CS 317 Automata

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CS 317 Automata

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Title: CS 317 Automata

1
CS 317 - Automata Formal Languages Exam 2
Review

Fall 2009Sarah Mocas

Exam 2 Tuesday 11/3
I expect you to know most basic definitions
I will look back at the homework problems
Always answer partial credit will be given
If unsure, ask
If still unsure, write down your assumptions
I will try to discern what you know based on what
you write

3
Class Info
Read these links if you need additional sources

Formal languages (sets of strings) and there
relationship to our theoretical machines
Regular languages
Can be recognized with finite automata
Can be described with regular expressions
Useful for string pattern matching
Useful for lexical analysis (initial phase of a
compiler that scans the input source)
Finite Automata (state machines)
Context-free languages
Can be recognized with nondeterministic pushdown
automata
Can be described with context-free grammars
Useful for describing the syntax of programming
languages
Pushdown Automata (state machines with a stack)

Dont forget everything from the first section
4
Overview

Old Material
Chapters 1, 2, 3, 4, 5, 6, 7, 8, 9
Notes 1, 2, 3, 4
This material is still important especially
As it relates to the pumping lemma for Regular
languages
Closure properties for Regular languages
Properties about NFAs and DFAs
I will not test on the methods in these sections

Definition 1 (Strings) Basic notions
An alphabet is a finite set S of symbols.
A string x over S is a finite-length sequence of
elements of S. Concatenation xy or simply xy.
The set of all strings over S is denoted S.
A language over S is a subset of S.
The length of a string x is denoted x.
The empty string is denoted e and e 0.
x is a prefix of y if xz y for some z.
Concatenation
(xy)z x(yz)
xe ex
aman amn

Basics
6

Definition (Sets) Basic notions
Set membership x ? A
Set union A U B x x ? A or x ? B
Set intersection A n B x x ? A and x ? B
String set concatenation AB xy x ?A and y
?B
AU(BnC) (AUB)n (AUC) etc.
(A U B) A n B De Morgan laws
Set difference (complement) A x ? S x ? A
Powers An (recursively defined) A0 e and
An1 AAn
A stared A U An for n 0
A (no 0) and
? e
An is different from an
?, e and e are different. What is ?, ? and
e?

Know the definitions
7
Chapter 3 Finite Automata Regular Sets

Definition 3 (DFA)
A deterministic finite automaton is a structure
M (Q, S, d, s, F) where
Q is a finite set of states
S is an input alphabet
d Q X S ? Q - transition function
(Induces d-hat Q X S ? Q. See next slides)
s ? Q - initial state or start state
F ? Q - final states
An FA is specified by a set of transitions
(function) or table or graph

8
Chapter 3 Finite Automata Regular Sets

M accepts x if d-hat(s, x) ? F.
(processing of string ends in a final state)
M rejects x if d-hat(s, x) ? F.
The language accepted by M is
L(M) x ? S M accepts x .
A language is called regular if it is accepted by
some DFA.
A set is called regular if it is accepted by some
DFA.

To show that a set is regular either give an FA
or give a regular expression for it.
9
Chapter 3 Finite Automata Regular Sets

An FA is specified by a set of transitions or
table or graph
What tuples (qi, a) ? qj give the transition
function?
What is the table for the transition function?

a
F q4 (q1, a) ? q2 (q1, b) ? q3 (q2, a) ? q1

b
a
q2
q4
q1
a, b
b
q3
a, b
a
q3 can be viewed as a dead state and can be left
out of the diagram without changing the set of
strings that are good.
b
a
q2
q4
q1
3 ways to specify and DFA, NFA
10
Closure properties of regular sets

Set union A U B x x ? A or x ? B
Set intersection
A n B x x ? A and x ? B
Set difference (complement)
A x ? S x ? A
Set concatenation
AB xy x ? A and y ? B
A stared A U An for n 0

This means that if A and B are regular than so
are the sets formed by union, intersection,
difference and the star operator
11
Chapter 4 Regular Sets

Closure properties of regular languages.
Regular languages are closed under complement.
Regular languages are closed under intersection.
Regular languages are closed under union.
Regular languages are closed under set
concatenation.
Regular languages are closed under A.

Know
12
Chapter 5 Nondeterministic FA

Nondeterminism is strange stuff.
NFA can have more than one start state
For each symbol in S a NFA can have zero, one ,
or more than one transition out of any state
There is no mechanical way to determine the
action of a NFA
A NFA, M accepts a string x if there exists at
least one computation path that accepts.
A NFA, M rejects a string x if all computation
paths do not accepts.
Model Guess and Verify

Know this definition
13
Chapter 6 Nondeterministic FA

Definition (NFA)
A nondeterministic finite automaton is a
structure M (Q, S, ?, S, F) where
Q is a finite set of states.
S is an input alphabet.
? Q X S ? 2Q - transition function.
(2Q A A ? Q it is the power set of Q this
means it is the set of all subset of Q)
S ? Q - initial states (set of states)
F ? Q - final states.

14
Chapter 6 Nondeterministic FA

A NFA, N accepts x ? S if there exists an accept
state q (so q ? F) such that q is reachable from
a start state on input string x
so x ? S and ?-HAT(S, x) n F? ?
L(N) x ? S N accepts x
L(N) x ? S ? an accepting computation
path for N on input x
Note a NFA for a language L must accept all
strings in L and can not accept stings in L. But
if x is in L there may also be some computation
paths that do not end in a final state.

15
Chapter 5 Nondeterministic FA

Nondeterminism
Each symbol in S can have any number of
transition out of a state
No mechanical way to determine the action of a
NFA

b
a
q2
q1
a, e
a
q3
a
16
Chapter 5 Nondeterministic FA

Theorem The languages accepted by NFAs are
exactly the regular languages.
Every DFA is a NFA and every NFA can be turned
into a DFA (by subset construction).

Important Theorem - Know
17
Chapter 8 Pattern Matching Regular Expressions

Regular Expressions are all patterns made with
just
Atoms a ? S, e, ?
operators , ,
precedence of operators highest to lowest
, (concatenation), Also use ( and )
Theorem Let A ? S. The following statements are
equivalent
A is regular so A L(M) for some finite
automata M
A L(a) for some regular expression a

18
Chapter 9 Regular Expressions and Finite
Automata

Laws to simplify regular expressions
a (ß ?) (a ß) ?
a(ß?) (aß)?
a ß ß a
a ? a
a? ?a ?
a a a
ea ae a
a(ß ?) aß a?
(a ß)? a? ß?
e aa a
e aa a

19
Chapter 9 Regular Expressions and Finite
Automata
L(ß)

Definition subset order
a ß ? L(a) ? L(ß)
? L(a ß) L(ß)
? a ß ß
Two rules for a, ß where is subset order
ß a? ? ? aß ?
ß ?a ? ? ßa ?
Useful equations
(aß)a a(ßa)
(aß)a (ß a)
a(ßa) (ß a)
(e a) a
aa aa

L(a)
or L(a) L(ß)
20
Overview

New Material
Chapters 11, 12, 13, 14, 19, 20, 21, 23, 24, 25,
27 (pages 195 197)
Notes 5, 6, 7, 9

21
Chapter 11 Limitations of Finite Automata

Nonregular set
C ancbn n 0 e, acb, aacbb, aaacbbb,
B anbn n 0 e, ab, aabb, aaabbb,
D a2n n 0 a, a2, a4, a8, a16,
PAL w w x rev_x
Regular Set
A ambn m, n 0 e, a, b, aa, ab, bb,
aaa, aab,
Think of other examples for the exam (S, empty
set)

I will give some languages and ask you to say
which are regular and which are not.
22
Chapter 12 Using the Pumping Lemma

Which of these is regular?
A anb 2m m, n 0
B anbm n ? m
C xcx x ? a, b
D anb n 481 n 0
E anb m n m and m 481

I will give some languages and ask you to say
which are regular and which are not.
23
Chapter 11 Limitations of Finite Automata

Pumping Lemma Let A be a regular set. Then the
following property holds of A
(P) There exists k 0 such that
for any x, y, z with xyz ? A and
y k,
there exists strings u, v, w such that
y uvw, v ? e, and
for all i 0, the string xuviwz ? A

Know the statement of the Pumping Lemma
24
Chapter 11 Limitations of Finite Automata

Pumping Lemma (contra positive) Let A be a set
of strings. Suppose that the following property
holds of A
(P) For all k 0
there exists strings x, y, z such that xyz ? A,
y k and
for all strings u, v, w with
y uvw and v ? e
There exists i 0, such that xuviwz not in A
Then A is not regular

Know how to use this to do proofs
25
Chapter 11 Limitations of Finite Automata

To show that a set is regular either give an FA
or give a regular expression for it.
To show that a set is nonregular use Pumping
Lemma (contrapositive).

Important Facts
26

Pumping Lemma (contrapositive) Let A be a set of
strings. Suppose that the following property
holds of A
For all k 0
there exists strings x, y, z such that xyz ? A,
y k and
for all strings u, v, w with
y uvw and v ? e
There exists i 0, such that xuviwz not in A
To use Pumping Lemma
k is pre picked
YOU pick x, y, z, such that xyz ? A and y k
This must divide up into any u, v, w with y uvw
and v ? e
You pick i 0
You prove the language is not regular if xuviwz ?
A

Theorem A anbn n 0 is not regular.
Proof Assume A is regular.
k is pre picked and YOU pick x, y, z, such that
xyz ? A and y k
pick s akbk where x ak, y bk and z e
This must divide up into any u, v, w with y uvw
and v ? e
Suppose that k jmn and so u bj, v bm, w
bn, where m gt 0.
You pick i 0
Let i2.
You prove the language is not regular if xuviwz ?
A
Now, the string is xuv2w ak bjbmbmbn ak bkm
this is not in A since there are more bs than
as so A is not regular.

Theorem A anbn n 0 is not regular.
Proof Assume A is regular. So by the Pumping
Lemma there exists k 0 such that
for any x, y, z with xyz ? A and y k,
there exists strings u, v, w such that
y uvw, v ? e, and
for all i 0, the string xuviwz ? A
So pick s akbk and divide s into xyz so
x ak, y bk and z e.
Now y k and y is picked to be all bs.
This y must divide up into any u, v, w with y
uvw and v ? e. So, suppose that k jmn and
that
u bj, v bm, w bn, where m gt 0.
Let i2.
Now, the string is xuv2w ak bjbmbmbn ak bkm
but this is not in A since there are more bs
than as (because k lt km) so A is not regular.

Use as a model for your proofs
29

Theorem B anbm n m is not regular.
Proof Assume B is regular. So by the Pumping
Lemma there exists k 0 such that
for any x, y, z with xyz ? A and y k,
there exists strings u, v, w such that
y uvw, v ? e, and
for all i 0, the string xuviwz ? A
So pick s akbkb and divide s into xyz so
x e, y ak and z bkb.
Now y k and y is picked to have all as.
This y must divide up into any u, v, w with y
uvw and v ? e. So, suppose that k jmn and so
u aj, v am, w an, where m gt 0.
Let i3. Now, the string is
xuv3w ajamamaman bkb ak2m bkb
but this is not in B since there are fewer bs
than as (because k2m lt k1) so B is not regular.

Use as a model for your proofs
30
Chapter 12 Using the Pumping Lemma

A anbn n 0
B anbm n gt m
C ancbn n 0
D a2n n 0 a, a2, a4, a8, a16,
E an! n 0
PAL w w x rev_x

I will give you variations on languages like
these and the ones that you had for homework
31
Chapter 12 Using the Pumping Lemma

Trick using properties about regular languages to
prove a L is not regular.
Example
Prove that D x ? a,b a(x) b(x) is
not regular
Proof Suppose that D is regular. Consider the
set D n ab. This set will be regular too since
regular languages are closed under intersection.
But D n ab anbn which is not regular so D can
not be regular.

Know how to do this type of proof
32
Chapter 12 Using the Pumping Lemma

Trick using properties about regular languages
Example
Prove that B anbm n m is not regular
Proof Suppose that B is regular.
Fact If B is regular than so are sets
revB bman n m and B ambn n m.
Consider B n B. This set will be regular too
since regular languages are closed under
intersection. But B n B anbn which is not
regular so B can not be regular.

Know how to do this type of proof
33
Chapter 12

Ultimate Periodicity
Periodic behavior - In mathematics, a periodic
function is a function that repeats its values
after some definite period p has been added to
its independent variable n
Definition Let U be a subset of the natural
numbers N. The set U is Ultimately Periodic if
there exists numbers n 0 and p gt 0 so that for
all m n, m is in U if and only if mp is in U.
After m n then strings have period p

34
Chapter 12 Using the Pumping Lemma

Ultimate Periodicity
Example U 0, 1, 5, 8, 10, 13, 16, 19, 22
here n 10, p 3
Theorem Let A ? a (any one symbol alphabet).
Then A is regular if and only if the set m am
? A, the set of lengths of strings in A, is
ultimately periodic. (Proof idea a loop is the
period)
Corollary Let A be any regular set over any
finite alphabet, Then the set length A x
x ? A, of lengths of strings in A is ultimately
periodic.

Know these results What regular sets are
periodic?
35
Chapter 13 DFA State Minimization

Simplifying a DFA
Collapse equivalent states to make the smallest
DFA possible minimum DFA
Mechanical method to make a minimal DFA
Get rid of unreachable states
Collapse equivalent states

I main idea is there is a smallest DFA for a
regular language and we can always make
it. Contrast this with trying to guarantee that
you have the smallest C program for a problem
(not possible).
36
Chapter 13 DFA State Minimization

Mechanical method to make a minimal DFA
Get rid of unreachable states
Any q where there is no string x such that
d-HAT(s,x) q
Use depth first search on the transition graph
of the DFA
Mark every state that can be reached from s
Delete all unmarked states

Know how to do - this is the easier part to do
37
Graph Depth-first Search
GO DEEP The first time a node is visited it is
processed.
8
6
3
1
2
7
5
16
8
4
9
0
9
11
14

Graph Depth-first Search
Visit node v then recursively visit each
unvisited node connected to v
void traverse(int k)
link t
visit(k)
visitedk 1
for (t adjk t ! NULL t t-gtnext)
if (!visitedt-gtv) traverse(t-gtv)

38
Graph Depth-first Search
39
Chapter 14 DFA State Minimization

Mechanical method to make a minimal DFA
Collapse equivalent states
Use the Quotient Construction
Rule If there exists a string x so that
d-HAT(p,x) ? F and d-HAT(q,x) ? F
then p and q can not be collapsed, otherwise
they can be collapsed

Know how to do - this is the harder part to do
40
Chapter 14 A Minimization Algorithm

Algorithm for computing minimum DFA
Write a table of all pairs p,q of states of M
Mark p,q if p ? F and q ? F or vise versa
Repeat until no more occur
If there exists an unmarked pair p,q such that
d(p,a),d(q,a) is marked for some a ? S then
mark p,q
q p iff p,q is not marked.

Know how to do - this is the harder part to do
41
Chapter 14 A Minimization Algorithm

Example
0
U 1
U U 2
U U U 3
U U U U 4
U U U U U 5
After Step 2
0
M 1
M U 2
U M M 3
U M M U 4
M U U M M 5

Unmarked pairs after Step 2 1,2 stays
unmarked 0,3 stays unmarked 0,4 stays
unmarked 3,4 stays unmarked 1,5 mark 2,5
mark After first pass Step 3 0 M 1 M U 2 U M M
3 U M M U 4 M M M M M 5
d(q1,a),d(q5,a) 3,5 marked d(q2,b),d(q5,b)
3,5 marked
If there exists an unmarked pair p,q such
that d(p,a),d(q,a) is marked for some a ? S
then mark p,q
Know how to do - this is the harder part to do
42
Chapter 14 A Minimization Algorithm
Unmarked pairs after Step 2 1,2 stays
unmarked 0,3 stays unmarked 0,4 stays
unmarked 3,4 stays unmarked 1,5 mark 2,5
mark After first pass Step 3 0 M 1 M U 2 U M M
3 U M M U 4 M M M M M 5

After second pass Step 3
0
M 1
M U 2
M M M 3
M M M U 4
M M M M M 5
Unmarked pairs after
Step 2 Second pass
1,2 stays unmarked
0,3 mark
0,4 mark
3,4 stays unmarked so 1 2 and 3 4

43
Chapter 14 A Minimization Algorithm
Unmarked pairs after Step 2 1,2 stays
unmarked 0,3 stays unmarked 0,4 stays
unmarked 3,4 stays unmarked 1,5 mark 2,5
mark After first pass Step 3 0 M 1 M U 2 U M M
3 U M M U 4 M M M M M 5

After second pass Step 3
0
M 1
M U 2
M M M 3
M M M U 4
M M M M M 5
Unmarked pairs after
Step 2 Second pass
1,2 stays unmarked
0,3 mark
0,4 mark
3,4 stays unmarked
so 1 2 and 3 4

q p iff p,q is not marked
44
(No Transcript)
45
Chapter 19 Context-Free Grammars Languages

A Context Free Grammar (CFG) is a set of
recursive rewriting rules used to generate
strings.
A CFG (N, S, P,S) where
N is a finite set of non-terminal symbols
S is a finite set of terminal symbols disjoint
from N
P is a finite set of productions, N X (N ? S)
rules for replacing (or rewriting) non-terminal
symbols (on the left side of the production) in a
string with other non-terminal or terminal
symbols (on the right side of the production).
S in N is a start symbol

Know the definition
46

Capital symbols for non-terminal A, B,
Non-Caps for terminal a, b, c,
Production rules are written A ? a and is
shorthand for OR
A ? a1, A ? a2 can be written A ? a1 a2
If a, ß are in (N U S) we say a ? ß means ß is
derivable from a (also derived in n steps from G)
A string in (N U S) derivable from the start
symbol S is in sentential form. It is a sentence
if there are no nonterminals in it.
A leftmost derivation is one where the
productions are always applied to the left most
nonterminal in the sentential form.

Know the definitions above
47

Example L anbn n 0
CFG (S, S a, b, P S ? aSb, S ? e, S)
non-terminal S
terminal a, b
Production rules S ? aSb e
If a, ß are in (N U S) we say a ? ß means ß is
derivable from a
aaSbb ? aaabbb means aaabbb derivable from aaSbb
A string in (N U S) derivable from the start
symbol S is in sentential form.

48
Chapter 19 Context-Free Grammars Languages

To generate a string of terminal symbols from a
CFG
Begin with a start symbol
Apply one of the productions with the start
symbol on the left hand side, replacing the start
symbol with the right hand side of the
production
Repeat the process of selecting non-terminal
symbols in the string, and replacing them with
the right hand side of some corresponding
production, until all non-terminals have been
replaced by terminal symbols.

Know how to do this
49

Example L anbn n 0
(S, S a, b, P S ? aSb, S ? e, S)
Production rules S ? aSb e
Derivation of aaaabbbb
S ? aSb
? aaSbb
? aaaSbbb
? aaaabbbb

sentential form
50

Example
L x x ? a,b and the number of as is
even
(S, S a, b, P listed below, S)
Production rules
S ? AB A B
B ? bB bA e
A ? aBa aaB aaA aAa e
Derivation of bbaabbbaba
S ? B
? bB
? bbA
? bbaaB
? bbaabB

? bbaabB ? bbaabbB ? bbaabbbA ? bbaabbbaBa ?
bbaabbbabBa ? bbaabbbaba
51

The language generated by a context-free grammar
G, denoted L(G), is the set of all strings
derived from G and containing only terminals.
A language is Context-Free if it is generated by
a context-free grammar G.
All regular set are context-free.
L(G) x in S x is derivable from G
starting from S
L is a CFL if it is L(G) for a context-free
grammar G.

Know these - Important
52
To prove this use the pumping lemma

Some CFL are not regular.
L1 anbn n 0
L2 x ? a, b x revx palindromes
L3 balanced string of parentheses
Some languages are not CFL.
L1 anbncn n 0
Definition A grammar is unambiguous if there is
only one derivation for each string in the
language of the grammar.

Fact not yet proved
53

Some CFL are not regular.
L1 anbn n 0
S ? aSb e
L2 x in a, b x rev_x palindromes
S ? aSa bSb a b e
L3 balanced string of parentheses
S ? (S) SS e

To prove a language is regular give a regular
expression or FA
To prove a language is not regular use the
pumping lemma
To prove a language is CF give a CFG or PDA

54
Chapter 19 Context-Free Grammars Languages

The set of real numbers in Pascal
ltrealgt ? ltdigitgt ltdigitsgt ltdecimal partgt ltexpgt
ltdigitsgt ? ltdigitgt ltdigitsgt e
ltdecimal partgt ? '.' ltdigitgt ltdigitsgt e
ltexpgt ? 'E' ltsigngt ltdigitgt ltdigitsgt e
ltsigngt ? '' '-' e
ltdigitgt ? 0 1 2 3 4 5 6 7 8 9
What is the derivation of 435.39E3

the letter E or e represents times ten raised to
the power -7.443E3 means X103
55
Chapter 19 Context-Free Grammars Languages

L(G) x in S S ? x
G
This means the string x is derivable from G via a
series of moves (application of the production
rules to the sentential forms)
A subset B of S is a context-free language if B
L(G) for some context-free grammar.

56
Chapter 21 Normal Forms

Know how to create CNF from a CFG
Know how to read/identify GNF

Special form easy for proofs
CNF Chomsky Normal Form
All productions are of the forms
A ? BC or A ? a
where A, B, C are in N and a is in S
GNF Greibach Normal Form
All productions are of the forms
A ? aB1B2 Bk if k0 then A ? a
where A, B1, B2 Bk are in N and a is in S

Examples for S ? S SS e
CNF S ? AB AC SS
C ? SB
A ?
B ?
GNF S ? B SB BS SBS
B ?
Theorem For any CFG G there is a CNF G and a
GNF G so that
L(G) L(G) L(G) e

Theorem For any CFG G there is a CNF G and a
GNF G so that L(G) L(G) L(G) e
Proof Get rid of A ? e and unit productions A ?
B
Lemma For any CFG G there is a CNG G with no e
or unit productions so that L(G) L(G) e
Method to remove
If A ? aBß and B ? e then replace with A ? aß
If A ? B and B ? ? then replace with A ? ?

Know how to remove
unit productions
e productions

59
Chapter 21 Normal Forms
Know how to do

CNF Chomsky Normal Form
After removing e and unit productions
For each terminal a in S make a production A ? a
and replace every a on right hand side with A
Now all productions are A ? B1B2 Bk and A ?
a
For A ? B1B2 Bk with k gt 2 replace this with A
? B1C and C ? B2 Bk. Repeat until all
productions have length at most 2

A context-free grammar for the language of all
strings over a,b containing an unequal number
of a's and b's
Context-free grammar - Wikipedia, the free
encyclopedia
S ? U V
U ? TaU TaT
V ? TbV TbT
T ? aTbT bTaT e
Eliminate e-productions to get
S ? U V
U ? TaU TaT aU aT Ta a
V ? TbV TbT bV bT Tb b
T ? aTbT bTaT abT aTb baT bTa ab ba

nonterminal T can generate all strings with the
same number of a's as b's, nonterminal U
generates all strings with more a's than b's
nonterminal V generates all strings with fewer
a's than b's.
61

S ? TaU TaT aU aT Ta a TbV TbT bV
bT Tb b
U ? TaU TaT aU aT Ta a
V ? TbV TbT bV bT Tb b
T ? aTbT bTaT abT aTb baT bTa ab ba
CNF step 1, replace terminals
S ? TAU TAT AU AT TA a
TBV TBT BV BT TB b
U ? TAU TAT AU AT TA a
V ? TBV TBT BV BT TB b
T ? ATBT BTAT ABT ATB BAT BTA AB BA
A ? a
B ? b

S ? TAU TAT AU AT TA a TBV TBT BV
BT TB b
U ? TAU TAT AU AT TA a
V ? TBV TBT BV BT TB b
T ? ATBT BTAT ABT ATB BAT BTA AB BA
A ? a
B ? b
CNF step 2, divide up long rules
S ? TC TD AU AT TA TE TF BV BT
TB a b
U ? TC TD AU AT TA a
V ? TE TF BV BT TB b
T ? DF FD AF DB BD FA AB BA
C ? AU
D ? AT
E ? BV

F ? BT A ? a B ? b
63
Chapter 21 Normal Forms

CNF try these
L1 anbn n 0
S ? aSb e
L2 x in a, b x rev_x
S ? aSa bSb a b e
One more
S? aSa bSb aAb bAa
A ? aAa bAb a b e

64
Chapter 21 Normal Forms

Good example
CS-311 HANDOUT Greibach Normal Form (GNF)
GNF Greibach Normal Form
All productions are of the forms
A ? aB1B2 Bk if k0 then A ? a
where A, B1, B2 Bk are in N and a is in S
Every CFG can be rewritten in Greibach Normal Form

Know how to read/understand GNF but you will not
need to convert a CFG to GNF
65

Example 2 G (S a, b, NS, A, B, P, S)
S ? a A S a
A ? S b A S S ba
B ? SA a
Start S, Terminal a, b
Nonterminals S, A, B rename to A1, A2, A3
A1 ? a A2 A1 a
A2 ? A1 b A2 A1 A1 ba
A3 ? A1 A2 a

66
A1 ? a A2 A1 a A2 ? A1 b A2 A1 A1 ba A3 ?
A1 A2 a

Next
A2 ? A1 b A2 A1 A1 ba
A1 ? a A2 A1 a
A2 ? a A2 A1 b A2
a A2 A1 A1
a b A2
a A1 ba
A3 ? a A2 A1 A2
a A2 a

Step 1 loop starts
A3 ? A1 A2
G is now
A1 ? a A2 A1 a
A2 ? A1 b A2
A1 A1 ba
A3 ? a A2 A1 A2
a A2 a

Last
Replaced b a by B2
A1 ? a A2 A1 a
A2 ? a A2 A1 B1
a A2 A1 A1 a B1
a A1 b B2
A3 ? a A2 A1 A2
a A2 a
B1 ? b A2
B2 ? a

Next
Replaced b A2 by B1
A1 ? a A2 A1 a
A2 ? a A2 A1 B1
a A2 A1 A1 a B1
a A1 ba
A3 ? a A2 A1 A2
a A2 a
B1 ? b A2

A1 ? a A2 A1 a A2 ? A1 b A2 A1 A1 ba A3 ?
A1 A2 a
68
Chapter 23 Pushdown Automata

Pushdown Automata (PDAs)
Like an FA but has a pushdown stack
Stack is unbounded memory
Stack is LIFO
Stack operations are PUSH and POP
Input head still read only, and moves right

69
Chapter 23 Pushdown Automata

Definition A nondeterministic PDA (Pushdown
Automata) is
M (Q, S, G, d, s, -, F) where
Q a finite set of states
S a finite set - input alphabet
G a finite set - stack alphabet
d ? (Q x (S U e) X G) X (Q X G) transitions
S start state in Q
- initial stack symbol to mark bottom
F ? Q the final states

Know the definition
70
Chapter 23 Pushdown Automata

NPDA M (Q, S, G, d, s, -, F)

Input Tape left to right, read only
Push/pop
Top of the Stack
Q Finite Control d
Stack
-
LIFO stack
71

Example L anbn n 0
(S, S a, b, P S ? aSb, S ? e, S)
Production rules S ? aSb e
Process aaaabbbb
Read a and Push A
Read a and Push A
Read a and Push A
Read a and Push A
Read b and Pop A
Read b and Pop A

Top of Stack
72

NPDA M (Q, S, G, d, s, -, F)
Two operations PUSH and POP the top symbol
((p,a,A), (q, B1B2 Bk)) ? d
In state p, reading a on the tape, pop A from the
top of the stack, push B1B2 Bk with Bk first
and B1 last, go to state q, move the tapes head
left 1 symbol
e-transitions allowed a PUSH or POP can happen
but the read head does not move to the next input
symbol
((p,e,A), (q, B1B2 Bk)) ? d
In state p, ignore the tape, pop A from the top
of the stack, push B1B2 Bk and go to state q

Know the definition
73
Chapter 23 Pushdown Automata

Configuration Gives the complete state of the
PDA
(p, baaabba, ABAC -)
(current state, tape input string, stack)
Left of the tape head can be not included
Start config. (s,w,-)
Nondeterministic so more than one possible move
from a given
(state, tape symbol, stack top)

Know the definition
74
Chapter 23 Pushdown Automata

Definition Since the PDAs are nondeterministic,
the machine accepts its input string if there
exists an accepting sequence of transitions.
Two Definitions of accept
Final State The PDA halts after having read the
entire input string and is in a final state
Empty Stack The PDA halts after having read the
entire input and after applying the last
transition the stack is empty - only has -

Know the definition
75
Chapter 23 Pushdown Automata

Example NPDA for PARENS
M (Q q, S ,, G -, , d, s q, -,
F)
d is
((q, , - ), (q, - )) push
((q, , ), (q, )) push
((q, , ), (q, e)) pop
((q, e , - ), (q, e)) stop with empty stack

Know how to make a PDA for a language or CFG
76
Chapter 23 Pushdown Automata

Examples L ww w ? a,b is not CF
The complement L a,b - ww w ? a,b
is CF
Idea Guess a point in the first half that does
not match the same point in the last half
suppose the input is aabbaaabaa
Then the 4th symbol in the 1st half does not
match the 4th symbol in the second half.

77
Chapter 23 Pushdown Automata

Example L x x wcwr and w in 0,1
M (q1, q2, 0, 1, c, -, B, G, d, q1, -, Ø)
((q1, 0, -), (q1, B-)) ((q1, 1, -), (q1, G-))
((q1, 0, B), (q1, BB)) ((q1, 1, B), (q1, GB))
((q1, 0, G), (q1, BG)) ((q1, 1, G), (q1, GG))
((q1, c, G), (q2, G)) ((q1, c, B), (q2, B))
((q1, c, -), (q2, -))
((q2, 0, B), (q2, e))
((q2, 1, G), (q2, e))
((q2, e, -), (q2, e))

Know how to make a PDA for a language or CFG
78
Chapter 23 Pushdown Automata

Extra Rules
The symbol - can be pushed and popped
Transition ((p,a,A), (q, B1 Bk)) and ((p,e,A),
(q, B1 Bk)) do not work if A is not on the top
of the stack.
If the top of the stack is empty then the PDA
breaks
Everything must be popped to accept with an empty
stack.

L a,b - ww w ? a,b is CF
((q1, a, - ), (q1, A- )) push a
((q1, b, - ), (q1, B- )) push b
((q1, a, ß), (q1, Aß)) push a ß is the top of
the stack
((q1, b, ß), (q1, Bß)) push b ß is the top of
the stack
((q1, a, ß), (q2, CACß)) push CAC go to q2 guess
mismatch
((q1, b, ß), (q2, CBCß)) push CBC go to q2 guess
mismatch
((q2, a, ß), (q2, Aß)) push a
((q2, b, ß), (q2, Bß)) push b
((q2, a, ß), (q3, e)) pop the top symbol ß go
to q3 - guess middle
((q2, b, ß), (q3, e)) pop the top symbol ß go
to q3 - guess middle
((q3, a, ß), (q3, e)) pop the top symbol ß
((q3, b, ß), (q3, e)) pop the top symbol ß
((q3, a, CAC), (q4, e)) pop CAC go to q4 reject
((q3, b, CBC), (q4, e)) pop CBC go to q4 reject
((q3, a, CBC), (q5, e)) pop CBC go to q5
accept on empty stack
((q3, b, CAC), (q5, e)) pop CAC go to q5
accept on empty stack

80
Chapter 27 very end, pages 195 - 197
Important - know

Closure Properties of CFLs
The Context-free languages are closed under union
If L(G1) L1 and L(G2) L2 then L1U L2 L3
will also be a CFL
Proof Idea S3 ? S1 S2
The Context-free languages are closed under
concatenation
If L(G1) L1 and L(G2) L2 then L3 will also
be CFL where L3 xy x ? L1 and y ? L2
Proof Idea S3 ? S1S2

81
Chapter 27 very end, pages 195 - 197
Important - know

Closure Properties of CFLs
The Context-free languages are closed under
astrate
If L(G1) L1 then L1 will also be a CFL
Proof Idea S2 ? S1S2 e
The Context-free languages are closed under
intersection with Regular sets
If L(G1) L1 is CFL and L(G2) L2 is a Regular
language then L1 n L2 L3 will also be a CFL
Proof Idea (harder) Intersection of the PDA and
DFA

82
Chapter 27 very end, pages 195 - 197
Important - know

Closure Properties of CFLs
The Context-free languages are NOT closed under
intersection
If L(G1) L1 is CFL and L(G2) L2 is CFL then
L1 n L2 L3 may not be CFL
Proof
anbncm ngt0 n ambncn ngt0 anbncn
ngt0
is not a CFL

83
Chapter 27 very end, pages 195 - 197
Important - know

Closure Properties of CFLs
The Context-free languages are NOT closed under
complementation
If L(G1) L1 is CFL then L1 may not be.
Proof
L ww w is in a, b is not a CFL but
its complement is a CLF

84
Important - know

Closure Properties of CFLs
The Context-free languages are NOT closed under
complementation
If L(G1) L1 is CFL then L1 may not be.
Second Proof (by contradiction)
Assume that the CLF are closed under
complementation. Then if L1and L2 are CFL so are
L1 and L2. Fact the Context-free languages are
closed under union. Now, by properties of sets L1
n L2 will be a CFL since
(L1 n L2) L1U L2 (De Morgans law)
and L1 n L2 ( L1U L2)
But the Context-free languages are NOT closed
under intersection so the assumption is wrong.

85
Chapter 27 very end, pages 195 - 197
Important - know

Closure Properties of CFLs
The context-free language have a proper subset of
languages accepted by deterministic PDAs.
These PDAs can not use nondeterministic moves.
L a, b - ww w is in a, b
is a CFL but is NOT deterministic CLF

86
The Language/Machine World so far

CFL NPDA
Deterministic CFL DPDA
REGULAR FA NFA
Have examples And know subset relationships
87
The Language/Machine World so far

If L1 and L2 are both regular (or CF) what about
L3?

Important - know
88
Chapter 24 PDAs and CFG

The languages accepted by nondeterministic
pushdown automata are the context-free languages

Important fact, dont worry about the proof.
89
Chapter 24 PDAs and CFG

Start with CFG (N, S, P,S) and construct NPDA M
so that L(M) L(G)
Assume that all productions are in GNF so
A ? aB1B2 Bk if k0 then A ? a with a ? S U e
Construct M (Qq, S, N, d, q, S, ?)
q is the only state
S the set of terminals in G is the input
alphabet to M
N the set of nonterminals in G is the stack
alphabet
q is the start state
S the start symbol of G is the initial stack
symbol
F ? because M accepts by empty stack
d for each production A ? cB1B2 Bk add to the
transitions of M ((q,c,A),(q,B1B2 Bk ))

Know that a PDA can be made from a CFG in GNF but
I will give you the rules above if needed.
90
For each production A ? cB1B2 Bk add to the
transitions of the PDA ((q,c,A),(q,B1B2 Bk
))

Example Set PAREN is generated by CFG
S ? BS ((q, , S), (q, BS))
S ? B ((q, , S), (q, B))
S ? SB ((q, , S), (q, SB))
S ? SBS ((q, , S), (q, SBS))
B ? ((q, , B), (q, e))

Know that a PDA can be made from a CFG in GNF but
I will give you the rules above is needed.
91

Example
S ? BS ((q, , S), (q, BS))
? S ((q, , B), (q, e))
? SBS ((q, , S), (q, SBS))
? BBS ((q, , S), (q, B))
? BS ((q, , B), (q, e))
? S ((q, , B), (q, e))
? SB ((q, , S), (q, SB))
? BSB
? BSB
? SB
? BB
? B
?

S ? BS B SB SBS B ? ((q, , S),
(q, BS)) ((q, , S), (q, B)) ((q, , S), (q,
SB)) ((q, , S), (q, SBS)) ((q, , B), (q,
e)) List of sentential forms in a leftmost
derivation of w is the list of the PDAs
configurations for a computation on input
w. Very nondeterministic!
92

Very nondeterministic! Why?
A language L is Context-Free if it is generated
(derivable) by a context-free grammar G.
This means that
Every string in the language L is derivable from
the grammar G (there exists at least one
derivation of the string)
No string in L is derivable from G

93
For the Test and the Future