Chapter 20Query ProcessingTransparenciesownerNoclient

Chapter 20 - ObjectivesObjectives of query processing and optimization.Static versus dynamic query optimization.How a query is decomposed and semantically analyzed.How to create a R.A.T. to represent a query.Rules of equivalence for RA operations.How to apply heuristic transformation rules to improve efficiency of a query.

Chapter 20 - ObjectivesTypes of database statistics required to estimate cost of operations.Different strategies for implementing selection.How to evaluate cost and size of selection. Different strategies for implementing join.How to evaluate cost and size of join. Different strategies for implementing projection.How to evaluate cost and size of projection.

Chapter 20 - ObjectivesHow to evaluate the cost and size of other RA operations.How pipelining can be used to improve efficiency of queries.Difference between materialization and pipelining.Advantages of left-deep trees.

IntroductionIn network and hierarchical DBMSs, low-level procedural query language is generally embedded in high-level programming language.Programmers responsibility to select most appropriate execution strategy. With declarative languages such as SQL, user specifies what data is required rather than how it is to be retrieved. Relieves user of knowing what constitutes good execution strategy.

IntroductionAlso gives DBMS more control over system performance.

Two main techniques for query optimization:heuristic rules that order operations in a query; comparing different strategies based on relative costs, and selecting one that minimizes resource usage.

Disk access tends to be dominant cost in query processing for centralized DBMS.

Query Processing Activities involved in retrieving data from the database.

Aims of QP:transform query written in high-level language (e.g. SQL), into correct and efficient execution strategy expressed in low-level language (implementing RA);execute strategy to retrieve required data.

Query OptimizationActivity of choosing an efficient execution strategy for processing query.

As there are many equivalent transformations of same high-level query, aim of QO is to choose one that minimizes resource usage. Generally, reduce total execution time of query. May also reduce response time of query. Problem computationally intractable with large number of relations, so strategy adopted is reduced to finding near optimum solution.

Example 20.1 - Different StrategiesFind all Managers who work at a London branch.

SELECT *FROM Staff s, Branch bWHERE s.branchNo = b.branchNo AND (s.position = Manager AND b.city = London);

Example 20.1 - Different StrategiesThree equivalent RA queries are:(1) (position='Manager') (city='London') (Staff.branchNo=Branch.branchNo) (Staff X Branch) (2) (position='Manager') (city='London')(Staff Staff.branchNo=Branch.branchNo Branch)(3) (position='Manager'(Staff)) Staff.branchNo=Branch.branchNo (city='London' (Branch))

Example 20.1 - Different StrategiesAssume:1000 tuples in Staff; 50 tuples in Branch;50 Managers; 5 London branches;no indexes or sort keys;results of any intermediate operations stored on disk;cost of the final write is ignored;tuples are accessed one at a time.

Example 20.1 - Cost Comparison Cost (in disk accesses) are:

(1) (1000 + 50) + 2*(1000 * 50) = 101 050 (2) 2*1000 + (1000 + 50) = 3 050 (3) 1000 + 2*50 + 5 + (50 + 5) = 1 160

Cartesian product and join operations much more expensive than selection, and third option significantly reduces size of relations being joined together.

Phases of Query ProcessingQP has four main phases:

decomposition (consisting of parsing and validation);optimization;code generation;execution.

Phases of Query Processing

Dynamic versus Static OptimizationTwo times when first three phases of QP can be carried out:dynamically every time query is run;statically when query is first submitted. Advantages of dynamic QO arise from fact that information is up to date. Disadvantages are that performance of query is affected, time may limit finding optimum strategy.

Dynamic versus Static OptimizationAdvantages of static QO are removal of runtime overhead, and more time to find optimum strategy. Disadvantages arise from fact that chosen execution strategy may no longer be optimal when query is run. Could use a hybrid approach to overcome this.

Query DecompositionAims are to transform high-level query into RA query and check that query is syntactically and semantically correct. Typical stages are:analysis, normalization, semantic analysis, simplification, query restructuring.

AnalysisAnalyze query lexically and syntactically using compiler techniques. Verify relations and attributes exist. Verify operations are appropriate for object type.

Analysis - ExampleSELECT staff_noFROM StaffWHERE position > 10;

This query would be rejected on two grounds:staff_no is not defined for Staff relation (should be staffNo).Comparison >10 is incompatible with type position, which is variable character string.

AnalysisFinally, query transformed into some internal representation more suitable for processing. Some kind of query tree is typically chosen, constructed as follows:Leaf node created for each base relation.Non-leaf node created for each intermediate relation produced by RA operation.Root of tree represents query result.Sequence is directed from leaves to root.

Example 20.1 - R.A.T.

NormalizationConverts query into a normalized form for easier manipulation. Predicate can be converted into one of two forms:

Conjunctive normal form:(position = 'Manager' salary > 20000) (branchNo = 'B003')

Disjunctive normal form:(position = 'Manager' branchNo = 'B003' ) (salary > 20000 branchNo = 'B003')

Semantic AnalysisRejects normalized queries that are incorrectly formulated or contradictory. Query is incorrectly formulated if components do not contribute to generation of result. Query is contradictory if its predicate cannot be satisfied by any tuple. Algorithms to determine correctness exist only for queries that do not contain disjunction and negation.

Semantic AnalysisFor these queries, could construct:A relation connection graph. Normalized attribute connection graph.

Relation connection graph Create node for each relation and node for result. Create edges between two nodes that represent a join, and edges between nodes that represent projection. If not connected, query is incorrectly formulated.

Semantic Analysis - Normalized Attribute Connection GraphCreate node for each reference to an attribute, or constant 0. Create directed edge between nodes that represent a join, and directed edge between attribute node and 0 node that represents selection. Weight edges a b with value c, if it represents inequality condition (a b + c); weight edges 0 a with -c, if it represents inequality condition (a c).If graph has cycle for which valuation sum is negative, query is contradictory.

Example 20.2 - Checking Semantic CorrectnessSELECT p.propertyNo, p.streetFROM Client c, Viewing v, PropertyForRent pWHERE c.clientNo = v.clientNo ANDc.maxRent >= 500 AND c.prefType = Flat AND p.ownerNo = CO93;

Relation connection graph not fully connected, so query is not correctly formulated. Have omitted the join condition (v.propertyNo = p.propertyNo) .

Example 20.2 - Checking Semantic CorrectnessRelation Connection graph

Normalized attribute connection graph

Example 20.2 - Checking Semantic CorrectnessSELECT p.propertyNo, p.streetFROM Client c, Viewing v, PropertyForRent pWHERE c.maxRent > 500 AND c.clientNo = v.clientNo AND v.propertyNo = p.propertyNo AND c.prefType = Flat AND c.maxRent < 200;

Normalized attribute connection graph has cycle between nodes c.maxRent and 0 with negative valuation sum, so query is contradictory.

SimplificationDetects redundant qualifications, eliminates common sub-expressions, transforms query to semantically equivalent but more easily and efficiently computed form. Typically, access restrictions, view definitions, and integrity constraints are considered. Assuming user has appropriate access privileges, first apply well-known idempotency rules of boolean algebra.

Transformation Rules for RA OperationsConjunctive Selection operations can cascade into individual Selection operations (and vice versa).

pqr(R) = p(q(r(R)))

Sometimes referred to as cascade of Selection. branchNo='B003' salary>15000(Staff) = branchNo='B003'(salary>15000(Staff))

Transformation Rules for RA OperationsCommutativity of Selection.

p(q(R)) = q(p(R))

For example:

branchNo='B003'(salary>15000(Staff)) = salary>15000(branchNo='B003'(Staff))

Transformation Rules for RA OperationsIn a sequence of Projection operations, only the last in the sequence is required.

LM N(R) = L (R)

For example:

lNamebranchNo, lName(Staff) = lName (Staff)

Transformation Rules for RA OperationsCommutativity of Selection and Projection.

If predicate p involves only attributes in projection list, Selection and Projection operations commute: Ai, , Am(p(R)) = p(Ai, , Am(R)) where p {A1, A2, , Am}For example:fName, lName(lName='Beech'(Staff)) = lName='Beech'(fName,lName(Staff))

Transformation Rules for RA OperationsCommutativity of Theta join (and Cartesian product).R p S = S p RR X S = S X RRule also applies to Equijoin and Natural join. For example:Staff staff.branchNo=branch.branchNo Branch = Branch staff.branchNo=branch.branchNo Staff

Transformation Rules for RA OperationsCommutativity of Selection and Theta join (or Cartesian product).

If selection predicate involves only attributes of one of join relations, Selection and Join (or Cartesian product) operations commute:p(R r S) = (p(R)) r Sp(R X S) = (p(R)) X Swhere p {A1, A2, , An}

Transformation Rules for RA OperationsIf selection predicate is conjunctive predicate having form (p q), where p only involves attributes of R, and q only attributes of S, Selection and Theta join operations commute as:

p q(R r S) = (p(R)) r (q(S))p q(R X S) = (p(R)) X (q(S))

Transformation Rules for RA OperationsFor example:

position='Manager' city='London'(Staff Staff.branchNo=Branch.branchNo Branch) = (position='Manager'(Staff)) Staff.branchNo=Branch.branchNo (city='London' (Branch))

Transformation Rules for RA OperationsCommutativity of Projection and Theta join (or Cartesian product).

If projection list is of form L = L1 L2, where L1 only has attributes of R, and L2 only has attributes of S, provided join condition only contains attributes of L, Projection and Theta join commute:

L1L2(R r S) = (L1(R)) r (L2(S))

Transformation Rules for RA OperationsIf join condition contains additional attributes not in L (M = M1 M2 where M1 only has attributes of R, and M2 only has attributes of S), a final projection operation is required:

L1L2(R r S) = L1L2( (L1M1(R)) r (L2M2(S)))

Transformation Rules for RA OperationsFor example:

position,city,branchNo(Staff Staff.branchNo=Branch.branchNo Branch) = (position, branchNo(Staff)) Staff.branchNo=Branch.branchNo (city, branchNo (Branch))

and using the latter rule:position, city(Staff Staff.branchNo=Branch.branchNo Branch) = position, city ((position, branchNo(Staff)) Staff.branchNo=Branch.branchNo ( city, branchNo (Branch)))

Transformation Rules for RA OperationsCommutativity of Union and Intersection (but not set difference).

R S = S RR S = S R

Transformation Rules for RA OperationsCommutativity of Selection and set operations (Union, Intersection, and Set difference).

p(R S) = p(S) p(R)p(R S) = p(S) p(R)p(R - S) = p(S) - p(R)

Transformation Rules for RA OperationsCommutativity of Projection and Union.

L(R S) = L(S) L(R)

Associativity of Union and Intersection (but not Set difference).

(R S) T = S (R T)(R S) T = S (R T)

Transformation Rules for RA OperationsAssociativity of Theta join (and Cartesian product).Cartesian product and Natural join are always associative:(R S) T = R (S T)(R X S) X T = R X (S X T)If join condition q involves attributes only from S and T, then Theta join is associative:(R p S) q r T = R p r (S q T)

Transformation Rules for RA OperationsFor example:

(Staff Staff.staffNo=PropertyForRent.staffNo PropertyForRent) ownerNo=Owner.ownerNo staff.lName=Owner.lName Owner =

Staff staff.staffNo=PropertyForRent.staffNo staff.lName=lName (PropertyForRent ownerNo Owner)

Example 20.3 Use of Transformation RulesFor prospective renters of flats, find properties that match requirements and owned by CO93.

SELECT p.propertyNo, p.streetFROM Client c, Viewing v, PropertyForRent pWHERE c.prefType = Flat AND c.clientNo = v.clientNo AND v.propertyNo = p.propertyNo ANDc.maxRent >= p.rent AND c.prefType = p.type AND p.ownerNo = CO93;

Example 20.3 Use of Transformation Rules

Example 20.3 Use of Transformation Rules

Example 20.3 Use of Transformation Rules

Heuristical Processing Strategies Perform Selection operations as early as possible.Keep predicates on same relation together.Combine Cartesian product with subsequent Selection whose predicate represents join condition into a Join operation.Use associativity of binary operations to rearrange leaf nodes so leaf nodes with most restrictive Selection operations executed first.

Heuristical Processing StrategiesPerform Projection as early as possible.Keep projection attributes on same relation together.Compute common expressions once.If common expression appears more than once, and result not too large, store result and reuse it when required. Useful when querying views, as same expression is used to construct view each time.

Cost Estimation for RA OperationsMany different ways of implementing RA operations. Aim of QO is to choose most efficient one. Use formulae that estimate costs for a number of options, and select one with lowest cost.Consider only cost of disk access, which is usually dominant cost in QP.Many estimates are based on cardinality of the relation, so need to be able to estimate this.

Database StatisticsSuccess of estimation depends on amount and currency of statistical information DBMS holds. Keeping statistics current can be problematic. If statistics updated every time tuple is changed, this would impact performance. DBMS could update statistics on a periodic basis, for example nightly, or whenever the system is idle.

Typical Statistics for Relation RnTuples(R) - number of tuples in R.

bFactor(R) - blocking factor of R.

nBlocks(R) - number of blocks required to store R:

nBlocks(R) = [nTuples(R)/bFactor(R)]

Typical Statistics for Attribute A of Relation RnDistinctA(R) - number of distinct values that appear for attribute A in R.minA(R),maxA(R)minimum and maximum possible values for attribute A in R.SCA(R) - selection cardinality of attribute A in R. Average number of tuples that satisfy an equality condition on attribute A.

Statistics for Multilevel Index I on Attribute AnLevelsA(I) - number of levels in I.

nLfBlocksA(I) - number of leaf blocks in I.

Selection OperationPredicate may be simple or composite. Number of different implementations, depending on file structure, and whether attribute(s) involved are indexed/hashed. Main strategies are:Linear Search (Unordered file, no index).Binary Search (Ordered file, no index). Equality on hash key.Equality condition on primary key.

Selection OperationInequality condition on primary key.Equality condition on clustering (secondary) index. Equality condition on a non-clustering (secondary) index. Inequality condition on a secondary B+-tree index.

Estimating Cardinality of SelectionAssume attribute values are uniformly distributed within their domain and attributes are independent.

nTuples(S) = SCA(R)For any attribute B A of S, nDistinctB(S) =nTuples(S)if nTuples(S) < nDistinctB(R)/2nDistinctB(R)if nTuples(S) > 2*nDistinctB(R)[(nTuples(S) + nDistinctB(R))/3]otherwise

Linear Search (Ordered File, No Index)May need to scan each tuple in each block to check whether it satisfies predicate. For equality condition on key attribute, cost estimate is:

[nBlocks(R)/2]For any other condition, entire file may need to be searched, so more general cost estimate is:

nBlocks(R)

Binary Search (Ordered File, No Index)If predicate is of form A = x, and file is ordered on key attribute A, cost estimate:[log2(nBlocks(R))]Generally, cost estimate is:[log2(nBlocks(R))] + [SCA(R)/bFactor(R)] - 1First term represents cost of finding first tuple using binary search. Expect there to be SCA(R) tuples satisfying predicate.

Equality of Hash KeyIf attribute A is hash key, apply hashing algorithm to calculate target address for tuple. If there is no overflow, expected cost is 1. If there is overflow, additional accesses may be necessary.

Equality Condition on Primary KeyCan use primary index to retrieve single record satisfying condition. Need to read one more block than number of index accesses, equivalent to number of levels in index, so estimated cost is:

nLevelsA(I) + 1

Inequality Condition on Primary KeyCan first use index to locate record satisfying predicate (A = x). Provided index is sorted, records can be found by accessing all records before/after this one. Assuming uniform distribution, would expect half the records to satisfy inequality, so estimated cost is:

nLevelsA(I) + [nBlocks(R)/2]

Equality Condition on Clustering IndexCan use index to retrieve required records. Estimated cost is:

nLevelsA(I) + [SCA(R)/bFactor(R)]

Second term is estimate of number of blocks that will be required to store number of tuples that satisfy equality condition, represented as SCA(R).

Equality Condition on Non-Clustering IndexCan use index to retrieve required records. Have to assume that tuples are on different blocks (index is not clustered this time), so estimated cost becomes:

nLevelsA(I) + [SCA(R)]

Inequality Condition on a Secondary B+-Tree IndexFrom leaf nodes of tree, can scan keys from smallest value up to x (< or or >=). Assuming uniform distribution, would expect half the leaf node blocks to be accessed and, via index, half the file records to be accessed. Estimated cost is :

nLevelsA(I) + [nLfBlocksA(I)/2 + nTuples(R)/2]

Composite Predicates - Conjunction without DisjunctionMay consider following approaches:- If one attribute has index or is ordered, can use one of above selection strategies. Can then check each retrieved record.- For equality on two or more attributes, with composite index (or hash key) on combined attributes, can search index directly.- With secondary indexes on one or more attributes (involved only in equality conditions in predicate), could use record pointers if exist.

Composite Predicates - Selections with DisjunctionIf one term contains an (OR), and term requires linear search, entire selection requires linear search. Only if index or sort order exists on every term can selection be optimized by retrieving records that satisfy each condition and applying union operator. Again, record pointers can be used if they exist.

Join OperationMain strategies for implementing join:

Block Nested Loop Join.Indexed Nested Loop Join.Sort-Merge Join.Hash Join.

Estimating Cardinality of JoinCardinality of Cartesian product is:nTuples(R) * nTuples(S)More difficult to estimate cardinality of any join as depends on distribution of values. Worst case, cannot be any greater than this value.

Estimating Cardinality of JoinIf assume uniform distribution, can estimate for Equijoins with a predicate (R.A = S.B) as follows:If A is key of R: nTuples(T) nTuples(S)If B is key of S: nTuples(T) nTuples(R)Otherwise, could estimate cardinality of join as:nTuples(T) = SCA(R)*nTuples(S) ornTuples(T) = SCB(S)*nTuples(R)

Block Nested Loop JoinSimplest join algorithm is nested loop that joins two relations together a tuple at a time. Outer loop iterates over each tuple in R, and inner loop iterates over each tuple in S. As basic unit of reading/writing is a disk block, better to have two extra loops that process blocks.Estimated cost of this approach is:

nBlocks(R) + (nBlocks(R) * nBlocks(S))

Block Nested Loop JoinCould read as many blocks as possible of smaller relation, R say, into database buffer, saving one block for inner relation and one for result. New cost estimate becomes:

nBlocks(R)+[nBlocks(S)*(nBlocks(R)/(nBuffer-2))]If can read all blocks of R into the buffer, this reduces to:

nBlocks(R) + nBlocks(S)

Indexed Nested Loop JoinIf have index (or hash function) on join attributes of inner relation, can use index lookup. For each tuple in R, use index to retrieve matching tuples of S. Cost of scanning R is nBlocks(R), as before. Cost of retrieving matching tuples in S depends on type of index and number of matching tuples. If join attribute A in S is PK, cost estimate is:

nBlocks(R) + nTuples(R)*(nlevelsA(I) + 1)

Sort-Merge JoinFor Equijoins, most efficient join is when both relations are sorted on join attributes. Can look for qualifying tuples merging relations. May need to sort relations first. Now tuples with same join value are in order. If assume join is *:* and each set of tuples with same join value can be held in database buffer at same time, then each block of each relation need only be read once.

Sort-Merge JoinCost estimate for the sort-merge join is:

nBlocks(R) + nBlocks(S)If a relation has to be sorted, R say, add:

nBlocks(R)*[log2(nBlocks(R)]

Hash JoinFor Natural or Equijoin, hash join may be used. Idea is to partition relations according to some hash function that provides uniformity and randomness. Each equivalent partition should hold same value for join attributes, although it may hold more than one value. Cost estimate of hash join as:

3(nBlocks(R) + nBlocks(S))

Projection OperationTo implement projection need following steps:Removal of attributes that are not required.Elimination of any duplicate tuples produced from previous step. Only required if projection attributes do not include a key. Two main approaches to eliminating duplicates: sorting;hashing.

Estimating Cardinality of ProjectionWhen projection contains key, cardinality is:

nTuples(S) = nTuples(R)If projection consists of a single non-key attribute, estimate is:

nTuples(S) = SCA(R)Otherwise, could estimate cardinality as:

nTuples(S) min(nTuples(R), im=1(nDistinctai(R)))

Duplicate Elimination using SortingSort tuples of reduced relation using all remaining attributes as sort key. Duplicates will now be adjacent and can be removed easily. Estimated cost of sorting is:

nBlocks(R)*[log2(nBlocks(R))].Combined cost is:

nBlocks(R) + nBlocks(R)*[log2(nBlocks(R))]

Duplicate Elimination using HashingTwo phases: partitioning and duplicate elimination. In partitioning phase, for each tuple in R, remove unwanted attributes and apply hash function to combination of remaining attributes, and write reduced tuple to hashed value. Two tuples that belong to different partitions are guaranteed not to be duplicates. Estimated cost is: nBlocks(R) + nB

Set OperationsCan be implemented by sorting both relations on same attributes, and scanning through each of sorted relations once to obtain desired result. Could use sort-merge join as basis. Estimated cost in all cases is:

nBlocks(R) + nBlocks(S) + nBlocks(R)*[log2(nBlocks(R))] + nBlocks(S)*[log2(nBlocks(S))]Could also use hashing algorithm.

Estimating Cardinality of Set OperationsAs duplicates are eliminated when performing union, difficult to estimate cardinality, but can give an upper and lower bound as:

max(nTuples(R), nTuples(S)) nTuples(T) nTuples(R) + nTuples(S)For set difference, can also give upper and lower bound:

0 nTuples(T) nTuples(R)

Aggregate OperationsSELECT AVG(salary)FROM Staff;

To implement query, could scan entire Staff relation and maintain running count of number of tuples read and sum of all salaries. Easy to compute average from these two running counts.

Aggregate OperationsSELECT AVG(salary)FROM StaffGROUP BY branchNo;

For grouping queries, can use sorting or hashing algorithms similar to duplicate elimination. Can estimate cardinality of result using estimates derived earlier for selection.

PipeliningMaterialization - output of one operation is stored in temporary relation for processing by next. Could also pipeline results of one operation to another without creating temporary relation. Known as pipelining or on-the-fly processing.Pipelining can save on cost of creating temporary relations and reading results back in again. Generally, pipeline is implemented as separate process or thread.

Types of Trees

PipeliningWith linear trees, relation on one side of each operator is always a base relation.However, as need to examine entire inner relation for each tuple of outer relation, inner relations must always be materialized.This makes left-deep trees appealing as inner relations are always base relations.Reduces search space for optimum strategy, and allows QO to use dynamic processing.Not all execution strategies are considered.

Query Optimization in OracleOracle supports two approaches to query optimization: rule-based and cost-based.

Rule-based15 rules, ranked in order of efficiency. Particular access path for a table only chosen if statement contains a predicate or other construct that makes that access path available. Score assigned to each execution strategy using these rankings and strategy with best (lowest) score selected.

QO in Oracle Rule-BasedWhen 2 strategies produce same score, tie-break resolved by making decision based on order in which tables occur in the SQL statement.

QO in Oracle Rule-based: Example SELECT propertyNoFROM PropertyForRentWHERE rooms > 7 AND city = LondonSingle-column access path using index on city from WHERE condition (city = London). Rank 9.Unbounded range scan using index on rooms from WHERE condition (rooms > 7). Rank 11. Full table scan - rank 15.Although there is index on propertyNo, column does not appear in WHERE clause and so is not considered by optimizer. Based on these paths, rule-based optimizer will choose to use index based on city column.

QO in Oracle Cost-BasedTo improve QO, Oracle introduced cost-based optimizer in Oracle 7, which selects strategy that requires minimal resource use necessary to process all rows accessed by query (avoiding above tie-break anomaly). User can select whether minimal resource usage is based on throughput or based on response time, by setting the OPTIMIZER_MODE initialization parameter. Cost-based optimizer also takes into consideration hints that the user may provide.

QO in Oracle StatisticsCost-based optimizer depends on statistics for all tables, clusters, and indexes accessed by query. Users responsibility to generate these statistics and keep them current. Package DBMS_STATS can be used to generate and manage statistics.Whenever possible, Oracle uses a parallel method to gather statistics, although index statistics are collected serially. EXECUTE DBMS_STATS.GATHER_SCHEMA_STATS(Manager);

QO in Oracle HistogramsPreviously made assumption that data values within columns of a table are uniformly distributed. Histogram of values and their relative frequencies gives optimizer improved selectivity estimates in presence of non-uniform distribution.

QO in Oracle Histograms(a) shows uniform distribution of rooms and (b) the actual non-uniform distribution. First can be stored compactly as low value (1) and a high value (10), and as total count of all frequencies (in this case, 100).

QO in Oracle HistogramsHistogram is data structure that can improve estimates of number of tuples in result. Two types of histogram:width-balanced histogram, which divides data into a fixed number of equal-width ranges (called buckets) each containing count of number of values falling within that bucket;height-balanced histogram, which places approximately same number of values in each bucket so that end points of each bucket are determined by how many values are in that bucket.

QO in Oracle Histograms(a) width-balanced for rooms with 5 buckets. Each bucket of equal width with 2 values (1-2, 3-4, etc.)(b) height-balanced height of each column is 20 (100/5).

QO in Oracle Viewing Execution Plan