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Midterm will be given in Week 11’s lecture (2 hurs) Hash Tables 1 Part E Hash Tables 0 1 2 3 4 Hash Tables 025-612-0001 981-101-0002 451-229-0004 2 Hash Tables • Questions: how to design a data structure, such that the following operations are all O(1)? – – – – Deletion Search Insertion Replace Hash Tables 3 Motivations of Hash Tables • We have n items, each contains a key and value (k, value). – The key uniquely determines the item. • Each key could be anything, e.g., a number in [0, 232], a string of length 32, etc. • How to store the n items such that given the key k, we can find the position of the item with key= k in O(1) time. – Another constraint: space required is O(n). • Linked list? Space O(n) and Time O(n). • Array? Time O(1) and space: too big, e.g., • If the key is an integer in [0, 2 32], then the space required is 2 32. • if the key is a string of length 30, the space required is 26 30. • Hash Table: space O(n) and time O(1). Hash Tables 4 Basic ideas of Hash Tables • A hash function h maps keys of a given type with a wide range to integers in a fixed interval [0, N - 1], where N is the size of the hash table such that • if k≠k’ then h(k)≠h(k’) ….. (1) . Problem: It is hard to design a function h such that (1) holds. What we can do: • We can design a function h so that with high chance, (1) holds. • i.e., (1) may not always holds, but (1) holds for most of the n keys. Hash Tables 5 Hash Functions • 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 • the goal is to store item (k, o) at index i = h(k) Hash Tables 6 Example Hash Tables 0 1 2 3 4 025-612-0001 981-101-0002 451-229-0004 … • We design a hash table storing entries as (HKID, Name), where HKID 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 • Need a method to handle collision. 9997 9998 9999 200-751-9998 7 0 1 2 3 4 Example 50xxxx01 50xxxx02 51xxxx02 86xxxx04 … • We design a hash table storing entries as (studentID, Name), 97 where studentID is a eight51xxxx98 98 digit positive integer 99 Task: to store the n items such • Our hash table uses an array that given the key k, we can of size N = 100 for our class of find the position of the item n= 49 students and the hash with key= k in O(1) time. As long as the chance for collision function is low, we can achieve this h(x) = last two digits of x goal. • Need a method to handle Setting N=1000 and looking at collision. the last three digits will reduce the chance of collision. Hash Tables 8 How to design a Hash Function • A hash function is usually • The hash code is specified as the composition of applied first, and the two functions: compression function is applied next on the Hash code: result, i.e., h1: keys integers h(x) = h2(h1(x)) – key could be anything, e.g., your name, an object, etc. • The goal of the hash Compression function: function is to “disperse” h2: integers [0, N - 1] the keys in an – The size of the array N cannot be apparently random way too large in order to save space. so that in most cases (1) – Trade-off between space and time. holds. Hash Tables 9 Integer cast: – We reinterpret the bits of the key as an integer Example: characters is mapped to its ASCII code. A=65=01000001, B=66=01000010, a=97, b=98, ,=44, .=46. – Suitable for keys of length less than or equal to the number of bits of the integer type (e.g., byte, short, int and float in Java) Hash Tables 10 Component sum: – We partition the bits of the key into components of fixed length (e.g., 16 or 32 bits) a0 a1 … an-1 and we sum the components (ignoring overflows) a0 + a1 + a2 + … +an-1 – Example 1: AB=0100000101000010 h(AB)= 01000001 + 01000010 10000011. Example 2: h(100000011000001001001000)= 10000001 10000010 01001000 01001011 (ignore 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 in Java) Hash Tables 11 Compression Functions • Division: – 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 – Example: keys: {200, 205, 210, 215, 220, 600}. If N=100, 200 and 600 have the same code, i.e., 0. It is better to choose N=101. Hash Tables 12 Compression Functions • Multiply, Add and Divide (MAD): – h2 (y) = (ay + b) mod N – a >0 and b>0 are nonnegative integers such that a mod N 0 and N is a prime number. Example: Keys={200, 205, 210, 215, 220, 600}. N=101. a=3 and b=7. h(200)=(600+7) mod 101 = 607 mod 101=1. H(205)=(615+7) mod 101 = 622 mod 101=16. h(210)=(630+7) mod 101 = 637 mod 101=31. h(215)=(645+7) mod 101 = 652 mod 101=46. h(220)=(660+7) mod 101 = 667 mod 101=61. H(600)=(3600 +7) mod 101=3607 mod 101=72. Hash Tables 13 Collision Handling • Collisions occur when different elements are mapped to the same cell • Separate Chaining: let each cell in the table point to a linked list of entries that map there 0 1 2 3 4 025-612-0001 451-229-0004 981-101-0004 • Separate chaining is simple, but requires additional memory outside the table Hash Tables 14 Open Addressing • The colliding item is placed in a different cell of the table • Load factor: n/N, where n is the number of items to store and N the size of the hash table. • n/N≤1. • To get a reasonable performance, n/N<0.5. Hash Tables 15 Linear Probing • Linear probing handles collisions by placing the colliding item in the next (circularly) available table cell • 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 0 1 2 3 4 5 6 7 8 9 10 11 12 41 18 44 59 32 22 31 73 0 1 2 3 4 5 6 7 8 9 10 11 12 Hash Tables 16 Search with Linear Probing • Consider a hash table A that uses linear probing • get(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 – To ensure the efficiency, if k is not in the table, we want to find an empty cell as soon as possible. The load factor can NOT be close to 1. Algorithm get(k) i h(k) p0 repeat c A[i] if c = return null else if c.key () = k return c.element() else i (i + 1) mod N pp+1 until p = N return null Hash Tables 17 Linear Probing • Search for key=20. • Example: – h(x) = x mod 13 – Insert keys 18, 41, 22, 44, 59, 32, 31, 73, 12, 20 in this order – h(20)=20 mod 13 =7. – Go through rank 8, 9, …, 12, 0. • Search for key=15 – h(15)=15 mod 13=2. – Go through rank 2, 3 and return null. 0 1 2 3 4 5 6 7 8 9 10 11 12 20 41 18 44 59 32 22 31 73 12 0 1 2 3 4 5 6 7 8 9 10 11 12 Hash Tables 18 Updates with Linear Probing • To handle insertions and deletions, we introduce a special object, called AVAILABLE, which replaces deleted elements • remove(k) • put(k, o) – We search for an entry with key k – If such an entry (k, o) is found, we replace it with the special item AVAILABLE and we return element o – Else, we return null – Have to modify other methods to skip available cells. Hash Tables – 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 entry (k, o) in cell i 19 Updates with Linear Probing Algorithm put(k,o) i h(k); p 0; av = -1 repeat • Example: c A[i] – h(x) = x mod 13 if av == -1 and c= – Insert keys 18, 41, 22, 44, A[i].key()=k 59, 32, 31, 73, 20, 12 in this A[i].element=o order return – Ti insert 12, we look at rank else if A[i].key()=k 12 and then rank 0. A[i].element=o return else if av > -1 and c= A[av].key()=k 0 1 2 3 4 5 6 7 8 9 10 11 12 A[av].element=o return else if av == -1 then av =i 12 41 18 44 59 32 22 31 73 20 i (i + 1) mod N 0 1 2 3 4 5 6 7 8 9 10 11 12 pp+1 until p = N Hash Tables 20 if av >-1 then A[av].key()=k; A[av].element=o A complete example • Example: – h(x) = x mod 13 – Insert keys 18, 41, 22, 44, 59, 32, 31, 73, 20, 12 in this order – Remove(): 20, 12 – Get(11): check the cell after AVAILABLE cells. – Insert keys 10, 11. 10 is at rank 12 and 11 is at rank 0. The Available cells are hard to deal with. Separate Chaining approach is simpler. 0 1 2 3 4 5 6 7 8 9 10 11 12 A A 12 41 18 44 59 32 22 31 73 20 0 1 2 3 4 5 6 7 8 9 10 11 12 Hash Tables 21 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 map 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 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: Hash Tables – small databases – compilers – browser caches 22