Analysis Of Algorithms - Hashing

Analysis of Algorithms
CS 477/677
Hashing
Thanks: George Bebis
(Chapter 11)

2
The Search Problem
• Find items with keys matching a given search key
– Given an array A, containing n keys, and a search key
x, find the index i such as x=A[i]
– As in the case of sorting, a key could be part of a
large record.

3
Applications
• Keeping track of customer account information at
a bank
– Search through records to check balances and perform
transactions
• Keep track of reservations on flights
– Search to find empty seats, cancel/modify reservations
• Search engine
– Looks for all documents containing a given word

4
Special Case: Dictionaries
• Dictionary = data structure that supports mainly
two basic operations: insert a new item and
return an item with a given key
• Queries: return information about the set S:
– Search (S, k)
– Minimum (S), Maximum (S)
– Successor (S, x), Predecessor (S, x)
• Modifying operations: change the set
– Insert (S, k)
– Delete (S, k) – not very often

5
Direct Addressing
• Assumptions:
– Key values are distinct
– Each key is drawn from a universe U = {0, 1, . . . , m - 1}
• Idea:
– Store the items in an array, indexed by keys
• Direct-address table representation:
– An array T[0 . . . m - 1]
– Each slot, or position, in T corresponds to a key in U
– For an element x with key k, a pointer to x (or x itself) will be placed
in location T[k]
– If there are no elements with key k in the set, T[k] is empty,
represented by NIL

6
Direct Addressing (cont’d)

7
Operations
Alg.: DIRECT-ADDRESS-SEARCH(T, k)
return T[k]
Alg.: DIRECT-ADDRESS-INSERT(T, x)
T[key[x]] x←
Alg.: DIRECT-ADDRESS-DELETE(T, x)
T[key[x]] NIL←
• Running time for these operations: O(1)

8
Comparing Different Implementations
• Implementing dictionaries using:
– Direct addressing
– Ordered/unordered arrays
– Ordered/unordered linked lists
Insert Search
ordered array
ordered list
unordered array
unordered list
O(N)
O(N)
O(N)
O(N)
O(1)
O(1)
O(lgN)
O(N)
direct addressing O(1) O(1)

9
Examples Using Direct Addressing
Example 2:
Example 1:

10
Hash Tables
• When K is much smaller than U, a hash table
requires much less space than a direct-address
table
– Can reduce storage requirements to |K|
– Can still get O(1) search time, but on the average
case, not the worst case

11
Hash Tables
Idea:
– Use a function h to compute the slot for each key
– Store the element in slot h(k)
• A hash function h transforms a key into an index in a
hash table T[0…m-1]:
h : U {0, 1, . . . , m - 1}→
• We say that k hashes to slot h(k)
• Advantages:
– Reduce the range of array indices handled: m instead of |U|
– Storage is also reduced

12
Example: HASH TABLES
U
(universe of keys)
K
(actual
keys)
0
m - 1
h(k3)
h(k2) = h(k5)
h(k1)
h(k4)
k1
k4 k2
k5
k3

14
Do you see any problems
with this approach?
U
(universe of keys)
K
(actual
keys)
0
m - 1
h(k3)
h(k2) = h(k5)
h(k1)
h(k4)
k1
k4 k2
k5
k3
Collisions!

15
Collisions
• Two or more keys hash to the same slot!!
• For a given set K of keys
– If |K| ≤ m, collisions may or may not happen,
depending on the hash function
– If |K| > m, collisions will definitely happen (i.e., there
must be at least two keys that have the same hash
value)
• Avoiding collisions completely is hard, even with
a good hash function

16
Handling Collisions
• We will review the following methods:
– Chaining
– Open addressing
• Linear probing
• Quadratic probing
• Double hashing
• We will discuss chaining first, and ways to
build “good” functions.

17
Handling Collisions Using Chaining
• Idea:
– Put all elements that hash to the same slot into a
linked list
– Slot j contains a pointer to the head of the list of all
elements that hash to j

18
Collision with Chaining - Discussion
• Choosing the size of the table
– Small enough not to waste space
– Large enough such that lists remain short
– Typically 1/5 or 1/10 of the total number of elements
• How should we keep the lists: ordered or not?
– Not ordered!
• Insert is fast
• Can easily remove the most recently inserted elements

19
Insertion in Hash Tables
Alg.: CHAINED-HASH-INSERT(T, x)
insert x at the head of list T[h(key[x])]
• Worst-case running time is O(1)
• Assumes that the element being inserted isn’t
already in the list
• It would take an additional search to check if it
was already inserted

20
Deletion in Hash Tables
Alg.: CHAINED-HASH-DELETE(T, x)
delete x from the list T[h(key[x])]
• Need to find the element to be deleted.
• Worst-case running time:
– Deletion depends on searching the corresponding list

21
Searching in Hash Tables
Alg.: CHAINED-HASH-SEARCH(T, k)
search for an element with key k in list T[h(k)]
• Running time is proportional to the length of the
list of elements in slot h(k)

22
Analysis of Hashing with Chaining:
Worst Case
• How long does it take to
search for an element with a
given key?
• Worst case:
– All n keys hash to the same slot
– Worst-case time to search is
Θ(n), plus time to compute the
hash function
0
m - 1
T
chain

23
Analysis of Hashing with Chaining:
Average Case
• Average case
– depends on how well the hash function
distributes the n keys among the m slots
• Simple uniform hashing assumption:
– Any given element is equally likely to hash
into any of the m slots (i.e., probability of
collision Pr(h(x)=h(y)), is 1/m)
• Length of a list:
T[j] = nj, j = 0, 1, . . . , m – 1
• Number of keys in the table:
n = n0 + n1 +· · · + nm-1
• Average value of nj:
E[nj] = α = n/m
n0 = 0
nm – 1 = 0
T
n2
n3
nj
nk

24
Load Factor of a Hash Table
• Load factor of a hash table T:
α = n/m
– n = # of elements stored in the table
– m = # of slots in the table = # of linked lists
∀ α encodes the average number of
elements stored in a chain
∀ α can be <, =, > 1
0
m - 1
T
chain
chain
chain
chain

25
Case 1: Unsuccessful Search
(i.e., item not stored in the table)
Theorem
An unsuccessful search in a hash table takes expected time
under the assumption of simple uniform hashing
(i.e., probability of collision Pr(h(x)=h(y)), is 1/m)
Proof
• Searching unsuccessfully for any key k
– need to search to the end of the list T[h(k)]
• Expected length of the list:
– E[nh(k)] = α = n/m
• Expected number of elements examined in an unsuccessful search is α
• Total time required is:
– O(1) (for computing the hash function) + α 
(1 )αΘ +
(1 )αΘ +

27
Analysis of Search in Hash Tables
• If m (# of slots) is proportional to n (# of
elements in the table):
• n = O(m)
• α = n/m = O(m)/m = O(1)
⇒ Searching takes constant time on average

28
Hash Functions
• A hash function transforms a key into a table
address
• What makes a good hash function?
(1) Easy to compute
(2) Approximates a random function: for every input,
every output is equally likely (simple uniform hashing)
• In practice, it is very hard to satisfy the simple
uniform hashing property
– i.e., we don’t know in advance the probability
distribution that keys are drawn from

29
Good Approaches for Hash Functions
• Minimize the chance that closely related keys
hash to the same slot
– Strings such as pt and pts should hash to
different slots
• Derive a hash value that is independent from
any patterns that may exist in the distribution
of the keys

30
The Division Method
• Idea:
– Map a key k into one of the m slots by taking
the remainder of k divided by m
h(k) = k mod m
• Advantage:
– fast, requires only one operation
• Disadvantage:
– Certain values of m are bad, e.g.,
• power of 2
• non-prime numbers

31
Example - The Division Method
• If m = 2p
, then h(k) is just the least
significant p bits of k
– p = 1 ⇒ m = 2
⇒ h(k) = , least significant 1 bit of k
– p = 2 ⇒ m = 4
⇒ h(k) = , least significant 2 bits of k
 Choose m to be a prime, not close to a
power of 2
 Column 2:
 Column 3:
{0, 1}
{0, 1, 2, 3}
k mod 97
k mod 100
m
97
m
100

32
The Multiplication Method
Idea:
• Multiply key k by a constant A, where 0 < A < 1
• Extract the fractional part of kA
• Multiply the fractional part by m
• Take the floor of the result
h(k) = = m (k A mod 1)
• Disadvantage: Slower than division method
• Advantage: Value of m is not critical, e.g., typically 2p
fractional part of kA = kA - kA

33
Example – Multiplication Method

34
Universal Hashing
• In practice, keys are not randomly distributed
• Any fixed hash function might yield Θ(n) time
• Goal: hash functions that produce random
table indices irrespective of the keys
• Idea:
– Select a hash function at random, from a designed
class of functions at the beginning of the execution

35
Universal Hashing
(at the beginning
of the execution)

36
Definition of Universal Hash Functions
H={h(k): U(0,1,..,m-1)}

37
How is this property useful?
Pr(h(x)=h(y))=

38
Universal Hashing – Main Result
With universal hashing the chance of collision
between distinct keys k and l is no more than the
1/m chance of collision if locations h(k) and h(l)
were randomly and independently chosen from
the set {0, 1, …, m – 1}

39
Designing a Universal Class
of Hash Functions
• Choose a prime number p large enough so that every
possible key k is in the range [0 ... p – 1]
Zp = {0, 1, …, p - 1} and Zp
*
= {1, …, p - 1}
• Define the following hash function
ha,b(k) = ((ak + b) mod p) mod m,
∀ a ∈ Zp
*
and b ∈ Zp
• The family of all such hash functions is
Hp,m = {ha,b: a ∈ Zp
*
and b ∈ Zp}
• a , b: chosen randomly at the beginning of execution
The class Hp,m of hash
functions is universal

40
Example: Universal Hash Functions
E.g.: p = 17, m = 6
ha,b(k) = ((ak + b) mod p) mod m
h3,4(8) = ((3⋅8 + 4) mod 17) mod 6
= (28 mod 17) mod 6
= 11 mod 6
= 5

41
Advantages of Universal Hashing
• Universal hashing provides good results on
average, independently of the keys to be stored
• Guarantees that no input will always elicit the
worst-case behavior
• Poor performance occurs only when the random
choice returns an inefficient hash function – this
has small probability

42
Open Addressing
• If we have enough contiguous memory to store all the keys
(m > N) ⇒ store the keys in the table itself
• No need to use linked lists anymore
• Basic idea:
– Insertion: if a slot is full, try another one,
until you find an empty one
– Search: follow the same sequence of probes
– Deletion: more difficult ... (we’ll see why)
• Search time depends on the length of the
probe sequence!
e.g., insert 14

43
Generalize hash function notation:
• A hash function contains two arguments now:
(i) Key value, and (ii) Probe number
h(k,p), p=0,1,...,m-1
• Probe sequences
<h(k,0), h(k,1), ..., h(k,m-1)>
– Must be a permutation of <0,1,...,m-1>
– There are m! possible permutations
– Good hash functions should be able to
produce all m! probe sequences
insert 14
<1, 5, 9>
Example

44
Common Open Addressing Methods
• Linear probing
• Quadratic probing
• Double hashing
• Note: None of these methods can generate
more than m2
different probing sequences!

45
Linear probing: Inserting a key
• Idea: when there is a collision, check the next available
position in the table (i.e., probing)
h(k,i) = (h1(k) + i) mod m
i=0,1,2,...
• First slot probed: h1(k)
• Second slot probed: h1(k) + 1
• Third slot probed: h1(k)+2, and so on
• Can generate m probe sequences maximum, why?
probe sequence: < h1(k), h1(k)+1 , h1(k)+2 , ....>
wrap around

46
Linear probing: Searching for a key
• Three cases:
(1) Position in table is occupied with an
element of equal key
(2) Position in table is empty
(3) Position in table occupied with a
different element
• Case 2: probe the next higher index
until the element is found or an
empty position is found
• The process wraps around to the
beginning of the table
0
m - 1
h(k3)
h(k2) = h(k5)
h(k1)
h(k4)

47
Linear probing: Deleting a key
• Problems
– Cannot mark the slot as empty
– Impossible to retrieve keys inserted
after that slot was occupied
• Solution
– Mark the slot with a sentinel value
DELETED
• The deleted slot can later be used
for insertion
• Searching will be able to find all the
keys
0
m - 1

48
Primary Clustering Problem
• Some slots become more likely than others
• Long chunks of occupied slots are created
⇒ search time increases!!
Slot b:
2/m
Slot d:
4/m
Slot e:
5/m
initially, all slots have probability 1/m

49
Quadratic probing
i=0,1,2,...

50
Double Hashing
(1) Use one hash function to determine the first slot
(2) Use a second hash function to determine the
increment for the probe sequence
h(k,i) = (h1(k) + i h2(k) ) mod m, i=0,1,...
• Initial probe: h1(k)
• Second probe is offset by h2(k) mod m, so on ...
• Advantage: avoids clustering
• Disadvantage: harder to delete an element
• Can generate m2
probe sequences maximum

51
Double Hashing: Example
h1(k) = k mod 13
h2(k) = 1+ (k mod 11)
h(k,i) = (h1(k) + i h2(k) ) mod 13
• Insert key 14:
h1(14,0) = 14 mod 13 = 1
h(14,1) = (h1(14) + h2(14)) mod 13
= (1 + 4) mod 13 = 5
h(14,2) = (h1(14) + 2 h2(14)) mod 13
= (1 + 8) mod 13 = 9
79
69
98
72
50
0
9
4
2
3
1
5
6
7
8
10
11
12
14

52
Analysis of Open Addressing
a
1 a−
(load factor)
k=0

53
Analysis of Open Addressing (cont’d)
Example (similar to Exercise 11.4-4, page 244)
Unsuccessful retrieval:
a=0.5 E(#steps) = 2
a=0.9 E(#steps) = 10
Successful retrieval:
a=0.5 E(#steps) = 3.387
a=0.9 E(#steps) = 3.670

Analysis Of Algorithms - Hashing

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