Chapter%2013A%20Week%203

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Chapter 13AWeek 3

• Records
• Files
• Hashing

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Record

• Analogous to a struct
• Collection of variables of different datatypes
• struct student
• char name30
• char ssn9
• char major4
• int age

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File

• Sequence of records
• Three types
• Fixed length records
• Variable length records
• Contains records of different types

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File Types

• Fixed length records are easiest to deal with
• File size is a multiple of record size
• We know where each record starts by its offset
  from the beginning
• Variable length records
• Variable length fields (e.g., name)
• Repeating groups (e.g., projects stored with
  employee)
• Optional fields (department supervised)
• Requires field separators
• Different record types in a single file
• Related records of different types are clustered
  (dependent record follows employee record)
• Requires record terminators

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Blocking

• Block is a unit of data transfer
• Blocking Factor
• Number of complete records per block
• Bf floor(B/R)
• where B number of blocks and
• R number of records
• If the blocking factor is not a multiple of
  record size we have wasted space
• W B Bf R
• The waste can be reclaimed if we let records span
  blocks

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File Operations

• File Organization
• A way of arranging records on a file
• Heap files
• Relative files
• Ordered files
• Hash files
• Access Methods
• Set of operations
• Open, Close
• Reset file pointer
• Scan All
• Find
• FindNext
• Read Current
• Delete
• Modify
• Insert
• FindAll
• Find N records that satisfy a condition

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Heap File

• Records are stored in the order in which they are
  inserted
• Scan All O(N)
• Find O(N/2) on average only N/2 records must be
  examined
• Find in a range O(N) since all records must be
  examined
• Insert O(1) fetch last block, insert, write back
  
• Delete O(N/2) Find record to delete. Mark it as
  deleted. Remove and reclaim space during batch
  reorganization.

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Relative File

• Records are of fixed length
• Records dont cross block boundaries
• Records are contiguous on disk
• So, records can be accessed by position
• Ex Bf 10
• Find(rec 13)
• Record is in ceiling(13/10) or 2nd block
• Record is 13 10 or 3rd record in that block

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Relative File Access Method Complexity

• Scan All O(N)
• Find O(1)
• Find in range O(N)
• Delete O(1)
• Insert O(1)

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Ordered Files

• Usually file itself is not sorted
• Rather an index on that file is maintained
• Advantages
• Reads are more efficient because block size of
  indexed file is smaller
• Can have multiple indexes

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Ordered File Access Method Complexity

• Scan All O(N)
• Find O(lgN) uses binary search
• Find in range O(lgN) O(R)
• First term finds the key
• R is the number of records that meet the search
  criteria

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Continued

• Insert
• In theory
• Find insertion spot O(lgN)
• Move all records down a position O(N/2)
• In Practice
• Maintain a sequential (heap) overflow file
• Insert records in the order in which they appear
• Reorganize primary file as a batch run
• Slows down find (because of the need for a
  sequential search)
• Speeds up insert

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Continued

• Delete
• Find Record and mark it for deletion O(lgN)
• Reorganize the file as a batch run

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Hashed Files

• Position of a record is based on a computation
• hash(key) ? position
• Scan All O(N)
• Find O(1)
• Find in range O(N) because it is necessary to
  scan the entire file
• Insert O(1)
• Delete O(1)

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Summary of Strengths and Weakness of File
Organizations

• Heap (sequential)
• Insertion is good
• Find is bad
• Easy to maintain
• Relative
• All access methods are fast
• But contiguity requirement is hard to maintain
• Ordered
• Find is good
• Insertion and deletion require reorganization
• Hash
• Find is good
• Scan in order is bad

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Hashing

• Two types
• Internal (done in RAM)
• External (done in secondary storage)
• Internal should be a review

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Internal Hashing

• Problem
• Given a key value, store records so that each one
  is directly accessible
• That is, speed up find, insert, delete
• Requires
• Hash key field of record upon which search is
  based
• Hash function function which when applied to key
  yields address in memory (of disk block) in which
  a record is stored

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Internal Hashing Example

• Suppose
• Key is SSN
• Task make find O(N)
• Solution 1 create an array so that each SS is
  an offset into the array
• Wastes 109 number of SSs
• Solution 2
• Use hash function h(k) k m
• Where
• k is the key
• m is the table size

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Example Continued

• Suppose we have 18 keys and a table size of 23
• 019, 392, 179, 359, 663, 262, 639, 321, 097, 468,
  814, 720, 260, 802, 364, 976, 774, 566
• h(019) 19
• h(392) 01
• h(179) 18
• h(359) 14
• H(663) 19
• But position 19 is occupied
• We need a collision resolution policy

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Possible Policies

• Linear Probing
• Double Hashing
• Chaining
• Linear Probing
• If there is a collision, insert key in the next
  open position, wrapping when you get to the end
  of the table
• Heres the hash table

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• Position Key Probes
• 0 802 4
• 1 392 1
• 2 364 7
• 3
• 4
• 5 097 1
• 6
• 7 720 1
• 8 468 1
• 9 262 1
• 10 814 2
• 11 260 5
• 12 976 3
• 13
• 14 359 1
• 15 774 1
• 16 566 3
• 17

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Find

• Find 360
• H(360) 15 occupied
• 16 occupied
• 17 empty
• Return Not Found

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Probe Sequence

• Path a key takes as it searches for an empty spot
• Notice
• Linear probing gives 1 probe sequence
• Each key enters it at a (possibly) different
  place
• We can improve performance by
• Improving probe sequence
• Increasing the size of the table

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General Probe SequenceReferred to as Open
Addressing

• (h(k) ip(k)) m for i 0, 1, , m-1
• Notice that under linear probing, p(k) 1
• Heres how the insertion of 802 worked
• h(802) 20 occupied
• 21 occupied
• 22 occupied
• 0 insert
• Using the general formula
• (h(802) 01) 23 20 occupied
• (h(802) 01) 23 21 occupied
• (h(802) 01) 23 22 occupied
• (h(802) 01) 23 0 insert

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Probe Sequence

• With linear probing there is a single probe
  sequence, different keys join it along the way
• Example
• Suppose k 802, m 13
• S(802) (802 13 01) mod 13 9
• S(802) (802 13 11) mod 13 10
• 11
• 12
• 0

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Continued

• Now k 11, m 13
• S(802) (11 13 01) mod 13 11
• 12
• 0

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Double Hashing

• Alter p(k) so that only keys hashing to the same
  address have the same probe sequence
• Make p(k) a function of h(k)
• P(k) (h(k) 4) m

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Continued

• K 137, m 13
• S(137) (137 13 0 p(k)) 13 7
• S(137) (137 13 1 (7 4) 13) 13 5
• The next values, in order, are 3, 1, 12, 8, 6,
  4, 2, 0, 11, 9
• But S(258) produces this sequence 11, 0, 2,
• Notice that a collision at 11 results in a
  different probe sequence

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Continued

• Result
• Only keys hashing to same address have the same
  probe sequence.
• This breaks up primary clusters, but produces
  secondary clusters
• These could be eliminated if p(k) were chosen
  independently of h(k)

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Continued

• But
• You must take care to generate a valid probe
  sequence
• Criteria for a key randomly chosen, each slot
  has equal probability of being hit and youll
  cycle through all slots in the table

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Chaining

• Each location in the hash table is the head of a
  linked list of keys that collide at that
  location.
• K 19, m 5, h(19) 4

Linear probing length of chain is length of
primary clusters Double Hash length of chain is
length of secondary clusters
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Deletion from a Hash Table

• Chaining
• Hash to location, traverse list until end or key
  is found. Delete
• Open addressing
• Mark location as deleted
• Recapture deleted locations on subsequent
  inserts
• If deleted locations build up so that search
  degrades, it will be necessary to rebuild the
  table