Hashing. Why Hashing? Applications of Hashing

Transcription

1 12 Hashing

2 Why Hashing? Hashing A Search algorithm is fast enough if its time performance is O(log 2 n) For 1 5 elements, it requires approx 17 operations But, such speed may not be applicable in real-world problems For instance, searching for a home address in the case of 119 call Hashing does not require any comparison operation; thus, O(1) complexity is pursued regardless of the number of Keys (n) Applications of Hashing Data Dictionary, Spelling Checker Thesaurus, Symbol tables (Compiler) Encryption

3 Hashing What & How Hashing? Key Hash Function Address Keys are stored in a fixed memory called a Hash Table A function f is used for deciding the address of key x in the Hash Table By using f(x), each key x is mapped into a certain number in the range to (HashTableSize 1) Hash Table U: Universe of Keys 1 = f(k 4 ) k 1 2 = f(k 1 ) A: actual Keys k 4 k 3 k 2 3 = f(k 3 ) = f(k 2 ) m-1

4 Example: Hashing Student Information in University Rapidly Store (Find) Student information into (from) a university database Key: Student Number Examples of Key - Student Number - Social Security Number - Telephone Number Hash Function Address M C Kang 2 G H Park 3 4 J W Lee 5 6 J H Oh 7 J H Kwon 99 J Y Jang

5 Hash Table Table size = m No keys = n Loading Factor; α = n / m Loading Factor where n : the number of keys & m : the size of Hash table Ex) Let n = 1,, m = 15; then, α = 1/15 = 2/3 (67%) Note Hash function f might map several different keys into the same position In Hashing, there must be some empty spaces As a rule of thumb, the loading fact should not exceed 75% 1 k a 1 Empty Space Available! 2 k b 2 3 k c 3 m-1 m-1

6 Collision/Overflow Collision Two keys k 1 and k 2 are synonyms with a hash function f if f(k 1 ) = f(k 2 ) where k 1 k 2 A collision occurs when we hash keys into the same position Overflow Note A overflow occurs when we hash a certain key into a full hash table Collision & Overflow occur simultaneously when k a k b k c 1 2 m-1 Collision Overflow Hash Table

7 Collision Collision occurs when f(k i ) = f(k j ) where k i k j Hash Table U: Universe of Keys A: actual Keys k 4 k 1 k 3 k x m-2 m-1 = f(k 4 ) = f(k 1 ) Collision = f(k 2 ) = f(k 5 ) k 5 = f(k 3 )

8 Hashing : Issues How to choose a good hash function? How to handle collision? How to decide the size of Hash table?

9 Hash Function What is a good Hash function? 1 Easy to Compute 2 Distribute Uniformly (ie, Unbiased) Key x random (,m-1) Prob[f(x)=i] = 1/m - If we randomly choose a key x, the probability that f(x) = i is 1/m (ie, an equal chance of hashing into any position in Hash table) 1 2 i m-1 Hash Table Examples: Good vs Bad Easy & Uniform Phone Numbers bad : the first 3 digits: better : the last 3 digits Key Hash Function Address Date of Birth bad : birth year better : birthday

10 Hash Function : Modular Division Modular Division Divide key x by a number M and use the remainder as hash address for x Hash function: h(x) = x % M M is the Hash table size The choice of a divisor (=M) is very important Key h(x) = x % 37 Address M C Kang J H Oh 35 D H Lim 36 G H Park H T Kim J Y Jang

11 Choice of a Divisor Hash Function : Modulo Division If M is an even number, even/odd-numbered keys are mapped into even/oddnumbered positions in Hash table Bad choice! Eg, M=1 ; 2%1 =, 215%1 = 15, 3%1 =, 45%1 = 5, 55%1 = 5, An odd number is better Generally, a Prime number is used as M! Eg, M = 11; 2%11 = 99, 215%11 = 13, 3%11 = 98, 45%11 = 1, 55%11 =, h(x) = x % G H Park 1 M C Kang J H Oh 98 D H Lim 99 J Y Choi 1

12 Collision Resolution Collision Resolution Inevitably, Collision occurs even if we use a good Hash function Generally, there are some (available) empty spaces Need to resolve the collisions by distributing homeless keys over empty spaces Two types of methods exist! 1) Open Addressing 2) Separate Chaining Key Hash Table k a k b 1 2 Empty Space Available! k a k b k c k c Open Addressing 3 m-1 Separate Chaining

13 Collision Resolution 1 Open Addressing If Collision occurs, permit the homeless keys to be positioned at other places Three methods exist! (1) Linear Probe (2) Quadratic Probe (3) Double Hashing 2 Separate Chaining Every key is located at its own address It is done by chaining the keys of the same address with a linked list

14 Linear Probe If a collision occurs when hashing a key x Examine the Hash table by the following rule until finding an empty space; h(x) = ( f(x) + i ) % M, where f(x) = x % M where M is Hash table size, i =, 1, 2,, & i M 1 If an empty space is found, then insert the key x into that position Resolve the collision by adding 1 repeatedly to current address! f(x) = x % G H Park 1 2 M C Kang 3 J Y Choi H T Kim 4 J H Oh Probe 1 Probe 2 ( f(x)+i ) % D H Lim 36

15 Primary cluster Linear Probe Problem of Linear Probe It suffers from Primary Cluster (ie, contiguous collection of many keys) Eg, Insert 64 into the following Hash table 63 f(x) = x % J Y Choi 2 M C Kang 3 H T Kim 4 J H Oh 5 D H Lim 6 G H Park ( f(x)+1) % 31 ( f(x)+2) % 31 ( f(x)+3) % 31 ( f(x)+4) % 31 ( f(x)+5) % 31 To resolve this problem, Quadratic prove has been considered! 29 D H Lim 3

16 Primary cluster Quadratic Probe If a collision occurs when hashing a key x Try to find an empty space in Hash table by a quadratic search h(x) = ( f(x) + i 2 ) % M If an empty space is found, then insert the key x into that position Resolve a collision by adding 1 2, 2 2, 3 2, sequentially to current address! 63 f(x) = x % J Y Choi 2 M C Kang 3 H T Kim 4 J H Oh 5 D H Lim 6 G H Park ( f(x)+1) % 31 ( f(x)+4) % 31 It doesn t suffer from Primary cluster! 1 D H Lim 3 ( f(x)+9) % 31

17 Secondary cluster Quadratic Probe Problem of Quadratic Probe It suffers from Secondary Cluster (ie, periodic, contiguous collection of keys) Eg, Insert 64 into the Hash table 1 J Y Choi 2 M C Kang D H Lim ( f(x)+1) % 31 ( f(x)+4) % f(x) = x % G H Park 1 D H Lim ( f(x)+9) % 31 To resolve this problem, Double hashing has been developed! D H Lim 3 ( f(x)+16) % 31

18 Double Hashing Double Hashing It resolves collision without suffering from Problems of Linear/Quadratic Probe Linear/quadratic probe: Probing sequence resorts to the regularities of the hashing function without considering Keys It eliminates some regularities of the probing rule Two Hash sub-functions are used; f 1 and f 2 f 1 (x) = x % M, f 2 (x) = x % U + 1 H(x) = ( f 1 (x) + if 2 (x) ) % M f 1 computes the address of a Key If collision, f 2 computes the interval size for the probing M and U (U<M) must be relatively prime numbers! eg, Relative prime numbers: M=14, U=9 What about M=14, U=12?

19 Secondary cluster Primary cluster Double Hashing Double Hashing Example f 1 (x) = x % 31, f 2 (x) = (x % 11)+1 H(x) = ( f 1 (x) + if 2 (x) ) % 31 1 J Y Choi 2 M C Kang 3 B T Shin Insert a Key 64 into a Hash table 4 M J Kim 5 D H Lim 6 G H Park 63 f 1 (x) = x % J H Lee 9 M J Kim 1 J K Jung ( f 1 (x)+9) % 31 1 st : H(63) = 1 2 nd : H(63) = (1+1*(8+1)) = 1 3 rd : H(63) = (1+2*(8+1)) = D C Park 19 J H Lee ( f 1 (x)+18) % 31 3