hash table theory and chaining. hash table: theory and chaining hash table formalism for hashing...
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Hash Table
Theory and chaining
Hash table: Theory and chaining
Hash Table
Formalism for hashing functions
Resolving collisions by chaining
FormFormαα∫∫ismism
Formalism for hashing functions
Hash table
Lets denote by U the set of possible values of the keys.
Lets denote by n the size of the hashing table.
h: U → {0, 1, …, n – 1}
Hashing function key position
U
h
1 m
Example: keys are words of 6 characters, and the hashing table has 13 cells.
h: 26^6 → 10
Size of the alphabet
If the hash table is at least as large as the number of different keys we’re expecting.
Formalism for hashing functions
h: U → {0, 1, …, n – 1}
Lets say that you were living in a really perfect world. What’s the
best feature of a hash function you could ever dream of?
The only problem that we have is when there is a collision: two keys
mapped on the same position.
h is perfect iff there is no collision
In which conditions can that happen?
If the table is exactly as large, h is a minimal perfect hash function.
Hash table
Formalism for hashing functions
If the table is exactly as large, h is a minimal perfect hash function.
Hash table
h: U → {0, 1, …, n – 1}
We are interested in hash tables smaller that the set S of possible keys.
S U, |S| > n⊆
Thus, we will have collisions.
What’s the best behaviour of the hash function we can hope for?
#of times the position gets hit
Position
Uniform distribution.
If x ≠ y then P[h(x) = h(y)] = 1/n
Formalism for hashing functions
Hash table
h: U → {0, 1, …, n – 1}
S U, |S| > n⊆ Uniform distribution.
Since there will be collisions, we will have to probe for empty spots.
<h (x), h (x), …, h (x)> is the probe sequence for a key x.1 2 n – 1
• What is the probability that T[h (x)] is already used?1
Lets say we have m elements in the table.
All cells have the same probability of being used.
The probability is m/n.
m/n
Formalism for hashing functions
Hash table
h: U → {0, 1, …, n – 1}
S U, |S| > n⊆ Uniform distribution.
Since there will be collisions, we will have to probe for empty spots.
<h (x), h (x), …, h (x)> is the probe sequence for a key x.1 2 n – 1
• What is the probability that T[h (x)] is already used?1 m/n
• If we have to probe more than once, what could be the cells targetted ...by the remaining sequence?
Any permutation of {0, 1, 2, …, n – 1} \ {h (x)}0
The expected number of probes is: E[T(n,m)] = 1 + * E[T(n–1,m–1)]mn
The number of times you have to probe is the complexity of a lookup.
Formalism for hashing functions
Hash table
h: U → {0, 1, …, n – 1}
S U, |S| > n⊆ Uniform distribution.
The expected number of probes is: E[T(n,m)] = 1 + * E[T(n–1,m–1)]mn
The number of times you have to probe is the complexity of a lookup.
If the hash table is empty, how many extra probes will we need?
0: first probe is the good one. So T(n,0) = 1.
The base case being proven, lets prove by recursion E[T(n,m)] ≤ n/(n–m )
mn
E[T(n,m)] = 1 + * E[T(n–1,m–1)]
≤ (n – 1)/(n – 1 – m + 1)≤ (n – 1)/(n – m)
≤m nn n – m1 + *
Because n – 1 ≤ 1
= n/(n–m)
Formalism for hashing functions
Hash table
h: U → {0, 1, …, n – 1}
S U, |S| > n⊆ Uniform distribution.
The expected number of probes is: E[T(n,m)] = 1 + * E[T(n–1,m–1)]mn
The number of times you have to probe is the complexity of a lookup.
If the hash table is empty, how many extra probes will we need?
0: first probe is the good one. So T(n,0) = 1.
The base case being proven, lets prove by recursion E[T(n,m)] ≤ n/(n–m )
The ratio of used cells m / n is called the load factor and denoted α.
E[T(n,m)] ≤ n/(n–m ) = 1 / (1 – α) ∈ O(1) because α is a constant.
Heuristics
Hash table
We’ve been wandering in the realms of somewhat pure mathematics.
By now you probably all love it.
But let’s come back to reality for a minute.
♥
α√π
♥
∑∂x²
♥
Heuristics
Hash table
The probe sequences that we generate are not totally random. We have:
• Linear probing: h (x) = (h(x) + i) mod ni
• Quadratic probing: h (x) = (h(x) + i²) mod ni
• Double hashing: h (x) = (h(x) + i.s(x)) mod niwhere s(x) is a secondary hashing function. Commonly q – (k % q).
ChainingChaining
Resolving collisions by chaining
Hash table
You’re back at the train station.
Instead of a seat number, the ticket is a compartment
number.
What do you do? You sit with the other people of your compartment.
Resolving collisions by chaining
Hash table
What do you do? You sit with the other people of your compartment.
0 1 2 3 4
Resolving collisions by chaining
Hash table
Each cell is now a data structure. Which one?
AVL!LinkedList!
HashTable!
Pretty much everything but an ArrayList.
Resolving collisions by chaining
Hash table
Each cell is now a data structure. Which one?
• With a LinkedList, access in O(1 + L(x)).
If L(x) denotes the length of the list at T[h(x)], then:
• With a balanced binary search tree, you access in O(1 + log L(x))
The second part depends on the complexity of the structure you use.
Pretty much everything but an ArrayList.