Hash Tables

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Hash Tables

Many applications require a dynamic set that only supports the dictionary operations: Insert, Delete, Search

A hash table is an efficient implementation of a dictionary

Worst case – same as linked list – O(n)

Under reasonable assumptions – O(1)

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Direct-address Tables

AssumptionsThe universe U of keys is reasonably small:U = { 0, 1, 2, …, m-1 }, for some small m

No two elements have the same key

ImplementationAllocate an array of size m

Insert the kth element into the kth slot in the array

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Direct-address Tables

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Direct-address Tables

Advantage O(1) time for all operations

DisadvantagesWasteful if the number of elements actually inserted is significantly smaller than the size of the universe (m)

Only applicable for small values of m,i.e. a limited range of keys

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Hash Tables

Performance is almost similar to that of a direct-address table, but without the limitations

The universe U may be very large

The storage requirement is O(|K|), where K is the set of keys actually used

Disadvantage – O(1) performance is now average case, not worst case

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Hash Tables

We need a hash function to map keys from the universe U into the hash tableh: U { 0, 1, …, m-1 }

For each key k, the hash function computes a hash value h(k)

If two keys hash to the same value:h(k1) = h(k2), we call this a collision

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Hash Tables

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Collisions

Can we avoid collisions altogether?

No. Since |U| > m, some keys must have the same hash value

A good hash function will be as ‘random’ as possible

Still, collisions must be resolved

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Collision Resolution

Chaining (also called open hash)Elements stored in their ‘correct’ slot

Collisions resolved by creating linked lists

Open addressing (also calledclosed hash)

All elements stored inside the table

Maybe rehashed if their slot is full

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Chaining – Open Hash

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Collision resolution – Chaining

All keys that have the same hash value are placed in a linked list

Insertion can be done at the beginning of the list in O(1) time

Searching is proportional to the length of the list – with a good hash function, will also be O(1)

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Hash Function Requirements

A hash function must be deterministic – the hash value generated for each key cannot change during the life of the hash table

Equal keys must always be mapped to the same hash value

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Hash Function Properties

Properties of a good hash functionEasy to evaluate – h(x) can be computed very quickly (not only in O(1), but also with a small constant)

Uniform distribution over all the table slots

Different keys are mapped to different slots (as much as possible)

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Simple Uniform HashingThe quality of the hash function strongly influences the efficiencyof the hash table

Simple Uniform Hashing assumption: The hash function will hash any keyinto any slot with equal probability

It is possible to define hash functions that almost satisfy this assumption

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Analysis

The load factor of a hash table is defined as the number of elements stored in the table, divided by the total number of slots:

A search will take under the assumption of simple uniform hashing

Therefore, all hash operations can be performed in O(1)

/n m

(1 )

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Computing Key Values

The first step is to represent the key as a natural integer number

For example if S is a string then we can interpret it as an integer value using the following formula:keylength

i

0

128 ( [ ])i

char key i

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The Division Method

Key k is mapped into one of m slots by taking the remainder of k divided by m:

Choosing the value of mPreferably prime

Not too close to a power of 2

( ) modh k k m

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Example – the Division Method

Let h be a hash table of 9 slots andh(k) = k mod 9. insert the elements:6, 43, 23, 62, 1, 13, 34, 55, 25

h(13) = 13 mod 9 = 4h(34) = 34 mod 9 = 7h(55) = 55 mod 9 = 1h(25) = 25 mod 9 = 7

h(6) = 6 mod 9 = 6h(43) = 43 mod 9 = 7h(23) = 23 mod 9 = 5h(62) = 62 mod 9 = 8h(1) = 1 mod 9 = 1

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Open AddressingEach element occupies a single slot in the hash table – no chaining is done

To insert an element, we probe the table according to the hash function until an empty slot is found

The hash function is now a function of both the key and the number of attempts in the insertion process: {0,1..... 1} {0,1..... 1}h U m m

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Linear Probing

A hash value is computed using any hash function h’, and then the number of the current attempt is added to it:

Slots are examined sequentially, until an empty one is found

( , ) ( '( ) )modh k i h k i m

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Linear Probing

Easy to implement but suffers from primary clustering

Clusters tend to grow:If an empty slot is preceded by i full slots, the probability that it will be the next one filled is (i+1)/m

If an empty slot is preceded by another empty slot, the probability is only 1/m

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Exercise

You are given a hash table H with 11 slots

Demonstrate inserting the following elements using linear probing and a hash function h(k) = k mod m

10, 22, 31, 4, 15, 28, 17, 88, 59

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Solutionh(10, 0) = (10 mod 11 + 0) mod 11 = 10

h(22, 0) = (22 mod 11 + 0) mod 11 = 0

h(31, 0) = (31 mod 11 + 0) mod 11 = 9

h(4, 0) = (4 mod 11 + 0) mod 11 = 4

h(15, 0) = (15 mod 11 + 0) mod 11 = 4

h(15, 1) = (15 mod 11 + 1) mod 11 = 5

h(28, 0) = (28 mod 11 + 0) mod 11 = 6

h(17, 0) = (17 mod 11 + 0) mod 11 = 6

0 1 2 3 4 5 6 7 8 9 10

22

4 15

28

31

10

25

Solutionh(17, 1) = (17 mod 11 + 1) mod 11 = 7

h(88, 0) = (88 mod 11 + 0) mod 11 = 0

h(88, 1) = (88 mod 11 + 1) mod 11 = 1

h(59, 0) = (59 mod 11 + 0) mod 11 = 4

h(59, 1) = (59 mod 11 + 1) mod 11 = 5

h(59, 2) = (59 mod 11 + 2) mod 11 = 6

h(59, 3) = (59 mod 11 + 3) mod 11 = 7

h(59, 4) = (59 mod 11 + 4) mod 11 = 8

0 1 2 3 4 5 6 7 8 9 10

22

88

4 15

28

17

59

31

10

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Quadratic Probing

In this case, the second attempt is a more complex function of i:

Tries to avoid primary clustering

However, suffers from secondary clustering

The entire probing sequence is determined by the initial probe:

21 2( , ) ( '( ) )modh k i h k c i c i m

1 2 1 2( , ) ( , ) ( , 1) ( , 1)h k i h k i h k i h k i

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Double Hashing

Given two hash functions

One of the best methods for open addressing collision resolution

Permutations are almost random

For the entire hash to be searched,m and h2(k) must be relatively prime

1 2,h h

1 2( , ) ( ( ) ( ))modh k i h k ih k m

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Double Hashing

Possible selections of h2(k)

Select m to be a power of 2, and design h2(k) to produce odd numbers

Select m to be prime, and m’ to be m-11

2

( ) mod

( ) 1 ( mod ')

h k k m

h k k m

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Double Hashing

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Issues in Open AddressingSearch may fail if items are deleted

Solution:Mark deleted items with a special symbol

Search treats this symbol as full, while insert treats it as empty

Table may be filled up

Solution:Rehashing (copy into a larger table)