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Lecture 6 Hashing. Motivating Example Want to store a list whose elements are integers between 1 and 5 Will define an array of size 5, and if the list.

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Presentation on theme: "Lecture 6 Hashing. Motivating Example Want to store a list whose elements are integers between 1 and 5 Will define an array of size 5, and if the list."— Presentation transcript:

1 Lecture 6 Hashing

2 Motivating Example Want to store a list whose elements are integers between 1 and 5 Will define an array of size 5, and if the list has element j, then j is stored in A[j-1], otherwise A[j-1] contains 0. Complexity of find operation is O(1)

3 Hash table The objective is to find an element in constant time ``on an average.’’ Supposing we know the elements belong to 1,2…U, and we are allowed an overall space of U, then this can be done as described before. But U can be very large. Space for storage is called ``hash table,’’ H

4 Assume that the hashtable has size M There is a hashfunction which maps an element to a value p in 0,….M-1, and the element is placed in position p in the hashtable. The function is called h[j], (the hash value for j is h[j]) If h[j] = k, then the element is added to H[k]. Suppose we want a list of integers, then an example hash function is h[j] = j modulo M.

5 M = 5 List contains 1, 3, 9,

6 We may want to store elements which are not numbers, e.g., names. Then we use a function to convert each element to an integer and hash the integer. We want to store string, abc Represent each symbol by the ASCII code, choose a number r, integer value for abc is ASCII(a)r 2 + ASCII(b)r + ASCII ( c )

7 Implementation Hashtables are arrays. Size of a hash table is normally a prime number Two different elements may hash to the same value (collision) Hashing needs collision resolution Hash functions are chosen so that the hash values are spread over 0,…..M-1, and there are only few collisions.

8 Separate Chaining Store all the elements mapped to the same position in a linked list. H[k] is the list of all elements mapped to k. To find an element j, compute h(j). Let h(j) = k. Then search in link list H[k] To insert an element j, compute h(j). Let h(j) = k. Then insert in link list H[k] To delete an element, delete from the link list M = 5

9 Insertion is O(1). Worst case searching complexity depends on the maximum length of a list H[p] O(q) if q is the maximum length. We are interested in average searching complexity M = 5 Search for 7, look in position 2 in the array, find it empty, conclude 2 not there Search for 13, look in position 3 in the array, search the link list, 13 not found, conclude that 13 is not there Insert 4, add it in the link list starting with 9

10 Load factor is the average size of a list. = number of elements in the hash table/ number of positions in the hash table(M) Average find complexity is 1 + Want to be approximately 1 To reduce worst case complexity we choose hash functions which distribute the elements evenly in the list.

11 Open Addressing Separate chaining requires manipulation of pointers and dynamic memory allocation which are expensive. Open addressing is an alternate scheme. Want to insert key (element) j Compute h(j) = k If H[k] is empty store in H[k], otherwise try H[k+1], H[k+2], etc. (increment in modulo size) Linear Probing

12 Every position in hash table contains one element each. Can always insert a key as long as the table is not full Finding may be difficult if the table is close to full M = 5 List contains 1, 3, 9, 8

13 The idea is to declare a hash table large enough so that it is never full. Initially, all slots are empty. Elements are inserted as described. When an element is deleted, the space is marked deleted (empty and deleted are different). During the find operation, one looks for element k starting from where it should be (H[h(k)]), till the element is found, or an empty slot is found. In the latter case, we conclude that the element is not in the list.

14 1389 M = 5 Looking for 13, Start from the position which has 3, then look at that with 9, then that with 8, next with 1, reach an empty slot, conclude not there Looking for 8, start from the position which has 3, then look at that with 9, then that with 8, conclude found Any problem if empty and deleted are not distinguished? Yes, may conclude that element not here even if it is 138 M = 5 Delete 9 Search for 8, start from the position with 3, go to next slot, finds nothing, concludes empty and thus 8 not there!

15 When we insert an element k, then start from H[h(k)] and move till an empty or deleted slot can be found. An element can be inserted as long as the hash-table is not full. If hash values are clustered, then even if hash table is relatively empty, finding may be difficult.

16 Quadratic Probing Alternative to linear probing. To insert key k, try slot h(k). If the slot is full try slot h(k) + 1, then h(k) + 4, then h(k) + 9 and so on. Advantage? Are we guaranteed to be able to insert as long as the hash table is not full? Not much of clustering M = 3, first two positions full, let h(k) = 0, h(k) + n 2 mod M is always 0 or 1. Prove it. Thus we never reach the third position which is the only empty one.

17 If size of hash table M is a prime number, then we can always insert a new element if the table is at most half full. We want to insert element k. h(k) = j. Let n =  M/2  If the locations j, j + 1, j + 4,…..,j + n 2 are all distinct modulo M, then we can insert an element in the hash table. Why? Now we show that j, j + 1, j + 4,…..,j + n 2 are distinct. Let that not be so, Suppose there is p, q, 0  p < q  n with j + p 2 = j + q 2 mod M If these are distinct, and still can not be inserted then the hashtable has n + 1 elements, i.e.,  M/2  + 1 elements, which is not possible as the hash table is half full.

18 p 2 = q 2 mod M (p – q)(p + q) = 0 mod M Then either p = q mod M or p + q = 0 mod M. Is that right? Since p and q are distinct and less than M/2, neither p = q mod M nor p + q = 0 mod M Yes, since M is prime.

19 Rehashing If the hash table is close to full, then a hash table of bigger size is used. The old hash table is copied into a new one. The old hash table is subsequently deleted. Should be done infrequently. Chapter 5 of Weiss


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