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Equality Join R X R.A=S.B S : : Relation R M PagesN Pages Relation S Pr records per page Ps records per page

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Simple Nested Loops Join For each tuple r in R do for each tuple s in S do if r.A == S.B then add to result Scan the outer relation R. For each tuple r in R, scan the entire inner relation S. Ignore CPU cost and cost for writing result to disk R has M pages with P R tuples per page. S has N pages with P S tuples per page. Nested loop join cost = M+M*P R *N. Cost to scan R. Memory Output 1 page for S 1 page for R

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Exercise What is the I/O cost for using a simple nested loops join? –Sailors: 1000 pages, 100 records/page, 50 bytes/record –Reserves: 500 pages, 80 records/page, 40 bytes/record Ignore the cost of writing the results Join selectivity factor: 0.1

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Exercise What is the I/O cost for using a simple nested loops join? –Sailors: 1000 pages, 100 records/page, 50 bytes/record –Reserves: 500 pages, 80 records/page, 40 bytes/record Ignore the cost of writing the results Join selectivity factor: 0.1 Answer If Sailors is the outer relation, –I/O cost = ,000*500 = 50,001,000 I/Os If Reserves is the outer relation, –I/O cost = ,000*1000 = 40,000,500 I/Os

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Block Nested Loops Join B-2 pages for R Memory Output buffer R Bring bigger relation S one page at a time. Cost=M+N Optimal if one of the relation can fit in the memory (M=B-2). For each r i of B-2 pages of R do For each s j of s in S do if r i.a == s j.a then add to result Cost=M+ B: Available memory in pages. Number of blocks of R for each retrieval 1 page for S 1 page * N

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Exercise What is the I/O cost for using a block nested loops join? –Sailors: 1000 pages, 100 records/page, 50 bytes/record –Reserves: 500 pages, 80 records/page, 40 bytes/record Ignore the cost of writing the results Memory available: 102 Blocks

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Exercise What is the I/O cost for using a block nested loops join? –Sailors: 1000 pages, 100 records/page, 50 bytes/record –Reserves: 500 pages, 80 records/page, 40 bytes/record Ignore the cost of writing the results Memory available: 102 Blocks Answer Sailors is the outer relation #times to bring in the entire Reserves I/O cost = *5=5500

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Indexed Nested Loops Join for each tuple r in R do for each tuple s in S do if r i. A == s j.B then add to result Use index on the joining attribute of S. Cost=M+M*P R *(Cost of retrieving a matching tuple in S). Depend on the type of index and the number of matching tuples. cost to scan R

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Exercise What is the I/O cost for using an indexed nested loops join? Ignore the cost of writing the results Memory available: 102 Blocks There is only a hash-based, dense index on sid using Alternative 2 Answer Sailors is the outer relation I/O Cost = ,000*(1.2+1)=88500 I/Os

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Grace Hash-Join Partition Phase Partition both relations using hash fn h: R tuples in partition i will only match S tuples in partition i. Join (probing) Phase v Read in a partition of R, hash it using h2 (<> h!). v Scan matching partition of S to search for matching tuples Partitions of R & S Input buffer for Si Hash table for partition Ri (k <= B-2 pages) B main memory buffers Disk Output buffer Disk Join Result hash fn h2 B main memory buffers Disk Original Relation OUTPUT 2 INPUT 1 hash function h B-1 Partitions 1 2 B-1...

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Cost of Grace Join Assumption: –Each partition fits in the B-2 pages –I/O cost for a read and a write is the same –Ignore the cost of writing the join results Disk I/O Cost –Partitioning Phase: I/O Cost: 2*M+2*N –Probing Phase I/O Cost: M+N –Total Cost 3*(M+N) Which one to use, block-nested loop or Grace join? –Block-nested loop : Cost=M+ * N

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What is the I/O cost for using Grace hash join? Ignore the cost of writing the results Assuming that each partition fits in memory. Answer 3*( )=4500 I/Os Exercise

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Ideally, each partition fits in memory –To increase this chance, we need to minimize partition size, which means to maximize #partitions Questions –What limits the number of partitions? –What is the minimum memory requirement? Partition Phase Probing phase Memory Requirement for Hash Join

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Partition Phase –To partition R into K partitions, we need at least K output buffer and one input buffer –Given B buffer pages, the maximum number of partition is B-1 –Assume a uniform distribution, the size of each R partition is equal to M/(B-1) B main memory buffers Disk Original Relation OUTPUT 2 INPUT 1 hash function h B-1 Partitions 1 2 B-1...

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Probing phase –# pages for the in-memory hash table built during the probing phase is equal to f*M/(B-1) f: fudge factor that captures the increase in the hash table size from the buffer size –Total number of pages since one page for input buffer for S and another page for output buffer Partitions of R & S Input buffer for Si Hash table for partition Ri (k <= B-2 pages) B main memory buffers Disk Output buffer Disk Join Result hash fn h2

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If memory is not enough to store a smaller partition Divide a partition of R into sub-partitions using another hash function h3 Divide a partition of S into sub-partitions using another hash function h3 Sub-partition j of partition i in R only matches sub-partition j of partition i in S

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Sort-Merge Join (R S) Sort R and S on the join attribute, then scan them to do a ``merge’’ (on join col.), and output result tuples. – Advance scan of R until current tuple R.i >= current tuple S.j, then advance scan of S until current S.j >= current R.i; do this until current R.i = current S.j. – At this point, all R tuples with same value in R.i (current R partition) and all S tuples with same value in S.j (current S partiton) match; – output for all pairs of such tuples. – Then resume scanning R and S. i=j i j : : Both are sorted

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Example of Sort-Merge Join I/O Cost for Sort-Merge Join Cost of sorting: TBD O(|R| log |R|)+O(|S| log |S|) ?? Cost of merging: M+N Could be up to O(M+N) if the inner-relation has to be scanned multiple times (very unlikely) Sailors Reserves

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Sort-Merge Join Attractive if one relation is already sorted on the join attribute or has a clustered index on the join attribute

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Exercise What is the I/O cost for using a sort-merge join? Ignore the cost of writing the results Buffer pool: 100 pages Answer Sort Reserves in 2 passes:2*( ) =4000 Sort Sailors in 2 passes: 2*( )=2000 Merge: ( ) =1500 Total cost = = 7500

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Cost comparison for Exercise Example Block nested loops join: 5500 Index nested loops join using a hash-index: Grace Hash Join: 4500 Sort-Merge Join: 7500

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Why Sort? Sort-merge join algorithm involves sorting. Problem: sort 1Gb of data with 1Mb of RAM. – why not virtual memory?

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2-Way Sort: Requires 3 Buffers Pass 1: Read a page, sort it, write it. – only one buffer page is used Pass 2, 3, …, etc.: – three buffer pages used. Main memory buffers INPUT 1 INPUT 2 OUTPUT Disk

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P1 INPUT 1 INPUT 2 OUTPUT P2 P1P2 P3 INPUT 1 INPUT 2 OUTPUT P4 P3P4 Example: Sorting 4 pages P1 and P2 are sorted individually P3 and P4 are sorted individidually P1 and P2 are sorted P3 and P4 are sorted

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P1 INPUT 1 INPUT 2 OUTPUT P2 P3P4 P3P2P1 The data in 4 pages are sorted

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Two-Way External Merge Sort Each pass we read + write each page in file. N pages in the file => the number of passes So total cost is: Idea: Divide and conquer: sort subfiles and merge Input file 1-page runs 2-page runs 4-page runs 8-page runs PASS 0 PASS 1 PASS 2 PASS 3 9 3,4 6,2 9,48,75,63,1 2 3,45,62,64,97,8 1,32 2,3 4,6 4,7 8,9 1,3 5,62 2,3 4,4 6,7 8,9 1,2 3,5 6 1,2 2,3 3,4 4,5 6,6 7,8

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Two-Phase Multi-Way Merge-Sort Phase 1: 1.Fill all available main memory with blocks from the original relation to be sorted. 2.Sort the records in main memory using main memory sorting techniques. 3.Write the sorted records from main memory onto new blocks of secondary memory, forming one sorted sublist. (There may be any number of these sorted sublists, which we merge in the next phase). * More than 3 buffer pages. How can we utilize them? File to be sorted MM Sorted f1 Sorted f2 Sorted fn : n = N/M, where N is the file size and P is the main memory in pages

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Phase 2: Multiway Merge-Sort Pointers to first unchosen records Select smallest unchosen key from the list for output Output buffer Input buffer, one for each sorted list Done in main memory Merge all the sorted sublists into a single sorted list. Partition MM into n blocks Load fi to block i

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Find the smallest key among the first remaining elements of all the lists. ( This comparison is done in main memory and a linear search is sufficient. Better technique can be used.) Move the smallest element to the first available position of the output block. If the output block is full, write it to disk and reinitialize the same buffer in main memory to hold the next output block. If the block from which the input smallest element was taken is now exhausted, read the next block from the same sorted sublist into the same buffer. If no block remains, leave its buffer empty and do not consider elements from that list. Phase 2: Multiway Merge-Sort

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Memory Requirement for Multi-Way Merge-Sort Partitioning Phase: Given M pages of main memory and a file of N pages of data, we can partition the file into N/M small files. File to be sorted MM Sorted f1 Sorted f2 Sorted fn : Merge Phase: At least M-1 pages of main memory are needed to merge N/M sorted lists, so we have M-1>=N/M, i.e., M(M-1)>=N. Main memory buffers INPUT 1 OUTPUT Disk INPUT 2 INPUT n

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1.pick one element in the array, which will be the pivot. 2.make one pass through the array, called a partition step, re- arranging the entries so that: a)the pivot is in its proper place. b)entries to the left of the pivot are smaller than the pivot c)entries to its right are larger than the pivot Detailed Steps: 1.Starting from left, find the item that is larger than the pivot, 2.Starting from right, find the item that is smaller than the pivot 3.Switch left and right 3.recursively apply quicksort to the part of the array that is to the left of the pivot, and to the part on its right. O(n log n) on average, worst case is O(n 2 ) In-memory Sort: Quick Sort left right 1st pass: pivot = 18

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Query Cost Estimation Select: No index unsorted data sorted data Index tree index hash-based index Join: R S Simple nested loop Block nested loop Grace Hash Sort-merge

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