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Query Evaluation: Join ,External SOrting

Query Evaluation: Join ,External SOrting. General Overview. Relational model - SQL Formal & commercial query languages Functional Dependencies Normalization Physical Design Indexing Query Processing and Optimization. Review: QP & O. SQL Query. Query Processor. Parser.

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Query Evaluation: Join ,External SOrting

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  1. Query Evaluation: Join ,External SOrting

  2. General Overview • Relational model - SQL • Formal & commercial query languages • Functional Dependencies • Normalization • Physical Design • Indexing • Query Processing and Optimization

  3. Review: QP & O SQL Query Query Processor Parser Query Optimizer Algebraic Expression Execution plan Evaluator Data: result of the query

  4. Review: QP & O Query Optimizer Algebraic Representation Query Rewriter Algebraic Representation Data Stats Plan Generator Query Execution Plan

  5. Review-Plan Generation Metadata: DBMS maintains statistics about each relation, attribute and index. Plan generation: • Generate many alternative plans • We saw many for selections, joins • Estimate cost for each and choose best Plans examined: Selection (exact match): Linear, binary, PI, SI Range: PI, SI Joins: NLJ, BNLJ, INLJ, SMJ, HJ

  6. Review-Plan Generation Depends upon a cost model For any query, must know its estimated cardinality its estimated cost (in # of I/Os) E.g.: A = K (R )  cardinality SC(A, R)  cost: depends on the plan, attribute: Linear Scan bR /2 bR Binary Search log2(bR) log2(bR)+sc(A, R)/fR -1 PI Scan HTi +1 HTi +sc(A, R) / fR

  7. Cost Model How do we predict the cost of a plan? Ans: Cost model • For each plan operator and each algorithm we have a cost formula • Inputs to formulas depend on relations, attributes • Database maintains statistics about relations for this (Metadata)

  8. Metadata • Given a relation r, DBMS likely maintains the following metadata: • Size (# of tuples) nr • Size (# of blocks) br • Block size (#tuples) fr (typicallybr =nr / fr ) • Tuple size (in bytes) sr • Attribute Variance (for each attribute r, # of different values) V(att, r) • Selection Cardinality (for each attribute in r, expected size of a selection: att = K (r ) ) SC(att, r)

  9. Cardinality of Joins in General Assume join: R S • If R, S have no common attributes: nr*ns • If R,S have attribute A in common: (take min) • If R, S have attribute A in common and: • A is a candidate key for R: ≤ ns • A is candidate key in R and candidate key in S : ≤ min(nr, ns) • A is a key for R, foreign key for S: = ns

  10. Nested-Loop Join Query: R S Algorithm 1: Nested Loop Join Idea: t1 u1 Blocks of... t2 u2 t3 u3 R S results Compare: (t1, u1), (t1, u2), (t1, u3) ..... Then: GET NEXT BLOCK OF S Repeat: for EVERY tuple of R

  11. Join Algorithms Query: R S Algorithm 2: Block Nested Loop Join Idea: t1 u1 Blocks of... t2 u2 t3 u3 R S results Compare: (t1, u1), (t1, u2), (t1, u3) (t2, u1), (t2, u2), (t2, u3) (t3, u1), (t3, u2), (t3, u3) Then: GET NEXT BLOCK OF S Repeat: for EVERY BLOCK of R

  12. Block Nested-Loop Join • Block Nested Loop Join for each block BRofR dofor each block BSof S do for each tuple trin BR do for each tuple usin Bsdo beginCheck if (tr,us) satisfy the join condition if they do (“match”), add tr• usto the result.

  13. Block Nested-Loop Join (Cont.) Cost: • Worst case estimate: br bs + br block accesses. • Improvements to nested loop and block nested loop algorithms for a buffer with M blocks: • In block nested-loop, use M — 2 disk blocks as blocking unit for outer relations, where M = memory size in blocks; use remaining two blocks to buffer inner relation and output • Cost = br / (M-2)  bs + br • If equi-join attribute forms a key on inner relation, stop inner loop on first match • Scan inner loop forward and backward alternately, to make use of the blocks remaining in buffer (with LRU replacement)

  14. Join Algorithms Query: R S Algorithm 3: Indexed Nested Loop Join Idea: t1 Blocks of... t2 t3 R S results (fill w/ blocks of S or index blocks) For each tuple ti of R if ti.A = K (A is the attribute R,S have in common) then use the index to compute att = K (S ) Demands: index on A for S

  15. Indexed Nested-Loop Join Indexed Nested Loop Join • For each tuple tRin the outer relation R, use the index to look up tuples in S that satisfy the join condition with tuple tR. • Worst case: buffer has space for only one page of R, and, for each tuple in R, we perform an index lookup on s. • Cost of the join: br + nr c • Where c is the cost of traversing the index and fetching all matching s tuples for one tuple from r • c can be estimated as cost of a single selection on s using the join condition. • If indices are available on join attributes of both R and S,use the relation with fewer tuples as the outer relation.

  16. Example of Nested-Loop Join Costs Query: depositor customer (cname, acct_no) (cname, ccity, cstreet) Metadata: customer: ncustomer = 10,000 fcustomer = 25 bcustomer = 400 depositor: ndepositor = 5000 fdepositor = 50 bdepositor = 100 V (cname, depositor) = 2500 i a primary index on cname (dense) for customer (fi = 20) Minimal buffer

  17. Plan generation for Joins Alternative 1: Block Nested Loop 1a: customer = OUTER relation depositor = INNER relation cost: bcustomer + bdepositor * bcustomer = 400 +(400 *100) = 40,400 1b: customer = INNER relation depositor = OUTER relation cost: bdepositor + bdepositor * bcustomer = 100 +(400 *100) = 40,100

  18. Plan generation for Joins Alternative 2: Indexed Nested Loop We have index on cname for customer. Depositor is the outer relation Cost: bdepositor + ndepositor * c = 100 +(5000 *c ) , c is the cost of evaluating a selection cname=K using index. What is c? Primary index on cname, cname a key for customer c = HTi +1

  19. Plan generation for Joins What is HTi ? cname a key for customer. V(cname, customer) = 10,000 fi = 20, i is dense LBi =  10,000/20 = 500 HTi ~ logfi(LBi) + 1 = log20 500 + 1 = 4 Cost of index nested loop is: = 100 + (5000 * (4+1)) = 25,100 BA (cheaper than NLJ)

  20. pR pS Another Join Strategy Query: R S Algorithm: Merge Join Idea: suppose R, S are both sorted on A (A is the common attribute) A A 2 2 3 5 1 2 3 4 ... ... Compare: (1, 2) advance pR (2, 2) match, advance pS  add to result (2, 2) match, advance pS  add to result (2, 3) advance pR (3, 3) match, advance pS add to result (3, 5) advance pR (4, 5) read next block of R

  21. Merge-Join GIVEN R, S both sorted on A • Initialization • Reserve blocks of R, S into buffer reserving one block for result • Pr= 1, Ps =1 • Join (assuming no duplicate values on A in R) WHILE !EOF( R) && !EOF(S) DO if BR[Pr].A == BS[Ps].A then output to result; Ps++ else if BR[Pr].A < BS[Ps].A then Pr++ else (same for Ps) if Pr or Ps point past end of block, read next block and set Pr(Ps) to 1

  22. Merge-Join (Cont.) • Each block needs to be read only once (assuming all tuples for any given value of the join attributes fit in memory) • Thus number of block accesses for merge-join is bR + bS • But.... What if one/both of R,S not sorted on A? Ans: May be worth sorting first and then perform merge join (Sort-Merge Join) Cost: bR + bS + sortR + sortS

  23. External Sorting Not the same as internal sorting Internal sorting:  minimize CPU (count comparisons)  best: quicksort, mergesort, .... External sorting:  minimize disk accesses (what we ‘re sorting doesn’t fit in memory!)  best: external merge sort WHEN used? 1) SORT-MERGE join 2) ORDER BY queries 3) SELECT DISTINCT (duplicate elimination)

  24. d e g m p r 31 16 24 3 2 16 External Sorting Idea: 1. Sort fragments of file in memory using internal sort (runs). Store runs on disk. 2. Merge runs. E.g.: a b c 19 14 33 a d g 19 31 24 sort a a b c d d d e g m p r g a d c b e r d m p d a 14 19 14 33 7 21 31 16 24 3 2 16 24 19 31 33 14 16 16 21 3 2 7 14 merge sort b c e 14 33 16 merge sort a d d 14 7 21 d m r 21 3 16 sort merge a d p 14 7 2

  25. External Sorting (cont.) Algorithm Let M = size of buffer (in blocks) 1. Sort runs of size M blocks each (except for last) and store. Use internal sort on each run. 2. Merge M-1 runs at a time into 1 and store. Merge for all runs. 3. if step 2 results in more than 1 run, goto step 2. Run m-1 Output Run 1 Run 2 ........ Run 3

  26. External Sorting (cont.) Cost: 2 bR * (logM-1(bR / M) + 1) Intuition: Step 1: create runs  every block read and written once  cost 2 bR I/Os Step 2: Merge  every merge iteration requires reading and writing entire file (2 bR I/Os) Total: logM-1(bR / M) Iteration # --------------- 1 2 3 ..... Runs Left to Merge ----------------------------

  27. What if we need to sort? Query: depositor customer Merge-sort Join Sorting depositor: bdepositor = 100 Sort depositor = 2 * 100 * (log2(100 / 3) + 1) = 1400 Similarly, for customer we get 7200 I/Os. Total: 100 + 400 + 1400 + 7200 = 9100 I/O’s! Still beats BNLJ (40K), INLJ (25K) Why not use SMJ always? Ans: 1) Sometimes inner relation can fit in memory 2) Sometimes index is small 3) SMJ only work for natural joins, “equijoins”

  28. Hybrid Merge Join • hybrid merge-join: If one relation is sorted, and the other has a secondary B+-tree index on the join attribute • Merge the sorted relation with the leaf entries of the B+-tree . • Sort the result on the addresses of the unsorted relation’s tuples • Scan the unsorted relation in physical address order and merge with previous result, to replace addresses by the actual tuples • Sequential scan more efficient than random lookup

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