Equality Join R XR AS B S Pr

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

Simple Nested Loops Join Memory For each tuple r in R do for each tuple s in S do if r. A == S. B then add <r, s> 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 1 page for R 1 page for S R has M pages with PR tuples per page. S has N pages with PS tuples per page. Nested loop join cost = M+M*PR*N. Cost to scan R. Output

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

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 = 1000 + 100, 000*500 = 50, 001, 000 I/Os • If Reserves is the outer relation, – I/O cost = 500 + 40, 000*1000 = 40, 000, 500 I/Os

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

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

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 = 500 + 1000*5=5500

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

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 = 500 + 40, 000*(1. 2+1)=88500 I/Os

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

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

Exercise • 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*(500+1000)=4500 I/Os

Memory Requirement for Hash Join • 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

• 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) Original Relation OUTPUT Partitions 1 INPUT . . . 1 2 2 hash function h B-1 Disk B main memory buffers Disk

• 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 Join Result hash fn Hash table for partition Ri (k <= B-2 pages) h 2 Input buffer for Si Disk Output buffer B main memory buffers Disk

If memory is not enough to store a smaller partition • Divide a partition of R into sub-partitions using another hash function h 3 • Divide a partition of S into sub-partitions using another hash function h 3 • Sub-partition j of partition i in R only matches sub-partition j of partition i in S

Sort-Merge Join (R • Sort R and S on the join attribute, then scan them to do a ``merge’’ (on join col. ), and output result tuples. i – – – 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 <r, s> for all pairs of such tuples. Then resume scanning R and S. S) i=j j : : Both are sorted

Example of Sort-Merge Join Sailors Reserves • 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)

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

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*(1000+1000) =4000 • Sort Sailors in 2 passes: 2*(500+500)=2000 • Merge: (1000+500) =1500 • Total cost = 4000+2000+1500 = 7500

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 88500

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

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. INPUT 1 OUTPUT INPUT 2 Disk Main memory buffers Disk

Example: Sorting 4 pages P 1 INPUT 1 OUTPUT P 2 P 1 and P 2 are sorted INPUT 1 OUTPUT P 4 P 2 INPUT 2 P 1 and P 2 are sorted individually P 3 P 1 P 3 P 4 INPUT 2 P 3 and P 4 are sorted individidually P 3 and P 4 are sorted

P 2 P 1 INPUT 1 OUTPUT P 4 P 3 P 2 P 1 INPUT 2 The data in 4 pages are sorted

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: 3, 4 6, 2 9, 4 8, 7 5, 6 3, 1 2 3, 4 2, 6 4, 9 7, 8 5, 6 1, 3 2 4, 7 8, 9 2, 3 4, 6 1, 3 5, 6 2 2 -page runs PASS 2 2, 3 4, 4 6, 7 8, 9 1, 2 3, 5 6 4 -page runs PASS 3 1, 2 2, 3 3, 4 • Idea: Divide and conquer: sort subfiles and merge Input file PASS 0 1 -page runs PASS 1 4, 5 6, 6 7, 8 9 8 -page runs

Two-Phase Multi-Way Merge-Sort * More than 3 buffer pages. How can we utilize them? Phase 1: 1. Fill 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). Sorted f 1 File to be sorted MM Sorted f 2 : Sorted fn n = N/M, where N is the file size and P is the main memory in pages

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

Phase 2: Multiway Merge-Sort • 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.

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. Sorted f 1 MM File to be sorted Sorted f 2 : 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. INPUT 1 INPUT 2 OUTPUT INPUT n Disk Main memory buffers Disk

In-memory Sort: Quick Sort 1. 2. 3. pick one element in the array, which will be the pivot. make one pass through the array, called a partition step, rearranging 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 recursively apply quicksort to the part of the array that is to the left of the pivot, and to the part on its right. 1 st pass: pivot = 18 18 5 6 19 3 1 5 6 11 21 33 23 left right O(n log n) on average, worst case is O(n 2)

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