ECE 569 Database System Engineering Spring 2004

Spring 2004ECE569 Lecture 04-2.1

ECE 569 Database System Engineering

Spring 2004

Yanyong Zhang www.ece.rutgers.edu/~yyzhang

Course URL www.ece.rutgers.edu/~yyzhang/spring04

http://www.ece.rutgers.edu/~yyzhang

http://www.ece.rutgers.edu/~yyzhang


Associative access The system is not asked to retrieve tuples based

on information about their storage location; rather, it has to find all tuples the attribute values of which fulfill certain conditions – associative access.

Associative access can be realized by sequential scanning, which happens for complicated queries.select R.x, S.y,from R,Swhere R.k = S.f and R.b < 12;

But for simple selection predicates, this is very slow (even for an in-memory database)


Access Path The class of algorithms and data structures

designed for translating attribute values into TID, or into other types of internal addresses of tuples having those attribute values, is called access paths.

Depending on what kind of selection predicate is to be supported, the techniques for associative access vary greatly.


Content addressability techniques Primary key access.

A tuple of a relation must be retrieved efficiently via the value of its primary (unique) key(s). e.g., key-sequenced files and hased files.

Point query vs. range query

Secondary key access A set of tuples are produced

Multi-table access Tuple access is often based on relationships between

different tuples. E.g., all orders placed by a given customer have to be found.


Associative access path techniques Hashing (key transformation)

Using the primary key value as a parameter to a function, which returns the storage location of the tuple.

Key comparison Maintaining a dynamic search structure on the set of

values in the key attribute. These values can be organized into tables, lists, trees, and so on

e.g, B+ tree


Operations on files (heap files) Assumptions

n = number of records in file R = number of records that can fit in block

Lookup – Given a key find corresponding record On average, n / (2R) block accesses.

Insertion – add record to file (allows duplicates) Read last block; it may need to allocate a new block.

Approximately, requires 2 accesses

Deletion – delete record look up record n / (2R) Write back to disk (1 access) Reorganize (unpinned) – move tuple from last page to utilize

space (2 disk accesses)


Hashed Files File is divided into B buckets

Hash function h maps elements of the key space to range [0, B)

Key space is large and unevenly distributed- SSNs as character strings- Each character takes on at most 10 of the possible 256

values Hash function h must map key values evenly among a

relatively small number of values.


Hash-based associative access

FOLDING HASHING

Ran

ge o

f pos

itive

inte

gers

tupleaddressspace

Range of PotentialKey Values(the shadedareas denoteused keyvalues)


Folding Convert arbitrary data types to a positive integer h can be

applied to. Reduce number of bits so that arithmetic is efficient. Example: Key is “Keefe” and 16803

Key value is the concatenation of byte representation of individual fields

Folded value of key is 0x4b 0x65 0x65 0x66 0x65 0x0 0x0 0x0 0x41 0xa3 Partition result into words and combine using XOR0x4b 0x65 0x65 0x660x65 0x0 0x0 0x00x41 0xa3 0x0 0x0

0x6f 0xc6 0x65 0x66 = 1875273062


Hashing goal of hashing

How to choose hash function if all the key values are uniformly distributed?

The critical issue is to produce 1:1 mapping

Collision: different inputs are mapped to the same output.

The criteria of a good hash function is to keep the collision as small as possible.


Static Hashing Input: folded key values

Output: bytes (relative to the beginning of the file), blocks ??

Bytes are not good because of the varying tuple size. A block/page is called a bucket.

H: {0 … 232-1} -> {0, B-1} Continuous allocation Fixed size: B pages are allocated at file creation time. Insert

- Determine the bucket- Check the bucket ( collision may happen)


How to find a good hash function Division / remainder (Congruential hashing)

H(Kb) = kb mod B where kb is folded key value and B is the number of buckets.

Nth power Compute kb

N, and from the resulting bit string (n x 31 bits) take log2B bits from the middle.

Base transformation

Polynomial division

Numerical analysis

encryption


Performance Assumption

Perfect hash function (tuples are uniformly distributed over B buckets)

Lookup ½ n/R 1/B To finish first match n/R 1/B If tuple does not exist

Insertion n/R 1/B + 1 Test for duplicates 1 Otherwise

Deletion ½ n/R 1/B delete first match


Collision Two keys collide if they hash to same value

A bucket with room for R tuples can accommodate R – 1 collisions before it overflows

Internal resolution: Place overflow blocks in another bucket- (h(K) + 1) mod B linear probing- (h2(h1(K)) multiple hashing


Collision - continued External resolution: Allocation overflow block, link to

overflow chain

buckets Overflowpages


Discussion What are the disadvantages of static hashing?

How do you limit the number of pages accessed when retrieving a tuple, for both external and internal resolution?


Trie The buckets will dynamically grow/shrink/balance

Fundamental: trie

A

B

C

0

0

1

1

(a)

A

B

C

0

0

1

1

0

1 D

(b)

000001010011100101110111

AD

B

C

(c)


Dynamic hashing function We need a hash function whose range of values

can change dynamically

One such hash function can be constructed using a series of functions hi(k), i = 0, 1, …., such that for any k, either hi(k) = hi-1(k) or hi(k) = hi-1(k)+2i-1.

Choose H(k), which maps the key space into random bit patterns of length m, for m sufficiently large. Then hi(k) may be defined as the integers formed by the last i bits of H(k).


Extendible Hashing The number of buckets can grow/shrink.

An intermediate data structure translates the hash results into page addresses. This data structure needs to be as compact as possible.

Hashes into an array of pointer to buckets (directory). The array is small enough to be kept in memory.


Directory Growth• To adapt to dynamically varying size of hash file-

modify directory size

• Assume a hash function h(Kb) that produces a bit string s.

• The directory is of size 2d. d is called the global depth and is initially 0.

• Use least significant d bits of s to determine bucket to access

• Each bucket has a corresponding local depth in the range [0, d] which indicates the difference between all the records in this bucket


Example

00x13,0x10 0

d = 0

00x13,0x10 0

d = 1

1

00x10,0x00 1

d = 1

1

0x13,0x07 1

000x10,0x00 1

d = 2

01

210

11

0x13,0x07 2

Insert 0x13, 0x10, 0x07, 0x00, 0x1fEach page can contain no more than 2 tuples


Example – insert 0x1f

000 0x10,0x00 1

d = 3

001

2 010

011

0x13 3

100

101

110

111

0x07,0x1f 3

local degree = global degree – log2(# of arrows pointing to this bucket)


Performance 2 steps for retrieving a tuple

If we can keep the directory in memory, each retrieval is one page access

Assuming 4 bytes per entry, 4KB pages, 1GB hash files, and we want to keep the entire directory in memory, what is the minimum buffer size?


Discussion How easy is it to keep the directory in the

memory?

How do we reduce the structure when the file shrinks?

How do you make the directory small, and increase space utilization? (deferred splitting)


Linear hashing

(a)00 01 10 11

(b)000 01 10 11

w

100

(c)000 001 10 11

w100 101

a b c d

a b c d

a b c d

x

(c)000 001 010 11

w100 101

a b c d

x

110

y

(d)000 001 010 011

w100 101

a b c d x110

y

111

d = 2

d = 2

d = 2

d = 2 d = 3


Addressing in linear hashing Which hash function should be used?

d is the degree, p is the address of the next page to split The algorithm is as follows: begin if (hd(k) >= p) then page:= hd(k)

else page := hd+1(k)

if necessary, chase the overflow chain end


Discuss Where to store the overflow?

Documents

ECE 569 Database System Engineering Spring 2004