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Chapter 7

Skip Lists and Hashing

Part 2: Hashing

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Sorted Linear Lists• For formula-based implementation

– Insert: O(n) comps & data moves

– Delete: O(n) comps & data moves

– Search: O(log(n)) comps

• For chained implementation: – Insert: O(n) comps

– Delete: O(n) comps

– Search: O(n) comps

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Sorted Chain

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Dictionary• A dictionary is a collection of elements, each element

has a field called key.• Key is unique for each element• Operations:

– Insert an element with a specified key value– Search the dictionary for an element with a specified key

value– delete an element with a specified key value

• The access mode for elements in a dictionary is random access (or direct access) mode: i.e. any element may be retrieved by performing a search on its key.

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Dictionary

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Ideal hashing

• Hash table: table used to store elements

• Hash function: function to map keys to positions: k => f(k)

• Search for an element with key k: if f(k) is not empty, found; otherwise, failed

• Insert: f(k) must be empty

• Delete: f(k) cannot be empty

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Example: Student record dictionary

• Use student ID (6 digit number) as the key

• ID range 951000 and 952000

• f(k) = k - 951000

• Table size: 1001 i.e. ht[0..1000]

• ht[i].key = 0 indicates an empty entry

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Evaluation: Ideal Hashing

• Initialize an empty dictionary: Θ(b) where b is the size of the table

• Search, insert, and delete: Θ(1)• Property: 1 key <=> 1 position • Problem: the range of the keys may be very

large resulting in large hash table, e.g. if the key is a 9 digit integer (ex SSN), the size of the table will be 109

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Hashing with linear open addressing

• Used when the size of the hash table (D) is smaller than the key range• f(k) = k % D• Positions in hash table are indexed 0..D-1• bucket - position in a hash table• If key values are not integral type, they need to be converted first.• two keys k1 and k2 map into the same bucket if f(k1) = f(k2) collision

• home bucket - position numbered f(k) is the home bucket for k• In general a bucket may contain space for more than one element.• An overflow occurs if there is not room in the home bucket for the

new element.• If bucket has space for only one element, collision and overflow are

the same.

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Collision, overflow and linear open addressing

80, 58, &35 map into home bucket ht(3).

In case of collision, insert in next available bucket in sequence.

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Search

• To search for an element with key k, begin at bucket f(k) and continue in successive bucket regarding the table as circular, until:– a bucket containing an element with k is found

(successful)– an empty bucket is reached (unsuccessful)– return to the home bucket (unsuccessful)

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deletion• After deletion, must move successive elements

until:– am empty bucket is reached– return to the bucket from which the deletion took

place

• To improve performance, use a NeverUsed field. May need reorganization when many buckets have their NeverUsed field set to false

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Class definition

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Constructor

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hSearch

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Search

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Insert

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Performance analysis

• b - the number of buckets in the hush table, b = D

• initialization - Θ(b)

• worst-case insert and search - Θ(n), where n is the number of elements in the table

• worst-case happens when all n keys have the same home bucket

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Performance analysis (continue)Average performance • Let α=n/b denote the loading factor

• Un and Sn - average number of buckets examined during and unsuccessful and successful search, respectively, then

]1

11[

2

1~

]1

1[2

1~

)1(2

S

U

n

n

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Performance analysis (continue)

• The performance of hashing with linear open addressing is superior:

– when α=0.5 table is half full

Un=2.5 and Sn=1.5

– when α=0.9 table is 90% full

Un=50.5 and Sn=5.5

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Determining D

• either a prime number or has no prime factors less than 20

• two methods:– begin with the largest possible value for b

– Then find the largest D (<= b) that is either a prime or has no factors smaller than 20

– e.g., when b = 530, then D = 23*23 = 529

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Determining D

Second method:– determine your accepted Un and Sn

– Estimate n

– determine α

– determine smallest b for the above α

– determine smallest integer D >= b that either prime or has no factor smaller than 20.

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Determining D

• n = 1000

• S 4 and U 50.5– S = 4 ==> α = 6/7

– U = 50.05 ==> α = 0.9

– α = min(6/7 , 0.9) = 6/7

– b = n/ α = 7000/6 = 1167

– note: 23*51 = 1173

• ==> select D = b = 1173

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Hashing with Chains

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Implementations

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An improved implementation

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Comparison with Linear Open Addressing

• Space complexity– Let s be the space required by an element– Let b and n denote the number of buckets and

number of elements, respectively– Linear open addressing: b(s+2) bytes (2 for an

element of empty array)– chaining: 2b+2n+ns bytes– when n < bs/(s+2), chaining takes less space

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Search time complexity• Worst-case time complexity= n occurs when all

elements map to same bucket (equal to that of linear open addressing)

• Average– average length of a chain is α=n/b– average number of nodes examined in an unsuccessful

search:

* if chain has i nodes, it may take 1, 2, 3, …,I examinations. Assuming equal probability, on average

search time = 2

1i

i2

)1i(ij

i

1i

1j

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Search time complexity Ctnd

2

1

2

)1(j

1

1jnU

If α=0, Un=0If α<1, Un<= α

If α>=1,

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Average time complexity for successful search

• Need to know the expected distance of each of the n elements from the head of its chain

• Without losing generality, we assume elements are inserted into the chain in increasing order

• When the ith element is inserted, the expected length of the chain is (i-1)/b; and the ith element is added into the end of the chain

• A search for this element will require examination of

1+(i-1)/b nodes

• Assuming n elements are searched for with equal probability, then

21~

2

11}/)1(1{

1

1

b

nbi

n

n

inS

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Comparison with linear open addressing

• The expected performance of chaining is superior, e.g.,– when α=0.9

– Chaining: Un=0.9, Sn=1.45

– Linear open addressing: Un=50.5, Sn=5.5

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Skip Lists

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20 24 30 40 807560

A sorted chain with head and tail nodes

20 24 30 40 807560

Pointers to middle are added

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20 24 30 40 807560

Pointers to every second node

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Skip List Implementation

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An application

• Text compression– compressor: file coding

• run-length coding: 1000 xs + 2000 ys => 1000x2000y• space needed: 3002 bytes (2 bytes for string ends) =>

12 bytes

– decompressor: decoding

• LZW Compression (Lempel, Ziv, and Welch)

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LZW Compression

• Try aaabbbbbbaabaaba

• encoded as: 0214537

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Input/Output

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Input/Output (continue)

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Dictionary organization

• Use code to represent the prefix of key

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Dictionary organization (continue)

• assume each code is 12 bits long. Hence there are at most 212=4096 codes

• Use hash table with divisor D = 4099

ChainHashTable<element, unsigned long> h(D)

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Output of codes

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Compression

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Compression (continue)

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Compression (continue)

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Headers and Function main

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Headers and Function main (continue)

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LZW Decompression• The dictionary is searched for an entry with a given code• The first code in the compressed file corresponds to a single

character• For all other codes p:

– Case 1: p is in the dictionary– Case 2: p is not in the dictionary

• If q is the code that precedes p in the compressed file, then pair (next code, test(q)fc(p)) is entered into dictionary, where fc(p) is the first character of text(p). This can only happen when text(p) = text(q)fc(q) and the current text segment is text(q)text(q)fc(q)

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Try

• Decode 0214537

• the result should be aaabbbbbbaabaaba

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Input/Output

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Input/Output (continue)

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Dictionary organization

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Input of Code

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Decompression

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Decompression (continue)

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Headers and Function main

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Headers and Function main (continue)

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End of Chapter 7