[PPT] - 16-HashTabl BitvectorReu.ec - Suppose we want to store some de # a PowerPoint Presentation

SLIDE 1

16-HashTabl

SLIDE 2

BitvectorReu.ec

Suppose we want to store

a set SE Lad] , for

some de #

A bitr¥r representation
f S is

a Boolean

array B

f

size dtt

sit

BLIJ

⇒ its

,

r

S

= { o sie d

: Bfi]

is true }

Eg

.

D= 20 , 5- { 3.7.93

:

B. = foIololohl4llotT4
Operations

member tx) , inserted

,

remove 4)

are all 04 ) .

. Only

practical

where

d. is small

^ Space inefficient if

1st Kd

Copy , Union, Intersection all

① (c)

SLIDE 3

HashFuncti

A hash function for a set D

is a function

h :D → M

where IM1E 1DI ,

ie

a map

to a smaller set.

Eg

h:[0, MAXINT] → Lo , 123 ,

h (x)

= x mod

13

( 1441=13 , ID 1=2,147,483,647)

There will be

values

ix.y ED stuffy but hcxthly) .

Notation :

Define

HCS)

= { y

'

y

= fix) and HES }

Eg

h (3) =3 ; h (7) = 4 ;

HC13)

O 's

HC15) = 2 's HC20)

= 7

h({ 3,7 , 13 , 15,203)

= {0,213,7 }

If hlxthcy

) for ix.yes ,

we call it

a collision leg 3,15)

We

will want hash functions h st .

ran

h

= [0 , m- I] for me#

( array indices)

h

tends to distribute S uniformly

ver L0 , m -I
m

= IM1 will be

prime

SLIDE 4

HashFunctiowtBitVec

Let h :D → Lo , m -D , B

a Boolean array of size M

For

a set

SED ,

set

BLIJ

= true iff

there is NED

sit . hlx)=i

er { i :BLil}

=

HCS)

Eg

:

S

= { 3,7 , 13,15 , 20}

h (x)

= x mod

13 ; m = 13

h (s)

= { 0,2 . 3,73

B

=

Fin

: { x : Bfhlx)] }

= { 0,2 , 3,7 , 13, 15 , 19,25 , 27, 31,

--3
Bfhtx))

= I

" suggests at 5

"

Bfh 4) 3=0

implies

x 45

.

e¥ there may

be false positives but

never false negatives

.

SLIDE 5

BIoomFiH

. Let

H

= {hi , ha , . . . he}

be a set of distinct

hash functions

for

a set D , each with range

Lam-D

.

For

S E D , set BLIJ

= true if

hlx)=i for some he H ;

Bfi)

= false

.w.TO

test for membership

in

S :

. if BLHCXIJ = true for all he H

, return true

-w . return false .

. We get

a false positive only when

hlx) is a

collision for every HEH

.

B

is

a Bloom¥s

If

m is large enough relative to 1st

and

the hi are good quality

, independent hash

functions

, then there will be few false positives

SLIDE 6

Hash Tables

Let h:D → M be a hash function for D with M= Lo, m-'I
Let A be an array of size IM1 and type D u E - 3

⇒

.

A : M → DUE-3

For

a set SED , we want

A [ had] = x ,

for

each XES A Lil

=

if

hcx) Fi for every des

.

Eg

.:S = { 2,12 , 17,213 , hlx) = x mod

13

h Is)={2,11 , 5,93 A

=

To check membership

in S , return

A Lhtx)]

.

A

is

a hashtabIef

But what if we have collisions ?
Need coHisionhang

.

We will look at

a few methods .

.

SLIDE 7

Hashingwithseparatechaining.lt

A be

a size - M array of

linked lists

Set A Lil to be a list of the elements {xesihlxt.is

.

T
test for membership in S :
Return true iff

x

is in the list

A Lhhd]

F⇒

.

5=21,5, 7,13 , 18 ,

A

=l

next x mod is

" }

It

¥?¥

T
insert1rem.

ve x : insertlremove x from Afhlx))

.

If

h distributes S

almost uniformly

ver M , the lists will

be

small

, and time will

be essentially 04) .

In the worst case ,

some lists

have length IN

,

and

performance degrades to that of

linked lists

: In) .

i)

.

SLIDE 8

Hashing with Probing

. (open Addressing)

> Let

A be an array of size M and type DUE-3 ,

£ a hash function h:D → Lo, m-7

. Let f be a function f. A -0N , that

has flo)=o

and is monotone increasing

leg . ixsy ⇒find > fly))

Define

, for it N

hi 4)

= Lhtx) tfli)) mod

m

> #

Ex. htx)

= x mod

13

,

fli ) = i

ho (3) = h (3) to

= 3

h , (3)

=

h (3) t I

= 4

hzl 3)

= h(3) t 2 = 5

To

resolve

collisions

, probe the sequence of

cells:

A Lino (a)3

. Ahh , CH3 , Afk Hl) ,

.

.

SLIDE 9

Hashingwithprobing-lopenh-ddressingl.tn

iCx)=(htx)tfmodm_

Toehechaformembershipofuxio

Examine the sequence of locations

A- tho

3

, A[him3 , Athe

.

. . . . stop at the first location containing

x

r

I.

return true if

x

was found, false otherwise .

T
insertu
Examine the sequence of locations

A- tho4)3

, AL4H)3 , ALWYN . . .

Stop

at thefirst location containing

and

store x there

.

Choice of

ft) determines properties

.

SLIDE 10

Hashingwithhinearprobing-let.fi

)

= i

F- The sequence of locations to probe is :

A LH4B , A LhHtD, A Chintz] , Afhlxtt 33 ,

. . . It

is mod m )

F¥

Suppose htt) = x mod 13 ,

5=12,9 ,18,363

( so hls) = { 25,9105) and A is

T
insert

5 :

. compute

HC5)=5 ;

see that

AL53 It

see that

ALGJ

= - ,

so set

AL63

= 5

Now: A = I
To check if

5 es :

. compute

HC5)=5 ;

see that

AL53 It , AC53 ¥5

seethat AL63

= 5 and return true

T
check if 31 ES :

. Compute HC31) = 5,

see that AL53 # 31

, AL53 #

i.

see that AL63 I 31 , AL6) I -

see that AL73 = -

and returnfalse

SLIDE 11

HashingwithQuadrakcprobing.hetfcif.it

→The sequence of

locations to

probe is :

A 1h43

, A Lhlxtt 13 , A-Lhtx)t 43 , ALhlx) to] , . . . It

is mod m )

Exi Suppose htt) = x mod 13 ,

5=12,9 , 18,363

( so h b) = { 25,9105) and A is

To insert 35

:

compute h(351=9

. see that AL93

F.

see that ACID

x.

see that AL13

= - and store 35 there .

Now : A is

T

check if 35Es :
compute h(351=9

. see that AL93 # - , AL93# 35 . see that Ado) ¥ - , All03 # 35 . see that AL13 = 35 and return true

T
check if 22 E S :

. compute h(22) =

9.

see that AG3A 4OI

,ALD , AL53 are

not

22

r

a.

see that AE23 = - and return false

SLIDE 12

Doublettashing.lt

flit = i. hash,4) ,

where hashzlx)

is

a hash function for D that

is

different from h , and

with

rain ( hash)

E Li , m]

The sequence of locations to probe

is :

( t

is mod m )

Afhlx)) , Afhlx) thashzlx)] , Afhlx)t 2. hash, HI ,

. . .

IE Suppose h 4) = x mod 13 , 8=12,9 , 18,36 }

( so h b) = { 25,9105) and A is

To insert 15 :

. compute htx) =

2.

see that A4J

F.

compute hashzlx) = 6

see that AL83 = , and store 15 there
Now : A is
To check if

15ES , check AL23 , then AL83 , andreturntrue

To check if 10 ES :

.Compute hGo) = to

see that A LID # 10 , A 403 #

i.

compute hashz(lot =

4.

see that AL13

= , and return false

SLIDE 13

Removal with Open Addressing

Suppose we have a hash table H for a set S containing x,

and want to remove x.

If

It uses separate chaining ,

we just delete x.

If

It uses open addressing

,

we cannot ,

because ix affects the probe sequence for other elements .

E¥

. Suppose htt) = x mod 13 ,

5=12 , 5,9 , 18,363

and A was obtained

as in

ur

Linear Probing

example : A =

Suppose we now delete

18 ,

so

A = I

I.

Now

, searching for 5 fails , because

ALHC5)I = - I

One

solution is to mask cells

where

we have deleted

elements

.

SLIDE 14

Removalwithopen-hddressing.FI

.

In the previous

example , to

remove

18 we replace

it with d. :

→

A =

Now

,

search 4 insert procedures

perform

as

if AL53

has

some

key that

we will

never

use .

Toremoueu
examine the sequence of

locations

A Lhote)), A Child

, A LhzHI ,

.

. . when

x

is found,

replace it with d.

Notice that search 4

insert work correctly

as they

are

Insert can

be modified to reclaim space : T

insert x :
Examine the sequence of prob locations
stop at the first one containing
r d

and store x there

.

NB : In implementation

, d and -

could

be special values

,

r A could be an

array of objects

r struts with "empty

" and "deleted " variablesfields .

SLIDE 15

↳adT ac

The loadfa
f

a hash table

H

is

. '

# of keys)t # of

elements marked d) (If It uses separate

A

= -

chaining

, there

m

are

no d's )

Good performance requires 1

not too large

.

For separate chaining

: I should not be much larger

than 1

,

so average list length is about 1

. " For

pen addressing , want 4<0.5 ,

so that

it

is not too

hard to find a place to make an insertion

.

SLIDE 16

somepropertieswithopenAddressing.li

nearprobing

\ Insertion always succeeds if 4L

I

Primatering

is

a serious problem .

Quadratic Probing
Avoids primary clustering
Exhibits secondary clustering
but less problematic
Insertion alway succeeds if At 0.5

,

but may

fail if I > 0.5 (even if there

is space)

. .Doublettash.me#

Requires design of a second suitable hash function

: Requires computing 2 hash functions whenever

probing beyond A[hotel)

is

needed

.

SLIDE 17

Rehashing

Rehashing

hash table It

means constructing

a completely

new hash table for the

contents of

It .

We may

want to

do it if

:

4

is too large ( close to 0.5 for

pen

addressing

,

much larger than 1 for separate chaining )

Performance has become

poor

( which may result from clustering

, from

long linked

lists ,

r from many removals)
Takes time

① Ln)

under the assumption that insert is Eli)

.

SLIDE 18

HashingPropert

Well
designed hash tables are effective in practice

,

with fast

insert

, member, remove operations

Require a good hash function for the domain of application
Operations 04)
narera→ , under assumptions that

may not hold

in practice :

all keys equally likey
hash function distributes keys uniformly
A

small

Do not support
perations

based

n order of keys ,

such as :

. enumerate

in order

min,

Max, range lookups

. Union

r intersection

(

These are efficient with

AVL Trees 4 B-Trees )

.