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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
Spring 2004ECE569 Lecture 04-2.2
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,S
where R.k = S.f and R.b < 12;
But for simple selection predicates, this is very slow (even for an in-memory database)
Spring 2004ECE569 Lecture 04-2.3
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.
Spring 2004ECE569 Lecture 04-2.4
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.
Spring 2004ECE569 Lecture 04-2.5
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
Spring 2004ECE569 Lecture 04-2.6
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)
Spring 2004ECE569 Lecture 04-2.7
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.
Spring 2004ECE569 Lecture 04-2.8
Hash-based associative access
FOLDING HASHING
Ran
ge
of
po
siti
ve in
teg
ers
tupleaddressspace
Range of PotentialKey Values(the shadedareas denoteused keyvalues)
Spring 2004ECE569 Lecture 04-2.9
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 XOR
0x4b 0x65 0x65 0x66
0x65 0x0 0x0 0x0
0x41 0xa3 0x0 0x0
0x6f 0xc6 0x65 0x66 = 1875273062
Spring 2004ECE569 Lecture 04-2.10
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.
Spring 2004ECE569 Lecture 04-2.11
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)
Spring 2004ECE569 Lecture 04-2.12
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
Spring 2004ECE569 Lecture 04-2.13
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
Spring 2004ECE569 Lecture 04-2.14
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
Spring 2004ECE569 Lecture 04-2.15
Collision - continued
External resolution: Allocation overflow block, link to overflow chain
buckets Overflow
pages
Spring 2004ECE569 Lecture 04-2.16
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?
Spring 2004ECE569 Lecture 04-2.17
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
1D
(b)
000
001
010
011
100
101
110
111
A
D
B
C
(c)
Spring 2004ECE569 Lecture 04-2.18
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).
Spring 2004ECE569 Lecture 04-2.19
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.
Spring 2004ECE569 Lecture 04-2.20
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
Spring 2004ECE569 Lecture 04-2.21
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
Spring 2004ECE569 Lecture 04-2.22
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)
Spring 2004ECE569 Lecture 04-2.23
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?
Spring 2004ECE569 Lecture 04-2.24
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)
Spring 2004ECE569 Lecture 04-2.25
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
Spring 2004ECE569 Lecture 04-2.26
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
Spring 2004ECE569 Lecture 04-2.27
Discuss
Where to store the overflow?