1. Dana Shapira Hash Tables
Data Structures
Dana Shapira
Hash Tables

2. Element Uniqueness Problem
Let
Determine whether there exist ij such that xi=xj
Sort Algorithm
Bucket Sort
for (i=0;i<m;i++)
T[i]=NULL;
for (i=0;i<n;i++){
if (T[xi]= = NULL)
T[xi]= i
else{
output (i, T[xi])
return; }
What happens when m is large or when we are dealing with real numbers??

3. Hash Tables U x1 x2 x3 x4 h(x1) h(x2) h(x4) h(x3) h Notations:
Set of array indices
Hash Tables
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). x1 x2 x3 x4
h(x1)
h(x2)
h(x4)
h(x3)

4. Example 81 62 53 17 19 1 2 3 7 9 4 5 6 8
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:
Chaining
Open addressing

5. Collision-Resolution by Chaining1 81 62 12 53 17 37 57 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. Analysis of Chaining
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)

7. Analysis of Chaining
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. Designing Good Hash Functions
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.

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

10. The Multiplication Method
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· … mod 1) =
= 10,000 ·(76, … mod 1) =
= 10,000 · … =   = 41

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

12. Example h(x) = m (cx mod 1)
x = , p = 14, m = 214 = 16384, w = 32,
Then sx = (76300 ⋅ 232)
The 14 most significant bits of are 67; that is, h(x) = 67
x =
s =
sx =
h(x) = = 67

13. Open Addressing
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.
h(x)

14. Linear Probing 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.

15. Quadratic Probing Hash function:
h(k, i) = (h'(k) + c1i + c2i2) 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.

16. Double Hashing Hash function:
h(k, i) = (h1(k) + ih2(k)) mod m, where h1 and h2 are two original hash functions.
h2(k) has to be prime w.r.t. m; that is, gcd(h2(k), m) = 1.
Two methods:
Choose m to be a power of 2 and guarantee that h2(k) is always odd.
Choose m to be a prime number and guarantee that h2(k) < m.
Benefits:
No more clustering
Drawback:
More complicated than linear and quadratic probing

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.
Ai = “there is an i-th probe, and it accesses a non-empty slot”

18. Analysis of Open Addressing – Cont.

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

20. Analysis of Open Addressing – Cont.
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.
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

21. Analysis of Open Addressing – Cont.

22. Universal Hashing
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].

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:
Yx = the number of keys ≠ x that hash to the same slot as x

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

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. Universal Family of Hash Functions
Theorem: The class  is universal.
Proof: Let
such that x ≠ y, w.l.o.g
For all a1,…,ar there exists a single a0 such that
For all there exists a single w such that z·w=1(mod m)
for mr values
The number of hash functions h in , for which h(k1) = h(k2) is at most mr/ mr+1 =1/m

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

28. Summary
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.