Data Structures

Hash Tables

Hashing Tables

Motivation: symbol tables A compiler uses a symbol table to relate symbols

to associated data Symbols: variable names, procedure names, etc. Associated data: memory location, call graph, etc.

For a symbol table (also called a dictionary), we care about search, insertion, and deletion

We typically don’t care about sorted order

Hash Tables

More formally: Given a table T and a record x, with key (= symbol) and

satellite data, we need to support: Insert (T, x) Delete (T, x) Search(T, x)

We want these to be fast, but don’t care about sorting the records

The structure we will use is a hash table Supports all the above in O(1) expected time!

Hashing: Keys

In the following discussions we will consider all keys to be (possibly large) natural numbers

How can we convert floats to natural numbers for hashing purposes?

How can we convert ASCII strings to natural numbers for hashing purposes?

Hashing a String

Let Ai represent the ascii code for a string character. Since ascii codes go from 0 to 127 we can perform the following calculation.

A3X3 + A2X2 + A1X1 +A0X0 = (((A3)X + A2)X + A1)X + A0

What is this called? Note that if X = 128 we can easily overflow. Solution:public static int hash(String key, int tableSize)

{

int hashVal = 0;

for(int i=0; i < key.length(); i++)

hashVal = ( hashVal * 128 + key.charAt(i))%tableSize;

return hashVal;

}

A better version

Hash functions should be fast and distribute the keys equitably! The 37 slows down the shifting of lower keys and we allow overflow to occur(sort of auto mod-ing) which can result in a neg.public static int hash(String key, int tableSize)

{

int hashVal = 0;

for(int i=0; i < key.length(); i++)

hashVal = hashVal * 37 + key.charAt(i);

hashVal %= tableSize;

if( hashVal < 0)hashVal += tableSize;// fix the sign

return hashVal;

}

Direct Addressing

Suppose: The range of keys is 0..m-1 Keys are distinct

The idea: Set up an array T[0..m-1] in which

T[i] = x if x T and key[x] = i T[i] = NULL otherwise

This is called a direct-address table Operations take O(1) time! So what’s the problem?

The Problem With Direct Addressing

Direct addressing works well when the range m of keys is relatively small

But what if the keys are 32-bit integers? Problem 1: direct-address table will have

232 entries, more than 4 billion Problem 2: even if memory is not an issue, the time to

initialize the elements to NULL may be Solution: map keys to smaller range 0..m-1 This mapping is called a hash function

Hash Functions

Next problem: collisionT

0

m - 1

h(k1)

h(k4)

h(k2) = h(k5)

h(k3)

k4

k2 k3

k1

k5

U(universe of keys)

K(actualkeys)

Resolving Collisions

How can we solve the problem of collisions? Solution 1: chaining Solution 2: open addressing

Open Addressing

Basic idea: To insert: if slot is full, try another slot, …, until an open

slot is found (probing) To search, follow same sequence of probes as would be

used when inserting the element If reach element with correct key, return it If reach a NULL pointer, element is not in table

Good for fixed sets (adding but no deletion) Example: spell checking

Table needn’t be much bigger than n

Open Addressing(Linear Probing)

Let H(k,p) denote the pth position tried for key K where p=0,1,...

Thus H(k,0)= h(k) where h is the basic hash function. H(k,p) represents the probe sequence we use to find the key.

If key K hashes to index i, and it is full, then try i+1, i+2, until an empty slot is found.

The Linear Probing sequence is therefore shown below.

H(K,0) = h(K)

H(K, p+1) = (H(K,p) + 1) mod m

Open Addressing(Double Hashing)

Here we use a second hash function h2 to determine the probe sequence. Note that linear probing is just h2(K)=1

H(K,0) = h(K)

H(K, p+1) = (H(K,p) + h2(K)) mod m

To ensure that the probe sequence visits all positions in the table h2(K) must be greater than zero and relatively prime to m for any K. Hence we usually let m be prime.

Mathematical Defn’s

Let us count as one probe each access to the data structure and n the size of the dictionary to be stored and m the size of the table. Let =n/m be called the load factor.

From a statistical stand point let S() be the expected number of probes to perform a LookUp on a key that is actually in the hash table, and U() be the expected number of probes in an unsuccessful LookUp.

Performance of Linear Probing

12 S. Adams 134 J. Adams 15 W. Floyd 26 T. Heyward 178 J. Hancock 1910 C.Braxton 111 J. Hart 112 J. Hewes 31314 C. Carroll 115 A. Clark 116 R. Ellery 217 B. Franklin 118 W. Hooper 5

key # probesThe names were inserted in alphabetical order. The third column shows the number of probes required to find a key in the table; this is the same as the number of slots that were inspected when it was inserted.

primary

clustering

S = 21/13=1.615U = 47/18=2.6load is 13/18=.72

Linear Probing

Knuth analyzed sequential probing and obtained the following.

1

11

2

1nS

2)1(

11

2

1

nU

=.9

SnUn

Linear

5.550.5

Double

2.5610.0

The small extra effort required to implement double hashing certainly pays off it seems.

Performance of Double Hashing

Assumption: each probe inot the hash table is independent and has a probability of hitting an occupied position exactly equal to the load factor. Let i = i/m for every in. This is then the probability of a collision on any probe after i keys have been inserted.

The expected number of probes in an unsuccessful search when n-1 items have been inserted is

Un-1 1(1- n-1)+2 n-1(1- n-1)+3 n-12 (1- n-1)+ . . .

= 1 + n-1 + n-12 + . . .

= 1/(1- n-1 )Prob. that third probe is empty and first 2 filled

Successful Searching(Double)

The # of probes in a successful search is the average of the number of probes it took to insert each of the n items.

The expected number of probes to insert the ith item is the expected # of probes in an unsuccessful search when i-1 items have been inserted.

n

i i

n

iin n

Un

S1 11

1 1

111

Successful Search Continued

nmm

n

i

n

i in HH

n

m

imn

m

nS

11 1 1

1

1

11

Where ii

H i ln1

3

1

2

11

So

nm

m

n

mnmm

n

mSn ln)ln(ln

nn

1

1ln

1

Observation(Double Hashing)

nnnS

1

1ln

1

When n = 20/31 we have Sn 1.606

Even when the table is 90% full Sn is 2.56

although Un 1/1-.9 = 10.0

Chaining

Chaining puts elements that hash to the same slot in a linked list:

——

——

——

——

——

——

T

k4

k2k3

k1

k5

U(universe of keys)

K(actualkeys)

k6

k8

k7

k1 k4 ——

k5 k2

k3

k8 k6 ——

——

k7 ——

Chaining

How do we insert an element?

——

——

——

——

——

——

T

k4

k2k3

k1

k5

U(universe of keys)

K(actualkeys)

k6

k8

k7

k1 k4 ——

k5 k2

k3

k8 k6 ——

——

k7 ——

Chaining

How do we delete an element?

——

——

——

——

——

——

T

k4

k2k3

k1

k5

U(universe of keys)

K(actualkeys)

k6

k8

k7

k1 k4 ——

k5 k2

k3

k8 k6 ——

——

k7 ——

Chaining

How do we search for a element with a given key?

——

——

——

——

——

——

T

k4

k2k3

k1

k5

U(universe of keys)

K(actualkeys)

k6

k8

k7

k1 k4 ——

k5 k2

k3

k8 k6 ——

——

k7 ——

Analysis of Chaining

Assume simple uniform hashing: each key in table is equally likely to be hashed to any slot

Given n keys and m slots in the table: the load factor = n/m = average # keys per slot

What will be the average cost of an unsuccessful search for a key?

Analysis of Chaining

Assume simple uniform hashing: each key in table is equally likely to be hashed to any slot

Given n keys and m slots in the table, the load factor = n/m = average # keys per slot

What will be the average cost of an unsuccessful search for a key? A: O(1+)

Analysis of Chaining

Assume simple uniform hashing: each key in table is equally likely to be hashed to any slot

Given n keys and m slots in the table, the load factor = n/m = average # keys per slot

What will be the average cost of an unsuccessful search for a key? A: O(1+)

What will be the average cost of a successful search?

Analysis of Chaining

Assume simple uniform hashing: each key in table is equally likely to be hashed to any slot

Given n keys and m slots in the table, the load factor = n/m = average # keys per slot

What will be the average cost of an unsuccessful search for a key? A: O(1+)

What will be the average cost of a successful search? A: O(1 + /2) = O(1 + )

Analysis of Chaining Continued

So the cost of searching = O(1 + ) If the number of keys n is proportional to the

number of slots in the table, what is ? A: = O(1)

In other words, we can make the expected cost of searching constant if we make constant

Choosing A Hash Function

Clearly choosing the hash function well is crucial What will a worst-case hash function do? What will be the time to search in this case?

What are desirable features of the hash function? Should distribute keys uniformly into slots Should not depend on patterns in the data

Hash Functions:The Division Method

h(k) = k mod m In words: hash k into a table with m slots using the slot

given by the remainder of k divided by m What happens to elements with adjacent

values of k? What happens if m is a power of 2 (say 2P)? What if m is a power of 10? Upshot: pick table size m = prime number not too

close to a power of 2 (or 10)

Hash Functions:The Multiplication Method

For a constant A, 0 < A < 1: h(k) = m (kA - kA)

What does this term represent?

Hash Functions:The Multiplication Method

For a constant A, 0 < A < 1: h(k) = m (kA - kA)

Choose m = 2P

Choose A not too close to 0 or 1 Knuth: Good choice for A = (5 - 1)/2

Fractional part of kA

Hash Functions: Worst Case Scenario

Scenario: You are given an assignment to implement hashing You will self-grade in pairs, testing and grading your

partner’s implementation In a blatant violation of the honor code, your partner:

Analyzes your hash function Picks a sequence of “worst-case” keys, causing your

implementation to take O(n) time to search

What’s an honest CS student to do?

Hash Functions: Universal Hashing

As before, when attempting to foil an malicious adversary: randomize the algorithm

Universal hashing: pick a hash function randomly in a way that is independent of the keys that are actually going to be stored Guarantees good performance on average, no

matter what keys adversary chooses

Universal Hashing

Let be a (finite) collection of hash functions …that map a given universe U of keys… …into the range {0, 1, …, m - 1}.

is said to be universal if: for each pair of distinct keys x, y U,

the number of hash functions h for which h(x) = h(y) is ||/m

In other words: With a random hash function from , the chance of a collision

between x and y (x y) is exactly 1/m

Universal Hashing

Theorem 12.3: Choose h from a universal family of hash functions Hash n keys into a table of m slots, n m Then the expected number of collisions involving a particular key

x is less than 1 Proof:

For each pair of keys y, z, let cyx = 1 if y and z collide, 0 otherwise

E[cyz] = 1/m (by definition)

Let Cx be total number of collisions involving key x

Since n m, we have E[Cx] < 1m

ncC

xyTy

xyx

1][E][E

A Universal Hash Function

Choose table size m to be prime Decompose key x into r+1 bytes, so that

x = {x0, x1, …, xr} Only requirement is that max value of byte < m Let a = {a0, a1, …, ar} denote a sequence of r+1 elements

chosen randomly from {0, 1, …, m - 1} Define corresponding hash function ha :

With this definition, has mr+1 members

r

iiia mxaxh

0

mod

A Universal Hash Function

is a universal collection of hash functions (Theorem 12.4)

How to use: Pick r based on m and the range of keys in U Pick a hash function by (randomly) picking the a’s Use that hash function on all keys

The End