S.K.H. Kei Hau Secondary School 2003-2004
Hashing
S.K.H. Kei Hau Secondary School
Department of Computer
AL Computer Studies
Hashing
1. Linear search
2. Hashing and hashing functions
1. Linear search
Suppose we have an array:
[1]
[2]
11
12
[3]
13
[4]
14
[5]
15
We may search an element by sequential searching method.
for i := 1 to 5 do
if number[i] = Target then
writeln('index is', i)
Performance:
For number of elements is N, searching an element must start from the beginning of
the list / array. So, the number of comparison is N.
2. Hashing and hashing functions
Question: How can a given collection of data be organized on a given key so that the
access of an item by the key involves the least amount of effort?
Consider,
Suppose a small business has an inventory file consisting of 100 data item and each
item has a unique eight-digit inventory number.
Then,
Size of "array of record" = 100 million (0..99999999)
Impossible!


The business has 100 million records?
The amount of memory is huge, waste space!
We can build a function that will accomplish the transformation: hashing function
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S.K.H. Kei Hau Secondary School 2003-2004
Hashing
Hashing function
The function that transforms the key (inventory number) into the corresponding array
index is called a hash function.
Example
H(k) = k DIV 10000
when k = 12347296, k = 1234
H(k) = k MOD 10000
when k = 12347296, k = 7296
However, it may cause many-to-one situation, collision occurs!
Resolving collisions by chaining
Records in the same slot are linked into a list.
T
0
i
49
86
52
m-1
h(49) = h(86) = h(52) = i
Analysis of chaining
We make the assumption of simple uniform hashing:

Each key k  K of keys is equally likely to be hashed to any slot of table T,
independent of where other keys are hashed.


Let n be the number of keys in the table
Let m be the number of slots

Define the load factor of T to be   n/ m , which is the average number of
keys per slot.
Search cost
Expected time to search for a record with a given key = O (1   ) :



1: apply hash function and access slot. (nearly to be direct access)
 : search the list
expected search time = O(1) if   O(1) , or equivalently, if n=O(m)
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S.K.H. Kei Hau Secondary School 2003-2004
Hashing
Choosing a hash function
The assumption of simple uniform hashing is hard to guarantee, but several common
techniques tend to work well in practice as long as their deficiencies (不足) can be
avoided.
Desirata:

A good hash function should distribute the keys uniformly into the slots of the
table.
 Regularity in the key distribution should not affect this uniformity.
 Quick to compute.
Note: non-uniform distribution may cause larger collision chance.
1. Division method
Assume all keys are integers, and define
h(k) = k mod m.


For decimal number: if m = 10r, the hash doesn't even depend on all bits of k.
Similarly, for binary number: if m = 2r, then the has doesn't even depend on all
the bits of k:
o If k = 1011000111011010, and r = 6, then h(k) = 011010

Pick m to be a prime not too close to power of 2 or 10 and not otherwise used
prominently in the computing environment.
2. Folding method
Divide key into sections and add together.
h(k) = p + q + r, where p, q, r are the digits of k


h(245690127) = 245+690+127
Subtract, multiply and divide are available
3. Middle-Squaring
Take some digits from the middle of the number and square them
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Hashing
4. Multiplication method (optional)
Assume that all keys are integers, m = 2r, and our computer has w-bit words. Define
h(k )  ( A  K mod 2w )rsh( w  r )
where rsh is the 'bit-wise right-shift" operator and A is an odd integer in the range
2w-1<A<2w.



Don't pick A too close to 2w
Multiplication modulo 2w is fast
The rsh operator is fast
Suppose that m = 8 = 23 and that our computer has w = 7-bit words:
1011001 = A
1101011 = k
X__________________________
10010100110011
h(k)
Resolving collisions by open addressing
No storage is used outside of the hash table itself.

Insertion systematically probes the table until an empty slot is found.

The hash function depends on both the key and probe number.
Example of open addressing
Insertion
Insert key k = 496:
T
0
0. Probe h(496,0)
586
133
204
collision
481
m-1
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Hashing
T
0
0. Probe h(496,0)
1. Probe h(496,1)
586
133
collision
204
481
m-1
T
0
0. Probe h(496,0)
1. Probe h(496,1)
586
133
2. Probe h(496,2)
204
496
481
Insertion
m-1
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S.K.H. Kei Hau Secondary School 2003-2004
Hashing
Searching for key k=496:
T
0
0. Probe h(496,0)
1. Probe h(496,1)
2. Probe h(496,2)
586
133
204
496
481
m-1
Note:
Search uses the same probe sequence, terminating successfully if it finds the key and
unsuccessfully if it encounters an empty slot.
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Hashing
Probing Strategies
1. Linear probing
Given an ordinary hash function h' (k ) , linear probing uses the hash function
h(k , i)  (h' (k )  i) mod m
This method, through simple, suffers from primary clustering, where long runs of
occupied slots build up, increasing the average search time. Moreover, the long runs
of occupied slots tend to get longer.
2. Double hashing
Given two ordinary hash functions h1(k) and h2(k), double hashing uses the function
h(k , i)  (h1 (k )  i  h2 (k )) mod m
This method generally produces excellent results, but h2(k) must be relatively prime
to m. One way is to make m a power of 2 and design h2(k) to produce only odd
numbers.
Analysis of open addressing
We make the assumption of uniform hashing:

Each key is equally likely to have any one of the m! permutations as its probe
sequence.
Theorem:
Given an open-addressed hash table with load factor   n / m  1, the expected
number of probes in an unsuccessful search is at most
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S.K.H. Kei Hau Secondary School 2003-2004
Hashing
Proof of the theorem (optional)


At least one probe is always necessary
With probability n/m, the first probe hits an occupied slot, and a second probe
is necessary.

With probability (n-1) / (m-1), the second probe hits an occupied slot, and a
third is necessary.

With probability (n-2) / (m-2), the third probe hits an occupied slot, etc.
Observe that
ni
n
   for i=1, 2, ..., n.
mi m
Therefore, the expected number of probes is
1
n 
n 1 
n2  
1
   
1 

1
...
1


...
m  m  1  m  2   m  n  1    
 1   (1   (1   (...(1   )...)))
 1     2   3  ...

  i
i 0

1
1
Implications of the theorem

If  is constant, then accessing an open-addressed hash table takes constant
time.


If the table is half full, then the expected number of probes is 1/(-0.5) = 2
If the table is 90% full, then the expected number of probes is 1/(1-0.9) = 10
Reference
1. Guy J.Hale, Richard J. Easton, Applied Data Structures Using Pascal 1987, D.C.
Heath and company.
2. Charles E. Leiserson, Note of Introduction to Algorithm 2001, MIT
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