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Hash Tables

Hash Tables. Chapter 12. Motivation. Many applications require only: Insert Search Delete Examples Symbol tables Memory management mechanisms. Direct-address tables. Assumption- unique keys are taken for a relatively small universe U={0,1,….,m-1} of keys

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Hash Tables

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  1. Hash Tables Chapter 12

  2. Motivation • Many applications require only: • Insert • Search • Delete • Examples • Symbol tables • Memory management mechanisms

  3. Direct-address tables • Assumption- unique keys are taken for a relatively small universe U={0,1,….,m-1} of keys • The set is represented by an array of references (direct access table) T[0..m-1] where entry T[j] may potentially point to an object with key=j

  4. Satellite data T U 2 0 1 3 9 4 8 2 5 3 5 6 7 8

  5. Pseudo-code • directAddressSearch(T,k) return T[k] • directAddressInsert(T,x) T[key(x)]=x • directAddressDelete(T,x) T[key(x)]=null

  6. Hash Tables • In many cases the universe U is large while the set K of keys actually stored is small. • Hash tables are data structures that require a table of size O(|K|) (instead of O(|U|)) only but still provide the O(1) (in average) access time.

  7. Hash function T U

  8. Collisions • A collision occurs when 2 different keys are mapped by h to the same slot in T • how to find hash functions that minimize collisions • Collisions are inevitable • Since |U|>m there must be two keys who have the same value • resolution by chaining • open addressing

  9. Chaining • All the elements that hash to the same slot are kept in a linked list. • chainedHashInsert(T,x) insert x at the head of list T[h(key(x))] • chainedHashSearch(T,k) search for an element with key k in list T[h(k)] • chainedHashDelete(T,x) delete x from the list T[h(key(x))]

  10. Analysis • Assume - n elements are stored in a hash table of size m • load factor • Worst case • If the hash function distributes the elements equally and can be computed in O(1) • average complexity • and if we choose m s.t n=O(m) the average complexity turns into O(1)

  11. Hash Functions • Assume that the keys are natural numbers. • Division method • How to choose m? • powers of 10 and 2 should be avoided (why) • good values for m are prime not to close to powers of 2 • ex- n=2000 m=701

  12. Hash Functions • Multiplication method • choose 0<A<1 • recommended value of A • ex k=123456 m=10000 h(k)=|10000*fract(123456*0.61803)| =|10000*fract(76300.004115)|=|10000*0.004115|=|41.151| =41

  13. Other type of keys • When the keys are not natural numbers we convert them. ex-strings s=“pt” ascii(‘p’) =112, ascii(‘t’)=116 nat(“pt”)= 112*256+116=28788

  14. Open Addressing • All the elements are stored in the table itself (m>n). • When inserting a new element, we successively probe the hash table until we find an empty slot. • The sequence of positions probed should depend on the key inserted (why?)

  15. Open addressing • The hash function is now a function that for each probe provide a slot in the table. • In order to ensure that every hash table position is considered we require probe sequence for every key to be a permutation of 0..m-1

  16. Pseudo Code • openHashInsert(T,x) j=0 k=key(x) repeat z=h(k,j) if T[z]=null or T[z]=Del then T[z]=x return z else j++ until j=m error “table overflow”

  17. Pseudo Code • openHashSearch(T,k) j=0 repeat z=h(k,j) if T[z]<> null and T[z]<>Del and key(T[z])=k then return z else j++ until T[z]=null or j=m return null

  18. Deleting • When deleting we must not replace the element with null (why?) • we use a special Del value to denote that the slot was once used. • openHashDelete(T,x) z = openHashSearch(T,x); if z<>null then T[z]=Del

  19. Hash Functions • We would like the probe sequences distributed equally among the m! possible permutations- Uniform Hashing • Uniform Hashing is difficult to implement • linear probing • quadratic probing • double hashing

  20. Linear Probing • Take an ordinary hash function then • Problem- the sequential quest for empty slots causes primary clustering • long run of occupied slots.

  21. Quadratic Probing • Let h’ be as above: • (mod m) • Adequate constants must be found to ensure that all m slots are probed. • Ex • Suffers from secondary clustering

  22. Double Hashing • The hash function is: • In order to probe all the table h’’(k) must be relatively prime to m. • choose m to be a power of 2 and h’’ to return a odd number • choose m to be prime and h’(k) = k mod m and h’’(k)=1+(k mod m’) where m’ is slightly less than m

  23. Analysis • Assuming a uniform hashing and load factor the average number of probes is at most: • unsuccessful search • successful search

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