Chapter 13B Week 4

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Chapter 13BWeek 4

• External Hashing

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External Hashing

• Hashing implemented in secondary storage

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Static Hashing

• Unix Example
• Each unix file has an inode that contains
• File ownership info
• Access rights
• Location of data

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Continued

• Have the start address contained in the inode
  point to a table that maps hash addresses to
  block addresses.
• Each block address is called a bucket.
• Each bucket contains multiple records

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The scheme looks like this
0 1 2 . . . M-1
inode
r1 r2 r3 r4 r5
Buckets are contiguous blocks. Can be one or
several. This one holds five records.
Block Address
Hash function maps key to here
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Problem Number of buckets is fixed at file
creation

• Maintain a linked list of overflow records for
  each bucket.

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What happens if a bucket is filled
inode
Record pointer contains both block address and
the relative record position within the block
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Trie-Based Hashing

• Another remedy to the overflow/fixed space
  problem
• Allow the number of allocated buckets to grow and
  shrink as needed
• Scheme
• H(K) produces an integer in binary
• Distributes records among buckets based on the
  values of the leading bits in their hash values
• Mimics a binary tree

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Insert

• Begin with a bucket of disk addresses

r1 r2 r3 r4
Overflow? Split the bucket (block) based on the
first binary digit of the hash address
0
1
r1 r3 r4 r5
r2
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Bucket 1 Overflows Again?

• Split on 2nd bit in the hash address
• Suppose we have
• Record Number Hash Address
• r1 0000
• r2 1000
• r3 0010
• r4 0100
• r5 0110
• Insert r6 0111 in the previous structure

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New Structure
0
1
0
1
r2
r1
r4
r3
r5
r6
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Find(K)

1. Compute h(k)
2. Traverse structure based on the binary value of
   the hash function
3. Do linear search of bucket for k
4. Go to the disk address contained there

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Issues

• Hashing function that produces n bit address can
  accommodate 2n buckets
• Total Records lt 2n NumRecs/bucket
• Hash function should distribute values uniformly
  to keep tree balanced
• Delete
• Suppose r4, r5, r6 deleted
• Remove pointer to their bucket
• Remove 00/01 internal node
• Reset the 0/1 internal node to point to bucket
  containing r1 r3

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Extendible Hashing

• Replaces the binary tree of dynamic hashing with
  a directory consisting of 2d bucket addresses
• d is called the global depth of the directory
• First d bits of the hash value is an index into
  the array
• d is called the local depth
• d lt d

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Continued

• Not necessarily 2d buckets allocated
• Several directory locations with the same first
  d bits for their hash values may point to the
  same bucket address when d lt d for a particular
  bucket

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Example
d3
Bucket for records whose hash values start with
000
000
001
b1
010

...
Bucket for records whose hash values start with 01
011
d 2

b2
111
d 3
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Continued

• Suppose bucket pointed to by 010,011 overflows
• Allocate a new bucket. The old b2 now contains
  records whose hash values begin with 010
• The new b3 contains records whose hash values
  begin 011
• d of each new bucket is 3

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Bucket splits
d3
Bucket for recs whose hash starts with 000
000
001
b1
010

...
Bucket for recs whose hash starts with 010
011
d 3

b2
111
Bucket for recs whose hash starts with 011
d3
b3
d 3
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Increase Table Size

• What happens when dd for some bucket and
  overflow occurs?
• We need an extra table entry to distinguish the
  new value
• E.g., 000 overflows we need 0000 and 0001
• Increase d by 1
• Doubles table size

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Continued

• 0000
• 0001
• 0010
• 0011
• 0100
• 0101
• 0110
• 0111
• .
• .
• .
• 111

d4r1, r2, , rn
d4r1, r2, , rn
d3r1, r2, , rn
d3r1, r2, , rn
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What if d gt d for all buckets?

• Means that at least two directory positions point
  to the same bucket for all buckets
• We can now halve the directory

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Continued

• 000 d 2
• 001
• 010 d 2
• 011
• 100 d 2
• 101
• 110 d 2
• 111
• Halve
• 00 d 2
• 01 d 2
• 10 d 2
• 11 d 2

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Linear Hashing

• Drawback of trie-based and extendible hashing is
  that each requires an auxiliary data structure
• Dynamic hashing? binary tree which may not be
  balanced
• Extendible hashing ? directory

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Task

• Design a scheme that adjusts size to the amount
  of data and requires no auxiliary data structures
• Linear Hashing Example