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1 Data Structures Dana Shapira Hash Tables. 2 Element Uniqueness Problem Let Determine whether there exist i j such that x i =x j Sort Algorithm Bucket.

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Presentation on theme: "1 Data Structures Dana Shapira Hash Tables. 2 Element Uniqueness Problem Let Determine whether there exist i j such that x i =x j Sort Algorithm Bucket."— Presentation transcript:

1 1 Data Structures Dana Shapira Hash Tables

2 2 Element Uniqueness Problem Let Determine whether there exist i j such that x i =x j Sort Algorithm Bucket Sort for (i=0;i { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/7/1649517/slides/slide_2.jpg", "name": "2 Element Uniqueness Problem Let Determine whether there exist i j such that x i =x j Sort Algorithm Bucket Sort for (i=0;i

3 3 Notations: U universe of keys of size |U|, K an actual set of keys of size n, T a hash table of size m Use a hash-function h:U {0,…,m-1}, h(x)=i that computes the slot i in array T where element x is to be stored, for all x in U. h(k) is computed in O(|k|) = O(1). x1x1 x2x2 x3x3 x4x4 h(x1)h(x1)h(x2)h(x2)h(x4)h(x4)h(x3)h(x3) U h Set of array indices Hash Tables

4 4 Example h:U {0,…, m-1} h(x)=x mod 10 (what is m?) input: 17,62,19,81,53 Collision: x y but h(x) = h(y). m « |U|. Solutions: 1. Chaining 2. Open addressing 81 62 53 17 19 1 2 3 7 9 0 4 5 6 8

5 5 Collision-Resolution by Chaining 181 6212 53 173757 19 Insert(T,x): Insert new element x at the head of list T[h(x.key)]. Delete(T,x):Delete element x from list T[h(x.key)]. Search(T,x):Search list T[h(x.key)].

6 6 Simple Uniform Hashing Any given element is equally likely to hash to any slot in the hash table. The slot an element hashes to is independent of where other elements hash. Load factor: α = n/m (elements stored in the hash table / number of slots in the hash table) Analysis of Chaining

7 7 Theorem: In a hash table with chaining, under the assumption of simple uniform hashing, both successful and unsuccessful searches take expected time Θ(1+α) on the average, where α is the hash table load factor. Proof: Unsuccessful Search: Under the assumption of simple uniform hashing, any key k is equally likely to hash to any slot in the hash table. The expected time to search unsuccessfully for a key k is the expected time to search to the end of list T[h(k)] which has expected length α. expected time - (1 + α ) including time for computing h(k). Successful search: The number of elements examined during a successful search is 1 more than the number of elements that appear before k in T[h(k)]. expected time - (1 + α ) Corollary: If m = (n), then Insert, Delete, and Search take expected constant time.

8 8 Example: Input = reals drawn uniformly at random from [0,1) Hash function: h(x) = mx Often, the input distribution is unknown. Then we can use heuristics or universal hashing. Designing Good Hash Functions

9 9 Hash function: h(x) = x mod m m = 2 k h(x) = the lowest k bits of x Heuristic: m = prime number not too close to a power of 2 The Division Method

10 10 Hash function: h(x) = m (cx mod 1), for some 0 < c < 1 Optimal choice of c depends on input distribution. Heuristic: Knuth suggests the inverse of the golden ratio as a value that works well: Example: x=123,456, m=10,000 h(x) = 10,000 ·(123,456·0.61803… mod 1) = = 10,000 ·(76,300.0041151… mod 1) = = 10,000 ·0.0041151… = 41.151... = 41 The Multiplication Method

11 11 Efficient Implementation of the Multiplication Method w bits p bits * Fractional part Integer part after multiplying by m = 2 p Let w be the size of a machine word Assume that key x fits into a machine word Assume that m = 2 p Restrict ourselves to values of c of the form c = s / 2 w Then cx = sx / 2 w sx is a number that fits into two machine words h(x) = p most significant bits of the lower word h(x) = m (cx mod 1)

12 12 x = 123456, p = 14, m = 2 14 = 16384, w = 32, Then sx = (76300 2 32 ) + 17612864 The 14 most significant bits of 17612864 are 67; that is, h(x) = 67 x = 0000 0000 0000 0001 1110 0010 0100 0000 s = 1001 1110 0011 0111 0111 1001 1011 1001 sx = 0000 0000 0000 0001 0010 1010 0000 1100 0000 0001 0000 1100 1100 0000 0100 0000 h(x) = 00 0000 0100 0011 = 67 Example h(x) = m (cx mod 1)

13 13 Open Addressing h(x)h(x) All elements are stored directly in the hash table. Load factor α cannot exceed 1. If slot T[h(x)] is already occupied for a key x, we probe alternative locations until we find an empty slot. Searching probes slots starting at T[h(x)] until x is found or we are sure that x is not in T. Instead of computing h(x), we compute h(x, i) i -the probe number.

14 14 Hash function: h(k, i) = (h'(k) + i) mod m, where h' is an original hash function. Benefits: Easy to implement Problem: Primary Clustering - Long runs of occupied slots build up as table becomes fuller. Linear Probing

15 15 Hash function: h(k, i) = (h'(k) + c 1 i + c 2 i 2 ) mod m, where h' is an original hash function. Benefits: No more primary clustering Problem: Secondary Clustering - Two elements x and y with h'(x) = h'(y) have same probe sequence. Quadratic Probing

16 16 Hash function: h(k, i) = (h 1 (k) + ih 2 (k)) mod m, where h 1 and h 2 are two original hash functions. h 2 (k) has to be prime w.r.t. m; that is, gcd(h 2 (k), m) = 1. Two methods: Choose m to be a power of 2 and guarantee that h 2 (k) is always odd. Choose m to be a prime number and guarantee that h 2 (k) < m. Benefits: No more clustering Drawback: More complicated than linear and quadratic probing Double Hashing

17 17 Analysis of Open Addressing Uniform hashing: The probe sequence h(k, 0), …, h(k, m – 1) is equally likely to be any permutation of 0, …, m – 1. Theorem: In an open-address hash table with load factor α < 1, the expected number of probes in an unsuccessful search is at most 1 / (1 – α ), assuming uniform hashing. Proof: Let X be the number of probes in an unsuccessful search. A i = there is an i-th probe, and it accesses a non-empty slot

18 18 Analysis of Open Addressing – Cont.

19 19 Corollary: The expected number of probes performed during an insertion into an open-address hash table with uniform hashing is 1 / (1 – α). Analysis of Open Addressing – Cont.

20 20 A successful search for an element x follows the same probe sequence as the insertion of element x. Consider the (i + 1)-st element x that was inserted. The expected number of probes performed when inserting x is at most Averaging over all n elements, the expected number of probes in a successful search is Theorem: Given an open-address hash table with load factor α < 1, the expected number of probes in a successful search is (1/ α ) ln (1 / (1 – α )), assuming uniform hashing and assuming that each key in the table is equally likely to be searched for. Analysis of Open Addressing – Cont.

21 21 Analysis of Open Addressing – Cont.

22 22 A family of hash functions is universal if for each pair k, l of keys, there are at most | | / m functions in such that h(k) = h(l). This means: For any two keys k and l and any function h chosen uniformly at random, the probability that h(k) = h(l) is at most 1/m. P= (| | / m )/ | | =1/m This is the same as if we chose h(k) and h(l) uniformly at random from [0, m – 1]. Universal Hashing

23 23 Analysis of Universal Hashing Theorem: For a hash function h chosen uniformly at random from a universal family, the expected length of the list T[h(x)] is α if x is not in the hash table and 1 + α if x is in the hash table. Proof: Indicator variables: Y x = the number of keys x that hash to the same slot as x

24 24 If x is not in T, then |{y T : x y}| = n. Hence, E[Y x ] = n / m = α. If x is in T, then |{y T : x y} = n – 1. Hence, E[Y x ] = (n – 1) / m < α. The length of list T[h(x)] is one more, that is, 1 + α. Analysis of Universal Hashing Cont.

25 25 Universal Family of Hash Functions Choose a prime p so that m = p. Let such For each define the hash function as follows: = | | = Example: m=p=253 a=[248,223,101] x=1025=[0,2,1].

26 26 Universal Family of Hash Functions Theorem: The class is universal. Proof: Let such that x y, w.l.o.g For all a 1,…,a r there exists a single a 0 such that For all there exists a single w such that z·w=1(mod m) for m r values The number of hash functions h in, for which h(k 1 ) = h(k 2 ) is at most m r/ m r+1 =1/m

27 27 Choose a prime p so that m < p. For any 1 a < p and 0 b < p, we define a function h a,b (x) = ((ax + b) mod p) mod m. Let H p,m be the family H p,m = {h a,b : 1 a < p and 0 b < p}. Theorem: The class H p,m is universal. Universal Family of Hash Functions

28 28 Hash tables are the most efficient dictionaries if only operations Insert, Delete, and Search have to be supported. If uniform hashing is used, the expected time of each of these operations is constant. Universal hashing is somewhat complicated, but performs well even for adversarial input distributions. If the input distribution is known, heuristics perform well and are much simpler than universal hashing. For collision-resolution, chaining is the simplest method, but it requires more space than open addressing. Open addressing is either more complicated or suffers from clustering effects. Summary


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