Module 12
• Computation and Configurations
– Formal Definition
– Examples
1
Definitions
• Configuration
– Functional Definition
• Given the original program and the current
configuration of a computation, someone should
be able to complete the computation
– Contents of a configuration for a C++ program
• current instruction to be executed
• current value of all variables
• Computation
– Complete sequence of configurations
2
Computation 1
1 int main(int x,y) {
2 int r = x % y;
3 if (r== 0) goto 8;
4 x = y;
5 y = r;
6 r = x % y;
7 goto 3;
8 return y; }
Input: 10 3
•
•
•
•
•
•
•
•
•
•
Line 1, x=?,y=?,r=?
Line 2, x=10, y=3,r=?
Line 3, x=10, y=3, r=1
Line 4, x=10, y=3, r=1
Line 5, x= 3, y=3, r=1
Line 6, x=3, y=1, r=1
Line 7, x=3, y=1, r=0
Line 3, x=3, y=1, r=0
Line 8, x=3, y=1, r=0
Output is 1
3
Computation 2
int main(int x,y) {
2 int r = x % y;
3 if (r== 0) goto 8;
4 x = y;
5 y = r;
6 r = x % y;
7 goto 3;
8 return y; }
Input: 53 10
•
•
•
•
•
•
•
•
•
Line 1, x=?,y=?,r=?
Line 2, x=53, y=10, r=?
Line 3, x= 53, y=10, r=3
Line 4, x=53, y=10, r=3
Line 5, x=10, y=10, r=3
Line 6, x=10, y=3, r=3
Line 7, x=10, y=3, r=1
Line 3, x=10, y=3, r=1
...
4
Computations 1 and 2
•
•
•
•
•
•
•
•
•
•
Line 1, x=?,y=?,r=?
Line 2, x=10, y=3,r=?
Line 3, x=10, y=3, r=1
Line 4, x=10, y=3, r=1
Line 5, x= 3, y=3, r=1
Line 6, x=3, y=1, r=1
Line 7, x=3, y=1, r=0
Line 3, x=3, y=1, r=0
Line 8, x=3, y=1, r=0
Output is 1
•
•
•
•
•
•
•
•
•
Line 1, x=?,y=?,r=?
Line 2, x=53, y=10, r=?
Line 3, x= 53, y=10, r=3
Line 4, x=53, y=10, r=3
Line 5, x=10, y=10, r=3
Line 6, x=10, y=3, r=3
Line 7, x=10, y=3, r=1
Line 3, x=10, y=3, r=1
...
5
Observation
int main(int x,y) {
2 int r = x % y;
3 if (r== 0) goto 8;
4 x = y;
5 y = r;
6 r = x % y;
7 goto 3;
8 return y; }
• Program and current
configuration
– Together, these two pieces of
information are enough to
complete the computation
– Are they enough to determine
what the original input was?
• No!
• Both previous inputs, 10 3 as
well as 53 10 eventually
reached the same
configuration (Line 3, x=10,
y=3, r=1)
• Line 3, x= 10, y=3, r=1
6
Module 13
• Studying the internal structure of REC, the
set of solvable problems
– Complexity theory overview
– Automata theory preview
• Motivating Problem
– string searching
7
Studying REC
Complexity Theory
Automata Theory
8
Current picture of all languages
All Languages
REC RE-REC
Solvable
Half
Solvable
All languages - RE
Not even half solvable
Which language class should be studied further?
9
Complexity Theory
REC RE - All languages
REC
- RE
• In complexity theory, we differentiate problems
by how hard a problem is to solve
– Remember, all problems in REC are solvable
• Which problem is harder and why?
– Max:
• Input: list of n numbers
• Task: return largest of the n numbers
– Element
• Input: list of n numbers
• Task: return any of the n numbers
10
Resource Usage *
• How do we formally measure the hardness
of a problem?
– We measure the resources required to solve
input instances of the problem
• Typical resources are?
• Need a notion of size of an input instance
– Obviously larger input instances require more resources to
solve
11
Poly Language
Class *
Poly
REC RE - All languages
REC
- RE
Rest of REC
Informal Definition: A problem L1 is easier than problem L2
if problem L1 can be solved in less time than problem L2.
Poly: the set of problems which can be solved in polynomial time
(typically referred to as P, not Poly)
Major goal: Identify whether or not a problem belongs to Poly
12
Working with Poly
Poly
Rest of REC
• How do you prove a problem L is in Poly?
• How do you prove a problem L is not in
Poly?
– We are not very good at this.
– For a large class of interesting problems, we
have techniques (polynomial-time answerpreserving input transformations) that show a
problem L probably is not in Poly, but few
which prove it.
13
Examples
• Shortest Path Problem
• Input
– Graph G
– nodes s and t
• Task
– Find length of shortest
path from s to t in G
Poly
Rest of REC
• Longest Path Problem
• Input
– Graph G
– nodes s and t
• Task
– Find length of longest
path from s to t in G
Which problem is provably solvable in polynomial time?
14
Automata Theory
REC RE - All languages
REC
- RE
• In automata theory, we will define new models of
computation which we call automata
– Finite State Automata (FSA)
– Pushdown Automata (PDA)
• Key concept
– FSA’s and PDA’s are restricted models of computation
• FSA’s and PDA’s cannot solve all the problems that C++
programs can
– We then identify which problems can be solved using
FSA’s and PDA’s
15
New language
classes
REC RE - All languages
REC
- RE
• REC is the set of solvable languages when we
start with a general model of computation like
C++ programs
• We want to identify which problems in REC can
be solved when using these restricted automata
Solvable Solvable
by FSA’s by
and PDA’s PDA’s
Rest of REC
16
Recap *
• Complexity Theory
– Studies structure of the set of solvable problems
– Method: analyze resources (processing time) used to
solve a problem
• Automata Theory
– Studies structure of the set of solvable problems
– Method: define automata with restricted capabilities
and resources and see what they can solve (and what
they cannot solve)
– This theory also has important implications in the
development of programming languages and compilers
17
Motivating Problem
String Searching
18
String Searching
• Input
– String x
– String y
• Tasks
– Return location of y in
string x
– Does string y occur in
string x?
• Can you identify
applications of this
type of problem in real
life?
• Try and develop a
solution to this
problem.
19
String Searching II
• Input
– String x
– pattern y
• Tasks
– Return location of y in
string x
– Does pattern y occur in
string x?
• Pattern
– [anything].html
– $EN4$$
• Try and develop a
solution to this
problem.
20
String Searching
• We will show an easy way to solve these
string searching problems
• In particular, we will show that we can
solve these problems in the following
manner
– write down the pattern
– the computer automatically turns this into a
program which performs the actual string
search
21
Module 14
• Regular languages
– Inductive definitions
– Regular expressions
• syntax
• semantics
22
Regular Languages
(Regular Expressions)
23
Regular Languages
• New language class
– Elements are languages
• We will show that this language class is identical
to LFSA
– Language class to be defined by Finite State Automata
(FSA)
– Once we have shown this, we will use the term “regular
languages” to refer to this language class
24
Inductive Definition of Integers *
• Base case definition
– 0 is an integer
• Inductive case definition
– If x is an integer, then
• x+1 is an integer
• x-1 is an integer
• Completeness
– Only numbers generated using the above rules are
integers
25
Inductive Definition of Regular Languages
• Base case definition
–
–
–
–
Let S denote the alphabet
{} is a regular language
{l} is a regular language
{a} is a regular language for any character a in S
• Inductive case definition
– If L1 and L2 are regular languages, then
• L1 union L2 is a regular language
• L1 concatenate L2 is a regular language
• L1* is a regular language
• Completeness
– Only languages generated using above rules are regular languages
26
Proving a language is regular *
• Prove that {aa, bb} is a regular language
– {a} and {b} are regular languages
• base case of definition
– {aa} = {a}{a} is a regular language
• concatenation rule
– {bb} = {b}{b} is a regular language
• concatenation rule
– {aa, bb} = {aa} union {bb} is a regular language
• union rule
• Typically, we will not go through this process to prove a language is
regular
27
Regular Expressions
• How do we describe a regular language?
– Use set notation
• {aa, bb, ab, ba}*
• {a}{a,b}*{b}
– Use regular expressions R
• Inductive def of regular languages and regular
expressions on page 72
• (aa+bb+ab+ba)*
• a(a+b)*b
28
R and L(R) *
• How we interpret a regular expression
– What does a regular expression R mean to us?
• aaba represents the regular language {aaba}
• f represents the regular language {}
• aa+bb represents the regular language {aa, bb}
– We use L(R) to denote the regular language
represented by regular expression R.
29
Precedence rules
• What is L(ab+c*)?
– Possible answers:
•
•
•
•
{a}({b} union {c}*}
({a}{b,c})*
({ab} union {c})*
{ab} union {c}*
– Must know precedence rules
• * first, then concatenation, then +
30
Precedence rules continued
• Precedence rules similar to those for
arithmetic expressions
– ab+c2
• (a times b) + (c times c)
• exponentiation first, then multiplication, then
addition
• Think of Kleene closure as exponentiation,
concatenation as multiplication, and union as
addition and the precedence rules are identical
31
Regular expressions are strings *
• Let L be a regular language over the
alphabet S
– A regular expression R for L is just a string
over the alphabet S union {(, ), +, *, f, l}
which follows certain syntactic rules
• That is, the set of legal regular expressions is itself a
language over the alphabet S union {(, ), +, *}
– f, a*aba are strings in the language of legal reg. exp.
– )(, *a* are strings NOT in the language of legal reg. exp.
32
Semantics *
• We give a regular expression R meaning when we
interpret it to represent L(R).
– aaba is just a string
– we interpret it to represent the language {aaba}.
• We do similar things with arithmetic expressions
– 10+72 is just a string
– We interpret this string to represent the number 59
33
Key fact *
• A language L is a regular language iff there
exists a reg. exp. R such that L(R) = L
– When I ask for a proof that a language L is
regular, rather than going through the inductive
proof we saw earlier, I expect you to give me a
regular expression R s.t. L(R) = L
34
Summary
• Regular expressions are strings
– syntax for legal regular expressions
– semantics for interpreting regular expressions
• Regular languages are a new language class
– A language L is regular iff there exists a regular
expression R s.t. L(R) = L
• We will show that the regular languages are
identical to LFSA
35
Module 15
• FSA’s
– Defining FSA’s
– Computing with FSA’s
• Defining L(M)
– Defining language class LFSA
– Comparing LFSA to set of solvable languages
(REC)
36
Finite State Automata
New Computational Model
37
Tape
• We assume that you have already seen
FSA’s in CSE 260
– If not, review material in reference textbook
• Only data structure is a tape
– Input appears on tape followed by a B character
marking the end of the input
– Tape is scanned by a tape head that starts at
leftmost cell and always scans to the right
38
Data type/States
• The only data type for an FSA is char
• The instructions in an FSA are referred to as
states
• Each instruction can be thought of as a
switch statement with several cases based
on the char being scanned by the tape head
39
Example program
1 switch(current tape cell) {
case a: goto 2
case b: goto 2
case B: return yes
}
2 switch (current tape cell) {
case a: goto 1
case b: goto 1
case B: return no;
}
40
New model of computation
1 switch(current tape cell) {
case a: goto 2
case b: goto 2
case B: return yes
}
2 switch (current tape cell) {
case a: goto 1
case b: goto 1
case B: return no;
}
• FSA M=(Q,S,q0,A,d)
– Q = set of states = {1,2}
– S = character set = {a,b}
• don’t need B as we see below
– q0 = initial state = 1
– A = set of accepting (final) states
= {1}
• A is the set of states where we
return yes on B
• Q-A is set of states that return no
on B
– d = state transition function
41
Textual representations of d *
1 switch(current tape cell) {
case a: goto 2
case b: goto 2
case B: return yes
d(1,a) = 2, d(1,b)=2,
d(2,a)=1, d(2,b) = 1
{(1,a,2), (1,b,2), (2,a,1), (2,b,1)}
}
2 switch (current tape cell) {
case a: goto 1
case b: goto 1
case B: return no;
}
1
2
a
b
2
1
2
1
42
Transition diagrams
1 switch(current tape cell) {
case a: goto 2
case b: goto 2
case B: return yes
}
2 switch (current tape cell) {
case a: goto 1
case b: goto 1
case B: return no;
}
a,b
2
1
a,b
Note, this transition diagram
represents all 5 components
of an FSA, not just the transition
function d
43
Exercise
• FSA M = (Q, S, q0, A, d)
–
–
–
–
–
Q = {1, 2, 3}
S = {a, b}
q0 = 1
A = {2,3}
d: {d(1,a) = 1, d(1,b) = 2, d(2,a)= 2, d(2,b) = 3,
d(3,a) = 3, d(3,b) = 1}
• Draw this FSA as a transition diagram
44
Transition Diagram
a
a
1
b
2
b
b
3
a
45
Computing with FSA’s
46
Computation Example *
a
a
1
b
2
b
b
3
a
Input: aabbaa
47
Computation of FSA’s in detail
• A computation of an FSA M on an input x is
a complete sequence of configurations
• We need to define
– Initial configuration of the computation
– How to determine the next configuration given
the current configuration
– Halting or final configurations of the
computation
48
Initial Configuration
a
a
b
1
2
b
b
3
FSA M
a
• Given an FSA M and an
input string x, what is the
initial configuration of the
computation of M on x?
– (q0,x)
– Examples
•
•
•
•
•
•
x = aabbaa
(1, aabbaa)
x = abab
(1, abab)
x=l
(1, l)
49
Definition of ├M
• (1, aabbaa) ├M (1, abbaa)
a
– config 1 “yields” config 2 in
one step using FSA M
3
• (1,aabbaa) ├ M (2, baa)
a
b
2
– config 1 “yields” config 2 in 3
steps using FSA M
• (1, aabbaa) ├* M (2, baa)
1
b
b
3
FSA M
a
– config 1 “yields” config 2 in 0
or more steps using FSA M
• Comment:
– ├ M determined by
transition function d
– There must always be one and
only one next configuration
• If not, M is not an FSA
50
Halting Configurations *
a
• Halting configuration
a
b
1
2
b
b
3
FSA M
a
– (q, l)
– Examples
• (1, l)
• (3, l)
• Accepting Configuration
– State in halting configuration is
in A
• Rejecting Configuration
– State in halting configuration is
not in A
51
a
a
FSA M on x
b
b
FSA M
b
a
• Two possibilities for M running on x
– M accepts x
• M accepts x iff the computation of M on x ends up in an
accepting configuration
• (q0, x) ├*M (q, l) where q is in A
– M rejects x
• M rejects x iff the computation of M on x ends up in a rejecting
configuration
• (q0, x) ├M (q, l) where q is not in A
*
– M does not loop or crash on x
• Why?
52
a
a
Examples
b
b
FSA M
b
a
– For the following input strings, does M
accept or reject?
• l
• aa
• aabba
• aab
• babbb
53
Definition of d*(q, x)
a
b
1
b
a
• Notation from the book
• d(q, c) = p
• dk(q, x) = p
2
– k is the length of x
b
3
FSA M
a
• d*(q, x) = p
• Examples
–
–
–
–
–
d(1, a) = 1
d(1, b) = 2
d4(1, abbb) = 1
d*(1, abbb) = 1
d*(2, baaaaa) = 3
54
a
L(M) and
LFSA *
• L(M) or Y(M)
a
b
b
FSA M
b
a
– The set of strings M accepts
• Basically the same as Y(P) from previous unit
– We say that M accepts/decides/recognizes/solves L(M)
• Remember an FSA will not loop or crash
– What is L(M) (or Y(M)) for the FSA M above?
• N(M)
– Rarely used, but it is the set of strings M rejects
• LFSA
– L is in LFSA iff there exists an FSA M such that L(M) = L.
55
LFSA Unit Overview
• Study limits of LFSA
– Understand what languages are in LFSA
• Develop techniques for showing L is in LFSA
– Understand what languages are not in LFSA
• Develop techniques for showing L is not in LFSA
• Prove Closure Properties of LFSA
• Identify relationship of LFSA to other
language classes
56
Comparing language classes
Showing LFSA is a subset of REC,
the set of solvable languages
57
LFSA subset REC
• Proof
– Let L be an arbitrary language in LFSA
– Let M be an FSA such that L(M) = L
• M exists by definition of L in LFSA
–
–
–
–
Construct C++ program P from FSA M
Argue P solves L
There exists a C++ program P which solves L
L is solvable
58
Visualization
•Let L be an arbitrary language in LFSA
•Let M be an FSA such that L(M) = L
•M exists by definition of L in LFSA
•Construct C++ program P from FSA M
•Argue P solves L
•There exists a program P which solves L
•L is solvable
L
L
LFSA
M
FSA’s
REC
P
C++ Programs
59
Construction
FSA
M
Program
Construction
P
Algorithm
• The construction is an algorithm which
solves a problem with a program as input
– Input to A: FSA M
– Output of A: C++ program P such that P solves
L(M)
– How do we do this?
60
Comparing computational
models
• The previous slides show one method for
comparing the relative power of two different
computational models
– Computational model CM1 is at least as general or
powerful as computational model CM2 if
• Any program P2 from computational model CM2 can be
converted into an equivalent program P1 in computational
model CM1.
– Question: How can we show two computational models
are equivalent?
61
Module 16
• Distinguishability
– Definition
– Help in designing/debugging FSA’s
62
Distinguishability
63
Questions
• Let L be the set of strings over {a,b} which end with aaba.
• Let M be an FSA such that L(M) = L.
• Questions
– Can aaba and aab end up in the same state of M? Why or why
not?
– How about aa and aab?
– How about l or a?
– How about b or bb?
– How about l or bbab?
64
Definition *
• String x is distinguishable from string y
with respect to language L iff there exists a
string z such that
– xz is in L and yz is not in L OR
– xz is not in L and yz is in L
• When reviewing, identify the z for pair of
strings on the previous slide
65
Questions
• Let L be the set of strings over {a,b} that have
length 2 mod 5 or 4 mod 5.
• Let M be an FSA such that L(M) = L.
• Questions
– Are aa and aab distinguishable with respect to L? Can
they end up in the same state of M?
– How about aa and aaba?
– How about l and a?
– How about b and aabbaa?
66
Design an FSA to accept L
• L = set of strings x
over {a,b} such that
length of x is 2 or 4
mod 5
• One design method
– Is l in L?
• Implication?
– Is a distinguishable
from l wrt L?
• Implication?
– Is b distinguishable
from l wrt L?
• Implication?
– Is b distinguishable
from a wrt L?
• Implication?
67
Design an FSA to accept L
• L = set of strings x
over {a,b} such that
length of x is 2 or 4
mod 5
• Design continued
– Is aa distinguishable
from l wrt L?
• Implication?
– Is aa distinguishable
from a wrt L?
• Implication?
68
Design an FSA to accept L
• L = set of strings x
over {a,b} such that
length of x is 2 or 4
mod 5
• Design continued
– What strings would we
compare ab to?
– What results do we
get?
– Implications?
– How about ba?
– How about bb?
69
Design an FSA to accept L
• L = set of strings x
over {a,b} such that
length of x is 2 or 4
mod 5
• Design continued
– We can continue in this
vein, but it could go on
forever
– Now lets try something
different
– Consider string l.
• What set of strings are
indistinguishable from
it wrt L?
• Implications?
70
Design an FSA to accept L
• L = set of strings x
over {a,b} such that
length of x is 2 or 4
mod 5
• Design continued
– Consider string a.
• What set of strings are
indistinguishable from
it wrt L?
• Implications?
– Consider string aa.
• What set of strings are
indistinguishable from
it wrt L?
• Implications?
71
Debugging an FSA
• Do essentially the same thing
– Identify some strings which end up in each state
– Try and generalize each state to describe the
language of strings which end up at that state.
72
Example 1
a,b
I
a
a
II
III
b
IV
a
V
b
b
a
b
VI
a,b
73
Example 2
b
I
a
a
a
II
III
a,b
b
IV
a
V
b
b
74
Example 3
a,b
a
II
b
I
a
IV
a
b
III
b
75
Module 17
• Closure Properties of Language class LFSA
– Remember ideas used in solvable languages
unit
– Set complement
– Set intersection, union, difference, symmetric
difference
76
LFSA is closed under set
complement
• If L is in LFSA, then Lc is in LFSA
• Proof
– Let L be an arbitrary language in LFSA
– Let M be the FSA such that L(M) = L
• M exists by definition of L in LFSA
–
–
–
–
Construct FSA M’ from M
Argue L(M’) = Lc
There exists an FSA M’ such that L(M’) = Lc
Lc is in LFSA
77
Visualization
L
Lc
•Let L be an arbitrary language in LFSA
•Let M be the FSA such that L(M) = L
•M exists by definition of L in
LFSA
•Construct FSA M’ from M
•Argue L(M’) = Lc
•Lc is in LFSA
LFSA
M
M’
FSA’s
78
Construct FSA M’ from M
• What did we do when we proved that REC, the set
of solvable languages, is closed under set
complement?
• Construct program P’ from program P
• Can we translate this to the FSA setting?
79
Construct FSA M’ from M
• M = (Q, S, q0, A, d)
• M’ = (Q’, S’, q’, A’, d’)
– M’ should say yes when M says no
– M’ should say no when M says yes
– How?
• Q’ = Q
• S’ = S
• q’ = q0
• d’ = d
• A’ = Q-A
80
Q’ = Q
S’ = S
q’ = q0
d’ = d
A’ = Q-A
Example
a
a
1
b
2
b
b
a
a
1
b
2
b
3
FSA M
a
b
3
FSA M’
a
81
Construction is
an algorithm *
FSA
M
FSA
Construction
M’
Algorithm
• Set Complement Construction
– Algorithm Specification
• Input: FSA M
• Output: FSA M’ such that L(M’) = L(M)c
– Comments
• This algorithm can be in any computational model.
– It does not have to be (and typically is not) an FSA
• These set closure constructions are useful.
– More on this later
82
Specification of
the algorithm
FSA
M
FSA
Construction
M’
Algorithm
• Your algorithm must give a complete specification of M’ in
terms of M
– Example:
• Let input FSA M = (Q, S, q0, A, d)
• Output FSA M’ = (Q’, S’, q’, A’, d’) where
– Q’ = Q
– S’ = S
– q’ = q0
– d’ = d
– A’ = Q-A
• When I ask for such a construction algorithm specification, this type
of answer is what I am looking for. Further algorithmic details on
how such an algorithm would work are unnecessary.
83
LFSA closed under Set
Intersection Operation
(also set union, set difference, and symmetric difference)
84
LFSA closed under set
intersection operation *
• Let L1 and L2 be arbitrary languages in LFSA
• Let M1 and M2 be FSA’s s.t. L(M1) = L1, L(M2) =
L2
– M1 and M2 exist by definition of L1 and L2 in LFSA
•
•
•
•
Construct FSA M3 from FSA’s M1 and M2
Argue L(M3) = L1 intersect L2
There exists FSA M3 s.t. L(M3) = L1 intersect L2
L1 intersect L2 is in LFSA
85
Visualization
•Let L1 and L2 be arbitrary languages in
LFSA
•Let M1 and M2 be FSA’s s.t. L(M1) = L1,
L(M2) = L2
•M1 and M2 exist by definition of L1
and L2 in LFSA
•Construct FSA M3 from FSA’s M1 and M2
•Argue L(M3) = L1 intersect L2
•There exists FSA M3 s.t. L(M3) = L1
intersect L2
•L1 intersect L2 is in LFSA
L1
L1 intersect L2
L2
LFSA
M1
M3
M2
FSA’s
86
Algorithm Specification
• Input
– Two FSA’s M1 and M2
• Output
– FSA M3 such that L(M3) = L(M1) intersection L(M2)
FSA M1
FSA M2
Alg
FSA M3
87
Use Old Ideas
FSA M1
FSA M2
Alg
FSA M3
• Key concept: Try ideas from previous closure
property proofs
• Example
– How did the algorithm that was used to prove that REC is
closed under set intersection work?
– If we adapt this approach, what should M3 do with
respect to M1, M2, and the input string?
88
Run M1 and M2
Simultaneously
0
l
Alg
FSA M3
1
1
0
FSA M1
FSA M2
0
1
0
1
0
2
l,A
0,A
1,A
2,A
l,B
0,B
1,B
2,B
1
1
A
M1
0
M2
0,1
B
M3
What happens when M1 and M2 run on input string 11010?
89
0
Construction *
• Input
– FSA M1 = (Q1, S1, q1, d1, A1)
– FSA M2 = (Q2, S2, q2, d2, A2)
l
1
0
0
1
1
1
1
1
M1
0,1
A
0
B
M2
0
0
2
• Output
– FSA M3 = (Q3, S3, q3, d3, A3)
– What is Q3?
• Q3 = Q1 X Q2 where X is cartesian product
• In this case, Q3 = {(l,A), (l,B), (0,A), (0,B), (1,A), (1,B), (2,A), (2,B)}
– What is S3?
• S3 = S1 = S 2
• In this case, S3 = {0,1}
90
0
Construction *
• Input
– FSA M1 = (Q1, S1, q1, d1, A1)
– FSA M2 = (Q2, S2, q2, d2, A2)
l
1
0
0
1
1
1
1
1
M1
0,1
A
0
B
M2
0
0
2
• Output
– FSA M3 = (Q3, S3, q3, d3, A3)
– What is q3?
• q3 = (q1, q2)
• In this case, q3 = (l,A)
– What is A3?
• A3 = {(p, q) | p in A1 and q in A2}
• In this case, A3 = {(0,B)}
91
0
Construction
• Input
– FSA M1 = (Q1, S1, q1, d1, A1)
– FSA M2 = (Q2, S2, q2, d2, A2)
l
1
0
0
1
1
1
1
1
M1
0,1
A
0
B
M2
0
0
2
• Output
– FSA M3 = (Q3, S3, q3, d3, A3)
– What is d3?
• For all p in Q1, q in Q2, a in S, d3((p,q),a) = (d1(p,a),d2(q,a))
• In this case,
– d3((0,A),0) = (d1(0,0),d2(A,0))
–
= (0,B)
– d3((0,A),1) = (d1(0,1),d2(A,1))
–
= (1,A)
92
Example Summary
0
l
1
0
1
1
0
1
0
1
0
2
l,A
1
1
A
0,A
0
M1
0
M2
1
1,A
0
2,A
0
1
0
0,1
B
1
l,B
0
0,B
1
1
1,B
0
2,B
M3
93
Observation
• Input
– FSA M1 = (Q1, S1, q1, d1, A1)
– FSA M2 = (Q2, S2, q2, d2, A2)
• Output
– FSA M3 = (Q3, S3, q3, d3, A3)
– What is A3?
• A3 = {(p, q) | p in A1 and q in A2}
• What if operation were different?
– Set union, set difference, symmetric difference
94
Observation continued *
• Input
– FSA M1 = (Q1, S1, q1, d1, A1)
– FSA M2 = (Q2, S2, q2, d2, A2)
• Output
– FSA M3 = (Q3, S3, q3, d3, A3)
– What is A3?
•
•
•
•
Set intersection: A3 = {(p, q) | p in A1 and q in A2}
Set union: A3 = {(p, q) | p in A1 or q in A2}
Set difference: A3 = {(p, q) | p in A1 and q not in A2}
Symmetric difference: A3 = {(p, q) | (p in A1 and q not in A2)
or (p not in A1 and q in A2) }
95
Observation conclusion
• LFSA is closed under
–
–
–
–
set intersection
set union
set difference
symmetric difference
• The constructions used to prove these closure
properties are essentially identical
96
Comments *
• You should be able to execute this algorithm
– Convert two FSA’s into a third FSA with the correct
properties.
• You should understand the idea behind this
algorithm
– The third FSA essentially runs both input FSA’s
simultaneously on any input string
– How we set A3 depending on the specific set operation
• You should understand how this algorithm can be
used to simplify design of FSA’s
• You should be able to construct new algorithms for
new closure property proofs
97
Comparison *
L1
L1 intersect L2
L
L
L2
LFSA
REC
LFSA
M1
M3
M2
M
FSA’s
P
C++ Programs
FSA’s
98
Module 18
• NFA’s
– nondeterministic transition functions
• computations are trees, not paths
– L(M) and LNFA
• LFSA subset of LNFA
– Comparing computational models
99
Nondeterministic Finite State
Automata
NFA’s
100
Change: d is a relation
• For an FSA M, d(q,a) results in one and
only one state for all states q and characters
a.
– That is, d is a function
• For an NFA M, d(q,a) can result in a set of
states
– That is, d is now a relation
– Next step is not determined (nondeterministic)
101
Example NFA
a,b
a,b
a
a
b
a
•Why is this only an NFA and not an FSA? Identify
as many reasons as you can.
102
Computing with NFA’s
• Configurations: same as they are for FSA’s
• Computations are different
– Initial configuration is identical
– However, there may be several next configurations or
there may be none.
• Computation is no longer a “path” but is now a “graph” (often
a tree) rooted at the initial configuration
– Definition of halting, accepting, and rejecting
configurations is identical
– Definition of acceptance must be modified
103
a,b
Computation
Graph (Tree)
a,b
a
a
b
a
Input string aaaaba
(1, aaaaba)
(1, aaaba)
(2, aaaba)
(1, aaba)
(2, aaba)
(3, aaba)
(1, aba)
(2, aba)
(3, aba)
(1, ba)
(2, ba)
(1, a)
(1, l)
crash
(3, ba)
(4, a)
(2, l)
(5, l)
104
a,b
Definition of ├
unchanged *
a
a
b
a
Input string aaaaba
(1, aaaaba)├ (1, aaaba)
(1, aaaaba)
(1, aaaba)
a,b
(1, aaaaba)├ (2, aaaba)
(2, aaaba)
(1, aaaaba)├3 (1, aba)
(1, aaba)
(2, aaba)
(1, aaaaba)├3 (3, aba)
(3, aaba)
(1, aaaaba)├* (2, aba)
(1, aba)
(2, aba)
(1, ba)
(2, ba)
(3, aba)
crash
(1, aaaaba)├* (3, aba)
(1, aaaaba)├* (1, l)
(3, ba)
(1, aaaaba)├* (5, l)
(1, a)
(1, l)
(4, a)
(2, l)
(5, l)
105
a,b
a
Acceptance and
Rejection
(1, aaba)
(1, aba)
(2, aba)
(1, ba)
(2, ba)
(1, a)
(1, l)
b
a
M accepts string x if one of the
configurations reached is an
accepting configuration
(2, aaaba)
(2, aaba)
a
Input string aaaaba
(1, aaaaba)
(1, aaaba)
a,b
(q0, x) ├* (f, l),f in A
(3, aaba)
(3, aba)
crash
(3, ba)
M rejects string x if all
configurations reached are
either not halting
configurations or are rejecting
configurations
(4, a)
(2, l)
(5, l)
106
Comparison
b
a
a
a
a,b
b
a
b
b
FSA
a,b
a,b
a
a
b
a
NFA
107
Defining L(M) and LNFA
• M accepts string x if one
of the configurations
reached is an accepting
configuration
– (q0, x) |-* (f, l),f in A
• M rejects string x if all
configurations reached are
either not halting
configurations or are
rejecting configurations
• L(M) (or Y(M))
• N(M)
• LNFA
– Language L is in language class
LNFA iff
108
Comparing language classes
LFSA subset of LNFA
109
LFSA subset LNFA
• Let L be an arbitrary language in LFSA
• Let M be the FSA such that L(M) = L
– M exists by definition of L in LFSA
•
•
•
•
Construct an NFA M’ such that L(M’) = L
Argue L(M’) = L
There exists an NFA M’ such that L(M’) = L
L is in LNFA
– By definition of L in LNFA
110
Visualization
•Let L be an arbitrary language in LFSA
•Let M be an FSA such that L(M) = L
•M exists by definition of L in LFSA
•Construct NFA M’ from FSA M
•Argue L(M’) = L
•There exists an NFA M’ such that L(M’)
=L
•L is in LNFA
L
LFSA
M
FSA’s
L
LNFA
M’
NFA’s
111
Construction
• We need to make M into an NFA M’ such
that L(M’) = L(M)
• How do we accomplish this?
112
Module 19
• LNFA subset of LFSA
– Theorem 4.1 on page 131 of Martin textbook
– Compare with set closure proofs
• Main idea
– A state in FSA represents a set of states in
original NFA
113
LNFA subset LFSA
• Let L be an arbitrary language in
• Let M be
– M exists by definition of
•
•
•
•
Construct an M’ such that L(M’)
Argue L(M’) =
There exists an M’ such that L(M’) =
L is in
– By definition of
114
Visualization
•Let L be an arbitrary language in LNFA
•Let M be an NFA such that L(M) = L
•M exists by definition of L in LNFA
•Construct FSA M’ from NFA M
•Argue L(M’) = L
•There exists an FSA M’ such that L(M’) =
L
•L is in LFSA
L
LNFA
M
NFA’s
L
LFSA
M’
FSA’s
115
Construction Specification
• We need an algorithm which does the
following
– Input: NFA M
– Output: FSA M’ such that L(M’) = L(M)
116
a,b
a,b
a
Difficulty *
(2, aaaba)
(1, aaba)
(2, aaba)
(3, aaba)
(1, aba)
(2, aba)
(3, aba)
(1, ba)
(2, ba)
(1, a)
(1, l)
b
a
Input string aaaaba
• An NFA can be in
several states after
processing an input
string x
(1, aaaaba)
(1, aaaba)
a
crash
(3, ba)
(4, a)
(2, l)
(5, l)
117
a,b
a,b
a
Observation *
(2, aaaba)
(1, aaba)
(2, aaba)
(3, aaba)
(1, aba)
(2, aba)
(3, aba)
(1, ba)
(2, ba)
(1, a)
(1, l)
b
a
Input string aaaaba
• All strings which end up
in the set of states {1,2,3}
are indistinguishable with
respect to L(M)
(1, aaaaba)
(1, aaaba)
a
crash
(3, ba)
(4, a)
(2, l)
(5, l)
118
a,b
a,b
a
Idea
a
b
a
Input string aaaaba
• Given an NFA M = (Q,S,q0,d,A), the equivalent FSA M’
will have state set 2Q (one state for each subset of Q)
• Example
– In this case there are 5 states in Q
– 2Q, the set of all subsets of Q, has 25 elements including {} and Q
– The FSA M’ will have 25 states
• What strings end up in state {1,2,3} of M’?
– The strings which end up in states 1, 2, and 3 of NFA M.
– In this case, strings which do not contain aaba and end with aa
such as aa, aaa, and aaaa.
119
a,b
a,b
a
Idea Illustrated
(1,aaaaba)
a
b
Input string aaaaba
({1}, aaaaba)
(1, aaaba)
(2, aaaba)
(1, aaba)
(2, aaba)
(3, aaba)
({1,2,3}, aaba)
(1, aba)
(2, aba)
(3, aba)
({1,2,3}, aba)
(1, ba)
(2, ba)
(3, ba)
({1,2,3}, ba)
(1, a)
(1, l)
a
(2, l)
({1,2}, aaaba)
(4, a)
({1,4}, a)
(5, l)
({1,2,5}, l)
120
a,b
Construction
a,b
NFA M
1
a
2
a
3
• Input NFA M = (Q, S, q0, d, A)
• Output FSA M’ = (Q’, S’, q’, d’, A’)
– What is Q’?
• all subsets of Q including Q and {}
• In this case, Q’ =
– What is S’?
• We always make S’ = S
• In this case, S’ = S = {a,b}
– What is q’?
• We always make q’ = {q0}
• In this case q’ =
121
a,b
Construction
a,b
NFA M
1
a
2
a
3
• Input NFA M = (Q, S, q0, d, A)
• Output FSA M’ = (Q’, S’, q’, d’, A’)
– What is A’?
• Suppose a string x ends up in states 1 and 2 of the NFA M above.
– Is x accepted by M?
– Should {1,2} be an accepting state in FSA M’?
• Suppose a string x ends up in states 1 and 2 and 3 of the NFA M above.
– Is x accepted by M?
– Should {1,2,3} be an accepting state in FSA M’?
• Suppose p = {q1, q2, …, qk} where q1, q2, …, qk are in Q
• p is in A’ iff at least one of the states q1, q2, …, qk is in A
• In this case, A’ =
122
a,b
Construction
a,b
NFA M
1
a
2
a
3
• Input NFA M = (Q, S, q0, d, A)
• Output FSA M’ = (Q’, S’, q’, d’, A’)
– What is d’?
• If string x ends up in states 1 and 2 after being processed by the
NFA above, where does string xa end up after being processed by
the NFA above?
• Figuring out d’(p,a) in general
– Suppose p = {q1, q2, …, qk} where q1, q2, …, qk are in Q
– Then d’(p,a) = d(q1,a) union d(q2,a) union … union d(qk,a)
» Similar to 2 FSA to 1 FSA construction
– In this case
» d’({1,2},a) =
123
Construction
Summary
a,b
a,b
NFA M
1
a
2
a
3
• Input NFA M = (Q, S, q0, d, A)
• Output FSA M’ = (Q’, S’, q’, d’, A’)
– Q’ = all subsets of Q including Q and {}
• In this case, Q’ = {{}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}}
–
S’ = S
• In this case, S’ = S = {a,b}
– q’ ={q0}
• In this case, q’ = {1}
– A’
• Suppose p = {q1, q2, …, qk} where q1, q2, …, qk are in Q
• p is in A’ iff at least one of the states q1, q2, …, qk is in A
– d’
• Suppose p = {q1, q2, …, qk} where q1, q2, …, qk are in Q
• Then d’(p,a) = d(q1,a) union d(q2,a) union … union d(qk,a)
124
Example Summary
a
b
a,b
1
a,b
a
2
a
3
a
{1}
b
b
{1,2} a {1,2,3}
a,b
NFA M
{}
b
a
{1,3}
a,b
b
{2}
a
{3}
a,b
{2,3}
FSA M’
125
Example Summary Continued
a
b
a,b
1
a,b
a
2
a
3
a
{1}
b
b
{1,2} a {1,2,3}
a,b
NFA M
{}
b
a
{1,3}
a,b
b
{2}
a
{3}
a,b
{2,3}
FSA M’
These states cannot be reached from initial state and are unnecessary.
126
Example Summary Continued
a
b
a,b
1
a,b
a
2
NFA M
a
3
a
{1}
b
b
b
{1,2} a {1,2,3}
a
{1,3}
Smaller FSA M’
By examination, we can see that state {1,3} is unnecessary.
However, this is a case by case optimization.
It is not a general technique or algorithm.
127
Example 2
a,b
A
a
B
b
C
NFA M
Step 1: name the three states of NFA M
128
a,b
Step 2:
transition table
A
a
B
b
C
NFA M
{A}
{B}
{}
{B,C}
a
{B}
{B}
b
{}
{B,C}
{}
{}
{B} {B,C}
d’({B,C},a) = d(B,a) U d(C,a)
= {B} U {}
= {B}
d’({B,C},b) = d(B,b) U d(C,b)
= {B,C} U {}
= {B,C}
129
a,b
Step 3:
accepting states
A
a
B
b
C
NFA M
{A}
{B}
{}
{B,C}
a
{B}
{B}
b
{}
{B,C}
{}
{}
{B} {B,C}
Which states should be accepting?
Any state which includes an
accepting state of M, in this case, C.
A’ = {{B,C}}
130
a,b
Step 4: Answer
A
a
B
b
C
NFA M
{A}
{B}
{}
{B,C}
a
{B}
{B}
b
{}
{B,C}
Initial state is {A}
Set of final states A’ = {{B,C}}
{}
{}
{B} {B,C}
This is sufficient. You do NOT need to turn this into a diagram.
131
a,b
Step 5:
Optional
a
A
a
b
{B}
b
C
NFA M
b
a
{A}
B
b
a
{B,C}
a,b
{}
FSA M’
132
Comments
• You should be able to execute this algorithm
– You should be able to convert any NFA into an equivalent FSA.
• You should understand the idea behind this algorithm
– For an FSA M’, strings which end up in the same state of M’ are
indistinguishable wrt L(M’)
– For an NFA M, strings which end up in the same set of states of M
are indistinguishable wrt L(M)
133
Comments
• You should understand the importance of this algorithm
– Design tool
• We can design using NFA’s
• A computer will convert this NFA into an equivalent FSA
– FSA’s can be executed by computers whereas NFA’s cannot (or at least
cannot easily be run by computers)
– Chaining together algorithms
• Perhaps it is easy to build NFA’s to accept L1 and L2
• Use this algorithm to turn these NFA’s to FSA’s
• Use previous algorithm to build FSA to accept L1 intersect L2
• You should be able to construct new algorithms for new
closure property proofs
134
Module 20
• NFA’s with l-transitions
– NFA-l’s
• Formal definition
• Simplifies construction
– LNFA-l
– Showing LNFA-l is a subset of LNFA (extra
credit)
• and therefore a subset of LFSA
135
Defining NFA-l’s
136
Change: l-transitions
• We now allow an NFA M to change state
without reading input
• That is, we add the following categories of
transitions to d
– d(q,l) is allowed
137
Example *
a,b
l
l
a,b
a,b
a
a
b
a
b
a
a
b
a,b
138
Defining L(M) and LNFA-l
• M accepts string x if one
of the configurations
reached is an accepting
configuration
– (q0, x) |-* (f, l),f e A
• M rejects string x if all
configurations reached are
either not halting
configurations or are
rejecting configurations
• L(M) or Y(M)
• N(M)
• LNFA-l
– Language L is in
language class LNFAl iff
139
LNFA-l subset LFSA
• Recap of what we already know
– Let M be any NFA
– There exists an algorithm A1 which constructs an FSA
M’ such that L(M’) = L(M)
• New goal
– Let M be any NFA-l
– There exists an algorithm A2 which constructs an FSA
M’ such that L(M’) = L(M)
140
Visualization
• Goal
– Let M be any NFA-l
– There exists an algorithm A2 which constructs an FSA
M’ such that L(M’) = L(M)
NFA-l M
A2
FSA M’
141
Modified Goal
NFA-l M
A2
FSA M’
• Question
– Can we use any existing algorithms to simplify the task
of developing algorithm A2?
• Yes, we can use algorithm A1 which converts an NFA M1 into
an FSA M’ such that L(M’) = L(M1)
NFA-l M
A2’
NFA
M1
A1
FSA M’
Algorithm A2
142
New Goal
(extra credit)
A2’
NFA-l M
NFA M1
• Difficulty
– NFA-l M can make transitions on l
– How can the NFA M1 simulate these l-transitions?
a
1
l
2
b
3
l
4
l
5
b
6
143
Basic Idea
A2’
NFA-l M
NFA M1
• For each state q of M and each character a of S, figure out which states
are reachable from q taking any number of l-transitions and exactly one
transition on that character a.
• In the NFA-d M1, directly connect q to each of these states using an arc
labeled with a.
a
1
l
2
b
3
l
4
b
2
3
5
b
6
b
b
1
l
4
5
6
144
Process State 2
NFA-d M1
A2’
NFA-l M
a
1
l
2
b
3
l
4
b
1
2
b
b
5
b
6
b
b
3
l
4
5
6
b
145
Process State 3
NFA-d M1
A2’
NFA-l M
a
1
l
2
b
3
l
4
b
1
2
b
a
b
5
b
6
b
a
b
3
l
4
5
6
b
b
146
Final Picture
NFA-d M1
A2’
NFA-l M
a
1
l
2
b
3
l
4
b
1
l
a
2
b
b
3
a
b
4
a
5
b
a
5
b
6
b
6
b
b
b
147
a
Construction
1
l
2
b
3
l
4
l
5
b
• Input NFA-l M = (Q, S, q0, d, A)
• Output NFA M1 = (Q1, S1, q1, d1, A1)
– What is Q1?
• Q1 = Q
• In this case, Q1 = {1,2,3,4,5,6}
– What is S1?
• S1 = S
• In this case, S1 = S = {a,b}
– What is q1?
• We always make q1 = q0
• In this case q1 = 1
148
6
a
Construction
1
l
2
b
3
l
4
l
5
b
6
• Input NFA-l M = (Q, S, q0, d, A)
• Output NFA M1 = (Q1, S1, q1, d1, A1)
– What is d1?
• d1(q,a) = the set of states reachable from state q in M taking any number
of l-transitions and exactly one transition on the character a
– More on this later
• In this case
– d1(1,a) = {}
– d1(1,b) = {3,4,5}
– What is A1?
• A1 = A with one minor change
– If an accepting state is reachable from q0 using only l-transitions, then we
make q1 an element of A1
• In this case, using only l-transitions, no accepting state is reachable
from q0, so A1 = A
149
Computing
d1(q,a)
•
a
1
l
2
b
3
l
4
l
5
b
d1(q,a) = the set of states reachable from state q in M taking
0 or more l-transitions and exactly one transition on the
character a
– Break this down into three steps
• First compute all states reachable from q using 0 or more l-transitions
– We call this set of states L(q)
• Next, compute all states reachable from any element of L(q) using the
character a
– We can denote these states as d(L(q),a)
• Finally, compute all states reachable from states in d(L(q),a) using 0 or
more l-transitions
– We denote these states as L(d(L(q),a))
– This is the desired answer
150
6
a
Example
•
1
l
2
b
3
l
4
l
5
b
d1(1,b) = {3,4,5}
– Compute L(1), all states reachable from state 1 using 0 or more ltransitions
• L(1) = {1,2}
– Compute d(L(1),b), all states reachable from any element L(1) of
using the character b:
• d(L(1),b) = d({1,2},b)
•
= d(1,b) U d(2,b)
•
= {} U {3} = {3}
– Compute L(d(L(1),b)), all states reachable from states in d(L(1),b)
using 0 or more l-transitions
• L(d(L(1),b)) = L(3)
•
= {3,4,5}
151
6
Comments
• For extra credit, you should be able to execute this
algorithm
– Convert any NFA-l into an equivalent NFA.
• For extra credit, you should understand the idea behind
this algorithm
– Why the transition function is computed the way it is
– Why A1 may need to include q1 in some cases
• You should understand the importance of this algorithm
– Compiler role again
– Use in combination with previous algorithm for converting any
NFA into an equivalent FSA to create a new algorithm for
converting any NFA-l into an equivalent FSA
152
LNFA-l = LFSA
• Implications
– Let us primarily use the term LFSA to refer to
this language class
– Given a language L is in LFSA
• We know there exists an FSA M s.t. L(M) = L
• We know there exists an NFA M s.t. L(M) = L
– To show a language L is in LFSA
• Show there exists an FSA M s.t. L(M) = L
• Show there exists an NFA-l M s.t. L(M) = L
153
Module 21
• Closure Properties for LFSA using NFA’s
– From now on, when I say NFA, I mean any
NFA including an NFA-l unless I add a specific
restriction
– union (second proof)
– concatenation
– Kleene closure
154
LFSA closed under set union
(again)
155
LFSA closed under set union
• Let L1 and L2 be arbitrary languages in LFSA
• Let M1 and M2 be NFA’s s.t. L(M1) = L1, L(M2) = L2
– M1 and M2 exist by definition of L1 and L2 in LFSA and the fact
that every FSA is an NFA
•
•
•
•
Construct NFA M3 from NFA’s M1 and M2
Argue L(M3) = L1 union L2
There exists NFA M3 s.t. L(M3) = L1 union L2
L1 union L2 is in LFSA
156
Visualization
•Let L1 and L2 be arbitrary languages in LFSA
•Let M1 and M2 be NFA’s s.t. L(M1) = L1, L(M2) =
L2
•M1 and M2 exist by definition of L1 and L2
in LFSA and the fact that every FSA is an
NFA
•Construct NFA M3 from NFA’s M1 and M2
•Argue L(M3) = L1 union L2
•There exists NFA M3 s.t. L(M3) = L1 union L2
•L1 union L2 is in LFSA
L1
L1 union L2
L2
LFSA
M1
M3
M2
NFA’s
157
Algorithm Specification
• Input
– Two NFA’s M1 and M2
• Output
– NFA M3 such that L(M3) = ?
NFA M1
NFA M2
A
NFA M3
158
Use l-transition
NFA M1
NFA M2
a
A
NFA M3
a
l
M1
a,b
a,b
a,b
l
a,b
a,b
a,b
M2
M3
159
General Case *
M1
NFA M1
NFA M2
A
NFA M3
l
l
M2
M3
160
Construction *
NFA M1
NFA M2
A
NFA M3
• Input
– NFA M1 = (Q1, S1, q1, d1, A1)
– NFA M2 = (Q2, S2, q2, d2, A2)
• Output
– NFA M3 = (Q3, S3, q3, d3, A3)
– What is Q3?
• Q3 =
– What is S3?
•
S3 = S1 = S 2
– What is q3?
• q3 =
161
Construction
NFA M1
NFA M2
A
NFA M3
• Input
– NFA M1 = (Q1, S1, q1, d1, A1)
– NFA M2 = (Q2, S2, q2, d2, A2)
• Output
– NFA M3 = (Q3, S3, q3, d3, A3)
– What is A3?
• A3 =
– What is d3?
•
d3 =
162
Comments
• You should be able to execute this algorithm
• You should understand the idea behind this algorithm
• You should understand how this algorithm can be used to
simplify design
• You should be able to design new algorithms for new
closure properties
• You should understand how this helps prove result that
regular languages and LFSA are identical
– In particular, you should understand how this is used to construct
an NFA M from a regular expression r s.t. L(M) = L(r)
– To be seen later
163
LFSA closed under set
concatenation
164
LFSA closed under set
concatenation
• Let L1 and L2 be arbitrary languages in LFSA
• Let M1 and M2 be NFA’s s.t. L(M1) = L1, L(M2) = L2
– M1 and M2 exist by definition of L1 and L2 in LFSA and the fact
that every FSA is an NFA
•
•
•
•
Construct NFA M3 from NFA’s M1 and M2
Argue L(M3) = L1 concatenate L2
There exists NFA M3 s.t. L(M3) = L1 concatenate L2
L1 concatenate L2 is in LFSA
165
Visualization
• Let L1 and L2 be arbitrary
languages in LFSA
• Let M1 and M2 be NFA’s s.t.
L(M1) = L1, L(M2) = L2
– M1 and M2 exist by definition
of L1 and L2 in LFSA and the
fact that every FSA is an NFA
• Construct NFA M3 from NFA’s
M1 and M2
• Argue L(M3) = L1 concatenate L2
• There exists NFA M3 s.t. L(M3) =
L1 concatenate L2
• L1 concatenate L2 is in LFSA
L1
L1 concatenate L2
L2
LFSA
M1
M3
M2
NFA’s
166
Algorithm Specification
• Input
– Two NFA’s M1 and M2
• Output
– NFA M3 such that L(M3) =
NFA M1
NFA M2
A
NFA M3
167
Use l-transition
a
NFA M1
NFA M2
A
NFA M3
a
M1
l
a,b
a,b
a,b
a,b
a,b
a,b
M2
M3
168
General Case
M1
NFA M1
NFA M2
A
NFA M3
l
l
M2
M3
169
Construction
NFA M1
NFA M2
A
NFA M3
• Input
– NFA M1 = (Q1, S1, q1, d1, A1)
– NFA M2 = (Q2, S2, q2, d2, A2)
• Output
– NFA M3 = (Q3, S3, q3, d3, A3)
– What is Q3?
• Q3 =
– What is S3?
•
S3 = S1 = S 2
– What is q3?
• q3 =
170
Construction
NFA M1
NFA M2
A
NFA M3
• Input
– NFA M1 = (Q1, S1, q1, d1, A1)
– NFA M2 = (Q2, S2, q2, d2, A2)
• Output
– NFA M3 = (Q3, S3, q3, d3, A3)
– What is A3?
• A3 =
– What is d3?
•
d3 =
171
Comments
• You should be able to execute this algorithm
• You should understand the idea behind this algorithm
• You should understand how this algorithm can be used to
simplify design
• You should be able to design new algorithms for new
closure properties
• You should understand how this helps prove result that
regular languages and LFSA are identical
– In particular, you should understand how this is used to construct
an NFA M from a regular expression r s.t. L(M) = L(r)
– To be seen later
172
LFSA closed under Kleene
Closure
173
LFSA closed under Kleene
Closure
• Let L be arbitrary language in LFSA
• Let M1 be an NFA s.t. L(M1) = L
– M1 exists by definition of L1 in LFSA and the fact that every FSA
is an NFA
•
•
•
•
Construct NFA M2 from NFA M1
Argue L(M2) = L1*
There exists NFA M2 s.t. L(M2) = L1*
L1* is in LFSA
174
Visualization
• Let L be arbitrary language
in LFSA
• Let M1 be an NFA s.t. L(M1)
=L
L1
– M1 exists by definition of L1
in LFSA and the fact that
every FSA is an NFA
• Construct NFA M2 from
NFA M1
• Argue L(M2) = L1*
• There exists NFA M2 s.t.
L(M2) = L1*
• L1* is in LFSA
L1*
LFSA
M1
M2
NFA’s
175
Algorithm Specification
• Input
– NFA M1
• Output
– NFA M2 such that L(M2) =
NFA M1
A
NFA M2
176
Use l-transition
NFA M1
NFA M2
l
a
M1
A
l
a
M2
177
General Case *
NFA M1
A
NFA M2
l
l
M1
l M
2
178
Construction
NFA M1
A
NFA M2
• Input
– NFA M1 = (Q1, S1, q1, d1, A1)
• Output
– NFA M2 = (Q2, S2, q2, d2, A2)
– What is Q2?
– What is S2?
•
S2 = S1
– What is q2?
• q2 =
179
Construction
NFA M1
A
NFA M2
• Input
– NFA M1 = (Q1, S1, q1, d1, A1)
• Output
– NFA M2 = (Q2, S2, q2, d2, A2)
– What is A2?
• A2 =
– What is d2?
•
d2 =
180
Comments
• You should be able to execute this algorithm
• You should understand the idea behind this algorithm
– Why do we need to make an extra state p?
• You should understand how this algorithm can be used to
simplify design
• You should be able to design new algorithms for new
closure properties
• You should understand how this helps prove result that
regular languages and LFSA are identical
– In particular, you should understand how this is used to construct
an NFA M from a regular expression r s.t. L(M) = L(r)
181
– To be seen later
Module 22
• Regular languages are a subset of LFSA
– algorithm for converting any regular expression
into an equivalent NFA
– Builds on existing algorithms described in
previous lectures
182
Regular languages are a subset of
LFSA
183
Reg. Lang. subset LFSA
• Let L be an arbitrary
• Let R be the
– R exists by definition of
• Construct an
– M is constructed from
• Argue
• There exists an
• L is in
– By definition of
184
Visualization
L
L
Regular
Languages
R
Regular
Expressions
LFSA
M
NFA-l’s
185
Algorithm Specification
• Input
– Regular expression R
• Output
– NFA M such that L(M) =
Regular expression R
A
NFA-l M
186
Recursive Algorithm
• We have an inductive definition for regular
languages and regular expressions
• Our algorithm for converting any regular
expression into an equivalent NFA is
recursive in nature
– Base Case
– Recursive or inductive Case
187
Base Case
• Regular expression R has zero operators
– No concatenation, union, Kleene closure
– For any alphabet S, only |S| + 2 regular
languages can be depicted by any regular
expression with zero operators
• The empty language f
• The language {l}
• The |S| languages consisting of one string {a} for all
a in S
188
Table lookup
• Finite number of base cases means we can
use table lookup to handle them
f
l
a
b
189
Recursive Case
• Regular expression R has at least one
operator
– This means R is built up from smaller regular
expressions using the union, Kleene closure, or
concatenation operators
– More specifically, there are 3 cases:
• R = R1+R2
• R = R1R2
• R = R1*
190
Recursive Calls
1) R = R1 + R2
2) R = R1 R2
3) R = R1*
• The algorithm recursively calls itself to generate
NFA’s M1 and M2 which accept L(R1) and L(R2)
• The algorithm applies the appropriate construction
– union
– concatenation
– Kleene closure
to NFA’s M1 and M2 to produce an NFA M such
that L(M) = L(R)
191
Pseudocode Algorithm
_____________ RegExptoNFA(_____________) {
regular expression R1, R2;
NFA M1, M2;
Modify R by removing unnecessary enclosing parentheses
/* Base Case */
If R = a, return (NFA for {a}) /* include l here */
If R = f, return (NFA for {})
/* Recursive Case */
Find “last operator O” of regular expression R
Identify regular expressions R1 (and R2 if necessary)
M1 = RegExptoNFA(R1)
M2 = RegExptoNFA(R2) /* if necessary */
return (OP(M1, M2)) /* OP is chosen based on O */
}
192
Example
A: R = (b+a)a*
Last operator is concatenation
R1 = (b+a)
R2 = a*
Recursive call with R1 = (b+a)
B: R = (b+a)
Extra parentheses stripped away
Last operator is union
R1 = b
R2 = a
Recursive call with R1 = b
193
Example Continued
C: R = b
Base case
NFA for {b} returned
B: return to this invocation of procedure
Recursive call where R = R2 = a
D: R = a
Base case
NFA for {a} returned
B: return to this invocation of procedure
return UNION(NFA for {b}, NFA for {a})
A: return to this invocation of procedure
Recursive call where R = R2 = a*
194
Example Finished
E: R = a*
Last operator is Kleene closure
R1 = a
Recursive call where R = R1 = a
F: R = a
Base case
NFA for {a} returned
E: return to this invocation of procedure
return (KLEENE(NFA for {a}))
A: return to this invocation of procedure
return CONCAT(NFA for {b,a}, NFA for {a}*)
195
Pictoral View
concatenate
(b|a)a*
a
l
l
l
b
(b|a) b
lunion
a
b
a
l
l
a
l
l Kleene
a* Closure
l
a
l
aa
196
Parse Tree
We now present the “parse” tree for regular expression (b+a)a*
concatenate
union
b
Kleene closure
a
a
197
Module 23
• Regular languages review
– Several ways to define regular languages
– Two main types of proofs/algorithms
• Relative power of two computational models
proofs/constructions
• Closure property proofs/constructions
– Language class hierarchy
• Applications of regular languages
198
Defining regular languages
199
Three definitions
• LFSA
– A language L is in LFSA iff there exists an FSA M s.t. L(M) = L
• LNFA
– A language L is in LNFA iff there exists an NFA M s.t. L(M) = L
• Regular languages
– A language L is regular iff there exists a regular expression R s.t.
L(R) = L
• Conclusion
– All these language classes are equivalent
– Any language which can be represented using any one of these
models can be represented using either of the other two models
200
Two types of
proofs/constructions
201
Relative power proofs
• These proofs work between two language classes
and two computational models
• The crux of these proofs are algorithms which
behave as follows:
– Input: One program from the first computational model
– Output: A program from the second computational
model that is equivalent in function to the first program
202
Closure property proofs
• These proofs work within a single language class
and typically within a single computational model
• The crux of these proofs are algorithms which
behave as follows:
– Input: 1 or 2 programs from a given computational
model
– Output: A third program from the same computational
model that accepts/describes a third language which is a
combination of the languages accepted/described by the
two input programs
203
Comparison
L
L1
L
L1 intersect L2
L2
LNFA
LFSA
LFSA
M1
M
NFA’s
M3
M’
FSA’s
M2
FSA’s
204
Language class hierarchy
?
H
H
regular
REC
RE
All languages over alphabet S
205
Three remaining topics
• Myhill-Nerode Theorem
– Provides technique for proving a language is not regular
– Also represents fundamental understanding of what a regular
language is
• Decision problems about regular languages
– Most are solvable in contrast to problems about recursive
languages
• Pumping lemma
– Provides technique for proving a language is not regular
206
Module 24
• Myhill-Nerode Theorem
–
–
–
–
distinguishability
equivalence classes of strings
designing FSA’s
proving a language L is not regular
207
Distinguishability
208
Distinguishable and
Indistinguishable
• String x is distinguishable from string y with
respect to language L iff
– there exists a string z such that
• xz is in L and yz is not in L OR
• xz is not in L and yz is in L
• String x is indistinguishable from string y with
respect to language L iff
– for all strings z,
• xz and yz are both in L OR
• xz and yz are both not in L
209
Example
• Let EVEN-ODD be the set of strings over
{a,b} with an even number of a’s and an
odd number of b’s
– Is the string aa distinguishable from the string
bb with respect to EVEN-ODD?
– Is the string aa distinguishable from the string
ab with respect to EVEN-ODD?
210
Equivalence classes of strings
211
Definition of equivalence classes
• Every language L partitions S* into equivalence
classes via indistinguishability
– Two strings x and y belong to the same equivalence
class defined by L iff x and y are indistinguishable w.r.t
L
– Two strings x and y belong to different equivalence
classes defined by L iff x and y are distinguishable w.r.t.
L
212
Example
How does EVEN-ODD partition {a,b}* into equivalence classes?
Strings with an
EVEN number of a’s
and an
EVEN number of b’s
Strings with an
ODD number of a’s
and an
EVEN number of b’s
Strings with an
EVEN number of a’s
and an
ODD number of b’s
Strings with an
ODD number of a’s
and an
ODD number of b’s
213
Second Example
Let 1MOD3 be the set of strings over {a,b} whose length mod 3 = 1.
How does 1MOD3 partition {a,b}* into equivalence classes?
Length mod 3 = 0
Length mod 3 = 1
Length mod 3 = 2
214
Designing FSA’s
215
Designing an FSA
for EVEN-ODD
l
a
b
b
Even
Even
Odd
Even
Even
Odd
Odd
Odd
a
b
a
ab
216
Length mod 3 = 0
Designing an
FSA for 1MOD3
Length mod 3 = 1
Length mod 3 = 2
a,b
l
a,b
a
a,b
aa
217
Proving a language is not regular
218
Third Example
• Let EQUAL be the set of strings x over {a,b} s.t.
the number of a’s in x = the number of b’s in x
• How does EQUAL partition {a,b}* into
equivalence classes?
• How many equivalence classes are there?
• Can we construct a finite state automaton for
EQUAL?
219
Myhill-Nerode Theorem
220
Theorem Statement
• Two part statement
– If L is regular, then L partitions S* into a finite
number of equivalence classes
– If L partitions S* into a finite number of
equivalence classes, then L is regular
• One part statement
– L is regular iff L partitions S* into a finite
number of equivalence classes
221
Implication 1
• Method for constructing FSA’s to accept a
language L
–
–
–
–
Identify equivalence classes defined by L
Make a state for each equivalence class
Identify initial and accepting states
Add transitions between the states
• You can use a canonical element of each
equivalence class to help with building the transition
function d
222
Implication 2
• Method for proving a language L is not
regular
– Identify equivalence classes defined by L
– Show there are an infinite number of such
equivalence classes
• Table format may help, but it is only a way to help
illustrate that there are an infinite number of
equivalence classes defined by L
223
Proving a language is not regular
revisited
224
Proving EQUAL is not regular
• Let EQUAL be the set of strings x over {a,b} s.t.
the number of a’s in x = the number of b’s in x
• We want to show that EQUAL partitions {a,b}*
into an infinite number of equivalence classes
• We will use a table that is somewhat reminiscent of
the table used for diagonalization
– Again, you must be able to identify the infinite number
of equivalence classes being defined by the table. They
ultimately represent the proof that EQUAL or whatever
language you are working with is not regular.
225
Table *
a
aa
aaa
aaaa
aaaaa
...
b
IN
OUT
OUT
OUT
OUT
...
bb
OUT
IN
OUT
OUT
OUT
...
bbb
OUT
OUT1
IN
OUT
OUT
...
bbbb
OUT
OUT
OUT
IN
OUT
...
bbbbb
OUT
OUT
OUT
OUT
IN
...
…
…
…
…
…
…
The strings being distinguished are the rows.
The tables entries indicate that the concatenation of the row
string with the column string is in or not in EQUAL.
Each complete column shows one row string is distinguishable
from all the other row strings.
226
Concluding EQUAL is
nonregular *
• We have shown that EQUAL partitions
{a,b}* into an infinite number of
equivalence classes
– In this case, we only identified some of the
equivalence classes defined by EQUAL, but
that is sufficient
• Thus, the Myhill-Nerode Theorem implies
that EQUAL is nonregular
227
Summary
• Myhill-Nerode Theorem and what it says
– It does not say a language L is regular iff L is finite
• Many regular languages such as S* are not finite
– It says that a language L is regular iff L partitions S*
into a finite number of equivalence classes
• Provides method for designing FSA’s
• Provides method for proving a language L is not
regular
– Show that L partitions S* into an infinite number of
equivalence classes
228
Two/Three Types of Problems
• Create a table that helps prove that a
specific language L is not regular
– You get to choose the “row” and “column”
strings
– I choose the “row” strings
• Identify the equivalence classes defined by
L as highlighted by a given table
229
Module 25
• Decision problems about regular languages
– Basic problems are solvable
• halting, accepting, and emptiness problems
– Solvability of other problems
• answer-preserving input transformations to basic
problems
230
Programs
• In this unit, our programs are the following
three types of objects
– FSA’s
– NFA’s
– regular expressions
• Previously, they were C++ programs
– Review those topics after mastering today’s
examples
231
Basic Decision Problems
(and algorithms for solving them)
232
Halting Problem
• Input
– FSA M
– Input string x to M
• Question
– Does M halt on x?
• Give an algorithm for solving the FSA halting problem.
233
Accepting Problem
• Input
– FSA M
– Input string x to M
• Question
– Is x in L(M)?
• Give an algorithm ACCEPT for solving the
accepting problem.
234
Empty Language Problem
• Input
– FSA M
• Question
– Is L(M)={}?
• Give an algorithm for solving the empty language problem.
– Don’t look ahead to the next slide.
235
Algorithms for solving empty
language problem
• Algorithm 1
– View FSA M as a directed graph (nodes, arcs)
– See if any accepting node is reachable from the start
node
• Algorithm 2
– Let n be the number of states in FSA M
– Run ACCEPT(M,x) for all input strings of length < n
– If any are accepted THEN no ELSE yes
• Why is algorithm 2 correct?
236
Solving Other Problems
(using answer-preserving input
transformations)
237
Complement Empty Problem
• Input
– FSA M
• Question
– Is (L(M))c = {}?
• Show how to use an answer-preserving input
transformation to help solve this problem
– Show that the Complement Empty problem transforms
to the Empty Language problem
– Don’t look at next two slides
238
Algorithm Description
• Convert input FSA M into an FSA M’ such that
L(M’) = (L(M))c
– We do this by applying the algorithm which we used to
show that LFSA is closed under complement
• Feed FSA M’ into algorithm which solves the
empty language problem
• If that algorithm returns yes THEN yes ELSE no
239
Input Transformation Illustrated
FSA M
Complement FSA M’
Construction
Algorithm for
solving empty
language problem
Yes/No
Algorithm for complement empty problem
The complement construction algorithm is the answer-pres. input transformation.
If M is a yes input instance of CE, then M’ is a yes input instance of EL.
If M is a no input instance of CE, then M’ is a no input instance of EL.
240
NFA Empty Problem
• Input
– NFA M
• Question
– Is L(M)={}?
• Show how to use answer-preserving input
transformations to help solve this problem
– Show that the NFA Empty problem transforms to the
Empty Language problem
241
Input Transformation
Yes/No
Algorithm for NFA empty problem
242
Equal Problem
• Input
– FSA’s M1 and M2
• Question
– Is L(M1) = L(M2)?
• Show how to use answer-preserving input transformations
to solve this problem
– Try and transform this problem to the empty language problem
– If L(M1) = L(M2), then what combination of L(M1) and L(M2)
must be empty?
243
Input Transformation Illustrated
Yes/No
Algorithm for Equal problem
244
Summary
• Decision problems with programs as inputs
• Basic problems
– You need to develop algorithms from scratch
based on properties of FSA’s
• Solving new problems
– You need to figure out how to combine the
various algorithms we have seen in this unit to
solve the given problem
245
Module 26
• Pumping Lemma
– A technique for proving a language L is NOT
regular
– What does the Pumping Lemma mean?
– Proof of Pumping Lemma
246
Pumping Lemma
How do we use it?
247
Pumping Condition
• A language L satisfies the pumping
condition if:
– there exists an integer n > 0 such that
– for all strings x in L of length at least n
– there exist strings u, v, w such that
•
•
•
•
x = uvw and
|uv| <= n and
|v| >= 1 and
For all k >= 0, uvkw is in L
248
Pumping Lemma
• All regular languages satisfy the pumping
condition
Regular languages
“Pumping Languages”
All languages over {a,b}
249
Pumping
Implications
Regular
• We can use the pumping lemma to prove a
language L is not regular
– How?
• We cannot use the pumping lemma to prove
a language is regular
– How might we try to use the pumping lemma to
prove that a language L is regular and why does
it fail?
250
Pumping Lemma
What does it mean?
251
Pumping Condition
• A language L satisfies the pumping condition if:
– there exists an integer n > 0 such that
– for all strings x in L of length at least n
– there exist strings u, v, w such that
•
•
•
•
x = uvw and
|uv| <= n and
|v| >= 1 and
For all k >= 0, uvkw is in L
252
v can be
pumped
1) x in L
2) x = uvw
3) For all k >= 0, uvkw is in L
• Let x = abcdefg be in L
• Then there exists a substring v in x such that v can be
repeated (pumped) in place any number of times and the
resulting string is still in L
– uvkw is in L for all k >= 0
• For example
– v = cde
• uv0w = uw = abfg is in L
• uv1w = uvw = abcdefg is in L
• uv2w = uvvw = abcdecdefg is in L
• uv3w = uvvvw = abcdecdecdefg is in L
• …
253
What the other parts mean
• A language L satisfies the pumping condition if:
– there exists an integer n > 0 such that
• defer what n is till later
– for all strings x in L of length at least n
• x must be in L and have sufficient length
– there exist strings u, v, w such that
• x = uvw and
• |uv| <= n and
– v occurs in the first n characters of x
• |v| >= 1 and
– v is not l
• For all k >= 0, uvkw is in L
254
Examples
• Example 1
– Let L be the set of even length strings over
{a,b}
– Let x = abaa
– Let n = 2
– What are the possibilities for v?
• abaa, abaa
• abaa
– Which one satisfies the pumping lemma?
255
Examples *
• Example 2
– Let L be the set of strings over {a,b} where the number
of a’s mod 3 is 1
– Let x = abbaaa
– Let n = 3
– What are the possibilities for v?
• abbaaa, abbaaa, abbaaa
• abbaaa, abbaaa
• abbaaa
– Which ones satisfy the pumping lemma?
256
Pumping Lemma
Proof
257
High Level Outline
• Let L be an arbitrary regular language
• Let M be an FSA such that L(M) = L
– M exists by definition of LFSA and the fact that
regular languages and LFSA are identical
• Show that L satisfies the pumping condition
– Use M in this part
• Pumping Lemma follows
258
First step: n+1 prefixes of x
• Let n be the number of states in M
• Let x be an arbitrary string in L of length at least n
– Let xi denote the ith character of string x
• There are at least n+1 distinct prefixes of x
–
–
–
–
–
–
–
length 0: l
length 1: x1
length 2: x1x2
...
length i: x1x2 … xi
...
length n: x1x2 … xi … xn
259
Example
• Let n = 8
• Let x = abcdefgh
• There are 9 distinct prefixes of x
–
–
–
–
–
length 0: l
length 1: a
length 2: ab
...
length 8: abcdefgh
260
Second step: Pigeon-hole
Principle
• As M processes string x, it processes each prefix of x
– In particular, each prefix of x must end up in some state of M
• Situation
– There are n+1 distinct prefixes of x
– There are only n states in M
• Conclusion
– At least two prefixes of x must end up in the same state of M
• Pigeon-hole principle
– Name these two prefixes p1 and p2
261
Third step: Forming u, v, w
• Setting:
– Prefix p1 has length i
– Prefix p2 has length j > i
• prefix p1 of length i: x1x2 … xi
• prefix p2 of length j: x1x2 … xi xi+1 … xj
• Forming u, v, w
–
–
–
–
Set u = p1 = x1x2 … xi
Set v = xi+1… xj
Set w = xj+1 … x|x|
x1x2 … xi xi+1… xj xj+1 … x|x|
u
v
w
262
Example 1
0
1
2
3
4
• Let M be a 5-state FSA that accepts all strings
over {a,b,c,…,z} whose length mod 5 = 3
• Consider x = abcdefghijklmnopqr, a string in L
• What are the two prefixes p1 and p2?
• What are u, v, w?
263
0
Example 2
0
1
1
1
0
2
• Let M be a 3-state FSA that accepts all strings
over {0,1} whose binary value mod 3 = 1
• Consider x = 10011, a string in L
• What are the two prefixes p1 and p2?
• What are u, v, w?
264
Fourth step: Showing u, v, w
satisfy all the conditions
• |uv| <= n
– uv = p2
– p2 is one of the first n+1 prefixes of string x
• |v| >= 1
– v consists of the characters in p2 after p1
– Since p2 and p1 are distinct prefixes of x, v is not l
• For all k >= 0, uvkw in L
– u=p1 and uv=p2 end up in the same state q of M
• This is how we defined p1 and p2
– Thus for all k >= 0, uvk ends up in state q
– The string w causes M to go from state q to an accepting state
265
Example 1 again
0
1
2
3
4
• Let M be a 5-state FSA that accepts all strings over
{a,b,c,…,z} whose length mod 5 = 3
• Consider x = abcdefghijklmnopqr, a string in L
• What are u, v, w?
– u=l
– v = abcde
– w = fghijklmnopqr
• |uv| = 5 <= 5
• |v| = 5 >= 1
• For all t>=0, (abcde)tfghijklmnopqr is in L
266
0
Example 2 again
0
1
1
1
0
2
• Let M be a 3-state FSA that accepts all strings over {0,1}
whose binary interpretation mod 3 = 1
• Consider x = 10011, a string in L
• What are u, v, w?
– u=1
– v = 00
– w = 11
• |uv| = 3 <= 3
• |v| = 2 >= 1
• For all k>=0, 1(00)k11 is in L
267
Pumping Lemma
• A language L satisfies the pumping condition if:
– there exists an integer n > 0 such that
– for all strings x in L of length at least n
– there exist strings u, v, w such that
•
•
•
•
x = uvw and
|uv| <= n and
|v| >= 1 and
For all k >= 0, uvkw is in L
• Pumping Lemma: All regular languages satisfy the
pumping condition
268
Module 27
• Applications of Pumping Lemma
– General proof template
• What is the same in every proof
• What changes in every proof
– Incorrect pumping lemma proofs
– Some rules of thumb
269
Pumping Lemma
Applying it to prove a specific
language L is not regular
270
How we use the Pumping Lemma
• We choose a specific language L
– For example, {ajbj | j > 0}
• We show that L does not satisfy the
pumping condition
• We conclude that L is not regular
271
Showing L “does not pump”
• A language L satisfies the
pumping condition if:
– there exists an integer n > 0
such that
– for all strings x in L of length
at least n
– there exist strings u, v, w such
that
•
•
•
•
x = uvw and
|uv| <= n and
|v| >= 1 and
For all k >= 0, uvkw is in L
• A language L does not satisfy
the pumping condition if:
– for all integers n of sufficient
size
– there exists a string x in L of
length at least n such that
– for all strings u, v, w such that
• x = uvw and
• |uv| <= n and
• |v| >= 1
– There exists a k >= 0 such that
uvkw is not in L
272
Example Proof
• A language L does not satisfy
the pumping condition if:
– for all integers n of sufficient
size
– there exists a string x in L of
length at least n such that
– for all strings u, v, w such that
• x = uvw and
• |uv| <= n and
• |v| >= 1
– There exists a k >= 0 such that
uvkw is not in L
• Proof that L = {aibi | i>0} does not
satisfy the pumping condition
• Let n be the integer from the
pumping lemma
• Choose x = anbn
• Consider all strings u, v, w s.t.
• x = uvw and
• |uv| <= n and
• |v| >= 1
• Argue that uvkw is not in L for
some k >= 0
– Argument must apply to all
possible u,v,w
– Continued on next slide
273
Example Proof Continued
• Proof that L = {aibi | i>0} does not
satisfy the pumping condition
• Let n be the integer from the
pumping lemma
• Choose x = anbn
• Consider all strings u, v, w s.t.
• x = uvw and
• |uv| <= n and
• |v| >= 1
• Argue that uvkw is not in L for
some k >= 0
– Argument must apply to all
possible u,v,w
– Continued on right
• uv0w = uw is not in L
– uv contains only a’s
• why?
– uw = an-|v|bn
• Follows from previous line
and uvw = x = anbn
– uw contains fewer a’s than b’s
• why?
– Therefore, uw is not in L
• Therefore L does not satisfy
the pumping condition
274
Alternate choice of k
• Proof that L = {aibi | i>0} does not
satisfy the pumping condition
• Let n be the integer from the
pumping lemma
• Choose x = anbn
• Consider all strings u, v, w s.t.
• x = uvw and
• |uv| <= n and
• |v| >= 1
• Argue that uvkw is not in L for
some k >= 0
– Argument must apply to all
possible u,v,w
– Continued on right
• uv2w = uvvw is not in L
– uv contains only a’s
• why?
– uvvw = an+|v|bn
• follows from previous line
and uvw = x = anbn
– uvvw contains more a’s than
b’s
• why?
– Therefore, uvvw is not in L
• Therefore L does not satisfy
the pumping condition
275
Pumping Lemma
Some bad applications of the
pumping lemma
276
Bad Pumping Lemma
Applications
• We now look at some examples of bad
applications of the pumping lemma
• We work with the language EQUAL
consisting of the set of strings over {a,b}
such that the number of a’s equals the
number of b’s
• We focus first on bad choices of string x
• We then consider another flawed technique
277
First bad choice of x
• A language L does not satisfy
the pumping condition if:
– Let n be the integer from the
pumping lemma
– there exists a string x in L of
length at least n such that
– for all strings u, v, w such that
• Let n be the integer from the
pumping lemma
• Choose x = a10b10
– What is wrong with this choice of
x?
• x = uvw and
• |uv| <= n and
• |v| >= 1
– There exists a k >= 0 such that
uvkw is not in L
278
Second bad choice of x
• A language L does not satisfy
the pumping condition if:
– Let n be the integer from the
pumping lemma
– there exists a string x in L of
length at least n such that
– for all strings u, v, w such that
• Let n be the integer from the
pumping lemma
• Choose x = anb2n
– What is wrong with this choice of
x?
• x = uvw and
• |uv| <= n and
• |v| >= 1
– There exists a k >= 0 such that
uvkw is not in L
279
Third bad choice of x
• A language L does not satisfy
the pumping condition if:
– Let n be the integer from the
pumping lemma
– there exists a string x in L of
length at least n such that
– for all strings u, v, w such that
• x = uvw and
• |uv| <= n and
• |v| >= 1
– There exists a k >= 0 such that
uvkw is not in L
• Let n be the integer from the
pumping lemma
• Choose x = (ab)n
– What is wrong with this choice
of x?
• The problem is there is a choice
of u, v, w satisfying the three
conditions such that for all k
>=0, uvkw is in L
• What is an example of such a u,
v, w?
280
Find the flaw in this proof
• A language L does not satisfy
the pumping condition if:
– Let n be the integer from the
pumping lemma
– there exists a string x in L of
length at least n such that
– for all strings u, v, w such that
• x = uvw and
• |uv| <= n and
• |v| >= 1
– There exists a k >= 0 such that
uvkw is not in L
• Let n be the integer from the
pumping lemma
• Choose x = anbn
• Let u = a2, v =a, w = an-3bn
– |uv| = 3 <= n
– |v| = 1
• Choose k = 2
• Argue uv2w is not in EQUAL
– uv2w = uvvw = a2aaan-3bn = an+1bn
– There is one more a than b in uv2w
– Thus uv2w is not in L
281
Pumping Lemma
Two rules of thumb
282
Two Rules of Thumb *
• Try to make the first n characters of x identical
– For EQUAL, choose x = anbn rather than (ab)n
• Simplifies case analysis as v only contains a’s
• Try k=0 or k=2
– k=0
• This reduces number of occurrences of that first character
– k=2
• This increases number of occurrences of that first character
283
Summary
• We use the Pumping Lemma to prove a
language is not regular
– Note, does not work for all nonregular
languages, though
• Choosing a good string x is first key step
• Choosing a good integer k is second key
step
• Must apply argument to all legal u, v, w
284