Admission in india 2015

AAddmmiissssiioonn iinn IInnddiiaa 22001155
By:
admission.edhole.com

FFiinniittee AAuuttoommaattaa aanndd
NNoonn DDeetteerrmmiinniissmm
http://cis.k.hosei.ac.jp/~yukita/

finite automaton
Q S
d
q F
A is a 5 - tupple ( , , , , ), where
states
Q
1. is a finite set called the ,
alphabet
S
transition function
Q ´S®
Q
d
3. : is the ,
start state
q Î
Q
4. is the , and
0
set of accept states
F Í
Q
5. is the .
0
DDeeffiinniittiioonn 11..11:: FFiinniittee AAuuttoommaattoonn
admission.edhole.com 3

0 1
1 0
SSttaattee DDiiaaggrraamm ffoorr MM11
4
qq 3 1 q2
0, 1

Q =
q q q
1. { , , }
1 2 3
S =
2. {0,1}
d
3. is described as
q
4. is the start state, and
1
F =
q
5. { }.
,
0 1
2
q q q
1 1 2
q q q
2 3 2
q q q
3 2 2
DDaattaa RReepprreesseennttaattiioonn ffoorr MM11

1. Implement M1 with your favorite
programming language.
2. GUI
• Two buttons for input 0 and 1
• State chart with the current state
highlighted
TTaasskk 0011
DDFFAA

M L M
language of machine
The , written as ( ),
is the set of all strings that machine accepts.
M A
We can say that .
1 L M w w
( ) { | contains at least one 1 and
an even number of 0s follow the last 1}
LLaanngguuaaggee ooff MM11
M
=
recognizes

0, <RESET> 0
2
SSttaattee DDiiaaggrraamm ffoorr MM55
8
q2
q1
q0
0
1
1
2
2, <RESET>
1, <RESET>

Q =
q q q
1. { , , }
0 1 2
S =
2. { RESET ,0,1,2}
d
3. is described as
R
q q q q q
0 0 0 1 2
q q q q q
1 0 1 2 0
q q q q q
2 0 2 0 1
q
4. is the start state, and
0
F =
q
5. { }.
,
0 1 2
0
DDaattaa RReepprreesseennttaattiioonn ffoorr MM55

M5 keeps a running count of the sum of
the numerical symbols it reads, modulo 3.
Every time it receives the <RESET>
symbol it resets the count to 0.
M5 accepts if the sum is 0, modulo 3.
IInnffoorrmmaall DDeessccrriippttiioonn ooff MM55

A language is called a regular language
if some finite automaton recognizes it.
DDeeffiinniittiioonn 11..77:: RReegguullaarr
LLaanngguuaaggee

E2 recognizes the regular language of all
strings that contain the string 001 as a
substring.
0010, 1001, 001, and 1111110011110
are all accepted,
but 11 and 0000 are not.
EExxaammppllee 11..99:: AA ffiinniittee aauuttoommaattoonn EE22

You
1. haven’t just seen any symbols of the
pattern,
2. have just seen a 0,
3. have just seen 00 or,
4. have just seen the entire pattern 001.
Assign the states q,q0,q00, and q001 to these
possibilities.
FFiinndd aa sseett ooff ssttaatteess ooff EE22

0 0, 1
1
1
DDrraaww aa SSttaattee DDiiaaggrraamm ffoorr EE22
14
q00 q001
1
q 0 0
0
q

A B
Let and be languages.
A È B = x x Î A x Î
B
: { | or }.
A B = xy x Î A y Î
B
: { | and }.
Union
Concatenation
*

A x x x k x A
k i = ³ Î
: { | 0 and each }.
1 2
binary

Star
The first two operations are operations,
unary
and the last one a operation.
RReegguullaarr OOppeerraattiioonnss oonn
LLaanngguuaaggeess

Let the alphabet be {a,b, , z}.
= =
Let {good, bad} and {boy, girl}. Then, we have
{good, bad, boy, girl},
A B
È =
{goodboy, goodgirl, badboy, badgirl}, and
=
A B
{ , good, bad, goodgood, goodbad, badbad,

*
goodgoodgood, good goodbad, goodbadgood,
goodbadbad, 
}.
EExxaammppllee 11..1111

= e
S
A
A B

TThheeoorreemm 11..1122 CClloosseeddnneessss ffoorr
UUnniioonn
The class of regular languages is closed under
the union operation.
In other words, if and are regular languages,
so is .
1 2
1 2
A A
A A
È

M A M = Q S
d
q F
Let recognize , where ( , , , , ),
1 1 1 1 1 1 1
M A M = Q S
d
q F
and recognize , where ( , , , , ).
2 2 2 2 2 2 2
M A È
A
Construct to recognize ,
M = Q S
d
q F
where ( , , , , ).
1 2
Then, check the correctness of the construction.
PPrrooooff ooff TThheeoorreemm 11..1122
admission.edhole.com Proof of Th 1.12 18

Q = Q ´ Q = r r r Î Q r Î
Q
1. {( , ) | and }.
1 2 1 2 1 1 2 2
M M
2. We can assume that and have the same
S
alphabet . (Why?)
1 2
r r Q a r r a r a r a
" Î " ÎS =
d d d
3. ( , ) , ; (( , ), ) ( ( , ), ( , ))
1 2 1 2 1 1 2 2
q =
q q
4. ( , ).
0 1 2
F = F ´ Q È Q ´ F = r r r Î F r Î
F
5. ( ) ( ) {( , ) | or }.
1 2 1 2 1 2 1 1 2 2
CCoonnssttrruuccttiioonn ooff MM

You should check the following.
1. For any string recognized by M1 is
recognized by M.
2. For any string recognized by M2 is
recognized by M.
3. For any string recognized by M is
recognized by M1 or M2.
CCoorrrreeccttnneessss

The class of regular languages is closed under concatenation
A A
operation. In other words, if and are regular languages,
A A
so is 
.
1 2
1 2
TThheeoorreemm 11..1133 CClloosseeddnneessss ffoorr
ccoonnccaatteennaattiioonn

To prove Theorem 1.13, we need
nondeterminism.
Nondeterminism is a generalization of
determinism. So, every deterministic
automaton is automatically a
nondeterministic automaton.
NNoonnddeetteerrmmiinniissmm

A nondeterministic finite automaton can
be different from a deterministic one in
that
◦ for any input symbol, nondeterministic one can
transit to more than one states.
◦ epsilon transition
NFA and DFA stand for nondeterministic
finite automaton and deterministic finite
automaton, respectively.
NNoonnddeetteerrmmiissttiicc FFiinniittee AAuuttoommaattaa

0,1
NNFFAA NN11
24
1 0,e q4
qq 3 1 q2
0,1
1

PPaarraalllleell wwoorrlldd aanndd NNFFAA
25
...
...
accept

0,1
1 0,1 q4
EExxaammppllee 11..1144 NNFFAA NN22
26
qq 3 1 q2
0,1
Let language A consist of all strings over {0,1} containing a 1 in the third
position from the end. N2 recognizes A.

0
0 0 q110
1 q111
AA DDFFAA eeqquuiivvaalleenntt ttoo NN22
27
qq 010 000 q100
qq 011 001 q101
1
0
1
0
0
1
1 0
1
1 0
1 admission.edhole.com

e
0
0
0
28
e
0
0
Let language A consist of all strings 0k , where k is a multiple of 2 or 3. N3
recognizes A.

0 q0
AA DDFFAA eeqquuiivvaalleenntt ttoo NN33
29
0 q 0 q 0 q0 qq１２ 3 4
5
0
q-1
1 1
1 1 1
1
0, 1

30
q1
q2 q3
a
b
a,b
e
a
N4 accepts e, a, baba, and baa. N4 does not accept b, nor babba.

nondeterministic finite automaton
A is a 5 - tupple ( , , , , ), where
states
Q
alphabet
transition function
Q
3. : 2 is the ,
start state
q Î
Q
4. is the , and
0
5. is the .
DDeeffiinniittiioonn 11..1177:: NNFFAA
0
set of accept states
F Q
Q q F
Q
Í
´S ®
S
S
e d
d

Q =
q q q q
1. { , , , },
1 2 3 4
2. {0,1}
3. is given as
0,1
q q q q
1 1 1 2
1 0,e
q q q
2 3 3
q q
Æ Æ
3 4
q
4. is the start state.
0 1
{ } { , }
{ } { }
{ }
{ } { }
EExxaammppllee 11..1188 NNFFAA NN5. { }. 11
4
1
4 4 4
F q
q q q
=
Æ
Æ
Æ
S =
e
d
32
qq 3 1 q2
0,1
q4
1

Von Neumann machines are deterministic.
However, there are many cases where
machine specification is all we need.
IInn wwhhaatt ssiittuuaattiioonn iiss NNoonn
DDeetteerrmmiinniissmm rreelleevvaanntt??

Every nondeterministic finite automaton
has an equivalent deterministic finite
automaton.
◦ Def. The two machines are equivalent is they
recognize the same language.
TThheeoorreemm 11..1199

N = Q S
d
q F A
Let ( , , , , ) be the NFA recognizing some language .
0
N
e
Let us assume first that has no arrows.
= ¢ S ¢ ¢ ¢
M Q d
q F
Construct ( , , , , ) such that
¢ =
Q
Q
1. 2 ,
¢ =
R a r a
d d
2. ( , ) ( , ),
¢ =
r Î
R
q q
3. { }, and
0 0
0
¢ = Î ¢ Ç ¹ Æ
F R Q R F
4. { | }.
PPrrooooff ooff TThh.. 11..1199

E R =
q q R
( ) { | can be reached from by traveling along zero or
more e
arrows.}
We modify d
¢
as follows.
¢ ( R , a ) = { q Î Q | q Î E ( ( r , a )) for some r Î
R
}.
We modify as follows.
d d
q
¢
0
¢ =
q E q
({ }).
0 0
We omit the correctness p
roof.
IInnccoorrppoorraattee e aarrrroowwss

A language is regular if and only if some
nondeterministic finite automaton
recognizes it.
CCoorroollllaarryy 11..2200

N =
a b d
Given {{1,2,3},{ , }, ,1,{3}}, we want to construct
4
D D
an equivalent DFA . The 's state set may be takes as
{1,2,3}
= Æ
2 { ,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}.
b
EExxaammppllee 11..2211 NNFFAA NN44 ttoo DDFFAA
38
1
2 3 a
a,b
e
a

1. Write a program that simulates N4.
2. GUI
◦ Three buttons for input 0, 1, and epsion.
◦ State chart that reflect the branching of the
world.
TTaasskk 0022
PPaarraalllleell WWoorrlldd

The start state is E({1}) = {1,3}.
The set of accept states is {{3},{1,3},{2,3},{1,2,3}}.
The state diagram is given as follows.
SSttaarrtt aanndd AAcccceepptt ssttaatteess

f {1} {2} {1,2}
{3} {1,3} {2,3} {1,2,3}
TThhee ssttaattee ddiiaaggrraamm ooff DD
41
a,b
a,b
a
a
a
a
a
a
b
b b b
b
b

TThheeoorreemm 11..2222 TThhee ccllaassss ooff rreegguullaarr
llaanngguuaaggeess iiss cclloosseedd uunnddeerr tthhee uunniioonn
ooppeerraattiioonn..
42
N1
N2
N
e
e

N = Q S
q F A
Let ( , , , , ) recognize , and
1 1 1 1 1 1
N Q q F A
( , , , , ) recognize .
2 2 2 2 2 2
N Q q F A A
Construct { , , , , ) to recognize .
Q q Q Q
1. { }
 
0 1 2
2. The state is the start state of
F F F
d
ì
ï ï
í
q a q Q
( , ) for
1 1
q a q Q
( , ) for
2 2
q q q q a
{ , } for and
ï ï
Î
Î
1 2 0
Æ = ¹
î
=
3.
= =
=
=
= S
= S
e
for and .
4. ( , )
0
1 2
0
0 1 2
e
d
d
d
d
d
q q a
q a
q N.



llaanngguuaaggeess iiss cclloosseedd uunnddeerr tthhee ccoonnccaatteennaattiioonn
44
N1 N2
N e
e

N = Q S
q F A
Let ( , , , , ) recognize , and
1 1 1 1 1 1
N Q q F A
( , , , , ) recognize .
2 2 2 2 2 2
N Q q F A A
1.
2. The start state is the same as that of .
3. The set of accept states is the same as that of
d
ì
ï ï
d e
í
d e
ï ï
î
q a q Q q F
( , ) for and
1 1 1
q a q F a
( , ) for and
1 1
q a q q F a
( , ) { } for and
1 2 1
Î Ï
Î ¹
Î =
Î
=
=
= S
= S
( , ) for .
4. ( , )
2 2
2
1 1
1 2
1 2 1 2
q a q Q
q a
N
q N
Q Q Q
d
d
d
d
d




llaanngguuaaggeess iiss cclloosseedd uunnddeerr tthhee ssttaarr
46
N1
N
e e
e

N = Q S
q F A
Let ( , , , , ) recognize .
1 1 1 1 1 1
N Q q F A
1. { }

2. The start state is the new start state.
F q F
ì
ï ï ï
q a q Q q F
q a q F a
d e
q a q q F a
d e
í
ï ï ï
Î Ï
( , ) for and
1 1 1
Î ¹
( , ) for and
1 1
Î =
( , ) { } for and
1 1 1
q q q a
{ } for and
1 0
Æ = ¹
î
=
3. { }
= =
=
=
= S
e
for and .
4. ( , )
0
0 1
0
0 1
*
0 1
e
d
d
d
d
q q a
q a
q
Q q Q



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