1. Dictionaries and Hash Tables
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Hash Tables  1 2 3  4
Dictionaries and Hash Tables

2. Hash ADT
The hash ADT models a searchable collection of key-element items
The main operations of a hash system (e.g., dictionary) are searching, inserting, and deleting items
Multiple items with the same key are allowed
Applications:
address book
credit card authorization
mapping host names (e.g., to internet addresses (e.g., )
Hash (dictionary) ADT operations:
find(k): if the dictionary has an item with key k, returns the position of this element, else, returns a null position.
insertItem(k, o): inserts item (k, o) into the dictionary
removeElement(k): if the dictionary has an item with key k, removes it from the dictionary and returns its element. An error occurs if there is no such element.
size(), isEmpty()
keys(), Elements()

3. Efficiency of Hash Tables
A dictionary is a hash table implemented by means of an unsorted sequence
We store the items of the dictionary in a sequence (based on a doubly-linked lists or a circular array), in arbitrary order
Performance:
insertItem takes O(1) time since we can insert the new item at the beginning or at the end of the sequence
find and removeElement take O(n) time since in the worst case (the item is not found) we traverse the entire sequence to look for an item with the given key

4. Hash Functions and Hash Tables
A hash function h maps keys of a given type to integers in a fixed interval [0, N - 1]
Example: h(x) = x mod N is a hash function for integer keys
The integer h(x) is called the hash value of key x
A hash table for a given key type consists of
Hash function h
Array (called table) of size N
When implementing a dictionary with a hash table, the goal is to store item (k, o) at index i = h(k)

5. Example
We design a hash table for a dictionary storing items (TC-no, Name), where TC-no (social security number) is a nine-digit positive integer
Our hash table uses an array of size N = 10,000 and the hash function h(x) = last four digits of x  1 2 3 4
9997
9998
9999 …

6. Hash Functions
A hash function is usually specified as the composition of two functions:
Hash code map: h1: keys  integers
Compression map: h2: integers  [0, N - 1]
The hash code map is applied first, and the compression map is applied next on the result, i.e., h(x) = h2(h1(x))
The goal of the hash function is to “disperse” the keys in an apparently random way

7. Hash Code Maps Memory address: Integer cast: Component sum:
We reinterpret the memory address of the key object as an integer
Good in general, except for numeric and string keys
Integer cast:
We reinterpret the bits of the key as an integer
Suitable for keys of length less than or equal to the number of bits of the integer type (e.g., char, short, int and float on many machines)
Component sum:
We partition the bits of the key into components of fixed length (e.g., 16 or 32 bits) and we sum the components (ignoring overflows)
Suitable for numeric keys of fixed length greater than or equal to the number of bits of the integer type (e.g., long and double on many machines)

8. Hash Code Maps (cont.) Polynomial accumulation:
We partition the bits of the key into a sequence of components of fixed length (e.g., 8, 16 or 32 bits) a0 a1 … an-1
We evaluate the polynomial
p(z) = a0 + a1 z + a2 z2 + … … + an-1zn-1
at a fixed value z, ignoring overflows
Especially suitable for strings (e.g., the choice z = 33 gives at most 6 collisions on a set of 50,000 English words)
Polynomial p(z) can be evaluated in O(n) time using Horner’s rule:
The following polynomials are successively computed, each from the previous one in O(1) time
p0(z) = an-1
pi (z) = an-i-1 + zpi-1(z) (i = 1, 2, …, n -1)
We have p(z) = pn-1(z)

9. Hash Functions Using Horner’s Rule
compute this by Horner’s Rule
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10. Horner’s Method: Compute a hash value
Compute the has value of string junk
A3X3+A2X2+A1X1+A0X0
Can be evaluated as
X = 128
(((A3)X+A2)X+A1 ) X+A0
‘j’ ‘u’ ‘n’ ‘k’
Apply mod (%) after each multiplication!

11. Compression Maps Division: Multiply, Add and Divide (MAD):
h2 (y) = y mod N
The size N of the hash table is usually chosen to be a prime
The reason has to do with number theory and is beyond the scope of this course
Multiply, Add and Divide (MAD):
h2 (y) = (ay + b) mod N
a and b are nonnegative integers such that a mod N  0
Otherwise, every integer would map to the same value b

12. Implementation of Hashtables
Dictionaries
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Implementation of Hashtables
Which is better, an array or a linked list?
Well, an array is better if we know how many
objects will be stored because we can access
each element of an array in O(1) time!!!
But, a linked list is better if we have to change the
number of objects stored dynamically and the
changes are significant, or if there are complex
relationships between objects that indicate
“neighbors”.
What to do?
Combine the advantages of both!!!
WELCOME HASHTABLES

13. Hashtables Wouldn't it be great if arrays could be made
Dictionaries
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Hashtables
Wouldn't it be great if arrays could be made
of infinite size without penalty? Then every
object would have a place in the array

14. Hashtables Wouldn't it be great if arrays could be made
Dictionaries
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Hashtables
Wouldn't it be great if arrays could be made
of infinite size without penalty? Then every
object would have a place in the array
...
453664
Object
453665
453666
...

15. Hashtables Wouldn't it be great if arrays could be made
Dictionaries
9/12/2018 4:33 AM
Hashtables
Wouldn't it be great if arrays could be made
of infinite size without penalty? Then every
object would have a place in the array
...
453664
Object
453665
453666
...
But this is impractical

16. Hashtables So we “fold” the array infinitely many times. In
Dictionaries
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Hashtables
So we “fold” the array infinitely many times. In
other words, the number of spaces for holding
objects is small but each space may hold lots of
objects.
...
Object 64
Object 65 66
Object
...
...

17. Hashtables But how do we maintain all the objects belonging
Dictionaries
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Hashtables
But how do we maintain all the objects belonging
to the same space?
...
Object 64
Object 65 66
Object
...
...

18. Hashtables But how do we maintain all the objects belonging
Dictionaries
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Hashtables
But how do we maintain all the objects belonging
to the same space? Answer: linked list
...
Object 64
Object 65 66
Object
...
...

19. Hashtables How do we locate an object? ... Object 64 65 Object 66
Dictionaries
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Hashtables
How do we locate an object?
...
Object 64
Object 65 66
Object
...
...

20. Hashtables How do we locate an object?
Dictionaries
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Hashtables
How do we locate an object?
1. Use a hash function to locate an array element
...
Object 64
Object 65 66
Object
...
...

21. Hashtables How do we locate an object?
Dictionaries
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Hashtables
How do we locate an object?
1. Use a hash function to locate an array element
2. Follow the linked list from that element to find
the object
...
Object 64
Object 65 66
Object
...
...

22. Hashtables How do we locate an object?
Dictionaries
9/12/2018 4:33 AM
Hashtables
How do we locate an object?
1. Use a hash function to locate an array element
2. Follow the linked list from that element to find
the object
...
Object 64
Object 65 66
Object
...
...

23. Hashtables How do we locate an object?
Dictionaries
9/12/2018 4:33 AM
Hashtables
How do we locate an object?
1. Use a hash function to locate an array element
2. Follow the linked list from that element to find
the object
...
Object 64
Object 65 66
Object
...
...

24. Hashtables How do we locate an object?
Dictionaries
9/12/2018 4:33 AM
Hashtables
How do we locate an object?
1. Use a hash function to locate an array element
2. Follow the linked list from that element to find
the object
...
Object 64
Object 65 66
Object
...
...

25. Hashtables Now the problem is how to construct the hash
Dictionaries
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Hashtables
Now the problem is how to construct the hash
function and the hash table size!!!

26. Hashtables Now the problem is how to construct the hash
Dictionaries
9/12/2018 4:33 AM
Hashtables
Now the problem is how to construct the hash
function and the hash table size!!!
Let's do the hash function first. The idea is to
take some object identity and convert it to some
random number.

27. Hashtables Now the problem is how to construct the hash
Dictionaries
9/12/2018 4:33 AM
Hashtables
Now the problem is how to construct the hash
function and the hash table size!!!
Let's do the hash function first. The idea is to
take some object identity and convert it to some
random number. Consider...
long hashvalue (void *object) {
long n = (long)valfun->value(object);
return (long)(n*357 % size); }

28. Table Size? What about the table size??? If the table size is
Dictionaries
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Table Size?
What about the table size??? If the table size is
N and the number of objects stored is M then
the average length of a linked list is M/N so if
M is about 2N the average length is 2 meaning that
to locate an object takes O(1) time!!!! Just like for arrays!!!

29. Collision Handling
Collisions occur when different elements are mapped to the same cell
Chaining: let each cell in the table point to a linked list of elements that map there  1 2 3 4
Chaining is simple, but requires additional memory outside the table

30. Izmir University of Economics
Handling Collisions
Separate Chaining
Open Addressing
Linear Probing
Quadratic Probing
Re-hashing
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31. Izmir University of Economics
Separate Chaining
Keep a list of elements that hash to the same value
New elements can be inserted at the front of the list
ex: x=i2 and hash(x)=x%10
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32. Performance of Separate Chaining
Load factor of a hash table, λ
λ = N/M (# of elements in the table/TableSize)
So the average length of list is λ
Search Time = Time to evaluate hash function + the time to traverse the list
Unsuccessful search= λ nodes are examined
Successful search=1 + ½* (N-1)/M (the node searched + half the expected # of other nodes) =1+1/2 *λ
Observation: Table size is not important but load factor is.
For separate chaining make λ=1  O(1)
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33. Separate Chaining: Disadvantages
Parts of the array might never be used.
As chains get longer, search time increases to O(N) in the worst case.
Constructing new chain nodes is relatively expensive (still constant time, but the constant is high).
Is there a way to use the “unused” space in the array instead of using chains to make more space?
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34. Linear Probing Example: h(x) = x mod 13
Open addressing: the colliding item is placed in a different cell of the table
Linear probing handles collisions by placing the colliding item in the next (circularly) available table cell
Each table cell inspected is referred to as a “probe”
Colliding items lump together, causing future collisions to cause a longer sequence of probes
Example:
h(x) = x mod 13
Insert keys 18, 41, 22, 44, 59, 32, 31, 73, in this order 1 2 3 4 5 6 7 8 9 10 11 12 41 18 44 59 32 22 31 73 1 2 3 4 5 6 7 8 9 10 11 12

35. Linear Probing: i = h(x) = key % tableSize49 49 49
Hash(89,10) = 9
Hash(18,10) = 8
Hash(49,10) = 9
Hash(58,10) = 8
Hash(9,10) = 9 1 58 58 2 9 3 4 5 6 7 8 18 18 18 18 9 89 89 89 89 89
After insert 89
After insert 18
After insert 49
After insert 58
After insert 9
Strategy: A[(i+1) %N], A[(i+2) %N], A[(i+3) %N],...
Employing wrap around

36. Quadratic Probing: i = h(x) = key % tableSize49 49 49
Hash(89,10) = 9
Hash(18,10) = 8
Hash(49,10) = 9
Hash(58,10) = 8
Hash(9,10) = 9 1 2 58 58 3 9 4 5 6 7 8 18 18 18 18 9 89 89 89 89 89
After insert 89
After insert 18
After insert 49
After insert 58
After insert 9
Strategy: A[(i+12) %N], A[(i+2 2) %N], A[(i+3 2) %N], ..., Employing wrap around

37. Search with Linear Probing
Consider a hash table A that uses linear probing
find(k)
We start at cell h(k)
We probe consecutive locations until one of the following occurs
An item with key k is found, or
An empty cell is found, or
N cells have been unsuccessfully probed
Algorithm find(k)
i  h(k)
p  0
repeat
c  A[i]
if c = 
return Position(null)
else if c.key () = k
return Position(c)
else
i  (i + 1) mod N
p  p + 1
until p = N

38. Updates with Linear Probing
To handle insertions and deletions, we introduce a special object, called AVAILABLE, which replaces deleted elements
removeElement(k)
We search for an item with key k
If such an item (k, o) is found, we replace it with the special item AVAILABLE and we return the position of this item
Else, we return a null position
insertItem(k, o)
We throw an exception if the table is full
We start at cell h(k)
We probe consecutive cells until one of the following occurs
A cell i is found that is either empty or stores AVAILABLE, or
N cells have been unsuccessfully probed
We store item (k, o) in cell i

39. Exercise: A hash function
#include<string.h>
unsigned int hash(const string &key, int tableSize){
unsigned int hashVal = 0;
for (int i= 0;i <key.length(); i++)
hashVal = (hashVal * key[i]) % tableSize;
return hashVal; }

40. Exercise: Another hash function
// A hash routine for string objects
// key is the string to hash.
// tableSize is the size of the hash table.
#include<string.h>
unsigned int hash(const string &key, int tableSize){
unsigned int hashVal = 0;
for (int i= 0;i <key.length(); i++)
hashVal = (hashVal * 37 + key[i]);
return hashVal % tableSize; }

41. Performance of Hashing
In the worst case, searches, insertions and removals on a hash table take O(n) time
The worst case occurs when all the keys inserted into the dictionary collide
The load factor a = n/N affects the performance of a hash table
Assuming that the hash values are like random numbers, it can be shown that the expected number of probes for an insertion with open addressing is 1 / (1 - a)
The expected running time of all the dictionary ADT operations in a hash table is O(1)
In practice, hashing is very fast provided the load factor is not close to 100%
Applications of hash tables:
small databases
compilers
browser caches

42. Exp: Find a prime >= n
#include<string.h>
// prime number must be at least as large as n
int nextPrime(int n) {
if (n % 2 == ) n++;
for ( ; !isPrime(n); n +=2)
; // NOP
return n; }

43. Universal Hashing
A family of hash functions is universal if, for any 0<i,j<M-1, Pr(h(j)=h(k)) < 1/N.
Choose p as a prime between M and 2M.
Randomly select 0<a<p and 0<b<p, and define h(k)=(ak+b mod p) mod N
Theorem: The set of all functions, h, as defined here, is universal.

44. Proof of Universality (Part 1)
Let f(k) = ak+b mod p
Let g(k) = k mod N
So h(k) = g(f(k)).
f causes no collisions:
Let f(k) = f(j).
Suppose k<j. Then
So a(j-k) is a multiple of p
But both are less than p
So a(j-k) = 0. I.e., j=k. (contradiction)
Thus, f causes no collisions.

45. Proof of Universality (Part 2)
If f causes no collisions, only g can make h cause collisions.
Fix a number x. Of the p integers y=f(k), different from x, the number such that g(y)=g(x) is at most
Since there are p choices for x, the number of h’s that will cause a collision between j and k is at most
There are p(p-1) functions h. So probability of collision is at most
Therefore, the set of possible h functions is universal.

46. Quadratic Probing With Prime Table Size
Theorem: If a Quadratic probing is used and the table size is prime, then
A new element can always be inserted if the table is at least half empty.
During the insertion, no cell is probed twice