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Previous Lecture Revision Hashing

Previous Lecture Revision Hashing. Searching : The Main purpose of computer is to store & retrieve Locating for a record is the most time consuming action Methods: Linear Search ( Search for a target element by element) Asymptotic notation is O(n)

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Previous Lecture Revision Hashing

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  1. Previous Lecture RevisionHashing • Searching : • The Main purpose of computer is to store & retrieve • Locating for a record is the most time consuming action • Methods: • Linear Search ( Search for a target element by element) • Asymptotic notation is O(n) • Binary Search ( Break the Group into two and search in one half) • Asymptotic notation is O(log 2 n) • Hashing ( Direct access )-Used in DBMS / File Systems • Asymptotic notation is O(1)

  2. Hashing Techniques • What is Hashing? : from the Key (INPUT) itself the index where it should be stored is derived. • Advantage :While Reading Back it can be READ IMMEDIATELY • Techniques: • Identity - Key itself becomes a index • Dis : Memory Should be LIMITLESS • Truncation - The Last digit is truncated and used as index • EG : Key 123456 So the index is 6 • Folding - Addition/Multiplication/Division is done on the key to obtain the index • EG : 123456 12 34 56 102 so index is 2

  3. Hashing Techniques • Modular Arithmetic : Use Some Arithmetic calculation on the Key to obtain the index • KEY % size EG : 123 % 10 • CAN WE USE 123 / 10 ? • What is a Collision ? • If the hashing Function gives out SAME INDEX for TWO keys then it is called a collision • EG : 123 % 10 index will be 3 • 223 % 10 index will be again 3 • We cannot store two values in the same location

  4. RESOLVING COLLISIONS

  5. How to resolve collisions • Three Methods are available, 1 . Resolving collisions by REPLACEMENT 2. Resolving collisions by OPEN ADDRESSING • Linear probing • Quadratic probing 3. Resolving collisions by CHAINING

  6. Resolving collisions by replacement • Working : we simply replace the old KEY with the new KEY when there is a collision • The Old Key is simply Lost • Or • It is combined with the new Key. • EG : 121 122 221 124 125 126 224 index = key % 10 • When it is used ? • Very Rarely used • Used only when the data are sets . Using UNION Operation old data is combined with the new data. 221 121 122 124 224 125 126

  7. Resolving collisions by Open addressing • We resolve the collision by putting the new Key in some other empty location in the table. • Two methods are used to locate the empty location 1. Linear Probing 2. Quadratic Probing • Linear Probing : • Start at the point where the collision occurred and do a sequential search through the table for an empty location. • Improvement : • Circular Probing : After reaching the end start probing from the first Example

  8. Example - Linear Probing Keys = 6 Table Size = 7 Function = key mod 7 key 12 15 21 36 84 96 Index 5 1 0 1 0 5 SOLUTION Index            0        1         2         3        4         5        6 21 15 36 84 12 96

  9. Resolving collisions by Open addressing • Quadratic Probing: If there is a collision at the address ‘h’, this method probes the table at locations h+I2 ( % hashsize )for I = 1,2,… Dis: It does not probe all locations in the table. Keys = 6 Table Size = 7 Function = key mod 7 key 12 15 21 36 84 96 Index 5 1 0 1 0 5index = h + I 2 % 7 I = 1 , 2 , 3 …… 36 21 15 84 12 96

  10. example Void main ( ) { int table[MAX],index,I target; for(I=1;I<=MAX;I++) table[I-1]=10*I; cin>>target; index = HASH(target); if (index!=1) {if table[index] == target) cout<<“Found at”<<index; else cout <<“Target Not found”;} else cout <<“Target Not found”;} # define MAX 20 int HASH(int key) { int index; index = key/10-1; if (index<MAX) return index; else return -1; }

  11. 21 84 0 15 1 36 2 3 4 12 5 96 6 Resolving collisions by Chaining • key 12 15 21 36 84 96 • Index 5 1 0 1 0 5 • Implemented using Linked List • Whenever a collision occurs a new node is created and the new value is stored and linked to the old value.

  12. 140145 137456 214562 Exercise Using modulo-division method and linear probing store the below in an array of 19 elements 224562, 137456, 214562,140145, 214576, 162145, 144467, 199645, 234534 Index is : 224562 % 19 = 1, 137456 % 19 =10 , 214562 % 19 =14,140145 % 19 =1 ( 2), 214576 % 19 =9, 162145 % 19 =18, 144467 % 19 =10 ( 11), 199645 % 19 =12, 234534 % 19 =17 224562 214576 162145 234534 199645 144467

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