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CSE 232 - Databases Systems

Lecture 1

Lecture 2 - Database Storage and Hardware Design

  • Different storage types used by databases
    • Typically cost is higher and amount of storage is lower the faster it is
    • CPU-level cache typically isn’t controlled by the software, so the DB can’t do much optimization around it
    • RAM is utilized by DB, but can’t guarantee transactions due to volatility
    • SSD, HDD, Tape, etc are non-volatile and can guarantee transactions

Peculiarities of storage and algorithms

  • most storage is accessible via block-based access (via kernel, i.e. block device)
  • not just pulling the bytes we need - taking big chunks.
  • clustering (for hard disks):
    • cost is lower for consecutive blocks.
  • sequential read times have increased immensely
    • minimum access time to first byte has not decreased propotionally

Ex: 2-phase merge sort

  • Pass 1: Read section by section into buffer, sort all data, write sorted section to file
  • 2nd pass: file pointers to all previously written files, read block from each file write in sorted order out to new file.
  • Assume RAM buffer of size \(M\)
  • Block size of $$B$
  • Number of block reads = \(M/B\)
  • File sections: \(N / M / B\)

With ram buffer of size \(M\) can sort up to \(M^2 / B\) records (at least one block from each ram buffer)

Horizontal Placement of SQL data in blocks

  • row oriented - tuples packed together in blocks
    • improve total table scan time
  • for deletions, clear the block, add a “mark” (bit?) for deletion
  • for additions to a sorted table by a key, leave a pointer on the block where it should exist to the new record (only applies on sorted files)

Lecture 3 - Indexing

Given condition on attribute, find the qualified records

  • Indexing is a trick to speed up the search for attributes with a particular value
    • Conditions may include operations like >, <, >=, <=, etc
  • Different types of indexes, each need evaluation
    • time for access, insert, delete, or disk space


  • Conventional (Traditional) Indexes
  • B-Trees
  • Hashing Schemes

Terms and Distinctions

  • Primary Index - index on the attributes which determines sequencing of the table.
    • can tell if value exists without accessing a file
    • easier access to overflow records (refer to previous lecture)
    • required for secondary index
  • Secondary Index - indexes on any other attribute
    • sparse index takes lest space.
    • better insertions
  • Dense Index - every value of the indexed attribute appears in the index
  • Sparse Index - many values do not appear

Can have multi-level indexes

  • index the index!
  • generally, two-level index is sufficient, more is rare

Pointers - a block pointer, value of record attribute, then a value representing the block and/or record location.

  • Block pointers cannot be record identifiers

What if a key isn’t unique?

  • Deletion from a dense primary index with no duplicate is the same way as a sequential file - leave blank space
  • Deletion from a sparse index, similar approach, but might not have to delete anything.
    • or, simply leave it, just change the pointer without changing any blocks
      • must assume the algorithm doesn’t assume a value exists just because it exists in the index.

Insertion in sparse index: if no new blocked is created for the key, do nothing

  • otherwise, create an overflow record
  • reorganize periodically

Duplicate values and secondary indexes

  • option 1: index entry to each tuple. not very efficient
  • option 2: index on unique values, then have blocks of “buckets” containing sequential pointers to duplicate values

Why bucket+record pointers is useful

  • enabled processing of queries working with pointers only
  • common technique in information retrieval
  • with buckets and record pointers, can simply do set intersection of record pointers in order to answer query
    • read less data blocks


  • Give up on sequential records, try to achieve balance instead
    • “node” and “block” are equivalent in lecture
    • node with three values \(j, k, l\), 4 pointers (\(n+1\))
      • \[p1 < j\]
      • \[j >= p2 > k\]
      • \[k >= p3 > l\]
      • \[l >= p4\]
  • non-root nodes must be at least half-full
    • non-leaf node \(\lceil \frac{n+1}{2} \rceil\) pointers
    • leaf node: \(\lfloor \frac{n+1}{2} \rfloor\).
  • bounded number of block access in the index based on the depth of the tree

Lecture 4 - Indexing Pt. 2

Hashing Schemes

  • hashing returns the address of bucket
  • In the case of key collisions, the bucket is a linked list
  • compute the table location as \(hash(key) % buckets\)
  • sorting in buckets only if cpu time for lookup is critical and inserts+deletes are not frequent
  • in DBs, hashing limited by data blocks to limit lookup accesses (different from array-based hash map in languages like Java)
    • entries in the buckets (blocks) have the key and a pointer to the corresponding data location.
    • if a bucket overflows too much, there is a pointer to an overflow block.
  • Rule of thumb: keep space utilization between 50% and 80% to limit overflow creations.
    • \(< 50%\) wasting space
    • \(> 80%\) slow down lookups due to overflow buckets
  • How to cope with growth?
    • Overflow blocks make it slow to lookup
    • re-hashing wastes lots of time

Extensible Hashing

  • Extensible Hashing: 2 ideas
    • use \(i\) of \(b\) bits which are output by the hash function.
    • use a directory to point to buckets
      • directory fits in main memory
      • points to buckets which contains full keys+pointer
  • When you overflow a bucket, increase the number of bits, \(i\) that the hashed values agree on, then create a new directory table without modifying the original blocks, and placing the new block from the block where records overflowed with the transferred records.
  • Strengths
    • Can handle growing files
      • less wasted space
      • no full reorganization
  • Weaknesses
    • Reorganizing table can be expensive
    • Indirection, not bad if everything in main memory
    • directories double in size

Linear Hashing

  • keep same idea of using low order bits from hash value
  • file grows in linear fashion
  • Variables
    • keys of \(b\) bits
    • hash bits of \(i\)
    • n keys in data blocks (again)
    • \(m\) to track maximum used block.
  • Rule

\(h(k)[i] \leq m \text{ bucket } h(k)[i]\)

  • else \(h(k)[i] - 2^{i-1}\)

  • When to expand the file?
    • keep track of \(\frac{\text{# used slots (incl. overflow)}}{\text{# total slots in primary buckets}}\)
    • when above a threshold, increase \(m\), as well as \(i\) when \(m = 2^i\)
  • Strengths
    • can handle growing files
    • less wasted space
    • no full reogranization

Lecture 5 - Indexing Pt. 3 - Hashing

  • Indexing vs hashing
    • Hashing is a good to probe for a specific key
    • Indexing (b+tree) is good for range searches
  • DBs tend to have tree-based indices because it makes it easy to implement predicates with greater, less, leq (less than eq.), geq
  • Traditional DBs don’t really let you define the types of indexes (b-tree/hash) in SQL
    • Some systems allow configuration parameters to change implementation
  • Multi-key indices to speed up queries with conditions on multiple attributes
    • Strategy for multi-condition queries with single index on each attribute
      • use one index and get tuples to satisfy cond1
        • iterate over list from above result to return correct set
      • another strategy
        • multi-index, find set of results satisfying both, then perform set intersection logic to get result (AND) condition.
    • Instead, structure keys as trees.
      • At the top level, sort by first key
        • all tuples underneath top-level attribute in a group, have same attribute value for top-level key.

Bitmap Indices

  • Heavily used in OLAP (online analytics)
  • Assume that tuples are in some order that each index sees globally
  • e.g. each index attribute value has a bitmap (string of 1’s and 0’s) e.g. given a table
name dep year
aaron suits 4
helen pens 3
jack PCs 4
yannis pens 1

then our bitmaps indices would be

dept bitmap
PCs 0010
pens 0101
suits 1000
year bitmap
1 0001
2 0000
3 0100
4 1010
  • unfortunate, bitmaps require a lot of space
  • increase in size w/ number of tuples
  • for 1B tuples, 1B / 8 bytes of space * number of values for the index
  • can try to compress bitmaps! (mostly 0’s)
  • Compression
    • naive solution for bitmap requires \(n*m\) bits
    • in reality, across the whole map there is only \(n\) 1’s
    • compress bitmap to \(2nlog(m)\)
      • \(n\) is the number of tuples
      • \(m\) is the number of distinct values.
    • e.g. 00011010 compresses to 301
      • 3 0’s until first 1
      • 0 0’s until 2nd 1
      • 1 0 until 3rd 1
    • further compress the encoding of the bitmap
      • 301 binary representation (digit-wise)
        • 11 (2 bits, 10 in binary)
        • 0 (1 bits)
        • 1 (1 bits)
        • end length scheme with 0, fill # of bits before - 1 with 1’s
          • eg. encode 10010 -> 1111010010
        • 0’s act as delimiters
      • Represent this as
        • 10110001

Bitmap example 2:

Pens bitmap: 01000001 Sequence: [1, 5] Encoding: 01110101

  • Handling Insertions and Deletions
    • Assertions, assume happens in order (nothing to do)
    • Deletions: do nothing, leave record and 0-out bitmap
    • Insertions: if tuple t with value v is inserted, add one more run in v’s sequence.

Proving 2*nlog(m)

  • n * m bits
  • will find exactly \(n\) runs. (Across all bitmaps)
  • why is the average run length \(2*log(m)\)?
    • \(n\) bits per bitmap - expect average \(\frac{n}{m}\)
    • n / (n / m) -> size of conceptual bitmap divided by the average run size -> \(m\)

Lecture 6 - Query Processing

  • A query processor turns user queries and data modification commands into a query plan
    • query plan is a sequence of operations on the database
  • Decisions taken by a query processor
    • which of the algebraically equivalent forms of a query will lead to the most efficient algorithm?
    • for each algebraic operator, what algorithm should we use to run the operator
    • how should the operators pass data from one to the other?


WHERE R.A = "c" and S.E = 2 AND R.C = S.C

How to execute this query?

  • idea 1:
    • scan relations
    • do a cartesian product
    • select tuples
    • do projection

Use relational algebra to describe the plan: e.g.

  • \[\Pi^{FLY}_{B,D}[\sigma^{FLY}_{R.A="c" \land S.E=2 \land R.C=S.C}(R^{SCAN} \times S^{SCAN})]\]

Above is very inefficient…

  • Plan 2: Instead of a full scan with a cartesian product, we can do something different:
    • scan each table and only include rows matching the predicates in the result
    • perform natural hash join on reduced tables
  • Plan 3: use indexes of each table
    • use R.A index to select tuples of R.A = "c"
    • for each R.C value found, use S.C index to find matching joins
    • eliminate join tuples where S.E != 2
    • projection
  • From query to optimal query plan
    • complex
    • algebra based logical and physical plans
    • transformations
    • may need to evaluate alternatives
  • Issues in query planning:
    • generating plans
      • employ efficient primitives to compute these operations
    • estimate cost of plans
    • “smart” search of possible plans

query optimization steps

Algebraic Operators

Predicates (\(\sigma\))

  • Select tuples of \(R\) that satisfy condition \(C\): \(\sigma^R_C\)
    • may contain multiple condition with AND/OR
    • may be of the for attr1 = value or attr1 = attr2
    • other operators like \(<,>,\neq,\text{LIKE}\)

Projections (SET vs BAG): \(\Pi\)

  • \(\Pi_{a_1...a_n}^{R}\):
    • returns a table that has only attributes \(a_1\),…,\(a_n\) of \(R\)
    • Set version: no duplicate tuples in the result
    • Bag version: allows duplicates

Cartesian Product: \(\times\)

  • \(\times\):
    • the schema is the combination of all attributes of \(R\) and \(S\)
    • for every tuple in \(R\) is matched to all tuples of \(S\)
    • if there are any attribute name collisions, prefix with the <Table_Name>.

Algebraic Operators: A Bag Version

  • Union of R and S: a tuple t is in the result as many times as the sum of the number of it is in R, plus the times it is in S.
  • Intersection of R and S: A tuple t is in the result the minimum number of time it is in R and S.
  • Difference of R and S A tuple t is in the result the number of times it is in R minus the number if times it is in S.
  • \(\delta(R)\) converts the bag \(R\) into a set.

The Extended Projection

  • Extend the relation projection \(\pi_A\):
    • the attribute list may include \(x\rightarrow y\) in the list to indicate the attribute x is renamed to y
    • Arithmetic, string operators, and scalar function on attributes are allowed
      • e.g. \(a+b\rightarrow x\) means teh sum of a and b is renamed to x.
      • e.g. \(c||d\rightarrow y\) concatenates teh result of c and d into a new attribute named y.
    • Note these are available only for scalar functions - not aggregations (like AVERAGE, COUNT).
    • results are generated by computing a new tuple with any renamed attributes and functions applied.


SELECT 2*A, as D, B, C as CPRIME

would be converted into

\[\pi_{2A\rightarrow D,B,C\rightarrow CPRIME} T\]

Cartesian Products -> Joins

  • Product of R and S (\(R\times S\))
    • if an attribute named a is found in both schemas, the rename by prefix
    • if a tuple r is found n times in R and a tuple s is found m times in S then the product contains nm instances of tuples rs.
  • Joins
    • Natural Joins: \(R \bowtie S\rightarrow \pi_{A}\sigma_{C}(R\times S)\)
      • C is a condition that equates common attributes
      • A is the concatenated list of attributes of R and S with no duplicates
    • Theta Join
      • arbitrary condition that involves multiple attributes

Example of a theta join


equates to

\[\pi_{T.A} (T \theta_{T.A=S.B} S)\]

Grouping and Aggregation: \(\gamma\)

  • \[\gamma_{\text{groupByList};\text{aggrFn1}\rightarrow \text{attr1}}\]
  • Conceptually, grouping leads to nested tables and is immediately followed by functions that aggregate the nested table.
  • example: \(\gamma_{\text{dept.;AVG(Salary)}\rightarrow\text{AvgSal},\text{SUM(Salary)}\rightarrow\text{SalaryExp}}\)

Example 2:

SELECT Dept, AVG(Salary) AS AvgSal
FROM Employee
HAVING SUM(Salary) > 100
\[\pi_{\text{Dept,AvgSal}} \sigma_{\text{SumSal > 100}}\gamma_{\text{Dept,AVG(SALARY)->AvgSal,SUM(Salary)->SumSal}}(\text{Employee})\]

Sorting and Lists

  • SQL and algebra results are ordered
  • Could be non-deterministic or dictated by SQL’s ORDER BY (\(\tau\))
  • A result of algebraic expression o(exp) is ordered if
    • o is \(\tau\)
    • if o retains ordering of exp and exp is ordered
      • depends on implementation !
    • if o creates ordering
  • consider leaf of tree may be scan (R)

Lecture 6 - Relational Algebra Optimization

  • Work with algebras - transform one expression to another equivalent
    • simple example \(x(y + z) = xy + xz\)
  • Need to determine which are “good” transformations

Commutativity and Associativity

  • Commutative: \(R \times S = S \times R\)
  • Associative: \(R \times (S \times T) = (R \times S) \times T\)
  • Natural join (\(\bowtie\)) is also commutative and associative

  • Commutative
    • cartesian product: \(\times\)
    • natural join/theta join: \(\bowtie\)
    • union: \(\cup\)
    • intersection: \(\cap\)
  • Associative
    • cartesian product: \(\times\)
    • natural join/theta join: \(\bowtie\)
    • union: \(\cup\)
    • intersection: \(\cap\)

Rewriting of algebraic expressions for logical connectives

instructor not clear on this section. refer to slides

Pushing Selection Through Binary Operators: Union and Difference

  • \[(R \cup S) \sigma_\text{cond} = \sigma_\text{cond} R \cup \sigma_\text{cond} S\]
  • \[(R - S) \sigma_\text{cond} = (\sigma_\text{cond} R) - (\sigma_\text{cond} S)\]

Pushing Selection Through Cartesian Product and Join

  • \[(R \times S) \sigma_\text{cond} = R \times (\sigma_\text{cond} S)\]
  • \[(R \bowtie S) \sigma_\text{cond} = R \bowtie (\sigma_\text{cond} S)\]

Pushing Projections Through Binary Operators

  • \[(R \cup S) \pi_A = (\pi_A R)\cup(\pi_A S)\]

Pushing Projections Through Join and Cartesian Product

  • \[(R\times S)\pi_A = ((R\pi_B) \times (S\pi_B))\pi_A\]
    • same for natural join

Deriving some rules

  • \[\sigma_{p\land q} (R \bowtie S) = \sigma_p R \bowtie \sigma_q S\]
  • see more examples in slides

What are always “good” transformations?

  • Some transformations are always good
  • others we do because it creates alternative (new) transformations which can also be efficient.


  • \[\sigma_{p1\wedge p2}(R) \rightarrow \sigma_{p1} (\sigma_{p2}(R))\]
    • not always good because conditions depend on index
  • \[\sigma_p(R\bowtie S) \rightarrow (\sigma_p (R)) \bowtie S\]
    • Usually good because joins are expensive, good to eliminate first.

The Bottom Line

  • No transformation is always good
  • Usualy good:
    • early selections
    • elimination of cartesian products
    • elimination of redundant subexpressions
  • many transformations can lead to “promising” plans
    • communicating/rearranging joins
    • in practive, too combinatorially explosive to handle a re-writing

Algorithms for Relation Operators

  • 3 Primary techniques
    • sorting
    • hashing
    • indexing
  • 3 degrees of difficulty
    • data small enough to fit in memory
    • too large to fit in main memory, but small enough to be handled by a two-pass algorithm
    • so large that two-pass methods have to be generalized to multi-pass
  • Dominant cost of operators running on disk
    • no. of disk block that must be read/written to execute query plan

Pipelining can greatly reduce cost.

  • each operator can scan through file with tuples
    • check condition (select) with sigma
    • project with pi
    • intermediate files between steps
    • open/close between each
    • (this is very very slow)
  • Instead, pipeline operators while opening data files
    • pass output from one operator into the next operator
    • no intermediate files

Iteration Join

  • \(R1 \bowtie R2\) on a common attribute C
  • Least efficient iteration join
for r in R1:
  for s in R2:
    if r.C = s.C, output (r, s)
  • merge join (conceptual)
    • if R1 and R2 not sorted by C, sort them
      • dual pointer algorithm
        • move smaller value pointer to next tuple
        • emitting pairs until hitting the end of the smaller table
        • however, slightly more complex, as you need to keep track of sets of equivalent values
          • need to output all combinations of tuples
    • O(n+m)

Lecture 7 - Joins with Indexes

  • Join with index (conceptual)
  • Hash join
    • hash function range \(0\rightarrow k\)
    • buckets for R1 into G buckets
    • buckets for R2 into H buckets
    • Algorithm
      • hash all R1 tuples into G buckets
      • hash all R2 tuples into H buckets
      • for i = 0 to k
        • match tuples in \(G_i = H_i\)
      • small joins on buckets
    • in postgres, “hash join” oriented is actually kind of like an index join with an ephemeral hash index

Disk Oriented Computation Model

  • \(M\) main memory buffers
    • each buffer is the size of a disk block
  • assume input of relation is read one block at a time
  • assume dominant factor is IO, and the number of blocks read directly correlates to time
    • assumption is fine for SSD
    • For HDD consecutive is very different from random
  • assume output buffers are not part of M buffers mentioned above
    • Pipelining allows the output buffers of an operator to be the input to the next operation
    • do not count the cost of writing output
  • Future cost notation
    • \(B(R) =\) number of blocks on disk that R occupies
    • \(T(R) =\) number of tuples in R
    • \(V(R,[a_1, a_2,\dots,a_n]) =\) number of distinct tuples in the projection of R on \(a_1, a_2,\dots,a_n\)
  • One-pass main memory algorithms for unary operators
    • Assumption: enough memory to keep entire relation in memory
    • Projection
      • scan R, apply operator to one tuple at a time
      • incremental cost of “on the fly” operators - 0 (no additional disk operators, aside from needing to read data at least once to produce the result)
    • Duplicate Elimination and Aggregation
      • Create one entry for each group, compute aggregated group value
        • hard to assume CPU cost is negligible here
        • require “complex” data structures
  • One pass nested loop join
    • cost is simply \(B(R) + B(S)\) because you simply loop over all tuples of each table once to join (since everything fits in memory)
  • Simple Sort-Mege Join
    • Assume R natural join on S on attribute C
    • Algorithm
      • Read and sort R on C
      • Read and sort S on C
      • Merge using two pointer algorithm
    • Cost assuming both tables not previous sorted and a 2-pass multiway merge sort is used
      • sort cost is \(4B(T)\rightarrow 4B(R) + 4B(S)\) (sort is 2x read, 2x write)
      • join cost is B(R) + B(S)
      • total cost is \(5B(R) + 5B(S)\)
    • Can we do better though?
      • Save two disk IOs per block by combining second pass sorting with merge
      • create sorted sublists of size M for R and S
      • bring first block of each sublist into a buffer
      • repeatedly find the leaste C value among the first tuples of each sublists. Identify all tuples with a join value c, and join them
        • when a buffer has no more tuples that has not already been considered, load another block into this buffer.
      • by joining on the final phase of merge sort, the cost is reduced \(3B(R)+3B(S)\)
  • Hash Join Algorithms
    • Assuming a natural join, use a hash function that is the same for both input arguments, and only uses join attributes
    • Algorithm:
      • Phase 1: Hash each tuple of R into one of the \(M-1\) buckets \(R_i\) and similarly for S
      • Phase 2: For i=1..M-1
        • load \(R_i\) and \(S_i\) in memory
        • join and save result to disk.
      • Cost: \((2B(R) + 2B(S)) + (B(R) + B(S))\)
  • Index-Based Join: Simple Version
    • Assume natural join of R and S where the index is on S
    • Algorithm:
      • read a block of R
        • for each tuple in the block use the index of S to find the tuples of S with value b
    • Assuming R is not sorted and S is indexed and clustered on B
      • Cost of \(B(R) + T(R)B(S)/V(S,B)\)
    • If R is sorted, find all tuples of R with a given b value, then issue index access

Lecture 8 - Estimating the Cost of a Query Plan


  • Esimating the size of results
  • Estimating the number of IOs

Estimating result size: need the following statistics

  • T(R): the number of tuples in R
  • S(R): number of bytes in each tuple of R
  • B(R): number of blocks to hold all R tuples
  • V(R, A): number of distinct values for attribute A in R


Given \(W = R1\times R2\)

  • \[T(W) = T(R1) \times T(R2)\]
  • \[S(W) = S(R1) + S(R2)\]

Size estimate for $$W = \sigma_{Z=val}(R)

  • S(W) = S(R)
  • T(W) = T(R) / V(R, Z)

Calculating Results with Inequalities

What about a query \(W = \sigma_{z \geq 15}R\)?

  • Keep statistics on the table, a histogram of frequencies over a number of buckets

Size estimate of a join…

  • query: \(W = R1\bowtie R2\)
  • Let X = attributes of R1
  • ket Y = attributes of R2

  • Join can have \([0, T(R1)T(R2)]\) tuples.
  • If we have some more information it can be helpful
    • Assume every value of A in R1 is in R2 (typically A is a foreign key of R1, R2.A is a primary key)

Computing T(W) when A of R1 is a foreign key of R2

  • If R2.A is a primary key, it occurs only once, so then the T(W) = T(R1)

  • Ok, but what happens when A is not a primary key (unique), then we need to multiple T(R1) by the average number of matches in R2. Thus

\[T(W) = \frac{T(R2)}{V(R2, A)}T(R1)\]
  • in the case of \(V(R1, A) \leq V(R2, A) \rightarrow T(W) = \frac{T(R2)T(R1)}{V(R2, A)}\)
  • in the case of \(V(R2, A) \leq V(R1, A) \rightarrow T(W) = \frac{T(R2)T(R1)}{V(R1, A)}\)

  • In general \(T(W) = \frac{T(R2)T(R1)}{max(V(R1, A), V(R2, A))}\)

Case: A 3-way join for 3 tables which share the same attribute?


R1, R2, R3

  • T(R1) = 10000
  • T(R2) = 20000
  • T(R3) = 5000
  • V(R1, A) = 1000
  • V(R2, A) = 20000
  • V(R3, A) = 500

Join: \(R1\bowtie R2\bowtie R3 \rightarrow (R1\bowtie R2) \bowtie R3\)

  • The intermediate result of \(R1\bowtie R2\) has \(T(R1\bowtie R2) = 10000\)
    • The intermediate of \(V(R1\bowtie R2, A) =\)1000, or 20,000?
      • Preserve probability of the larger table?

Lecture 9 - Plan Enumeration

  • Smart exhaustive algorithm for generating query plans
    • Textbook Section 16.6
  • INGRES Heuristic for enumeration

Practice problems

Given the relations R(A, B, C) and S(C, D, E) give 6 valid logical query plans

where R.C=S.C AND R.A=5;

Given the relations R, S, and T, Give an algebraic expression that only contains the join operator

where R.A=S.A AND S.B=T.B;

Given the plans from the above problem, give plans for executing the query.

  • To generate a query plan, pick one of the types of algorithms for join operators
    • e.g.
      • Hash Join (HJ)
      • Sort Sort Merge (SSM)
      • **M (merge only if sorted)
      • *SM (sort only one)
      • S*M (sort only one)

Arranging the Join Order: Wong-Yussefi Algorithm

  • Challenges with large natural join expressions
    • for simplicity, assume that in the query
      • all joins natural (equality conditions)
      • whenever two tables of the FROM clause have common attributes, we oin them
      • consider right-index only
        • assume that on join operator, the right-hand argument has an index key that maps directly to the index of the left-hand operand.
  • Wong-Yussefi assumptions and objectives
    • assumption 1 (weak): indexes on all join attributes (keys+foreign keys)
    • assumption 2 (strong): at least one selection creates a small relation
      • joins on small relations results in a small relation
    • objective: create a sequence of index-based joins such that all intermediate results are small.
  • Calculates cost via “hypergraph” data structure
    • nodes are attributes
    • relations are hyperedges
  • pick small relation (and conditions) to start the plan.
    • Remove the small relation (hypergraph reduction) and color as many small relations that join with the removed “small” relation
  • What happens if there are multiple foreign keys? multiple selections?
    • multiple options to follow in the hypergraph.
      • generally, pick smallest items before joining on larger tables.
  • Algorithm has generally good ideas, but does not really survive because it makes too many assumptions
    • new approach, use dynamic programming to enumerate plans.

Lecture 10 - Failure Recovery

  • Integrity and correctness of data is of utmost importance
  • Integrity+consistency constraints
    • some predicates DB must satisfy:
      • x is key of R
      • x –> Y holds in R
      • domain(x) = {R, G, B}
      • no employee should make more than twice the average salary
    • these types of consistency are based on business logic constraints.
    • DB builders are concerned with the view of data pre and post transaction are consistent.
  • Transaction: a collection of actions that preserve DB consistency
  • Working assumption for transactions:
    • if T starts with a consistent state, and T executes until completion and isolation, then T leaves a consistent state.
  • How to prevent and fix violations?
    • failure recovery: fixing violations due to failures
    • concurrency control: fixing violations due to concurrency and data sharing
    • a mix: fixing violations that stem from interaction of failures with sharing
  • What is not considered in this class
    • how to write correct transactions (buggy transactions can violate constraints)
    • how to write a correct dbms

Failure Model

  • Events
    • desired
    • undesired
      • expected
        • memory loss
        • cpu halt, crash, reset
      • unexpected (db crash)
        • disk data lost, memory lost without CPU halt, skynet CPU decides to wipe out programmers….

Storage Hierarchy

  • memory <–> disk
  • operations:
    • input(x) block with X –> memory
    • output(x) block with X –> disk
    • read(x, t): input (x) if necessary, t <– value of X in block
    • write(x, t): input(x) if necessary, value of X <– t
  • key problem: unfinished transactions
    • constraint: A=B
    • \[T_1 \leftarrow A\times 2 \rightarrow B\leftarrow B\times 2\]
  • if db crashes before all updates a written, on recovery we could have onconsistent state of disk.
  • need ability to execute “all or none” semantics
    • undo logs of unfinished transactions
  • use a commit log to log all changes to DB
    • complications
      • log first written to mem, not written to disk on every action (too expensive)
  • undo logging rules
    • for every action, generate an undo log record (containing old value)
    • before x is modified on disk, log records pertaining to x must be on disk (Write-ahead-log, WAL)
    • before commit is flushed to log, all writes of transaction must be reflected on disk.
  • recovery rules: undo logging:
    • let S = set of transactions that have a start log but no commit (or maybe an abort log)
    • for each transaction statement, in reverse order (latest to earliest)
      • if \(T_i\) in the set of transactions then (write(X, v), output(X))
    • for each transaction, write a transaction abort to log.
  • What about failure during recovery?
    • undo is idempotent, can replay again.
  • Redo logging
    • do transactions, but don’t output results until the transaction is committed in the log.
    • rules
      • for every action, generate a redo log record
      • before X is modified on disk, all log records for transaction that modified X (including final commit) must be on disk
      • flush log at commit.
    • recovery rules
      • Let S be a set of transactions with a commit entry in the log
      • for each transaction statement in forward order (earliest to latest) do
        • write(X, v)
        • Output(X)
    • recovery is slow
  • checkpoint to speed up
    • periodically:
      • do not accept new txns
      • wait until all txns finish
      • flush all log records
      • flush all buffers
      • write checkpoint record
      • resume txns
  • key drawbacks
    • undo logging:
      • cannot bring backup DB copies up to date, real writes at end of txn needed
    • redo logging:
      • need to keep all modified block in memory until commit.

Lecture 11 - Transactions Cont’d

  • how to address issues from last lecture?
    • undo/redo logging! (use both)
  • rules
    • page X can be flushed before or after commit \(T_i\)
    • log record can be flushed before corresponding updated page
    • flush at commit (log only)
  • problem: at boot up, need to replay committed transactions
    • can checkpoint so that the log doesn’t grow too large
    • problem with naive checkpoint is that working to establish a checkpoint needs database downtime.
    • solve problem with non-quiescent checkpoint
  • non-quiescent checkpoint
    • start checkpoint, but take note of active transactions.
    • on DB recovery, undo transaction statements that began before or after checkpoint, and didn’t complete
    • redo transaction blocks that were written after checkpoint start but where transaction already began
    • recovery process summary
    • backwards pass (end of log to latest checkpoint start)
      • construct set of S committed transactions
      • undo actions of transactions not in S
    • undo pending transactions
    • follow undo chains for transactions in (active checkpoint list) - S
    • forward pass (latest checkpoint start to end of log)
    • redo actions of S transactions.
  • real-world actions
    • execute real-world actions (e.g. subtracting from bank account balance) after commit
      • want actions to be idempotent
  • summary
    • transaction processing for keeping DB consistent
      • source of problem: DB/hardware failures
        • use logging + redundancy
      • next source of problems: concurrency and data sharing

Intro to Concurrency Control

  • DBs usually have many processes communicating with the DB with multiple Txns in parallel.
  • concurrency control: how to best interleave reads and writes to maintain consistency
  • want transaction schedules that are “good”
    • equivalent to serial execution regardless of
      • initial state
      • transaction semantics
    • only want to look at order of reads and writes.
  • create dependency graph based on reads and writes required by each transaction
    • if no cycles in graph, then the schedule of operations is equivalent to a serial schedule

Lecture 12 - Transaction Concurrency Control Cont’d

  • Transaction
    • a sequence of \(r_i(x)\) and \(w_i(x)\) actions
  • Conflicting actions
    • when \(i\) is different for read or write actions on the same piece of data.
    • When two write operations at different points operate on the same data.
  • Schedule
    • represents chronological order in which actions are executed
  • Serial Schedule
    • no interleaving of actions or transactions.
  • S1 and S2 are conflict equivalent schedules if S1 can be transformed into S2 by a series of swaps on non-conflicting actions.
  • a schedule is conflict serializable if it is conflict equivalent to a serial schedule.
  • A precedence graph \(P(S)\) where \(S\) is a schedule
    • nodes are transactions in S
    • edges: \(T_i\rightarrow T_j\) whenever
      • \(p_i(A), q_j(A)\) are actions in S
      • \(p_i(A) <_s q_j(A)\).
      • at least one of \(p_i,q_i\) is a write.

Exercise: What is P(S) for $$S=w_3(A)w_2(C)r_1(A)w_1(B)r_1(C)w_2(A)r_4(A)w_4(D)

  • edges:
    • T3->T1
    • T3->T4
    • T1->T2
    • T2->T4
    • T1->T2
  • conflict serializable if there is a cycle
  • Lemma: S1,S2 conflict equivalent if P(S1) = P(S2)
  • How to prove?
    • assume they are not equivalent. List all edges. They are not equivalent if edges are different.
  • if two graphs have equivalent precedence, doesn’t necessarily imply conflict equivalence
  • if P(S1) is acyclic (no cycles) –> S1 is conflict serializable
    • proof: Assume S1 is conflict serializable
      • therefore, another schedules \(S_s\) is conflict equivalent to S1
        • \(P(S_1) = P(S_s)\).
        • \(P(S_1)\) is acyclic since \(P(S_s)\) is acyclic
    • assume P(S1) is acyclic
      • transform S1 as follows
        • Take T1 to be transaction with no incident (incoming) arcs
        • move all T1 actions to the front.
        • repeat again discarding node T1
  • how to enforce serializable schedules?
    • option 1: run system, recording P(S). Check P(S) for cycles and declare if execution was good, or abort transactions as soon as they generate a cycle.
    • option 2: prevent P(S) cycles from occurring via a scheduler that receives all transactions.
      • requires a locking protocol to prevent bad concurrent transactions
        • \(l_i(A)\) - exclusive lock on A
        • \(u_i(A)\) - unlock A
      • scheduler reads lock table.
      • may only operate on items which have locks
      • still doesn’t fully solve the problem, because locks don’t prevent cycles
        • 2-phase locking for transactions required.
        • basic rule: once first unlock occurs, cannot unlock again.
  • Assume deadlocked transactions are rolled back
    • have no effect
    • do not appear in schedule
  • beyond a 2PL protocol, it is about improving performance and allowing concurrency -> other methods
    • shared locks
    • multiple granularity
    • inserts, deletes, phantoms
    • other types of concurrency control mechanisms
  • shared locks – allowed on reads. multiple transactions can lock an item at the same time.
    • transactions which read and write same object:
      • option 1: request exclusive lock
      • option 2: upgrade (need to read, but unsure about write)

Lecture 13 - Concurrency Control Cont’d

  • Concurrency Control judged on few aspects
    • how performant
    • allow as many transactions as possible to occur in parallel
  • Increment locks
    • atomic increment action
    • hardware ensures read and write executes atomically.
  Shared Exclusive Increment
Shared T F F
Exclusive F F F
Increment F F T
  • common table schema in practice: CREATE TABLE T(attr SERIAL PRIMARY KEY,..);
    • prevents needing to lock the counter on attr.
  • locking in practice
    • don’t trust transactions to request/release locks
    • hold all locks until transaction commits
  • scheduler uses a lock table for which transactions for determining who needs or wants locks.
  • lock table
    • table of every possible object that can be locked
      • null if unlocked, otherwise lock into for the object,
    • locking work in most cases, but should we lock at large or fine grain?
      • large objects (e.g. relations)
        • need new locks, low concurrency
      • lock small objects (tuples, fields)
        • more locks, more concurrency
        • high overhead
  • Share intentions about which locks to get, but don’t request them yet
    • IS (Intend Shared)
    • IX (Intend Exclusive)
    • SIX (Shared Intend Exclusive)
Parent Lock (table) Child Locked In (tuple)
S [S, IS] not necessary
X None


  • follow multiple granularity comp functions
  • lock root of tree first, any mode
  • node Q can be locked by a transaction in S or IS only if the parent is locked by IX or IS
  • Node Q can be locked by a transaction in X, SIX, or IX only if the parent is locked by IX or SIX
  • the transaction is a two phase locking process
  • transaction can unlock node Q only if none of Q’s children are locked by the transaction

Handling Inserts and Deletes

  • get exclusive lock on A before deleting
  • At insertion of A on transaction, given exclusive lock to A on insertion
  • Still have issue: Phantoms:
    • can have serialized transactions without proper locks update same attr twice (e.g. unique instead)
  • instead of locking on R, lock on an index of R
    • generalize to tree-based concurrency control
    • increase concurrency 0 don’t need to lock whole table
  • generalized tree-based concurrency control
    • all objects accessed through root
    • follow pointers
      • lock parent, then child, then grandchild, etc
  • rules: tree protocol
    • first lock by a transaction may be on any item
    • after that, Q may be locked by any transaction only if the parent of Q was locked by the transaction
    • items may be unlocked at any time
    • after a transaction unlocks Q, it cannot relock Q.

Lecture 14 - More Concurrency Control and Transactions

  • Transactions have 3 phases
    • read
      • all DB values read
      • writes to temp storage
      • no locking
    • validate
      • check is schedule so far is serializable
    • write
      • if validate ok, write to DB
  • key idea –> make sure validation is atomic
  • if T1, T2, T3, etc.. is validation order, then resulting scheduling will be conflict equivalent to S = (T1)(T2)(T3)

  • to implement transaction validation - system has two sets
    • FIN –> Transactions that have finished phase 3 (completed)
    • VAL –> transactions that have successfully finished phase 2 (validation)
  • validation rules for Tj
    • when Tj starts phase 1:
      • ignore(Tj) <– FIN
    • at Tj validation:
      • if check(Tj) then
        • V <– \(V \cup T_j\)
        • do write phase
        • FIN <– FIN \(\cup T_j\)
  • implementation of check(Tj)
    for Ti in VAL -IGNORE (Tj) DO
    IF ( WS(Ti) intersectes RS(Tj) != empty set
      OR Ti not in FIN ) THEN
        RETURN false
    RETURN true
  • This method is called “optimistic concurrency control”
    • assumes that conflicts are rare
    • system resources plentiful
    • have real-time constraints

Concurrency Control and Recovery

  • Two transactions, Ti and Tj write/read to same relation
    • Tj writes first
    • Ti reads after the write, and then commits.
    • Tj then aborts at the end.
  • without proper validation, would need a cascading rollback because Ti would have read data that technically wouldn’t have existed (the write from Tj)
  • a schedule is recoverable is each transaction commits only after all transactions from which it read are committed
  • A schedule avoids cascading rollback if each transaction may only read those values written by committed transactions -

Lecture 15 - Virtual and Materialized Views

Virtual view


Materialized View

  Virtual View Materialized View
When updating R Database does nothing Ideally, DB refreshes V to reflect changes of R
When querying V optimize and run query that view represents simply use the view as the base table
  • essentially, its up to the DB implementation to update a materialized view when its underlying tables are queried. This is called Incremental View Maintenance (IVM)
  • views only represent algebraic expressions and are simply re-evaluated upon every query. Kind of like an alias.

  • At the end of a transaction which updates the relation R that a materialized view depends on, must be reflected in any materialized views which depend on the relation R.
    • Two options
      • delete and recompute view
      • incrementally maintain view
  • Capturing IVM as a computation of the changes of views

  • neglect update commands
    • think of update as a delete - insert
  • Capture transaction as two delta tables \(\Delta R^+\) and \(\Delta R^-\)
  • compute tuples to be deleted from view as \(\Delta V^-\) and \(\Delta V^+\)
  • delete \(\Delta V^-\) from V, and insert \(\Delta V^+\) into V

  • IVM - Eager version
    • problem: find efficient view updates after every table update
  • IVM - deferred version
    • allows choice of how long after transaction the DB updates a view
  • IVM - self-maintaining
    • works for some views
    • typically, view change will require the delta of the R, the old view, and the new relation state.
    • self-maintaining views don’t need the current state of the table to compute the new view.
  • Basic IVM Algorithm
    • Rule for \(V = R\bowtie S\)
      • \[\Delta V^+ = ((\Delta R^+\bowtie S) \cup (R \bowtie \Delta S^+)) - (\Delta R^+ \bowtie \Delta S^+)\]
      • \[\Delta V^- = (()) - ()\]
    • Rule for \(V = \sigma_c R\)
      • \[\Delta V^+ = \sigma_c \Delta R^+\]
    • Compose the above rules to do IVM for more complex view queries
      • \[\Delta V^- = \sigma_c \Delta R^-\]
      • e.g. \(V = T \bowtie \sigma_{A > 5}W\)
      • dumb way - compute two separate views by decomposing the views, then combine the views (inefficient)
      • for above example
        • \[\Delta V^+ = ((\Delta T^+ \bowtie \sigma_{A>5} W) \cup (T \bowtie \sigma_{A>5}\Delta W^+)) - (\Delta T^+ \bowtie \sigma_{A>5}\Delta W^+)\]
  • IVM with caching
    • associate intermediate views (caches) with subexpressions
    • bottom up: from updating caches to reaching the materialized view
    • caches will typically need indices
    • caches may or may not pay off as they incur a maintenance cost.