Transcript Hashing

Hashing
Vishnu Kotrajaras, PhD
Nattee Niparnan, PhD
Flashback
Still recall something before midterm?
Recall the previous ADT
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List, Stack, Queue
BST
AVL
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We wish to do Insert, Delete and Find
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Do we happy with AVL tree?
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We want something FASTER!!!
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Possible with “hashing”
Overview
What is hash?
Hashing property
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Insert O(1)
Delete O(1)
Find O(1)
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Lacks order, traversal
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Hashing Idea
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What happen if all possible data is the date of
the month?
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Post Office analogy
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Addressing!!!
Hash Table
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Table (storage)
Key (address)
And possibly the Value (things to store)
Realization of Hash
Table
Put it into a practical use
What is the problem?
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What if the data is not only 1 – 31?
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What if the data is not a number ?
Hash function
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Mash key into something else
Hash function
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We use it to try to distribute values evenly
throughout our table. We may use:
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Key number % tableSize
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But if tableSize is 10, 20, 30, …we cannot use this
function.
What if keys are Strings?
Let’s see some example.
Hash function
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st
(1
example)
Sum the ASCII values of all alphabets
public static int hash(String key, int tableSize){
int hashVal = 0;
for(int i =0; i<key.length(); i++)
hashVal += key.charAt(i);
return hashVal%tableSize;
}
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The method in the last page is not good if the table is large:
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Whet if each key is short (e.g. 8 alphabets?)
An ASCII normally has a maximum value of 127.
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Therefore the sum of all 8 alphabets will not exceed 127*8.
If the table is big, data will not be distributed evenly.
The
10,000th
member
Indices will concentrate at the front.
Hash function
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
nd
(2
example)
Assume we have a big table, and each key is
made from at least 3 random alphabets.
We look at the first 3 alphabets only.
public static int hash(String key, int tableSize){
return (key.charAt(0) +27*key.charAt(1) +729*
key.charAT(2))%tableSize;
}
All alphabets, including space
27*27
This distributes well in a table of size 10000. (10007 is the first prime after
10000, we will use this number. You will see why).
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Wait, any actual key will never be random like
this:
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There will be a lot of repetition.
Hash function (3rd example)
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We calculate a polynomial function of 37, using
Horner’s Rule.
We can calculate k0 + 37k1+ 37*37k2 by using
[(k2*37)+k1]*37 +k0
Horner rule is to repeat this -> n times. In fact, it is a
calculation of:
KeySize1
 Key[ KeySize i 1] * 37
i
i 0
public static int hash(String key, int tableSize){
int hashVal = 0;
for(int i =0; i<key.length(); i++)
hashVal= 37*hashVal+key.charAt(i);
hashVal %= tableSize;
if(hashVal<0)
hashVal += tableSize;
Possible overflow
return hashVal;
}
What is “GOOD” hash function?
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Low cost
Determinism
Uniformity
Variable Range
Injective
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Perfect hash function?
Side Note
eMule, BitTorrent, all P2P
MD5
Still More Problem
Collision
Collision Resolution
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Separate Chaining
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Toolbox analogy
Open addressing
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Library shelf analogy
Separate Chaining
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Try to put it into the same position
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Use another “Data Structure”
Fixing collision: separate chaining
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Store repeated elements in a linked list.
If you want to search for an element, use hash function, then search in the list given
by that hash function.
If you want to insert an element,
 use hash function to find a list to put that element in.
 After that, check the list to see whether it already contains the element. If the list
does not have that element then insert the element at the front.
 Statistically, a newly inserted element is often accessed again soon after the
insertion.
Code for an object that has a hash
function.
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public interface Hashable
{
/**
* Compute a hash function for this object.
* @param tableSize the hash table size.
* @return (deterministically) a number between
* 0 and tableSize-1, distributed equitably.
*/
int hash( int tableSize );
}
How we use a Hashable object.
Public class Student implements Hashable{
private String name;
private double number;
private int year;
public int hash(int tableSize){
return SeparateChainingHashTable.hash(name, tableSize);
}
static method from our
HashTable class.
public boolean equals(Object rhs){
return name.equals(((Student)rhs).name);
}
}
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public class SeparateChainingHashTable
{
/**
* Construct the hash table.
*/
public SeparateChainingHashTable( )
{
this( DEFAULT_TABLE_SIZE );
}
/**
* Construct the hash table.
* @param size approximate table size.
*/
public SeparateChainingHashTable( int size )
{
theLists = new LinkedList[ nextPrime( size ) ];
for( int i = 0; i < theLists.length; i++ )
theLists[ i ] = new LinkedList( );
}
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/**
* Insert into the hash table. If the item is
* already present, then do nothing.
* @param x the item to insert.
*/
We use Student
public void insert( Hashable x )
{
LinkedList whichList = theLists[ x.hash( theLists.length ) ];
LinkedListItr itr = whichList.find( x );
if( itr.isPastEnd( ) )
whichList.insert( x, whichList.zeroth( ) );
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}
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/**
* Remove from the hash table.
* @param x the item to remove.
*/
public void remove( Hashable x )
{
theLists[ x.hash( theLists.length ) ].remove( x );
}
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/**
* Find an item in the hash table.
* @param x the item to search for.
* @return the matching item, or null if not found.
*/
public Hashable find( Hashable x )
{
return (Hashable)theLists[ x.hash( theLists.length ) ].find( x ).retrieve( );
}
/**
* Make the hash table logically empty.
*/
public void makeEmpty( )
{
for( int i = 0; i < theLists.length; i++ )
theLists[ i ].makeEmpty( );
}
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/**
* A hash routine for String objects.
* @param key the String to hash.
* @param tableSize the size of the hash table.
* @return the hash value.
*/
public static int hash( String key, int tableSize )
{
int hashVal = 0;
for( int i = 0; i < key.length( ); i++ )
hashVal = 37 * hashVal + key.charAt( i );
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hashVal %= tableSize;
if( hashVal < 0 )
hashVal += tableSize;
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return hashVal;
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}
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private static final int DEFAULT_TABLE_SIZE = 101;
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/** The array of Lists. */
private LinkedList [ ] theLists;
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/**
* Internal method to find a prime number at least as large as n.
* @param n the starting number (must be positive).
* @return a prime number larger than or equal to n.
*/
private static int nextPrime( int n )
{
if( n % 2 == 0 )
n++;
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for( ; !isPrime( n ); n += 2 )
;
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return n;
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}
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/**
* Internal method to test if a number is prime.
* Not an efficient algorithm.
* @param n the number to test.
* @return the result of the test.
*/
private static boolean isPrime( int n )
{
if( n == 2 || n == 3 )
return true;
if( n == 1 || n % 2 == 0 )
return false;
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for( int i = 3; i * i <= n; i += 2 )
if( n % i == 0 )
return false;
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return true;
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}
// Simple main
public static void main( String [ ] args )
{
SeparateChainingHashTable H = new SeparateChainingHashTable( );
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final int NUMS = 4000;
final int GAP = 37;
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System.out.println( "Checking... (no more output means success)" );
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for( int i = GAP; i != 0; i = ( i + GAP ) % NUMS )
H.insert( new MyInteger( i ) );
for( int i = 1; i < NUMS; i+= 2 )
H.remove( new MyInteger( i ) );
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for( int i = 2; i < NUMS; i+=2 )
if( ((MyInteger)(H.find( new MyInteger( i ) ))).intValue( ) != i )
System.out.println( "Find fails " + i );
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for( int i = 1; i < NUMS; i+=2 )
{
if( H.find( new MyInteger( i ) ) != null )
System.out.println( "OOPS!!! " + i );
}
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}
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}
Separate Chaining : More variation
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Can we use BST instead?
AVL?
 B-Tree?
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Some Analysis
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num berOfElem entsInTheTable
Load factor  
tableSize
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It is an average length of linked list.
Search time = time to do hashing + time to search list
= constant + time to search list
 Unsuccessful search
Search time == average list length == load factor
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Successful search
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In a list that we will search, there is one node that contains an
object that we want to find. There are other nodes too (0 or
more).
in a table, if we have N members, distributed into M lists.
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There are N-1 nodes that do not have what we want.
If we distribute these nodes evenly among the lists. Each list will have
(N-1)/M nodes.
= lambda- (1/M)
= lambda, because M is large.
On average, half the list will be searched before we find what we
want. That is, lambda/2 steps will be executed.
Therefore the average time to find the required element is 1 +
(lambda/2) steps.
The tableSize is not important. What really matters is the load factor.
Open Addressing
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Try to use another slot
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“Probing”
Try h0(x), h1(x), …
hi(x)=[hash(x)+f(i)]%tableSize, f(0)=0
 “i” is the collision count
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Use no extra space
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Load factor is very important
Open Addressing Technique
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Linear probing
Quadratic probing
Double hashing
Fixing collision by using Open
addressing
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No list.
If there is a collision, then keep calculating a
new index until an empty slot is found.
The new index is at h0(x), h1(x), …
 hi(x)=[hash(x)+f(i)]%tableSize, f(0)=0
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Every data must be put into our table. Therefore
the table must be large enough to distribute data.
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Load factor <=0.5
Open addressing: linear probing
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F is a linear function of i.
Normally we have -> f(i)=i
It is “looking ahead one slot at a time.”
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This may take time.
There will be consecutive filled slots, called
primary clustering. If a new collision takes
place, it will take some time before we can
find another empty slot.
Open addressing: quadratic
probing
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There is no primary clustering by this method.
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We usually have -> f(i)=i2
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hi(x)=[hash(x)+f(i)]%tableSize
a
if b collides with a, we add 12 to find a new empty slot.
If c also collides with a, we add 12 to find b.
We need to go further by adding 22 instead.
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However, if our table is more than half full or
the tableSize is not prime, this method does not
guarantee an empty slot.
But if the table is not yet half full and the
tableSize is prime, it is proven that we can always
find an empty slot for a new value.
Proof
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Let the tableSize be a prime number greater than
3.
Be 2 empty slot positions.
Let (h(x)+i2) mod tableSize
 tableSize
0  i, j  

2
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(h(x)+j2) mod tableSize
Prove by contradiction
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Assume both positions are the same and i !=j.
h( x )  i 2  h( x )  j 2
i2  j2
i2  j2  0
(i  j )(i  j )  0
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i-j =0 is impossible because we assumed they are not
equal.
 tableSize
0  i, j  

i+j=0 is also impossible,
2

Therefore our assumption that the two positions are
the same is wrong.
Thus the two positions are always different.
So there is always a slot for a new value, if the table is
not yet half full and the tableSize is prime.
Why prime?
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If not, the number of available slots will greatly
reduce.
Example: tableSize == 16. Assume a normal
hashing gives index ==0. (quadratic probing)
12
42
72
22
62
32
52
You can see that they fall in the same positions.
Deleting in open addressing
Open addressing implementation
class HashEntry {
Hashable element; // the element
boolean isActive; // false means -> deleted
public HashEntry( Hashable e ){
this( e, true );
}
public HashEntry( Hashable e, boolean i ){
element = e;
isActive = i;
}
}
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public class QuadraticProbingHashTable{
private static final int DEFAULT_TABLE_SIZE = 11;
/** The array of elements. */
private HashEntry [ ] array; // The array of elements
private int currentSize;
// The number of occupied cells
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public QuadraticProbingHashTable( ){
this( DEFAULT_TABLE_SIZE );
}
null active nonactive
/**
* Construct the hash table.
* @param size the approximate initial size.
*/
public QuadraticProbingHashTable( int size ){
allocateArray( size );
makeEmpty( );
}
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/**
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* Internal method to allocate array.
* @param arraySize the size of the array.
*/
private void allocateArray( int arraySize ){
array = new HashEntry[ arraySize ];
}
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/**
* Make the hash table logically empty.
*/
public void makeEmpty( ){
currentSize = 0;
for( int i = 0; i < array.length; i++ )
array[ i ] = null;
}
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/**
* Return true if currentPos exists and is active.
* @param currentPos the result of a call to findPos.
* @return true if currentPos is active.
*/
private boolean isActive( int currentPos ){
return array[ currentPos ] != null && array[ currentPos ].isActive;
}
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/**
* Method that performs quadratic probing resolution.
* @param x the item to search for.
* @return the position where the search terminates.
*/
private int findPos( Hashable x ) {
2=f(i-1)+2i-1
f(i)=i
/* 1*/ int collisionNum = 0;
/* 2*/ int currentPos = x.hash( array.length );
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/* 3*/
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while( array[ currentPos ] != null &&
!array[ currentPos ].element.equals( x ) ){
currentPos += 2 * ++collisionNum - 1; // Compute ith
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/* 4*/
probe
/* 5*/
/* 6*/
}
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if( currentPos >= array.length )
currentPos -= array.length;
/* 7*/
}
return currentPos;
// Implement the mod
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/**
* Find an item in the hash table.
* @param x the item to search for.
* @return the matching item.
*/
public Hashable find( Hashable x ){
int currentPos = findPos( x );
return isActive( currentPos ) ? array[ currentPos ].element : null;
}
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/**
* Insert into the hash table. If the item is
* already present, do nothing.
* @param x the item to insert.
*/
public void insert( Hashable x )
{
// Insert x as active
int currentPos = findPos( x );
if( isActive( currentPos ) )
return; //x is already inside, so do nothing
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array[ currentPos ] = new HashEntry( x, true );
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// Rehash; see Section 5.5
if( ++currentSize > array.length / 2 )
rehash( );
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}
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/**
* Remove from the hash table.
* @param x the item to remove.
*/
public void remove( Hashable x )
{
int currentPos = findPos( x );
if( isActive( currentPos ) )
array[ currentPos ].isActive = false;
}
hash, nextPrime, isPrime are the
same as before.
rehashing
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Rehash can be done due to 3 situations.
Do it immediately when the table is half full.
 Do it when our insert starts to fail.
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Do it when a load factor is up to some value (Does
not have to be 0.5)
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Do not forget that the more the load factor value, the
more difficult it is to insert.
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/**
* Expand the hash table.
*/
private void rehash( )
{
HashEntry [ ] oldArray = array;
O(N) because there are N
members to be rehashed.
This is not done often
because the table has to be
half filled first.
// Create a new double-sized, empty table
allocateArray( nextPrime( 2 * oldArray.length ) );
currentSize = 0;
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// Copy table over
for( int i = 0; i < oldArray.length; i++ )
if( oldArray[ i ] != null && oldArray[ i ].isActive )
insert( oldArray[ i ].element );
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return;
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}
recalculate index because
this is a new array.
Downside of quadratic probing
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Secondary clustering
Fixed by double hashing:
f(i) = i*hash2(x)
 We find hash2(x), 2 *hash2(x), …and so on.
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Must be careful when choosing a function.
If our array has 9 slots and hash2(x) = x%9 -> if we
insert 99, we will always get 0.
 hash2(x) must not give 0.
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Example of hash2
Assume hash(x) = x%tableSize
 hash2(x)=R-(x%R) , R is prime and
R< tableSize
 Let our tableSize be 16. We insert 9, 25, 26, 41,
42, 58 respectively.

26
9
25
25 collides, so we add 13-(25%13)=1
26 collides, so we add 13-(26%13)=13
41
26
9
25
42
41 collides, so we add 13-(41%13)=11
42 collides, so we add 13-(42%13)=10
but 42 still collides, so we add 2*10 from its
original index.
58
41
26
9
25
42
58 collides, so we add 13-(58%13)=7