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 has element j, then j is stored in A[j-1], otherwise A[j-1] contains 0. Complexity of find operation is O(1) Hash table
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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)
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
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.
Note down example from board
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)r2 + ASCII(b)r + ASCII ( c )
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.
Store all the elements mapped to the same position in a linked list.
Note down the illustration from the board.
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.
Insertion is O(1).
We are interested in average searching complexity.
Load factor names. 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.
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)
Note down example from board.
Can always insert a key as long as the table is not full
Finding may be difficult if the table is close to full.
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.
Any problem if empty and deleted are not distinguished? is never full.
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.
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.
Are we guaranteed to be able to insert as long as the hash table is not full?
If size of hash table M is a prime number greater than 3, 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 + n2 are all distinct modulo M, then we can insert an element in the hash table. Why?
Proof by contradiction.
Suppose there is p, q, 0 p < q n with
j + p2 = j + q2 mod M
p then we can always insert a new element if the table is at most half full.2 = q2 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
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