Collision Resolution: Open Addressing

1
Collision Resolution Open Addressing

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

2
Open Addressing Quadratic Probing

• Quadratic probing Attempts to avoid cluster
  buildup.
• In this method, c(i) is a quadratic function in
  i
• c(i) ai 2 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 i1,2,,(n-1)/2
• Thus the probe sequence is
• h(r)i2, h(r)- i2, for i1,2,,(n-1)/2

3
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

4
Example 1 Quadratic Probing (cont'd)
0 1 2 3 4 5 6
7 8 9 10 11 12

Husain
Yousuf
Radha
Amr
Musab
Adel
Husam
Louai
Ziyad
Khalid
5
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,
  otherwise
• Property 2 will not be satisfied.
• Ideally, table size, n, should be a prime,
  satisfying n4j3, for an integer j, which
  guarantees Property 2.

6
Quadratic Probing Concluding Notes (cont'd)

• A disadvantage colliding keys at a given address
  follow 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.

7
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) ihp(r)) mod n
• We see that c(i) ihp(r) satisfies Property 2
  as well,
• provided hp(r) and n are relatively prime.
• To guarantee Property 2, n must be a prime
  number.

8
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, j1hp(r), j2hp(r), j3hp(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.

9
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)).

10
Example 2 Animating Double Hashing
0 1 2 3 4 5 6
7 8 9 10 11 12

Husain
Yousuf
Radha
Amr
Musab
Adel
Husam
Louai
Ziyad
Khalid
11
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) (ac(i-1) 1) mod n, i1,..n-1
• We choose a to ensure 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.

12
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.

13
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 does not eliminates the problem of
  secondary clusters.

14
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.

15
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?