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Collision Resolution: Open AddressingPowerPoint Presentation

Collision Resolution: Open Addressing

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Collision Resolution: Open Addressing

- Quadratic Probing
- Double Hashing
- Random Probing
- Informal Analysis of Hashing

Open Addressing: Quadratic Probing

- Quadratic probing: Attempts to avoid cluster buildup.
- In this method, c(i) is a quadratic function in i:
c(i) = a i2 + bi + c

- Quadratic probing is usually done using c(i) = i2 .
- We use the following slight modification to c (i) :
c(i)=i2, c(i)= -i2, for i=1,2,…,(n-1)/2

- Thus the probe sequence is
h(r)+i2, h(r)- i2, for i=1,2,…,(n-1)/2

Example 2: Linear Probing

- Use the hash function h(r) = r.id % 13 to load the following records into an array of size 13.
Al-Otaibi Ziyad 1.73 985926

Al-Turki, Musab Ahmad Bakeer 1.60 970876

Al-Saegh, Radha Mahdi 1.58 980962

Al-Shahrani, Adel Saad 1.80 986074

Al-Awami, Louai Adnan Muhammad 1.73 970728

Al-Amer, Yousuf Jauwad 1.66 994593

Al-Helal, Husain Ali AbdulMohsen 1.70 996321

Then insert the following records quadratic probing to resolve collisions, if any.

Al-Najjar, Khaled Ziyad 1.69 987615

Al-Ali, Amr Ali Zaid 1.79 987630

Al-Ramadi, Husam Yahya 1.58 987602

Example 1: Quadratic Probing (cont'd)

0 1 2 3 4 5 6 7 8 9 10 11 12

Husain

Yousuf

Khalid

Louai

Ziyad

Amr

Radha

Husam

Musab

Adel

Quadratic Probing: Concluding Notes

- In general, quadratic probing improves on linear probing but does not avoid cluster buildup.
- Gives rise to secondary clusters which are less harmful than the primary clusters in linear probing.
- Hash table size should not be an even number, therwise
Property 2 will not be satisfied.

- Ideally, table size should be a prime, 4j+3, for an integer j, which guarantees Property 2.

Quadratic Probing: Concluding Notes (cont'd)

- A disadvantage: colliding keys at a given address tread the
same probe sequence.

- This sequence of locations is called a secondary cluster.
- Unlike primary clusters, secondary clusters cannot combine into
larger secondary clusters.

- To eliminate secondary clustering, synonyms must have
different probe sequences.

- Double hashing achieves this by having two hash functions that both depend on the hash key.

Open Addressing: Double Hashing

- Double hashing: Best addresses secondary clustering.
- Uses two hash functions, h and hp, with the usual probe
sequence:

hi(r) = (h(r) + i*hp(r)) mod n

- We see that c(i) = i*hp(r) satisfies Property 2 as well,
provided hp(r) and n are relatively prime.

- To guarantee Property 2, n must be a prime number.

Open Addressing: Double Hashing (cont'd)

- Using two hash functions can be expensive.
- In practice, a common definition for hp is :
hp(r)= 1 + (r mod (n-1)).

- Thus, the probe sequence for r hashing to position j is:
j, j+1*hp(r), j+2*hp(r), j+3*hp(r), …

- Notice that if hp(r)=1, the probe sequence for r is the same
as linear probing.

- But if hp(r)=2, the probe sequence examines every other
array location.

Example 2: Illustrating Double Hashing

- Use double hashing to load the following records into a hash table
Al-Otaibi Ziyad 1.73 985926

Al-Turki, Musab Ahmad Bakeer 1.60 970876

Al-Saegh, Radha Mahdi 1.58 980962

Al-Shahrani, Adel Saad 1.80 986074

Al-Awami, Louai Adnan Muhammad 1.73 970728

Al-Amer, Yousuf Jauwad 1.66 994593

Al-Helal, Husain Ali AbdulMohsen 1.70 996321

Al-Najjar, Khaled Ziyad 1.69 987615

Al-Ali, Amr Ali Zaid 1.79 987630

Al-Ramadi, Husam Yahya 1.58 987602

Use the following pair of hash functions

h(r) = r.id % 13

hp(r) = 1 + (r.id % (n-1)).

Example 2: Animating Double Hashing

0 1 2 3 4 5 6 7 8 9 10 11 12

Husain

Yousuf

Husam

Louai

Ziyad

Amr

Radha

Khalid

Musab

Adel

Open Addressing: Random Probing

- Random probing: Uses a pseudo-random function to "jump
around" in the hash table.

- Here, c(i) is defined so that it always produces a 'random‘
integer in the range 0, 1, ..., n - 1.

- The function c(i) can be defined recursively as follows
c(0) = 0

c(i+1) = (a*c(i) + 1) mod n

- We choose a to insure Property 2, too.
- This recursive definition of c() is a permutation of
0, 1, ..., n - 1 iff a - 1 is a multiple of every prime

divisor of n, where 4 is considered as prime.

Example 3: Random Probing

- Let n = 16, a = 3. We get the sequence for c(i) :
0 1 4 13 8 9 12 5 0 1 4 13 8 9 12 5 0 ...

which is not a permutation of 0,1,2, …, 15.

- The prime divisors of 16 are 2 and 4, so (a-1) must be a multiple of 4.
- If we choose a = 5, then, we get the sequence
0 1 6 15 12 13 2 11 8 9 14 7 4 5 10 3 0 1 6 ...

which is a permutation of 0, 1, 2, …, 15.

- We could also select a to be 9, 13, 17, etc.

Random Probing: Concluding Remarks

- For our students records example, n = 13 which has itself as its only prime divisor.
- We can therefore select a = 14. The probe sequence turns out to be: 0 1 2 3 4 5 6 7 8 9 10 11 12 0 1 ... which is the same as in linear probing.
- Note that the name "random probing" is exaggerative.
- Random probing eliminates the problem of secondary clusters.

Hashing: Informal Analysis

- In the best case, successful hash search requires one probe.
- In the worst case, hash search becomes a sequential search.
- For the average case, the number of probes depends on the load
factor.

- The load factor in open addressing is less than 1 while it
can be greater than 1 in chaining.

- Note that some methods are worse than others at high load
factors.

- Performance of open addressing methods about the same at
low load factors.

Exercises

1. If a hash table is 25% full what is its load factor?

2. Given that,

c(i) = i2,

for c(i) in quadratic probing, we discussed that this equation

does not satisfy Property 2, in general. What cells are missed by

this probing formula for a hash table of size 17? Characterise

using a formula, if possible, the cells that are not examined by

using this function for a hash table of size n.

3. It was mentioned in this session that secondary clusters are less

harmful than primary clusters because the former cannot combine

to form larger secondary clusters. Use an appropriate hash table

of records to exemplify this situation.

4.What value would you select for a given a hash table of size 100?

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