1. Jeffrey D. Ullman Stanford University

2. 2  Communication cost for a MapReduce job = the total number of key-value pairs generated by all the mappers.  For many applications, communication cost is the dominant cost, taking more time than the computation performed by the mappers and reducers.  Focus: minimizing communication cost for a join of three or more relations.

3. 3  Given: a collection of relations, each with attributes labeling their columns.  Find: Those tuples over all the attributes such that when restricted to the attributes of any relation R, that tuple is in R.

4. 4 A B 0 1 1 2 A C 1 3 2 1 B C 1 2 2 3 The join: A B C 1 2 3

5. 5  Suppose we want to compute R(A,B) JOIN S(B,C), using k reducers.  Remember: a “reducer” corresponds to a key of the intermediate file, so we are really asking how many keys we want to use.  R and S are each stored in a chunked file of a distributed file system.  Subtle point: when joining two relations, each tuple is sent to one reducer, so k can be as large as we like; in a multiway join, large k implies more communication.

6. 6  Use a hash function h from B-values to k buckets.  Earlier, we used h = identity function.  Map tasks take chunks from R and S, and send:  Tuple R(a,b) to reducer h(b).  Tuple S(b,c) to reducer h(b).  If R(a,b) joins with S(b,c), then both tuples are sent to reducer h(b).  Thus, their join (a,b,c) will be produced there and shipped to the output file.

7. 7  Consider a chain of three relations: R(A, B) JOIN S(B, C) JOIN T(C,D)  Example: R, S, and T are “friends” relations.  We could join any two by the 2-way MapReduce algorithm, then join the third with the resulting relation.  But intermediate joins are large.

8. 8  An alternative is to divide the work among k = m 2 reducers.  Hash both B and C to m values.  A reducer (key) corresponds to a hashed B- value and a hashed C-value.

9. 9  Each S-tuple S(b,c) is sent to one reducer: (h(b), h(c)).  But each tuple R(a,b) must be sent to m reducers, those whose keys are of the form (h(b), x).  And each tuple T(c,d) must be sent to m reducers of the form (y, h(c)).

10. 10 h(b)=0 h(b)=1 h(b)=2 h(b)=3 h(c) = 0 1 2 3 S(b, c) R(a, b) T(c, d)

11. 11  Thus, any joining tuples R(a,b), S(b,c), and T(c,d) will be joined at the reducer (h(b), h(c)).  Communication cost: s + mr + mt.  Convention: Lower-case letter is the size of the relation whose name is the corresponding upper- case letter.  Example: r is the size of R.

12. 12  Suppose for simplicity that:  Relations R, S, and T have the same size r.  The probability of two tuples joining is p.  The 3-way join has communication cost r(2m+1).  Two two-way joins have a communication cost:  3r to read the relations, plus  pr 2 to read the join of the first two.  Total = r(3+pr).

13. 13  3-way beats 2-way if 2m + 1 < 3 + pr.  pr is the multiplicity of each join.  Thus, the 3-way chain-join is useful when the multiplicity is high.  Example: relations are “friends”; pr is about 300. m 2 = k can be 20,000.  Example: relations are Web links; pr is about 15. m 2 = k can be 64.

14. 14 Share Variables and Their Optimization Special Case: Star Joins Special Case: Chain Joins Application: Skew Joins

15. 15  When we discussed the 3-way chain-join R(A, B) JOIN S(B, C) JOIN T(C,D), we used attributes B and C for the map-key (attributes that determined the reducers).  Why not include A and/or D?  Why use the same number of buckets for B and C?

16. 16  For the general problem, we use a share variable for each attribute.  The number of buckets into which values of that attribute are hashed.  Convention: The share variable for an attribute is the corresponding lower-case letter.  Example: the share variable for attribute A is always a.

17. 17  The product of all the share variables is k, the number of reducers.  The communication cost of a multiway join is the sum of the size of each relation times the product of the share variables for the attributes that do not appear in the schema of that relation.

18. 18  Consider the cyclic join R(A, B) JOIN S(B, C) JOIN T(A, C)  Cost function is rc + sa + tb.  Construct the Lagrangean remembering abc = k: rc + sa + tb – (abc – k)  Take the derivative with respect to each share variable, then multiply by that variable.  Result is 0 at minimum.

19. 19  d/da of rc + sa + tb – (abc – k) is s – bc.  Multiply by a and set to 0: sa – abc = 0.  Note: abc = k : sa = k.  Similarly, d/db and d/dc give: sa = tb = rc = k.  Solution: a = (krt/s 2 ) 1/3 ; b = (krs/t 2 ) 1/3 ; c = (kst/r 2 ) 1/3.  Cost = rc + sa + tb = 3(krst) 1/3.

20. 20  Certain attributes can’t be in the map-key.  A dominates B if every relation of the join with B also has A.  Example: R(A,B,C) JOIN S(A,B,D) JOIN T(A,E) JOIN U(C,E) Every relation with B Also has A

21. 21  Cost expression: rde + sce + tbcd + uabd  Since b appears wherever a does, if there were a minimum-cost solution with b > 1, we could replace b by 1 and a by ab, and the cost would lower. R(A,B,C) JOIN S(A,B,D) JOIN T(A,E) JOIN U(C,E)

22. 22  This rule explains why, in the discussion of the chain join R(A, B) JOIN S(B, C) JOIN T(C,D) we did not give dominated attributes A and D a share.

23. 23  Unfortunately, there are more complex cases than dominated attributes, where the equations derived from the Lagrangean imply a positive sum of several terms = 0.  We can fix, generalizing dominated attributes, but we have to branch on which attribute needs to be eliminated from the map-key.

24. 24  Solutions not in integers:  Drop an attribute with a share < 1 from the map-key and re-solve.  Round other nonintegers, and treat k as a suggestion, since the product of the integers may not be k.

25. 25  A star join combines a large fact table F(A 1,A 2,…,A n ) with tiny dimension tables D 1 (A 1,B 1 ), D 2 (A 2,B 2 ),…, D n (A n,B n ).  There may be other attributes not shown, each belonging to only one relation.  Example: Facts = sales; dimensions tell about buyer, product, day, time, etc.

26. 26 A1A1 A2A2 A3A3 A4A4 B1B1 B2B2 B3B3 B4B4

27. 27  Map-key = the A’s.  B’s are dominated.  Solution: d i /a i = k for all i.  That is, the shares are proportional to the dimension-table sizes.

28. 28  Fact/dimension tables are often used for analytics.  Aster Data approach: partition fact table among nodes permanently (shard the fact table); replicate needed pieces of dimension tables. Fact Shard 1 Fact Shard 2 Fact Shard k...

29.  Shard fact table by country.  Wins for distributing the Customer dimension table.  Each Customer tuple needed only at the shard for the country of the customer.  Loses big for other dimension tables, e.g., Item.  Pretty much every item has been bought by someone in each country, so Dimension tuples are replicated to every shard. 29

30. 30  Our solution lets you partition the fact table to k shards, one for each reducer.  Analogy: Think of sharding process as taking the join of the fact table and all dimension tables.  Shards = reducers.  Only dimension keys (the A’s) get shares, so each Fact tuple goes to exactly one shard/reducer.  Total copies of dimension tuples = communication cost, which is minimized.

31. 31  A chain join has the form R(A 0, A 1 ) JOIN R(A 1, A 2 ) JOIN … JOIN R(A n -1, A n )  Other attributes may appear, but only in one relation.  A 0 and A n are dominated; other attributes are in the map-key. AnAn A2A2 A3A3 A n -1 A1A1 A0A0...

32. 32  Illustrates strange behavior.  Even and odd n have very different distributions of the share variables.  Even n : a 2 = a 4 = … = a n -2 = 1; a 1 = a 3 = … = a n -1 = k 2/n

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34. 34  Even a’s grow exponentially.  That is, a 4 = a 2 2 ; a 6 = a 2 3 ; a 8 = a 2 4,…  The odd a’s form the inverse sequence.  That is, a 1 = a n-1 ; a 3 = a n-3 ; a 5 = a n-5 ;…

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36.  Suppose we want to join R(A,B) with S(B,C) using MapReduce, as in the introduction.  But half the tuples of R and S have value B=10.  No matter how many reducers we use, one reducer gets half the tuples and dominates the wall-clock time.  To deal with this problem, systems like PIG handle heavy-hitter values like B=10 specially. 36

37.  Pick one of the relations, say R.  Divide the tuples of R having B=10 into k groups, with a reducer for each.  Note: we’re pretending B=10 is the only heavy hitter.  Works for any value and > 1 heavy hitter.  Send each tuple of S with B=10 to all k reducers.  Other tuples of R and S are treated normally. 37

38.  Let R have r tuples with B=10 and S have s tuples with B=10.  Then communication cost = r + ks.  Can be minimum if s << r, but also might not be best. 38

39.  Let’s partition the tuples of R with B=10 into x groups and partition the tuples of S with B=10 into y groups, where xy = k.  Use one of k reducers for each pair (i, j) of a group for R and a group for S.  Send each tuple of R with B=10 to all y groups of the form (i, ?), and send each tuple of S with B=10 to all x reducers (?, j).  Communication = ry + sx. 39

40.  We need to minimize ry + sx, under the constraint xy = k.  Solution: x =  kr/s; y =  ks/r; communication = 2  krs.  Note: 2  krs is always < r + ks (the cost of the implemented skew join), often much less.  Example: if r = s, then new skew join takes 2r  k, while old takes r(k+1). 40

41. 41 1. Multiway joins can be computed by replicating tuples and distributing them to many compute nodes. 2. Minimizing communication requires us to solve a nonlinear optimization. 3. Multiway beats 2-way joins for star queries and queries on high-fanout graphs. 4. Exact solution for chain and star queries. 5. Simple application to optimal skew joins.

42. 1. We really don’t know the complexity of optimizing the multiway join in general.  The algorithm we offered in 2010 EDBT is exponential in the worst case.  But is it NP-hard? 2. The Skew-join technique borrows from the multiway join, but is a 2-way join.  Can you generalize the idea to all multiway joins with skew? 42