Lecture 17 April 11, 11 Chapter 5, Hashing dictionary operations general idea of hashing hash functions chaining closed hashing

Dictionary operations o search o insert o delete Applications: data base search books in a library patient records, GIS data etc. web page caching (web search) combinatorial search (game tree)

Dictionary operations o search o insert o delete ARRAY LINKED LIST sorted unsorted Search Insert delete O(log n) O(n) O(n) O(n) O(n) O(1) O(n) O(n) O(n) O(n) O(n) O(n) comparisons and data movements combined (Assuming keys can be compared with and = outcomes) Exercise: Create a similar table separately for data movements and for comparisons.

Performance goal for dictionary operations: O(n) is too inefficient. Goal O(log n) on average O(log n) in the worst-case O(1) on average Data structure that achieve these goals: (a) binary search tree (b) balanced binary search tree (AVL tree) (c) hashing. (but worst-case is O(n))

Hashing oAn important and widely useful technique for implementing dictionaries. oConstant time per operation (on the average). oWorst case time proportional to the size of the set for each operation (just like array and linked list implementation)

General idea U = Set of all possible keys: (e.g. 9 digit SS #) If n = |U| is not very large, a simple way to support dictionary operations is: map each key e in U to a unique integer h(e) in the range 0.. n – 1. Boolean array H[0.. n – 1] to store keys.

General idea

Ideal case not realistic U the set of all possible keys is usually very large so we can’t create an array of size n = |U|. Create an array H of size m much smaller than n. Actual keys present at any time will usually be smaller than n. mapping from U -> {0, 1, …, m – 1} is called hash function. Example: D = students currently enrolled in courses, U = set of all SS #’s, hash table of size = 1000 Hash function h(x) = last three digits.

Example (continued) Insert Student “Dan” SS# = h( ) = hash table buckets 871 Dan NULL

Example (continued) Insert Student “Tim” SS# = h( ) = 871, same as that of Dan. Collision hash table buckets 871 Dan NULL

Hash Functions If h(k 1 ) =  = h(k 2 ): k 1 and k 2 have collision at slot  There are two approaches to resolve collisions.

Collision Resolution Policies Two ways to resolve: (1) Open hashing, also known as separate chaining (2) Closed hashing, a.k.a. open addressing Chaining: keys that collide are stored in a linked list.

Previous Example: Insert Student “Tim” SS# = h( ) = 871, same as that of Dan. Collision hash table buckets 871 Dan NULL Tim

Open Hashing The hash table is a pointer to the head of a linked list All elements that hash to a particular bucket are placed on that bucket’s linked list Records within a bucket can be ordered in several ways by order of insertion, by key value order, or by frequency of access order

Open Hashing Data Organization D-1...

Implementation of open hashing - search bool contains( const HashedObj & x ) { list whichList = theLists[ myhash( x ) ]; return find( whichList.begin( ), whichList.end( ), x ) != whichList.end( ); } Find is a function in the STL class algorithm. Code for find is described below: template InputIterator find ( InputIterator first, InputIterator last, const T& value ) { for ( ;first!=last; first++) if ( *first==value ) break; return first; }

Implementation of open hashing - insert bool insert( const HashedObj & x ) { list whichList = theLists[ myhash( x ) ]; if( find( whichList.begin( ), whichList.end( ), x ) != whichList.end( ) ) return false; whichList.push_back( x ); return true; } The new key is inserted at the end of the list.

Implementation of open hashing - delete

Choice of hash function A good hash function should: be easy to compute distribute the keys uniformly to the buckets use all the fields of the key object.

Example: key is a string over {a, …, z, 0, … 9, _ } Suppose hash table size is n = (Choose table size to be a prime number.) Good hash function: interpret the string as a number to base 37 and compute mod h(“word”) = ? “w” = 23, “o” = 15, “r” = 18 and “d” = 4. h(“word”) = (23 * 37^ * 37^ * 37^1 + 4) % 10007

Computing hash function for a string Horner’s rule: (( … (a 0 x + a 1 ) x + a 2 ) x + … + a n-2 )x + a n-1 ) int hash( const string & key ) { int hashVal = 0; for( int i = 0; i < key.length( ); i++ ) hashVal = 37 * hashVal + key[ i ]; return hashVal; }

Computing hash function for a string int myhash( const HashedObj & x ) const { int hashVal = hash( x ); hashVal %= theLists.size( ); return hashVal; } Alternatively, we can apply % theLists.size() after each iteration of the loop in hash function. int myHash( const string & key ) { int hashVal = 0; int s = theLists.size(); for( int i = 0; i < key.length( ); i++ ) hashVal = (37 * hashVal + key[ i ]) % s; return hashVal % s; }

Analysis of open hashing/chaining Open hashing uses more memory than open addressing (because of pointers), but is generally more efficient in terms of time. If the keys arriving are random and the hash function is good, keys will be nicely distributed to different buckets and so each list will be roughly the same size. Let n = the number of keys present in the hash table. m = the number of buckets (lists) in the hash table. If there are n elements in set, then each bucket will have roughly n/m If we can estimate n and choose m to be ~ n, then the average bucket will be 1. (Most buckets will have a small number of items).

Analysis continued Average time per dictionary operation: m buckets, n elements in dictionary  average n/m elements per bucket n/m = is called the load factor. insert, search, remove operation take O(1+n/m) = O(1  time each (1 for the hash function computation) If we can choose m ~ n, constant time per operation on average. (Assuming each element is likely to be hashed to any bucket, running time constant, independent of n.)

Closed Hashing Associated with closed hashing is a rehash strategy: “If we try to place x in bucket h(x) and find it occupied, find alternative location h 1 (x), h 2 (x), etc. Try successively until all the cells have been probed. If this happens, then the hash table is full.” h(x) is called home bucket Simplest rehash strategy is called linear hashing h i (x) = (h(x) + i) % D In general, the collision resolution strategy is to generate a sequence of hash table addresses (probe sequence); test each slot until you find an empty one (probing)

Closed Hashing Example: m =8, keys a,b,c,d have hash values h(a)=3, h(b)=0, h(c)=4, h(d)= b a c Where do we insert d? 3 already filled Probe sequence using linear hashing: h 1 (d) = (h(d)+1)%8 = 4%8 = 4 h 2 (d) = (h(d)+2)%8 = 5%8 = 5* h 3 (d) = (h(d)+3)%8 = 6%8 = 6 Etc. Wraps around the beginning of the table d

Operations Using Linear Hashing Test for membership: search Examine h(k), h 1 (k), h 2 (k), …, until we find k or an empty bucket or home bucket case 1: successful search -> return true case 2: unsuccessful search -> false case 3: unsuccessful search and table is full If deletions are not allowed, strategy works! What if deletions?

Dictionary Operations with Linear Hashing What if deletions? If we reach empty bucket, cannot be sure that k is not somewhere else and empty bucket was occupied when k was inserted Need special placeholder deleted, to distinguish bucket that was never used from one that once held a value

Implementation of closed hashing Code slightly modified from the text. // CONSTRUCTION: an approximate initial size or default of 101 // // ******************PUBLIC OPERATIONS********************* // bool insert( x ) --> Insert x // bool remove( x ) --> Remove x // bool contains( x ) --> Return true if x is present // void makeEmpty( ) --> Remove all items // int hash( string str ) --> Global method to hash strings There is no distinction between hash function used in closed hashing and open hashing. (I.e., they can be used in either context interchangeably.)

template class HashTable { public: explicit HashTable( int size = 101 ) : array( nextPrime( size ) ) { makeEmpty( ); } bool contains( const HashedObj & x ) { return isActive( findPos( x ) ); } void makeEmpty( ) { currentSize = 0; for( int i = 0; i < array.size( ); i++ ) array[ i ].info = EMPTY; }

bool insert( const HashedObj & x ) { int currentPos = findPos( x ); if( isActive( currentPos ) ) return false; array[ currentPos ] = HashEntry( x, ACTIVE ); if( ++currentSize > array.size( ) / 2 ) rehash( ); // rehash when load factor exceeds 0.5 return true; } bool remove( const HashedObj & x ) { int currentPos = findPos( x ); if( !isActive( currentPos ) ) return false; array[ currentPos ].info = DELETED; return true; } enum EntryType { ACTIVE, EMPTY, DELETED };

private: struct HashEntry { HashedObj element; EntryType info; }; vector array; int currentSize; bool isActive( int currentPos ) const { return array[ currentPos ].info == ACTIVE; }

int findPos( const HashedObj & x ) { int offset = 1; // int offset = s_hash(x); /* double hashing */ int currentPos = myhash( x ); while( array[ currentPos ].info != EMPTY && array[ currentPos ].element != x ) { currentPos += offset; // Compute ith probe // offset += 2 /* quadratic probing */ if( currentPos >= array.size( ) ) currentPos -= array.size( ); } return currentPos; }

Performance Analysis - Worst Case Initialization: O(m), m = # of buckets Insert and search: O(n), n number of elements currently in the table – Suppose there are close to n elements in the table that form a chain. Now want to search x, and say x is not in the table. It may happen that h(x) = start address of a very long chain. Then, it will take O(c) time to conclude failure. c ~ n. No better than an unsorted array.

Example h(k) = k%11 = What if next element has home bucket 0?  go to bucket 3 Same for elements with home bucket 1 or 2! Only a record with home position 3 will stay.  p = 4/11 that next record will go to bucket 3 2. Similarly, records hashing to 7,8,9 will end up in Only records hashing to 4 will end up in 4 (p=1/11); same for 5 and 6 I II insert 1052 (h.b. 7) 1052 next element in bucket 3 with p = 8/11

Performance Analysis - Average Case Distinguish between successful and unsuccessful searches Delete = successful search for record to be deleted Insert = unsuccessful search along its probe sequence Expected cost of hashing is a function of how full the table is: load factor = n/m

Random probing model vs. linear probing model It can be shown that average costs under linear hashing (probing) are: Insertion: 1/2(1 + 1/(1 - ) 2 ) Deletion: 1/2(1 + 1/(1 - )) Random probing: Suppose we use the following approach: we create a sequence of hash functions h, h,… all of which are independent of each other. insertion: 1/(1 – ) deletion: 1/ log(1/ (1 – ))

Random probing – analysis of insertion (unsuccessful search) What is the expected number of times one should roll a die before getting 4? Answer: 6 (probability of success = 1/6.) More generally, if the probability of success = p, expected number of times you repeat until you succeed is 1/p. If the current load factor =, then the probability of success = 1 – since the proportion of empty slots is 1 –.

Improved Collision Resolution Linear probing: h i (x) = (h(x) + i) % D all buckets in table will be candidates for inserting a new record before the probe sequence returns to home position clustering of records, leads to long probing sequence Linear probing with increment c > 1: h i (x) = (h(x) + ic) % D c constant other than 1 records with adjacent home buckets will not follow same probe sequence Double hashing: h i (x) = (h(x) + i g(x)) % D G is another hash function that is used as the increment amount. Avoids clustering problems associated with linear probing.

Comparison with Closed Hashing Worst case performance is O(n) for both. Average case is a small constant in both cases when  is small. Closed hashing – uses less space. Open hashing – behavior is not sensitive to load factor. Also no need to resize the table since memory is dynamically allocated.

Random probing model vs. linear probing model It can be shown that average costs under linear hashing (probing) are: Insertion: 1/2(1 + 1/(1 - ) 2 ) ‏ Deletion: 1/2(1 + 1/(1 - )) ‏ Random probing: Suppose we use the following approach: we create a sequence of hash functions h, h,… all of which are independent of each other. insertion: 1/(1 – ) ‏ deletion: 1/ log(1/ (1 – )) ‏

Random probing – analysis of insertion (unsuccessful search) ‏ What is the expected number of times one should roll a die before getting 4? Answer: 6 (probability of success = 1/6.) More generally, if the probability of success = p, expected number of times you repeat until you succeed is 1/p. Probes are assumed to be independent. Success in the case of insertion involves finding an empty slot to insert.

Proof for the case insertion: 1/(1 – ) ‏ Recall: geometric distribution involves a sequence of independent random experiments, each with outcome success (with prob = p) or failure (with prob = 1 – p). We repeat the experiment until we get success. The question is: what is the expected number of trials performed? Answer: 1/p In case of insertion, success involves finding an empty slot. Probability of success is thus 1 –. Thus, the expected number of probes = 1/(1 – ) ‏

Improved Collision Resolution Linear probing: h i (x) = (h(x) + i) % D all buckets in table will be candidates for inserting a new record before the probe sequence returns to home position clustering of records, leads to long probing sequence Linear probing with increment c > 1: h i (x) = (h(x) + ic) % D c constant other than 1 records with adjacent home buckets will not follow same probe sequence Double hashing: h i (x) = (h(x) + i g(x)) % D G is another hash function that is used as the increment amount. Avoids clustering problems associated with linear probing.

Comparison with Closed Hashing Worst case performance is O(n) for both. Average case is a small constant in both cases when  is small. Closed hashing – uses less space. Open hashing – behavior is not sensitive to load factor. Also no need to resize the table since memory is dynamically allocated.

Another hash function - Multiplication Method We choose m to be power of 2 (m=2 p ) and For example, k=123456, m=512 then:

Multiplication Method: Implementation x w bits A 2 W key h(key) ‏ extract p bits product high order word low order word