2 tuples ![R2] Input: S3, S4, T1, T2, T3 Output: 2 tuples ![R3] Input: S4, S5, S6, T4, T5, T6 Output: 5 tuples !!Max-Reducer-Input: 6 Max-Reducer-Output: 5


• We see there could be many possible mappings from join matrix to reducers

• We will see in different cases, which mapping is (close to) optimal and algorithms to compute such mapping.

LemmaWe will be using the following lemma repeatedly to show how (close to) optimal each mapping is.

[ LEMMA 1 ] A reducer that is assigned to c cells of the join matrix M will receive at least input tuples

[ Proof ] Consider a reducer r that receives m records from T and n records from S. Then,

!!

2pc

mn � c2pmn � 2

pc

m+ n � 2pc

Cross Product• We first consider cross product, where all of

tuples from two datasets satisfy the join condition. The join matrix would look like the following:

5 5 6 8 8 105668810

Join condition: S X T

ST

Cross Product• Since all entries of the join matrix are true, we

can see that the maximum-reducer-output (MRO) . (Otherwise, there would be tuples not mapped to a reducer.)

• Along with Lemma 1, we have a lower bound for the maximum-reducer-input (MRI):

MRI

� |S||T |/r

� 2

r|S||T |

r

[ LEMMA 1 ] A reducer that is assigned to c cells of the join matrix M will receive at least input tuples 2

pc

Cross Product• Since all entries of the join matrix are true, we

can see that the maximum-reducer-output (MRO) . (Otherwise, there would be tuples not mapped to a reducer.

• Along with Lemma 1, we have a lower bound for the maximum-reducer-input (MRI):

MRI

� |S||T |/r

� 2

r|S||T |

r


pc

Cross Product• We will revisit these two properties frequently to

see the quality of join mappings:

� |S||T |/rMRO and MRI � 2

r|S||T |

r

p|S||T |/rCase 1: Suppose |S| and |T| are multiples of .

Namely, and .|S| = csp|S||T |/r |T | = cT

p|S||T |/r

Then, partitioning the join matrix with squares of size is an optimal mapping.p

|S||T |/r

Proof : is trivial. Each region mapped to a reducer !has output size: and input size: |S||T |/r 2

r|S||T |

r

Cross Product� |S||T |/rMRO and MRI � 2

r|S||T |

r

Properties


r|S||T |

r

Properties

5 5 6 8 8 105668810

ST

Suppose |S| = |T| = 6 and r = 9


r|S||T |

r

Properties

5 5 6 8 8 105668810

ST

Suppose |S| = |T| = 6 and r = 9

MRO = 4 = 2

r|S||T |

r

MRI = 4 = |S||T |/r

Case 2: Suppose the cardinality of one dataset is significantly greater than that of the other. (WLOG, assume ). Then, rectangle cover


r|S||T |

r

Properties

|S| < |T |/r |S|⇥ |T |/ris the optimal mapping.

Case 2: Suppose the cardinality of one dataset is significantly greater than that of the other. (WLOG, assume ). Then, rectangle cover


r|S||T |

r

Properties

|S| < |T |/r |S|⇥ |T |/ris the optimal mapping.

(e.g., |S| = 3, |T| = 20, r = 5)

Case 3: The remaining case where . !

Let , !

Then, covering M with squares is a mapping worse than an optimal mapping by a factor no greater than 4.


r|S||T |

r

Properties

|T |/r |S| |T |

CT =

$|T |/

r|S||T |

r

%CS =

$|S|/

r|S||T |

r

%

p|S||T |/r ⇥

p|S||T |/r

If |S| and/or |T| is not a multiple of , scale each !

side by and/or respectively to !

cover M. Given , we see that


r|S||T |

r

Properties

p|S||T |/r

✓1 +

1

CS

◆ ✓1 +

1

CT

◆

|T |/r |S| |T |✓1 +

1

CS

◆r|S||T |

r 2

r|S||T |

r

Hence, and


r|S||T |

r

Properties

Comparing these with the lower bounds given above, we see that the MRO and MRI produced by this mapping are at most 4 times (twice for MRI) the lower bounds.

MRI 4p

|S||T |/rMRO 4|S||T |/r

Implementation• Now we know how to (nearly) optimally partition

the join matrix. So let’s run it!!

• However, when a reducer is given a record (either from S or T), it does NOT have enough information where exactly in the dataset (in which row/col) the record belongs to.

• We could run another pre-process to get that info, but it can be avoided by running a randomized algorithm!

Mapping & Randomized Algorithm

Algorithm 1 : Map (Theta - Join) !Input : input tuple 1: if then 2: matrixRow = random(1,|S|) 3: for all regionID in lookup.getRegions(matrixRow) do 4: Output (regionID, (x, “S”) ) 5: else 6: matrixCol = random (1,|T|) 7: for all regionID in lookup.getRegions(matrixCol) do 8: Output (regionID, (x, “T”) )

x 2 S [ T

x 2 S


Algorithm 1 : Map (Theta - Join) !Input : input tuple 1: if then 2: matrixRow = random(1,|S|) 3: for all regionID in lookup.getRegions(matrixRow) do 4: Output (regionID, (x, “S”) ) 5: else 6: matrixCol = random (1,|T|) 7: for all regionID in lookup.getRegions(matrixCol) do 8: Output (regionID, (x, “T”) )

x 2 S [ T

x 2 S

1. Given a record ( WLOG ) 2. Get a row uniformly randomly 3. Get all the regions intersecting that row and output ( regID, (x, S) )

x 2 S


5 7 7 7 8 9577899

ST


(1) (2)

(3)

3 5 1 5 1 2

6 2 2 3 6 4

(1,S1) (2,S1) (3,S2) (1,S3) (2,S3) (3,S4) (1,S5) (2,S5) (1,S6) (2,S6) (2,T1) (3,T1) (1,T2) (3,T2) (1,T3) (3,T3) (1,T4) (3,T4) (2,T5) (3,T5) (2,T6) (3,T6)

Input Tuple

Random Row/Col Output

MapReducer 1 : key 1 (regID)Input: S1, S3, S5, S6, T2, T3, T4Output: (S3,T2) (S3,T3) (S3,T4)

Reducer 2 : key 2 (regID)Input: S1, S3, S5, S6, T1, T5, T6Output: (S1,T1) (S5,T6) (S6,T6)

Reducer 3 : key 3 (regID)Input: S2, S4, T1, T2, T3, T4, T5, T6Output: (S2,T2) (S2,T3) (S2,T4) (S4,T5)

Reduce

S1.A = 5 S2.A = 7 S3.A = 7 S4.A = 8 S5.A = 9 S6.A = 9 T1.A = 5 T2.A = 7 T3.A = 7 T4.A = 7 T5.A = 8 T6.A = 9

Cross Product… NOT!

• We have verified that 1 Bucket Theta algorithm is close to optimal when the join condition is cross product.

• How does 1 Bucket Theta algorithm perform when join condition is NOT cross product ?

• We will compare the quality of 1 Bucket Theta algorithm to any join algorithm

1BT vs ANY join algorithmLet . Any matrix to reducer mapping that has to cover at least of the cells of the join matrix, by Lemma 1, has MRI

1 � x > 0

x|S||T | |S||T |� 2

px|S||T |


pc

As we have seen, 1BT guarantees that MRI . !Hence,

4p|S||T |

MRI1BT

MRI

AnyJoinAlg

=4p

|S||T |/r2p

x|S||T |/r=

2px

1BT vs ANY join algorithm

1BT vs ANY join algorithm

When , the ratio < 3. !Hence,compared to ANY join algorithm that assigns more than 50% of its matrix cells to reducers, the MRI for 1BT is at most 3 times the MRI of that algorithm.

x = 0.5

M-Bucket-I• In the previous slide, we see that instead of

covering the entire matrix, mapping smaller regions would yield better MRI result.

• Ideally, we only want to map those satisfying the join condition, but it cannot be done before knowing input statistics and/or join condition.

• M-Bucket-I exploits statistics to improve over 1 Bucket Theta join algorithm

M-Bucket-I[ Step 1 ] Approximate Equi-Depth Histograms

1) With probability n /|S|, sample approx. n records from |S|

2) Build k-quantiles (k buckets), where k < n 3) Iterate through |S| and count the number of

records in each bucket 4) Do the same for |T| and build the join matrix

accordingly


S_id Value

1 7

2 2

3 4

4 2

5 1

6 9

7 10

8 2

9 5

10 3

Dataset S Dataset T

T_id Value

1 5

2 5

3 6

4 8

5 8

6 10

7 2

8 4

9 1

10 3


S_id Value

1 7

2 2

3 4

4 2

5 1

6 9

7 10

8 2

9 5

10 3

Dataset S Dataset T

T_id Value

1 5

2 5

3 6

4 8

5 8

6 10

7 2

8 4

9 1

10 3

Sample S 7, 2, 2, 9, 2, 3

Sample T 5, 6, 8, 2, 1, 3


S_id Value

1 7

2 2

3 4

4 2

5 1

6 9

7 10

8 2

9 5

10 3

Dataset S Dataset T

T_id Value

1 5

2 5

3 6

4 8

5 8

6 10

7 2

8 4

9 1

10 3

Sample S 7, 2, 2, 9, 2, 3

Sample T 5, 6, 8, 2, 1, 3

Samples


S_id Value

1 7

2 2

3 4

4 2

5 1

6 9

7 10

8 2

9 5

10 3

Dataset S Dataset T

T_id Value

1 5

2 5

3 6

4 8

5 8

6 10

7 2

8 4

9 1

10 3

Sample S 7, 2, 2, 9, 2, 3

Sample T 5, 6, 8, 2, 1, 3

Samples

Buckets

S

T

0 2 3 9

0 1 5 8 1

1


S_id Value

1 7

2 2

3 4

4 2

5 1

6 9

7 10

8 2

9 5

10 3

Dataset S Dataset T

T_id Value

1 5

2 5

3 6

4 8

5 8

6 10

7 2

8 4

9 1

10 3

Sample S 7, 2, 2, 9, 2, 3

Sample T 5, 6, 8, 2, 1, 3

Samples

Buckets

S

T

0 2 3 9

0 1 5 8 1

1

4 1 4 1

1 5 3 1


S S S S S S S S S STTTTTTTTTT




2 3 9

1

5

8




2 3 9

1

5

8


We now have candidate cells. How do we map these cells to reducers?

M-Bucket-I[ Step 2 ] M-Bucket-I Algorithm

Algorithm : M-Bucket-I !Input : maxInput, r, M 1: row = 0 2: while row < M.noOfRows do 3: (row,r) = CoverSubMatrix(row, maxInput, r, M) 4: if r < 0 then!5: return false 6: return true!

M-Bucket-I

Algorithm : CoverSubMatrix !Input : row_s, maxInput, r, M 1: maxScore = -1, rUsed = 0 2: for i = 1 to maxInput-1 do 3: R_i = CoverRows(row_s, row_s + i, maxInput, M) 4: area = totalCandidateArea(row_s, row_s + i, M) 5: score = area/R_i.size 6: if score >= maxScore then!7: bestRow = row_s + i 8: rUsed = R_i.size 9: r = r - rUsed 10: return (bestRow + 1, r)

[ Step 2 ] M-Bucket-I Algorithm

M-Bucket-I

Algorithm : CoverRows !Input : row_f, row_l, maxInput, M 1: Regions = 0; r = newRegion() 2: for all c_i in M.getColumns do 3: if r. cap < c_i.candidateInputCosts then!4: Regions = Regions U r 5: r = newRegion() 6: r.Cells = r.Cells U c_i.candidateCells 7: return Regions


M-Bucket-I

Run the algorithm with r = 6 maxInput = 5


M-Bucket-I


row : 0 cost : 4


M-Bucket-I


row : 0 cost : 4

row : 1 cost : 13/3 = 4.3


M-Bucket-I


row : 0 cost : 4

row : 1 cost : 13/3 = 4.3

row : 2 cost : 22/4 = 5.5


M-Bucket-I


row : 0 cost : 4

row : 1 cost : 13/3 = 4.3

row : 2 cost : 22/4 = 5.5

row : 3 cost : 31/7 = 4.428..


M-Bucket-I


row : 0 cost : 4

row : 1 cost : 13/3 = 4.3

row : 2 cost : 22/4 = 5.5

row : 3 cost : 31/7 = 4.428..

We choose the mapping with highest score!

(1) (2)(3) (4)


M-Bucket-I


row : 3 cost : 3

(1) (2)(3) (4) So on and so forth…


M-Bucket-I


Final mapping!


(1) (2)(3) (4)

(7)(6)(5)

(8) (9)(10)

(11) (12)(13)

M-Bucket-I


(1) (2)(3) (4)

However, we have mapped the candidate cells to > r reducers. !We do binary search until we get to the point where we a mapping to <= r reducers.(7)(6)(5)

(8) (9)(10)

(11) (12)(13)


M-Bucket-I[ Step 3 ] Binary Search

MaxInput = |S|+|T| = 20

Num.Reducers = 1

MaxInput = 5

Num.Reducers = 13



Num.Reducers = 1

MaxInput = 5

Num.Reducers = 13

MaxInput = 12

Num.Reducers = 3



Num.Reducers = 1

MaxInput = 5

Num.Reducers = 13

MaxInput = 12

Num.Reducers = 3

MaxInput = 8

Num.Reducers = 5

Since 7 reducers are required when MaxInput = 7, we stop the binary search here and output the mapping with MRI = 8.

Performance1 Bucket Theta Standard Equi Join

Data set Output size (billion)

Output Imbalance

Runtime (secs)

Output Imbalance

Runtime (secs)

Synth - 0 25.00 1.0030 657 1.0124 701

Synth - 0.4 24.99 1.0023 650 1.2541 722

Synth - 0.6 24.98 1.0033 676 1.7780 923

Synth - 0.8 24.95 1.0068 678 3.0103 1482

Synth - 1 24.91 1.0089 667 5.3124 2489

Skew

ed

Where Output Imbalance = MRI

Ave.RI

MRI

Ave.RI

Skew Resistance of 1 Bucket Theta

Performance

Step Number of Buckets

1 10 100 1000 10,000 100,000 1,000,000

M-Bucket-I cost details (seconds)

Quantiles 0 115 120 117 122 124 122

Histogram 0 140 145 147 157 167 604

Heuristic 74.01 9.21 0.84 1.50 16.67 118.03 111.27

Join 49384 10905 1157 595 548 540 536

Total 49,458.01 11,169.21 1,422.84 860.5 843.67 949.03 1,373.27

Theta join (M-bucket-I algorithm explained)

Engineering

Transcript of Theta join (M-bucket-I algorithm explained)