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B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

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Review Last time discussed File Organization –Unordered heap files –Sorted Files –Clustered Trees –Unclustered Trees –Unclustered Hash Tables Indexes –B-Trees – dynamic, good for changing data, range queries –Hash tables – fastest for equality queries, useless for range queries
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B-Trees, Part 2 Hash-Based Indexes R&G Chapter 10 Lecture 10
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Page 1: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

B-Trees, Part 2Hash-Based

Indexes R&G Chapter 10Lecture 10

Page 2: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Administrivia• The new Homework 3 now available

– Due 1 week from Sunday– Homework 4 available the week after

• Midterm exams available here

Page 3: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Review• Last time discussed File Organization

– Unordered heap files– Sorted Files– Clustered Trees– Unclustered Trees– Unclustered Hash Tables

• Indexes– B-Trees – dynamic, good for changing data,

range queries– Hash tables – fastest for equality queries,

useless for range queries

Page 4: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Review (2)

• For any index, 3 alternatives for data entries k*:– Data record with key value k– <k, rid of data record with search key value k>– <k, list of rids of data records with search key

k>– Choice orthogonal to the indexing technique

Page 5: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Today: • Indexes

– Composite Keys, Index-Only Plans

• B-Trees – details of insertion and deletion

• Hash Indexes– How to implement with changing data sets

Page 6: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Inserting a Data Entry into a B+ Tree• Find correct leaf L. • Put data entry onto L.

– If L has enough space, done!– Else, must split L (into L and a new node L2)

• Redistribute entries evenly, copy up middle key.• Insert index entry pointing to L2 into parent of L.

• This can happen recursively– To split index node, redistribute entries evenly, but

push up middle key. (Contrast with leaf splits.)• Splits “grow” tree; root split increases height.

– Tree growth: gets wider or one level taller at top.

Page 7: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Indexes with Composite Search Keys

• Composite Search Keys: Search on a combination of fields.– Equality query: Every field value is

equal to a constant value. E.g. wrt <sal,age> index:

• age=20 and sal =75– Range query: Some field value is

not a constant. E.g.:• age =20; or age=20 and sal > 10

• Data entries in index sorted by search key to support range queries.– Lexicographic order, or– Spatial order.

sue 13 75

bobcaljoe 12

10

208011

12

name age sal

<sal, age>

<age, sal> <age>

<sal>

12,2012,10

11,80

13,75

20,12

10,12

75,1380,11

11121213

10207580

Data recordssorted by name

Data entries in indexsorted by <sal,age>

Data entriessorted by <sal>

Examples of composite keyindexes using lexicographic order.

Page 8: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Composite Search Keys• To retrieve Emp records with age=30 AND

sal=4000, an index on <age,sal> would be better than an index on age or an index on sal.– Choice of index key orthogonal to clustering etc.

• If condition is: 20<age<30 AND 3000<sal<5000: – Clustered tree index on <age,sal> or <sal,age> is

best.• If condition is: age=30 AND 3000<sal<5000:

– Clustered <age,sal> index much better than <sal,age> index!

• Composite indexes are larger, updated more often.

Page 9: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Index-Only Plans• A number of

queries can be answered without retrieving any tuples from one or more of the relations involved if a suitable index is available.

SELECT D.mgrFROM Dept D, Emp EWHERE D.dno=E.dno

SELECT D.mgr, E.eidFROM Dept D, Emp EWHERE D.dno=E.dno

SELECT E.dno, COUNT(*)FROM Emp EGROUP BY E.dno

SELECT E.dno, MIN(E.sal)FROM Emp EGROUP BY E.dno

SELECT AVG(E.sal)FROM Emp EWHERE E.age=25 AND E.sal BETWEEN 3000 AND 5000

<E.dno><E.dno,E.eid>

Tree index!

<E.dno>

<E.dno,E.sal>Tree index!

<E. age,E.sal> or<E.sal, E.age>

Tree!

Page 10: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Index-Only Plans (Contd.)• Index-only plans

are possible if the key is <dno,age> or we have a tree index with key <age,dno>– Which is better?– What if we

consider the second query?

SELECT E.dno, COUNT (*)FROM Emp EWHERE E.age=30GROUP BY E.dno

SELECT E.dno, COUNT (*)FROM Emp EWHERE E.age>30GROUP BY E.dno

Page 11: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

B-Trees: Insertion and Deletion• Insertion

– Find leaf where new record belongs– If leaf is full, redistribute*– If siblings too full, split, copy middle key up– If index too full, redistribute*– If index siblings full, split, push middle key up

• Deletion– Find leaf where record exists, remove it– If leaf is less than 50% empty, redistribute*– If siblings too empty, merge, remove key above– If index node above too empty, redistribute*– If index siblings too empty, merge, move above key

down

Page 12: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

B-Trees: For Homework 2• Insertion

– Find leaf where new record belongs– If leaf is full, redistribute*– If siblings too full, split, copy middle key up– If index too full, redistribute*– If index siblings full, split, push middle key up

• Deletion– Find leaf where record exists, remove it– If leaf is less than 50% empty, redistribute*– If siblings too empty, merge, remove key above– If index node above too empty, redistribute*– If index siblings too empty, merge, move above key

downThis means that after deletion, nodes will often be < 50% full

Page 13: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

B-Trees: For Homework 2 (cont)• Splits

– When splitting nodes, choose the middle key– If there are even number of keys, choose

middle

• Your code must Handle Duplicate Keys– We promise that there will never be more

than 1 page of duplicate values.– Thus, when splitting, if the middle key is

identical to the key to the left, you must find the closest splittable key to the middle.

Page 14: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

B-Tree Demo

Page 15: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Hashing• Static and dynamic hashing techniques

exist; trade-offs based on data change over time

• Static Hashing– Good if data never changes

• Extendable Hashing – Uses directory to handle changing data

• Linear Hashing– Avoids directory, usually faster

Page 16: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Static Hashing• # primary pages fixed, allocated

sequentially, never de-allocated; overflow pages if needed.

• h(k) mod N = bucket to which data entry with key k belongs. (N = # of buckets)

h(key) mod N

hkey

Primary bucket pages Overflow pages

20

N-1

Page 17: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Static Hashing (Contd.)• Buckets contain data entries.

• Hash fn works on search key field of record r. Must distribute values over range 0 ... N-1.– h(key) = (a * key + b) usually works well.– a and b are constants; lots known about how to

tune h.

• Long overflow chains can develop and degrade performance. – Extendible and Linear Hashing: Dynamic techniques

to fix this problem.

Page 18: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Extendible Hashing• Situation: Bucket (primary page) becomes full.

• Why not re-organize file by doubling # of buckets?– Reading and writing all pages is expensive!

• Idea: Use directory of pointers to buckets, – double # of buckets by doubling the directory, – splitting just the bucket that overflowed!– Directory much smaller than file, doubling much

cheaper. – No overflow pages!– Trick lies in how hash function is adjusted!

Page 19: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Extendible Hashing Details• Need directory with pointer to each bucket

• Need hash function to incrementally double range– can just use increasingly more LSBs of h(key)

• Must keep track of global “depth”– how many times directory doubled so far

• Must keep track of local “depth”– how many time each bucket has been split

Page 20: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Example• Directory is array of size 4.• To find bucket for r, take

last `global depth’ # bits of h(r); we denote r by h(r).– If h(r) = 5 = binary

101, it is in bucket pointed to by 01.

Insert: If bucket is full, split it (allocate new page, re-distribute). If necessary, double the directory. (As we will see, splitting a bucket does not always require doubling; we can tell by comparing global depth with local depth for the split bucket.)

13*00

01

10

11

2

2

2

2

2

LOCAL DEPTH

GLOBAL DEPTH

DIRECTORY

Bucket A

Bucket B

Bucket C

Bucket D

DATA PAGES

10*

1* 21*

4* 12* 32* 16*

15* 7* 19*

5*

Page 21: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Insert h(r)=20 (Causes Doubling)

00011011

2 2

2

2

LOCAL DEPTH 2

DIRECTORY

GLOBAL DEPTHBucket A

Bucket B

Bucket C

Bucket D

1* 5* 21*13*

32*16*

10*

15* 7* 19*

4* 12*

19*

2

2

2

000001010

011100101

110111

3

3

3DIRECTORY

Bucket A

Bucket B

Bucket C

Bucket D

Bucket A2(`split image'of Bucket A)

32*

1* 5* 21*13*

16*

10*

15* 7*

4* 20*12*

LOCAL DEPTH

GLOBAL DEPTH

Page 22: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Now Insert h(r)=9 (Causes Split Only)

19*

3

2

2

000001010

011100101

110111

3

3

3

DIRECTORY

Bucket A

Bucket B

Bucket C

Bucket D

Bucket A2

32*

1* 9*

16*

10*

15* 7*

4* 20*12*

LOCAL DEPTH

GLOBAL DEPTH

3

Bucket B25* 21* 13*

19*

2

2

2

000001010

011100101

110111

3

3

3DIRECTORY

Bucket A

Bucket B

Bucket C

Bucket D

Bucket A2(`split image'of Bucket A)

32*

1* 5* 21*13*

16*

10*

15* 7*

4* 20*12*

LOCAL DEPTH

GLOBAL DEPTH

Page 23: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Points to Note• 20 = binary 10100. Last 2 bits (00) tell us r belongs

in A or A2. Last 3 bits needed to tell which.– Global depth of directory: Max # of bits needed to tell

which bucket an entry belongs to.– Local depth of a bucket: # of bits used to determine if

splitting bucket will also double directory

• When does bucket split cause directory doubling?– If, before insert, local depth of bucket = global depth.

• Insert causes local depth to become > global depth; • directory is doubled by copying it over and `fixing’ pointer to

split image page. • (Use of least significant bits enables efficient doubling via

copying of directory!)

Page 24: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Directory DoublingWhy use least significant bits in directory?

Allows for doubling via copying!

13*00

01

10

11

2

2

2

2

2

10*

1* 21*

4* 12* 32* 16*

15* 7* 19*

5* 13*

000

001

010

011

32

2

2

2

10*

1* 21*

4* 12* 32* 16*

15* 7* 19*

5*

100

101

110

111

Page 25: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Deletion• Delete: If removal of data entry makes bucket

empty, can be merged with `split image’. If each directory element points to same bucket as its split image, can halve directory.

13*00

01

10

11

2

2

2

2

2

10*

1* 21*

4* 12* 32* 16*

15*

5* 13*00

01

10

11

2

1

2

2

1* 21*

4* 12* 32* 16*

15*

5*Delete 10

Page 26: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Deletion (cont)• Delete: If removal of data entry makes bucket

empty, can be merged with `split image’. If each directory element points to same bucket as its split image, can halve directory.

13*

00

01

10

11

2 1

1

1* 21*

4* 12* 32* 16*

5*

Delete 15

13*00

01

10

11

2

1

2

2

1* 21*

4* 12* 32* 16*

15*

5*

13*

0

1

1 1

1

1* 21*

4* 12* 32* 16*

5*

Page 27: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Comments on Extendible Hashing

• If directory fits in memory, equality search answered with one disk access; else two.– 100MB file, 100 bytes/rec, 4K pages contains

1,000,000 records (as data entries) and 25,000 directory elements; chances are high that directory will fit in memory.

– Directory grows in spurts, and, if the distribution of hash values is skewed, directory can grow large.

• Biggest problem:– Multiple entries with same hash value cause problems!– If bucket already full of same hash value, will keep

doubling forever! So must use overflow buckets if dups.

Page 28: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Linear Hashing• This is another dynamic hashing scheme, an

alternative to Extendible Hashing.

• LH handles the problem of long overflow chains without using a directory, and handles duplicates.

• Idea: Use a family of hash functions h0, h1, h2, ...– hi(key) = h(key) mod(2iN); N = initial # buckets– h is some hash function (range is not 0 to N-1)– If N = 2d0, for some d0, hi consists of applying h and

looking at the last di bits, where di = d0 + i.– hi+1 doubles the range of hi (similar to directory doubling)

Page 29: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

• Let’s start with N = 4 Buckets– Start at “round” 0, “next” 0, have 2round buckets– Each time any bucket fills, split “next” bucket

– If (O ≤ hround(key) < Next), use hround+1(key) instead

Linear Hashing Example

13*

10*

1* 21*

4* 12* 32* 16*

15*

5*

Start

13*

10*

1* 21*

4* 12*

32* 16*

15*

5*

Add 9Nex

t9*Nex

t

13*

10*

1* 21*

4* 12*

32* 16*

15*

5*

Add 20

9*Next

20*

Nround Nround

Nround

Page 30: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

• Overflow chains do exist, but eventually get split• Instead of doubling, new buckets added one-at-a-time

Linear Hashing Example (cont)

13*

10*

1* 21*

4* 12*

32* 16*

15*

5*

Add 6

9*Next

20*

6*

13*

10*

1*

21*

4* 12*

32* 16*

15*

5*

Add 17

9*

Next

20*

6*

Nround Nround

Page 31: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Linear Hashing (Contd.)• Directory avoided in LH by using overflow pages,

and choosing bucket to split round-robin.– Splitting proceeds in `rounds’. Round ends when all

NR initial (for round R) buckets are split. Buckets 0 to Next-1 have been split; Next to NR yet to be split.

– Current round number also called Level.

– Search: To find bucket for data entry r, find hround(r):• If hround(r) in range `Next to NR’ , r belongs here.• Else, r could be hround(r) or hround(r) + NR;

– must apply hround+1(r) to find out.

Page 32: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Overview of LH File • In the middle of a round.

Levelh

Buckets that existed at thebeginning of this round:

this is the range of

NextBucket to be split

of other buckets) in this round

Levelh search key value )(

search key value )(

Buckets split in this round:If is in this range, must useh Level+1

`split image' bucket.to decide if entry is in

created (through splitting`split image' buckets:

Page 33: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Linear Hashing (Contd.)• Insert: Find bucket by applying hround / hround+1:

– If bucket to insert into is full:• Add overflow page and insert data entry.• (Maybe) Split Next bucket and increment Next.

• Can choose any criterion to `trigger’ split.

• Since buckets are split round-robin, long overflow chains don’t develop!

• Doubling of directory in Extendible Hashing is similar; switching of hash functions is implicit in how the # of bits examined is increased.

Page 34: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Another Example of Linear Hashing

• On split, hLevel+1 is used to re-distribute entries.

0hh

1

(This infois for illustrationonly!)

Round=0, N=4

00

01

10

11

000

001

010

011

(The actual contentsof the linear hashedfile)

Next=0PRIMARY

PAGES

Data entry rwith h(r)=5

Primary bucket page

44* 36*32*

25*9* 5*

14* 18*10*30*

31*35* 11*7*

0hh

1

Round=0

00

01

10

11

000

001

010

011

Next=1

PRIMARYPAGES

44* 36*

32*

25*9* 5*

14* 18*10*30*

31*35* 11*7*

OVERFLOWPAGES

43*

00100

Page 35: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Example: End of a Round

0hh1

22*

00

01

10

11

000

001

010

011

00100

Next=3

01

10

101

110

Round=0PRIMARYPAGES

OVERFLOWPAGES

32*

9*

5*

14*

25*

66* 10*18* 34*

35*31* 7* 11* 43*

44*36*

37*29*

30*

0hh1

37*

00

01

10

11

000

001

010

011

00100

10

101

110

Next=0

Round=1

111

11

PRIMARYPAGES

OVERFLOWPAGES

11

32*

9* 25*

66* 18* 10* 34*

35* 11*

44* 36*

5* 29*

43*

14* 30* 22*

31*7*

50*

Page 36: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

LH Described as a Variant of EH• The two schemes are actually quite similar:

– Begin with an EH index where directory has N elements.– Use overflow pages, split buckets round-robin.– First split is at bucket 0. (Imagine directory being

doubled at this point.) But elements <1,N+1>, <2,N+2>, ... are the same. So, need only create directory element N, which differs from 0, now.

• When bucket 1 splits, create directory element N+1, etc.• So, directory can double gradually. Also, primary

bucket pages are created in order. If they are allocated in sequence too (so that finding i’th is easy), we actually don’t need a directory! Voila, LH.

Page 37: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Summary• Hash-based indexes: best for equality searches,

cannot support range searches.

• Static Hashing can lead to long overflow chains.

• Extendible Hashing avoids overflow pages by splitting a full bucket when a new data entry is to be added to it. (Duplicates may require overflow pages.)– Directory to keep track of buckets, doubles periodically.– Can get large with skewed data; additional I/O if this

does not fit in main memory.

Page 38: B-Trees, Part 2 Hash-Based Indexes RG Chapter 10 Lecture 10.

Summary (Contd.)• Linear Hashing avoids directory by splitting

buckets round-robin, and using overflow pages. – Overflow pages not likely to be long.– Duplicates handled easily.– Space utilization could be lower than Extendible

Hashing, since splits not concentrated on `dense’ data areas.• Can tune criterion for triggering splits to trade-off

slightly longer chains for better space utilization.

• For hash-based indexes, a skewed data distribution is one in which the hash values of data entries are not uniformly distributed!


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