CSC 172 DATA STRUCTURES

SETS and HASHING Unadvertised in-store special: SETS! in JAVA, see Weiss 4.8 Simple Idea: Characteristic Vector HASHING...The main event.

Representation of Sets

ListSimple O(n) dictionary operations

Binary Search TreesO(log n) average timeRange queries, sorting

Characteristic Vector O(1) dictionary ops, but limited to small sets

Hash TableO(1) average for dictionary opsTricky to expand, no range queries

Characteristic Vectors

Boolean Strings whose position corresponds to the members of some fixed “universal” set

A “1” in a location means that the element is in the setA “0” means that it is not

MUSIC THEORY

A chord is a set of notes played at the same time. Represented by a 12 bit vector called a “pitch

class” {B,A#,A,G#,G,F#,F,E,D#,D,C#,C} 000010010001 represents C major 000010001001 represents C minor Rotation is “transposition” Bit reversal is “inversion”

UNIX file privileges

{user, group, others} x {read, write, execute}9 possible privilegesType “ls –l” on UNIX

total 142-rw-rw-r-- 1 pawlicki none 76 Jun 20 2000 PKG416.desc-rw-rw-r-- 1 pawlicki none 28906 Jun 20 2000 PKG416.pdf-rw-rw-r-- 1 pawlicki none 1849 Jun 20 2000 let.1-rw-rw-r-- 1 pawlicki none 0 Apr 2 13:03 out-rw-rw-r-- 1 pawlicki none 39891 Jun 20 2000 stapp.uu

UNIX files

The order is rwx for each of user (owner), group, and others

So, a protection mode of 110100000 means that the owner may read and write (but not execute), the group can read only and others cannot even read

GAMBLING A deck has 52 cards {2C,2H,2S,2D,3C, .... KD,AC,AH,AS,AD} Represent a “hand” as a vector of 52 bits 00000000000000000000000000000000000000000000

00000101 is a pair of aces In “Texas Hold'em” everyone gets two “hole” cards and

5 “board” cards We can use bitwise & to find “hands”

CV advantages

If the universal set is small, sets can be represented by bits packed 32 to a word

Insert, delete, and lookup are O(1) on the proper bit

Union, intersection, difference are implemented on a word-by-word basis

O(m) where m is the size of the setSmall constant factor (1/32)Fast, machine operations

Hashing

A cool way to get from an element x to the place where x can be found

An array [0..B-1] of bucketsBucket contains a list of set elementsB = number of buckets

A hash function that takes potential set elements and quickly produces a “random” integer [0..B-1]

Example

If the set elements are integers then the simplest/best hash function is usually h(x) = x % B or h(x) = x - (x%B), (never 0).

Suppose B = 6 and we wish to store the integers {70, 53, 99, 94, 83, 76, 64, 30}

They belong in the buckets 4, 5, 3, 4, 5, 4, 4, and 0Note: If B = 7 0,4,1,3,6,6,1,2

Pitfalls of Hash Function Selection

We want to get a uniform distribution of elements into buckets

Beware of data patterns that cause non-uniform distribution

Example

If integers were all even, then B = 6 would cause only buckets 0,2, and 4 to fill

If we hashed words in the UNIX dictionary into 10 buckets by length of word then 20% go into bucket 7

Dictionary Operations

LookupGo to head of bucket h(x)Search for bucket list. If x is in the bucket

Insertion: append if not foundDelete – list deletion from bucket list

Analysis

If we pick B to be new N, the number of elements in the set, then the average list is O(1) long

Thus, dictionary ops take O(1) timeWorst case: all elements go into one bucketO(n)

Managing Hash Table Size

If n gets as high as 2B, create a new hash table with 2B buckets

“Rehash” every element into the new tableO(n) time total

There were at least n inserts since the last “rehash”

All these inserts took time O(n)

Thus, we “amortize” the cost of rehashing over the inserts since the last rehash

Constant factor, at worst

So, even with rehashing we get O(1) time ops

Collisions

A collision occurs when two values in the set hash to the same value

There are several ways to deal with thisChaining (using a linked list or some secondary structure)Open AddressingDouble hashingLinear Probing

Chaining

4

5

6

3

2

1

0 70

99 64

83 76

94

53

30

Very efficientTime Wise

Other approachesUse less space

Open Addressing

When a collision occurs, if the table is not full find an available spaceLinear ProbingQuadratic ProbingDouble Hashing

Linear ProbingIf the current location is occupied, try the next table

locationLinearProbingInsert(K) {

if (table is full) error;probe = h(K);while (table[probe] is occupied)

probe = ++probe % M;table[probe] = K;

}

Walk along table until an empty spot is foundUses less memory than chaining (no links)Takes more time than chaining (long walks)Deleting is a pain (mark a slot as having been deleted)

Linear Probingh(K) = K % 13

181211109876543210

Insert: 18, 41, 22, 59, 32, 31, 73

h(K) : 5,

Linear Probingh(K) = K % 13

18411211109876543210

Insert: 18, 41, 22, 59, 32, 31, 73

h(K) : 5, 2,

Linear Probingh(K) = K % 13

2218411211109876543210

Insert: 18, 41, 22, 59, 32, 31, 73

h(K) : 5, 2, 9,

Linear Probingh(K) = K % 13

225918411211109876543210

Insert: 18, 41, 22, 59, 32, 31, 73

h(K) : 5, 2, 9, 7,

Linear Probingh(K) = K % 13

22593218411211109876543210

Insert: 18, 41, 22, 59, 32, 31, 73

h(K) : 5, 2, 9, 7, 6,

Linear Probingh(K) = K % 13

22593218411211109876543210

Insert: 18, 41, 22, 59, 32, 31, 73

h(K) : 5, 2, 9, 7, 6, 5,

Linear Probingh(K) = K % 13

22593218411211109876543210

Insert: 18, 41, 22, 59, 32, 31, 73

h(K) : 5, 2, 9, 7, 6, 5,

Linear Probingh(K) = K % 13

22593218411211109876543210

Insert: 18, 41, 22, 59, 32, 31, 73

h(K) : 5, 2, 9, 7, 6, 5,

Linear Probingh(K) = K % 13

2231593218411211109876543210

Insert: 18, 41, 22, 59, 32, 31, 73

h(K) : 5, 2, 9, 7, 6, 5,

Linear Probingh(K) = K % 13

2231593218411211109876543210

Insert: 18, 41, 22, 59, 32, 31, 73

h(K) : 5, 2, 9, 7, 6, 5, 8

Linear Probingh(K) = K % 13

2231593218411211109876543210

Insert: 18, 41, 22, 59, 32, 31, 73

h(K) : 5, 2, 9, 7, 6, 5, 8

73

Double HashingIf the current location is occupied, try another table locationUse two hash functionsIf M is prime, eventually will examine every location DoubleHashInsert(K) {

if (table is full) error;probe = h1(K);offset = h2(K);while (table[probe] is occupied)

probe = (probe+offset) % M;table[probe] = K;

}

Many of the same (dis)advantages as linear probingDistributes keys more evenly than linear probing

Quadratic Probing

Don't step by 1 each time. Add i2 to the h(x) hashed location (mod B of course) for i = 1,2,...

Double Hashingh1(K) = K % 13h1(K) = 8 - K % 8

1211109876543210

Insert: 18, 41, 22, 59, 32, 31, 73

h1(K) : 5, 2, 9, 7, 6, 5, 8

h2(K) : 6, 7, 2, 5, 8, 1, 7

Double Hashingh1(K) = K % 13h1(K) = 8 - K % 8

22593218411211109876543210

Insert: 18, 41, 22, 59, 32, 31, 73

h1(K) : 5, 2, 9, 7, 6, 5, 8

h2(K) : 6, 7, 2, 5, 8, 1, 7

31

Double Hashingh1(K) = K % 13h1(K) = 8 - K % 8

22593218411211109876543210

Insert: 18, 41, 22, 59, 32, 31, 73

h1(K) : 5, 2, 9, 7, 6, 5, 8

h2(K) : 6, 7, 2, 5, 8, 1, 7

3173

Theoretical Results

Double Hashing

Linear Probing

Chaining

FoundNot Found

1 +α 1+α2

12

+ 1

2 (1−α )212

+ 12 (1−α )

1(1−α )

1αln 1

(1−α )

Expected Probes

0.5 1.0

1.0

Linear Probing

Double Hashing

Chaining