CS 3240 – Chapter 3
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How would you delete all C++ files from a
directory from the command line?
How about all PowerPoint files that start with
the letter a?
PowerPoint file names that contain the string
3240?
CS 3240 - Regular Languages and Grammars
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*.cpp
a*.ppt
*3240*.ppt
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These are wildcard expressions
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 Not bona fide regular expressions
CS 3240 - Regular Languages and Grammars
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Language
Machine
Grammar
Regular
Finite Automaton
Regular Expression,
Regular Grammar
Context-Free
Pushdown Automaton
Context-Free
Grammar
Recursively
Enumerable
Turing Machine
Unrestricted PhraseStructure Grammar
CS 3240 - Introduction
4
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Text patterns that represent regular languages
 We’ll show shortly that for every regular
expression there is a finite automaton that
accepts that language
 And vice-versa
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The operators are:
 ()
*
+
 xy
(Grouping)
(Kleene Star)
(Union)
(Concatenation)
CS 3240 - Regular Languages and Grammars
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1) Specify base case(s)
2) Show how to generate other elements
 Rules that use what’s in the set already
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Example: Non-negative multiples of 5, F
 1) 0, 5 is in F
 2) For x, y in F, then x + y is in F
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Alternate definition:
 1) 0 is in F
 2) For x in F, so is x + 5
CS 3240 - Regular Languages and Grammars
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Base cases:
 The empty set: ∅ or ( )
 The empty string: λ
 Any letter in Σ
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Recursive rules: Given regular expressions r, r1, r2:
 (r)
 r*
 r1 + r2
 r1r2
(Grouping)
(Kleene Star)
(Union)
(Concatenation)
CS 3240 - Regular Languages and Grammars
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All strings beginning with a:
 a(a + b)*
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All strings containing aba:
 (a + b)*aba(a + b)*
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All strings of even length:
 ((a + b)(a + b))* = (aa + ba + ab + bb)* = ((a + b)2)*
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All strings of odd length:
 (a+b)((a + b)2)*
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Valid decimal integers in C:
 (1+2+3+4+5+6+7+8+9)(0+1+2+3+4+5+6+7+8+9)*
CS 3240 - Regular Languages and Grammars
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Put anything you want on an edge
Use an “else” branch as well
 [0-9] (if-branch)
 ~[0-9] or [^(0-9)] or else
(Decimal integers)
CS 3240 - Regular Languages and Grammars
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(b*ab*ab*ab* + b) *
 = b* (ab*ab*ab*) *
 = b* + (b*ab*ab*ab*) *

(a(a+bb) *) *

((a + b)a) *
CS 3240 - Regular Languages and Grammars
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L(∅) =∅
L(λ) = λ
L(c) = c, for c ∊ Σ
L((r)) = L(r)
L(r*) = L(r)*
L(r1 + r2) = L(r1) ∪ L(r2)
L(r1r2) = L(r1)L(r2)
CS 3240 - Regular Languages and Grammars
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r+s = s+r
(r+s)+t = r+(s+t)
r+r = r
r+∅=r
(rs)t = r(st)
rλ = λr = r
r ∅ = ∅r = ∅
r(s+t) = rs+rt
(r+s)t = rt+st
CS 3240 - Regular Languages and Grammars
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1.
For every regular expression there is an
associated NFA that accepts the same
language
 And therefore a DFA, by conversion
2.
For every FA (either NFA or DFA) there is a
regular expression that represents the same
language
CS 3240 - Regular Languages and Grammars
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We will show how to convert each element of
the definition of regular expressions to an
NFA
This is sufficient!
 And shows the convenience of recursive
definitions (review slide 7 now)
 because if we can give a machine for every case in
the definition of REs, we are done!
CS 3240 - Regular Languages and Grammars
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• Empty Language
• Empty String
CS 3240 - Regular Languages and Grammars
• Single Character
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CS 3240 - Regular Languages and Grammars
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Just draw the lambdas from a new start state
to the start states of each machine
Remove the start notation from the original
start states
 (No need to have a new final state)
CS 3240 - Regular Languages and Grammars
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CS 3240 - Regular Languages and Grammars
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1) Just draw a lambda from each final state of
the first machine to the start state of the
second machine
2) remove the acceptability of those final
states of the first machine
CS 3240 - Regular Languages and Grammars
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CS 3240 - Regular Languages and Grammars
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We need to do two things:
 1) Add the empty string, if needed
 2) Loop from each final state back to the start state
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Procedure:
 1) If the empty string is not accepted, create a new start
state which accepts, and connect to the original start state
with λ
 2) Add a λ-edge from each final state to the original (or the
new) start state
CS 3240 - Regular Languages and Grammars
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Draw NFAs for the REs on slides 8 and 9
CS 3240 - Regular Languages and Grammars
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First remove all jails
 Then, if needed, convert the DFA to an equivalent
NFA with

 A start state with no incoming edges
 A single final state with no outgoing edges
 Will need lambda transitions for this
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Then “eliminate” all but the start and final states
 Without changing the language accepted
 Using GTGs…
CS 3240 - Regular Languages and Grammars
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Allow regular expressions on the edges
Accepts a* + a*(a+b)c*
[Note: (c*)* = c*]
CS 3240 - Regular Languages and Grammars
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If the start state has an incoming edge (even if
it’s a loop), create a new start state with a
lambda transition to the old start state:
CS 3240 - Regular Languages and Grammars
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If there is more than one final state, or if the single
final state has an outgoing edge (even if it’s a loop),
create a new final state and link to it with a lambda
transition from each final state:
CS 3240 - Regular Languages and Grammars
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“Remove” each intermediate state, one at a
time:
1. Combine each incoming path with each outgoing
path (only “through” paths; not loops)
2. Determine the regular expression equivalent to
the combined path through the current state
3. Add an edge with that RE between the incoming
state and the outgoing state
4. Repeat until all intermediate states vanish
CS 3240 - Regular Languages and Grammars
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CS 3240 - Regular Languages and Grammars
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To eliminate 2:
• 1-2-1: af*b
• 1-2-3: af*c
• 3-2-1: df*b
• 3-2-3: df*c
CS 3240 - Regular Languages and Grammars
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To eliminate 1:
• 0-1-3: (e+af*b)*(h+af*c)
• 3-1-3: (i+df*b)(e+af*b)*(h+af*c)
CS 3240 - Regular Languages and Grammars
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Eliminate 3 (Final Result):
(e+af*b) *(h+af*c)(g+df*c+(i+df*b)(e+af*b) *(h+af*c))*
CS 3240 - Regular Languages and Grammars
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CS 3240 - Regular Languages and Grammars
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Find a regular expression for the language
containing all strings that do not contain the
substring aa
CS 3240 - Regular Languages and Grammars
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See bypass.doc
 Shows different possibilities by eliminating states
in different orders
 But the REs obtained are equivalent
▪ Meaning they represent the same language
CS 3240 - Regular Languages and Grammars
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Language
Machine
Grammar
Regular
Finite Automaton
Regular Expression,
Regular Grammar
Context-Free
Pushdown Automaton
Context-Free
Grammar
Recursively
Enumerable
Turing Machine
Unrestricted PhraseStructure Grammar
CS 3240 - Introduction
35
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There is a natural correspondence between
FAs and grammars
Right-linear Grammars
 “Linear” means there is at most one variable on
the right-hand side of the rule
 “Right-linear” means the variable occurs as the
last entry in the rule:
▪ A → abC
CS 3240 - Regular Languages and Grammars
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The variables represent states
The right-hand side contains the character(s) on the edge,
optionally followed by the target state
 The accepting states have a lambda rule
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A → aB | bC | λ
B → aA | bD
C → aD | bA
D → aC | bB
CS 3240 - Regular Languages and Grammars
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Go to an accepting state with no out-edges
A→b
CS 3240 - Regular Languages and Grammars
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S → aaS | bbS | abA | baA | λ
A → aaA | bbA | abS | baS
a GTG
CS 3240 - Regular Languages and Grammars
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Construct a regular grammar for the language
denoted by aab*a
1. First build a GTG
2. Then map to a right-linear grammar
CS 3240 - Regular Languages and Grammars
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S → Xa
X → Xb | aa
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How did I come up with this?
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CS 3240 - Regular Languages and Grammars
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If you have the single variable only at the left
ends, you have a left-linear grammar
This is also a regular grammar
We will show how to convert between rightlinear and left-linear grammars
We will use two facts to establish the process:
 If L is regular, so is LR (Section 2.3, exercise 12)
 L(GR) = L(G)R (obvious, but on next slide…)
CS 3240 - Regular Languages and Grammars
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GR means you reverse the right-hand sides of
each rule in a grammar, G
The language generated is L(G)R (the reverse of
L(G))
S → abS | X
X → bX | λ
S → Sba | X
X → Xb | λ
(ab)*b*
b*(ba)*
CS 3240 - Regular Languages and Grammars
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1.
2.
Convert the right-linear grammar to a GTG
“Reverse” the GTG (a la Section 2.3, #12)
 Ensure a single final state (use λ if needed)
 Interchange the role of the start and final states
 Reverse all arrows
3.
4.
Convert the reversed GTG to a right-linear
grammar
Reverse the right-hand sides of each rule to
obtain the left-linear grammar
CS 3240 - Regular Languages and Grammars
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A → aB
B → abA | b
(rev)
C → bB
B → aA
A → baB | λ
(rev)
ba(baa)*
CS 3240 - Regular Languages and Grammars
C → Bb
B → Aa
A → Bab | λ
(aab)*ab
45
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Reverse the grammar, G, obtaining rightlinear grammar, GR, for L(G)R
Convert to GTG
Reverse the GTG
Convert to Right-linear
CS 3240 - Regular Languages and Grammars
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CS 3240 - Regular Languages and Grammars
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