2. HASH TABLE

3. a group of people could be arranged in a database like this:
Hashing
is the transformation of a string of characters into a usually shorter fixed-length value or key that represents the original string.
Example
a group of people could be arranged in a database like this:
Allen, Jane
Moore, Sarah
Smith, Dan

4. H A S I N G Hash Table 7864 Allen, Jane Moore, Sarah 9802 1990
Smith, Dan
HASH VALUES
HASH KEYS
HASH FUNCTION

5. Hash Table
stores things and allows 3 operations: insert, search and delete.
associated with a set of records

6. Bob Miller 34
John Smith
Sally Wood 21 H
John Smith 29 5

7. Each slot of a hash table is called a bucket and hash values are called bucket indices.
7864
Allen, Jane
BUCKET
BUCKET
INDEX

8. HASH FUNCTION Mapping of the keys to indices of a hash table
2 compositions
Hash code map:
key
integer
Compression map:
integer
[0, N-1]

9. DIVISION Example: If table size m = 12 key k = 100
Map a key k into one of m slots by using this
function:
h(k) = k mod m
Example:
    If table size m = 12                 key k = 100
    than         h(100) = 100 mod 12                    = 4

10. Ex. k=3121 then 31212=9740641 thus h(3121)= 406 MID-SQUARE FUNCTION
The key is squared and the mid part is used as the address.
Ex.
k=3121
then 31212=
thus h(3121)= 406

11. Folding Key is divided into several parts 2 types 1. shift folding
2. boundary folding

12. Shift Folding Ex. (SSN) 123-45-6789 1. Divide into 3 parts:
123, 456 and 789.
2. Add them.
=1368
3. h(k)=k mod M
where M = 1000
h(1368) = 1368 mod 1000
= 368
1. Divide into five parts: 12, 34, 56, 78 and 9.
2. Add them.
= 189
3. h(k)=k mod M
where M = 1000
h(189) = 189 mod 1000
= 189

13. 1st 4 digits = 1234 Last 4 digits = 6789
Extraction
Only a part of the key is used to compute the address.
Ex. (SSN)
1st 4 digits = 1234
Last 4 digits = 6789
1st 2 combined with the last 2 = 1289(address)

14. Hash Method : Folding Chopping the Key in Two Parts
Add the Two Parts to Generate the Hash
Leading Digit will be Ignored
Example
Key
Parts
H(x)
Option Rotate the Second Digit
Parts
H(x)

15. Radix Transformation K is transformed into another number base M = 100
21210=2559
M = 100
H(k) = k mod M
H(255) = 255 mod 100
= 55

16. 212= 9(9(2)+ 5)+ 5 = 2(92)+ 5(9)+ 5. divide 212 by 9.
9 divides into times with remainder = 9(23)+ 5
9 divides into 23 twice with remainder = 9(2)+5
212= 9(9(2)+ 5)+ 5
= 2(92)+ 5(9)+ 5.

17. different keys happen to have same hash value
Hash Collision
different keys happen to have same hash value

18. Bob Miller 34
Jane Depp 18
Collision!
Sally Wood 21 2
John Smith 29

19. Collision Resolution There are two kinds of collision resolution:
1 – Chaining makes each entry a linked list so that when a collision occurs the new entry is added to the end of the list.
2 – Open Addressing uses probing to discover an empty spot.

20. Collision Resolution – Open Addressing
the table is probed for an open slot when the first one already has an element.
Linear probing in which the interval between probes is fixed — often at 1.
Quadratic probing in which the interval between probes increases linearly (hence, the indices are described by a quadratic function).
Double hashing in which the interval between probes is fixed for each record but is computed by another hash function.

21. H(x,i) = (H(x) + i)(mod M)
Linear Probing
is a scheme in resolving hash collisions of values of hash functions by sequentially searching the hash table for a free location
two values - one as a starting value and one as an interval between successive values 
newLocation = (startingValue + stepSize) % arraySize
H(x,i) = (H(x) + i)(mod M)

22. Linear Probing - Example1 2 3 4 5 6 7 8 9
empty
Insert 15, 17, 8 1 2 3 4 5 6 7 8 9
empty 15 17 8
H(8)=8 mod 10 = 8
H(15)=15 mod 10 = 5
H(17)=17 mod 10 = 7

23. 1 2 3 4 5 6 7 8 9
empty 75 15 35 17 8 25
Insert 25
Insert 35
H(1,8)=(1 + 8) mod 10 = 9
H(35)=35 mod 10 = 5
H(1,6)=(1 + 6) mod 10 = 7
H(1,7)=(1 + 7) mod 10 = 8
H(1,5)=(1 + 5) mod 10 = 6
H(25)=25 mod 10 = 5
H(1,5)=(1 + 5) mod 10 = 6
Insert 75
H(1,9)=(1+9) mod 10 = 0
H(1,8)=(1+8) mod 10 = 9
H(1,5)=(1+5) mod 10 = 6
H(75)=75 mod 10 = 5
H(1,6)=(1+6) mod 10 = 7
H(1,7)=(1+7) mod 10 = 8

24. Has anyone spotted the flaw in the linear probing technique
Has anyone spotted the flaw in the linear probing technique? Think about this: what would happen if we now inserted 85, then 95, then 55?

25. Each one would probe exactly the same positions as its predecessors
Each one would probe exactly the same positions as its predecessors. This is known as clustering. It leads to inefficient operations, because it causes the number of collisions to be much greater than it need be.

26. eliminates primary clustering p(K, i) = c1 i2 + c2i + c3
Quadratic Probing
eliminates primary clustering
p(K, i) = c1 i2 + c2i + c3 
p(K, i) = i2 (i.e., c1 = 1, c2 = 0, and c3 = 0)

27. Quadratic Probing - Example
Table Size is 11 (0..10)
Hash Function: h(x) = x mod 11
Insert keys:
20 mod 11 = 9
30 mod 11 = 8
2 mod 11 = 2
13 mod 11 = 2  2+12=3
25 mod 11 = 3  3+12=4
24 mod 11 = 2  2+12, 2+22=6
10 mod 11 = 10
9 mod 11 = 9  9+12, 9+22 mod 11,
9+32 mod 11 =7 1 2 3 13 4 25 5 6 24 7 9 8 30 20 10

28. not all hash table slots will be on the probe sequence
Using p(K, i) = i2 gives particularly inconsistent results
If all slots on that cycle happen to be full, this means that the record cannot be inserted at all!

29. Double Hashing P = (1 + P) mod TABLE_SIZE
 increment P, not by a constant but by an amount that depends on the Key. 
P = (1 + P) mod TABLE_SIZE
P = (P + INCREMENT(Key)) mod TABLE_SIZE

30. Double Hashing - Example
P = (P + INCR(Key)) mod TABLE_SIZE
Suppose INCR(Key) = 1 + (Key mod 7)
Adding 1 guarantees it is never 0!
Insert 15, 17, 8:

31. Insert 35:
P = H(35) = 5.
P = (5 + ( mod 7)) mod 10 = 6.
Insert 25:P = H(25) = 5.
P = (5 + ( mod 7)) mod 10 = 0

32. 10 3 2 1 4 9 5 8 6 7
Let’s try!
Insert 75:
P = (P + INCR(Key)) mod TABLE_SIZE
Suppose INCR(Key) = 1 + (Key mod 7)

33. Chaining/Separate Chaining
uses an array as the primary hash table
an array of lists of entries

34. Chaining nil nil nil : nil
One way to handle collision is to store the collided records in a linked list. The array now stores pointers to such lists. If no key maps to a certain hash value, that array entry points to nil. 1
nil 2
nil 3 4
nil 5 :
Key:
name: tom
score: 73
HASHMAX
nil

36. 29 16 14 99
127
129 16
127 99 29 14 29
129

37. is a collision resolution method that
Coalesced Hashing
is a collision resolution method that
uses pointers to connect the elements of a synonym
chain.
A hybrid of separate chaining and open addressing.
Linked lists within the hash table handle collisions.
This strategy is effective, efficient and very easy to
implement.

38. A5, A2, A3 B5, A9, B2 B9, C2 Insert: A2 A2 A2 A3 A3 A3 C2 A5 A5 A5 B9

39. Insert: A5, A2, A3 B5, A9,B2 B9,C2 A2 A2 A2 A3 A3 A3 A5 A5 A5 C2 A9 A9

40. using additional space
Bucket Addressing
using additional space
A bucket can be defined as a block of space that can be used to store multiple elements that hash to the same position.

41. Insert:
A5, A2, A3, B5, A9, B2, B9 A2 B2 A3 A5 B5 A9 B9

42. TOMBSTONE DELETION Deleting a record must not hinder later searches.
The search process must still pass through the newly emptied slot to reach records whose probe sequence passed through this slot. 
It should not mark the slot as empty.
Freed slot should be available to a future insertion.
TOMBSTONE

43. 28 83 25 75 35
Insert:
Collision Probing Sequence: 25 75
Delete: 35 83
Match Found!
TOMBSTONE 75 28

44. A1, A4, A2,B4,B1 A2 A4
Delete:
Insert: A1 B1 A2 B1 B4 A4 B4

45. Perfect Hash Functions
Quick to compute
Distributes keys uniformly throughout the table
Very rare(birthday paradox)
No collisions
Perfect hash functions are rare.

46. A Perfect Hash Function for Strings
R. J. Cichelli gave an algorithm for finding perfect hash functions for strings.
He proposes the hash function:
h(s)=size+g(s.charAt(0))+
g(s.charAt(size-1))%n
where size = s.length().
The function g is to be constructed so that h(s) is unique
for each string s.

51. Example 1: Illustrating Perfect Hashing
Use Cichelli's algorithm to build a minimal perfect hash
function for the following nine strings: DO
DOWNTO
ELSE
END IF IN
TYPE
VAR
WITH

52. Example 1: Solution
For Step 1 in the algorithm, we find the frequencies of the
first and last letter of each word to find:
D O E I F N T V R W H
Next we find the sum of the first and last letter of each word:
DO=5(D+0=3+2), DOWNTO=5, ELSE = 8, END=7,
IF=3, IN=3, TYPE=5, VAR=2,WITH=2
Sorting the keywords in decreasing frequency yields:
ELSE END DOWNTO DO TYPE IN IF VAR WITH
We are now at step 5 of the algorithm, the heart of the algorithm. We try the words in frequency order:

53. Example 1: Cichelli's Method (cont'd)
s = ELSE g(E)=0 h(s)= s.length()+g(E)+g(E)=4
s = END g(D)=0 h(s)= s.length()+g(E)+g(D)=3
s = DOWNTO g(O)=0 h(s)= s.length()+g(D)+g(O)=6
s = DO h(s)= s.length()+g(D)+g(O)=2
s = TYPE g(T)=0 h(s)= s.length()+g(T)+g(E)=4*
s = TYPE g(T)=1 h(s)= s.length()+g(T)+g(E)=5
s = IN g(I)=0,g(N)=0 h(s)=s.length()+g(I)+g(N)=2*
s = IN g(I)=1,g(N)=0 h(s)=s.length()+g(I)+g(N)=3*
s = IN g(I)=2,g(N)=0 h(s)=s.length()+g(I)+g(N)=4*
s = IN g(I)=3,g(N)=0 h(s)=s.length()+g(I)+g(N)=5*
s = IN g(I)=3,g(N)=1 h(s)=s.length()+g(I)+g(N)=6*
s = IN g(I)=3,g(N)=2 h(s)=s.length()+g(I)+g(N)=7
s = IF g(F)=0 h(s)=s.length()+g(I)+g(F)=5*
s = IF g(F)=1 h(s)=s.length()+g(I)+g(F)=6*
s = IF g(F)=2 hash(s)=s.length()+g(I)+g(F)=7*
s = IF g(F)=3 h(s)=s.length()+g(I)+g(F)=8

54. Example 1: Cichelli's Algorithm (cont'd)
VAR
WITH DO
END
ELSE
TYPR
DOWNTO IN IF
The hash table above is fully occupied with empty slots.
Note that if there are empty slots or there is a collision, then the g-value assignments are in error.

55. Extendible hashing
Hashing with buckets

56. Extendible Hashing - Class Example

57. Leaf Pages Directory d1 = local depth d = global depth rec 1 rec 4
splitting bucket
rec 1
rec 2
d1=0 1
splitting bucket
d = 0
record 3 = overflow!!
d = 1
rec 2
rec 3
record 5 = overflow!!
NEXT

58. rec 1 rec 4 00 01 rec 2 10 splitting bucket rec 3
d = 2
d1 = 2
d1 = 1 11 01
rec 1
rec 4
rec 2
splitting bucket
rec 3
record 7 = overflow!!
rec 5
rec 6
NEXT

59. 000 110 d = 3 111 001 010 011 100 101 rec 1 rec 4 splitting bucket
record 8 = overflow!!
d1 = 3
rec 2
rec 7
d1 = 3
rec 3
rec 5
rec 6
NEXT

60. rec 1 NEXT rec 4 rec 8 000 001 010 011 100 rec 2 101 rec 7 110 rec 3
d1 = 3
d1 = 2
rec 1
rec 4
rec 8
000
110
d = 3
111
001
010
011
100
101
d1 = 3
d1 = 2
rec 2
rec 3
rec 5
rec 6
rec 7
rec 9
splitting bucket
record 10 = overflow!!

61. NEXT
rec 7
rec 9
d1 = 3
rec 1
rec 4
000
110
d = 3
111
001
010
011
100
101
rec 2
rec 3
d1 = 2
rec 8
rec 11
rec 12
d1 = 3
rec 5
rec 6
splitting bucket
rec 10
record 13 = overflow!!

62. d1 = 4
d1 = 3
d1 = 2
0000
1110
d = 4
1111
0001
0010
0011
1100
1101
0100
0101
0110
0111
1010
1011
1000
1001
rec 1
rec 4
rec 2
rec 3
rec 5
rec 7
rec 8
rec 11
rec 12
rec 14
rec 15
rec 6
rec 10
rec 13

63. Hash Table Uses
driver's license record's
Internet search engines
telephone book databases
electronic library catalogs
implementing passwords for systems with multiple users. Hash Tables allow for a fast retrieval of the password which corresponds to a given username