Maps, Dictionaries, Hashing

Maps, Dictionaries, Hashing

Outline and Reading Map ADT (§9.1) Dictionary ADT (§9.5)
Hash Tables (§9.2) Ordered Maps (§9.3)

Map ADT The map ADT models a searchable collection of key-element items The main operations of a map are searching, inserting, and deleting items Multiple items with the same key are not allowed Applications: address book mapping host names (e.g., cs16.net) to internet addresses (e.g., ) Map ADT methods: find(k): if M has an entry with key k, return an iterator p referring to this element, else, return special end iterator. put(k, v): if M has no entry with key k, then add entry (k, v) to M, otherwise replace the value of the entry with v; return iterator to the inserted/modified entry erase(k) or erase(p): remove from M entry with key k or iterator p; An error occurs if there is no such element. size(), isEmpty()

Map - Direct Address Table
A direct address table is a map in which The keys are in the range {0,1,2,…,N-1} Stored in an array of size N - T[0,N-1] Item with key k stored in T[k] Performance: insertItem, find, and removeElement all take O(1) time Space - requires space O(N), independent of n, the number of items stored in the map The direct address table is not space efficient unless the range of the keys is close to the number of elements to be stored in the map, I.e., unless n is close to N.

Dictionary ADT The dictionary ADT models a searchable collection of key-element items The main difference from a map is that multiple items with the same key are allowed Any data structure that supports a dictionary also supports a map Applications: Dictionary that has multiple definitions for the same word Dictionary ADT methods: find(k): if the dictionary has an entry with key k, returns an iterator p to an arbitrary element findAll(k): Return iterators (b,e) s.t. that all entries with key k are between them insert(k, v): insert entry (k, v) into D, return iterator to it erase(k), erase(p): remove arbitrary entry with key k or entry referenced by iterator p. Error occurs if there is no such entry Begin(), end(): return iterator to first or just beyond last entry of D size(), isEmpty()

Map/Dictionary - Log File (unordered sequence implementation)
A log file is a dictionary implemented by means of an unsorted sequence We store the items of the dictionary in a sequence (based on a doubly-linked lists or a circular array), in arbitrary order Performance: insert takes O(1) time since we can insert the new item at the beginning or at the end of the sequence find and erase take O(n) time since in the worst case (item is not found) we traverse the entire sequence to find the item with the given key Space - can be O(n), where n is the number of elements in the dictionary The log file is effective only for dictionaries of small size or for dictionaries on which insertions are the most common operations, while searches and removals are rarely performed (e.g., historical record of logins to a workstation)

Map/Dictionary implementations
n - #elements in map/Dictionary Insert Find Space Log File O(1) O(n) Direct Address Table (map only) O(N)

Hash Tables Hashing Issues Hash table (an array) of size N, H[0,N-1]
Hash function h that maps keys to indices in H Issues Hash functions - need method to transform key to an index in H that will have nice properties. Collisions - some keys will map to the same index of H (otherwise we have a Direct Address Table). Several methods to resolve the collisions Chaining - put elements that hash to same location in a linked list Open addressing - if a collision occurs, have a method to select another location in the table

Hash Functions and Hash Tables
A hash function h maps keys of a given type to integers in a fixed interval [0, N - 1] Example: h(x) = x mod N is a hash function for integer keys The integer h(x) is called the hash value of key x A hash table for a given key type consists of Hash function h Array (called table) of size N When implementing a dictionary with a hash table, the goal is to store item (k, o) at index i = h(k)

Example We design a hash table for a dictionary storing items (SSN, Name), where SSN (social security number) is a nine-digit positive integer Our hash table uses an array of size N = 10,000 and the hash function h(x) = last four digits of x  1 2 3 4 9997 9998 9999 …

Collisions Collisions occur when different elements are mapped to the same cell collisions must be resolved Chaining (store in list outside the table) Open addressing (store in another cell in the table) Example with Modulo Method h(k) = k mod N If N=10, then h(k)=0 for k=0,10,20, … h(k)= 1 for k=1, 11, 21, etc …

Collision Resolution with Chaining
Collisions occur when different elements are mapped to the same cell Chaining: let each cell in the table point to a linked list of elements that map there  1 2 3 4 Chaining is simple, but requires additional memory outside the table

Exercise: chaining Assume you have a hash table H with N=9 slots (H[0,8]) and let the hash function be h(k)=k mod N. Demonstrate (by picture) the insertion of the following keys into a hash table with collisions resolved by chaining. 5, 28, 19, 15, 20, 33, 12, 17, 10

Collision Resolution in Open Addressing - Linear Probing
Open addressing: the colliding item is placed in a different cell of the table Linear probing handles collisions by placing the colliding item in the next (circularly) available table cell. So the i-th cell checked is: H(k,i) = (h(k)+i)mod N Each table cell inspected is referred to as a “probe” Colliding items lump together, causing future collisions to cause a longer sequence of probes Example: h(x) = x mod 13 Insert keys 18, 41, 22, 44, 59, 32, 31, 73, in this order 1 2 3 4 5 6 7 8 9 10 11 12

Open addressing: the colliding item is placed in a different cell of the table Linear probing handles collisions by placing the colliding item in the next (circularly) available table cell. So the i-th cell checked is: H(k,i) = (h(k)+i)mod N Each table cell inspected is referred to as a “probe” Colliding items lump together, causing future collisions to cause a longer sequence of probes Example: h(x) = x mod 13 Insert keys 18, 41, 22, 44, 59, 32, 31, 73, in this order 18 1 2 3 4 5 6 7 8 9 10 11 12

Open addressing: the colliding item is placed in a different cell of the table Linear probing handles collisions by placing the colliding item in the next (circularly) available table cell. So the i-th cell checked is: H(k,i) = (h(k)+i)mod N Each table cell inspected is referred to as a “probe” Colliding items lump together, causing future collisions to cause a longer sequence of probes Example: h(x) = x mod 13 Insert keys 18, 41, 22, 44, 59, 32, 31, 73, in this order 41 18 1 2 3 4 5 6 7 8 9 10 11 12

Open addressing: the colliding item is placed in a different cell of the table Linear probing handles collisions by placing the colliding item in the next (circularly) available table cell. So the i-th cell checked is: H(k,i) = (h(k)+i)mod N Each table cell inspected is referred to as a “probe” Colliding items lump together, causing future collisions to cause a longer sequence of probes Example: h(x) = x mod 13 Insert keys 18, 41, 22, 44, 59, 32, 31, 73, in this order 41 18 22 1 2 3 4 5 6 7 8 9 10 11 12

Open addressing: the colliding item is placed in a different cell of the table Linear probing handles collisions by placing the colliding item in the next (circularly) available table cell. So the i-th cell checked is: H(k,i) = (h(k)+i)mod N Each table cell inspected is referred to as a “probe” Colliding items lump together, causing future collisions to cause a longer sequence of probes Example: h(x) = x mod 13 Insert keys 18, 41, 22, 44, 59, 32, 31, 73, in this order 41 18 44 22 1 2 3 4 5 6 7 8 9 10 11 12

Open addressing: the colliding item is placed in a different cell of the table Linear probing handles collisions by placing the colliding item in the next (circularly) available table cell. So the i-th cell checked is: H(k,i) = (h(k)+i)mod N Each table cell inspected is referred to as a “probe” Colliding items lump together, causing future collisions to cause a longer sequence of probes Example: h(x) = x mod 13 Insert keys 18, 41, 22, 44, 59, 32, 31, 73, in this order 41 18 44 59 22 1 2 3 4 5 6 7 8 9 10 11 12

Open addressing: the colliding item is placed in a different cell of the table Linear probing handles collisions by placing the colliding item in the next (circularly) available table cell. So the i-th cell checked is: H(k,i) = (h(k)+i)mod N Each table cell inspected is referred to as a “probe” Colliding items lump together, causing future collisions to cause a longer sequence of probes Example: h(x) = x mod 13 Insert keys 18, 41, 22, 44, 59, 32, 31, 73, in this order 41 18 44 59 32 22 1 2 3 4 5 6 7 8 9 10 11 12

Open addressing: the colliding item is placed in a different cell of the table Linear probing handles collisions by placing the colliding item in the next (circularly) available table cell. So the i-th cell checked is: H(k,i) = (h(k)+i)mod N Each table cell inspected is referred to as a “probe” Colliding items lump together, causing future collisions to cause a longer sequence of probes Example: h(x) = x mod 13 Insert keys 18, 41, 22, 44, 59, 32, 31, 73, in this order 41 18 44 59 32 22 31 1 2 3 4 5 6 7 8 9 10 11 12

Open addressing: the colliding item is placed in a different cell of the table Linear probing handles collisions by placing the colliding item in the next (circularly) available table cell. So the i-th cell checked is: H(k,i) = (h(k)+i)mod N Each table cell inspected is referred to as a “probe” Colliding items lump together, causing future collisions to cause a longer sequence of probes Example: h(x) = x mod 13 Insert keys 18, 41, 22, 44, 59, 32, 31, 73, in this order 41 18 44 59 32 22 31 73 1 2 3 4 5 6 7 8 9 10 11 12

Search with Linear Probing
Consider a hash table A that uses linear probing find(k) We start at cell h(k) We probe consecutive locations until one of the following occurs An item with key k is found, or An empty cell is found, or N cells have been unsuccessfully probed Algorithm find(k) i  h(k) p  0 repeat c  A[i] if c =  return Position(null) else if c.key () = k return Position(c) else i  (i + 1) mod N p  p + 1 until p = N

Updates with Linear Probing
To handle insertions and deletions, we introduce a special object, called AVAILABLE, which replaces deleted elements removeElement(k) We search for an item with key k If such an item (k, o) is found, we replace it with the special item AVAILABLE and we return the position of this item Else, we return a null position insertItem(k, o) We throw an exception if the table is full We start at cell h(k) We probe consecutive cells until one of the following occurs A cell i is found that is either empty or stores AVAILABLE, or N cells have been unsuccessfully probed We store item (k, o) in cell i

Exercise: Linear Probing
Assume you have a hash table H with N=11 slots (H[0,10]) and let the hash function be h(k)=k mod N. Demonstrate (by picture) the insertion of the following keys into a hash table with collisions resolved by linear probing. 10, 22, 31, 4, 15, 28, 17, 88, 59

Open Addressing: Double Hashing
Double hashing uses a secondary hash function d(k) and handles collisions by placing an item in the first available cell of the series h(k,i) =(h(k) + i*d(k)) mod N for i = 0, 1, … , N - 1 The secondary hash function d(k) cannot have zero values The table size N must be a prime to allow probing of all the cells Common choice of compression map for the secondary hash function: d2(k) = q - (k mod q) where q < N q is a prime The possible values for d2(k) are 1, 2, … , q

Example of Double Hashing
Consider a hash table storing integer keys that handles collision with double hashing N = 13 h(k) = k mod 13 d(k) = 7 - (k mod 7) Insert keys 18, 41, 22, 44, 59, 32, 31, 73, in this order 1 2 3 4 5 6 7 8 9 10 11 12 31 41 18 32 59 73 22 44 1 2 3 4 5 6 7 8 9 10 11 12

Exercise: Double Hashing
Assume you have a hash table H with N=11 slots (H[0,10]) and let the hash functions for double hashing be h(k,i)=(h(k) + i*h2(k))mod N h(k)=k mod N h2(k)=1 + (k mod (N-1)) Demonstrate (by picture) the insertion of the following keys into H 10, 22, 31, 4, 15, 28, 17, 88, 59

Hash Functions A hash function is usually specified as the composition of two functions: Hash code map: h1: keys  integers Compression map: h2: integers  [0, N - 1] The hash code map is applied first, and the compression map is applied next on the result, i.e., h(x) = h2(h1(x)) The goal of the hash function is to “disperse” the keys in an apparently random way

Hash Code Maps Memory address: Integer cast: Component sum:
We reinterpret the memory address of the key object as an integer Good in general, except for numeric and string keys Integer cast: We reinterpret the bits of the key as an integer Suitable for keys of length less than or equal to the number of bits of the integer type (e.g., char, short, int and float on many machines) Component sum: We partition the bits of the key into components of fixed length (e.g., 16 or 32 bits) and we sum the components (ignoring overflows) Suitable for numeric keys of fixed length greater than or equal to the number of bits of the integer type (e.g., long and double on many machines)

Hash Code Maps (cont.) Polynomial accumulation: We partition the bits of the key into a sequence of components of fixed length (e.g., 8, 16 or 32 bits) a0 a1 … an-1 We evaluate the polynomial p(z) = a0 + a1 z + a2 z2 + … … + an-1zn-1 at a fixed value z, ignoring overflows Especially suitable for strings (e.g., the choice z = 33 gives at most 6 collisions on a set of 50,000 English words) Polynomial p(z) can be evaluated in O(n) time using Horner’s rule: The following polynomials are successively computed, each from the previous one in O(1) time p0(z) = an-1 pi (z) = an-i-1 + zpi-1(z) (i = 1, 2, …, n -1) We have p(z) = pn-1(z)

Compression Maps Division: Multiply, Add and Divide (MAD):
h2 (y) = y mod N The size N of the hash table is usually chosen to be a prime The reason has to do with number theory and is beyond the scope of this course Multiply, Add and Divide (MAD): h2 (y) = (ay + b) mod N a and b are nonnegative integers such that a mod N  0 Otherwise, every integer would map to the same value b

Performance of Hashing
In the worst case, searches, insertions and removals on a hash table take O(n) time The worst case occurs when all the keys inserted into the dictionary collide The load factor a = n/N affects the performance of a hash table Assuming that the hash values are like random numbers, it can be shown that the expected number of probes for an insertion with open addressing is 1 / (1 - a) The expected running time of all the dictionary ADT operations in a hash table is O(1) In practice, hashing is very fast provided the load factor is not close to 100% Applications of hash tables: small databases compilers browser caches

Uniform Hashing Assumption
The probe sequence of a key k is the sequence of slots that will be probed when looking for k In open addressing, the probe sequence is h(k,0), h(k,1), h(k,2), h(k,3), … Uniform Hashing Assumption: Each key is equally likely to have any one of the N! permutations of {0,1, 2, …, N-1} as is probe sequence Note: Linear probing and double hashing are far from achieving Uniform Hashing Linear probing: N distinct probe sequences Double Hashing: N2 distinct probe sequences

Performance of Uniform Hashing
Theorem: Assuming uniform hashing and an open-address hash table with load factor a = n/N < 1, the expected number of probes in an unsuccessful search is at most 1/(1-a). Exercise: compute the expected number of probes in an unsuccessful search in an open address hash table with a = ½ , a=3/4, and a = 99/100.

Maps/Dictionaries n = #elements in map/dictionary,
N=#possible keys (it could be N>>n) or size of hash table Insert Find Space Log File O(1) O(n) Direct Address Table (map only) O(N) Hashing (chaining) O(n/N) O(n+N) Hashing (open addressing) O(1/(1-n/N))

Ordered Map In an ordered Map, we wish to perform the usual map operations, but also maintain an order relation for the keys in the dictionary. Naturally supports Look-Up Tables - store dictionary in a vector by non-decreasing order of the keys Binary Search Ordered Dictionary ADT: In addition to the generic dictionary ADT, the ordered dictionary ADT supports the following functions: closestBefore(k): return the position of an item with the largest key less than or equal to k closestAfter(k): return the position of an item with the smallest key greater than or equal to k

Lookup Table A lookup table is a dictionary implemented by means of a sorted sequence We store the items of the dictionary in an array-based sequence, sorted by key We use an external comparator for the keys Performance: find takes O(log n) time, using binary search insertItem takes O(n) time since in the worst case we have to shift n/2 items to make room for the new item removeElement take O(n) time since in the worst case we have to shift n/2 items to compact the items after the removal The lookup table is effective only for dictionaries of small size or for dictionaries on which searches are the most common operations, while insertions and removals are rarely performed (e.g., credit card authorizations)

Example of Ordered Map: Binary Search
Binary search performs operation find(k) on a dictionary implemented by means of an array-based sequence, sorted by key similar to the high-low game at each step, the number of candidate items is halved terminates after a logarithmic number of steps Example: find(7) 1 3 4 5 7 8 9 11 14 16 18 19 m l h 1 3 4 5 7 8 9 11 14 16 18 19 m l h 1 3 4 5 7 8 9 11 14 16 18 19 m l h 1 3 4 5 7 8 9 11 14 16 18 19 l=m =h

Universal Hashing A family of hash functions is universal if, for any 0<i,j<M-1, Pr(h(j)=h(i)) < 1/N. Choose p as a prime between M and 2M. Randomly select 0<a<p and 0<b<p, and define h(k)=(ak+b mod p) mod N Theorem: The set of all functions, h, as defined here, is universal.

Proof of Universality (Part 1)
Let f(k) = ak+b mod p Let g(k) = k mod N So h(k) = g(f(k)). f causes no collisions: Let f(k) = f(j). Suppose k<j. Then So a(j-k) is a multiple of p But both are less than p So a(j-k) = 0. I.e., j=k. (contradiction) Thus, f causes no collisions.

Proof of Universality (Part 2)
If f causes no collisions, only g can make h cause collisions. Fix a number x. Of the p integers y=f(k), different from x, the number such that g(y)=g(x) is at most Since there are p choices for x, the number of h’s that will cause a collision between j and k is at most There are p(p-1) functions h. So probability of collision is at most Therefore, the set of possible h functions is universal.

Maps, Dictionaries, Hashing

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