CS 3240 – Chapter 4

Closure Properties

Algorithms for Elementary Questions:
 Is a given word, w, in L?
 Is L empty, finite or infinite?
 Are L1 and L2 the same set?

Detecting non-regular languages
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
Closure of operations
 If x and y are in the same set, is x op y also?
 Example: The integers are closed under addition
▪ They are not closed under division

Regular languages are closed under
everything!
 Typical set operations
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
Regular languages are closed under:
 Kleene Star (*)
 Union (+)
 Concatenation (xy)
 (By definition!)

They are also closed under:
 Complement (reverse state acceptability✓)
 Intersection
 Set difference
 Reversal (already proved in homework #12, 2.3✓)
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
Proof from set theory:
 L1 ∩ L2 = (L1’ ∪ L2’)’
 Since complement and union are closed,
intersection must be also! QED
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• Note how the intersection is never shaded
• L1’ ∪ L2’ shades everything but where they overlap
• Therefore, (L1’ ∪ L2’)’ is the overlap (intersection)
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
A – B:
 Everything that is in A but not in B

A – B = A ∩ B’
 We have already shown that regular languages
are closed under intersection and complement.
QED
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
Start with a composite start state:
 Consisting of the two start states

Follow all out-edges simultaneously
 As we did for NFA-to-DFA conversion

States containing any original final state is a final state
in the result for union
 Because one of the machines accepts there

States containing an original final state from each
original machine is a final state in the result for
intersection
 Because both of the machines accept there

¿How would you construct the difference machine?
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b
a,b
a
a
-x1
x2
+x3
b
b
 y1
a
Double-a
y2
b
a
a
a
EVEN-EVEN
b
y3
y4
b
9
xi, yi
a
b
x1, y1
x2, y3
x1, y2
x2, y3
x3, y1
x1, y4
x1, y2
x2, y4
x1, y1
x3, y1
x3, y3
x3, y2
x1, y4
x2, y2
x1, y3
x2, y4
x3, y2
x1, y3
x3, y3
x3, y1
x3, y4
x3, y2
x3, y4
x3, y1
x2, y2
x3, y4
x1, y1
x1, y3
x2, y1
x1, y4
x3, y4
x3, y2
x3, y3
x2, y1
x3, y3
x1, y2
For union: assign accepting
states where any original xi or
yi accept.
For intersection: assign
accepting states only where
both original xi or yi accept
simultaneously. No need to
compute (L1’ ∪ L2’)’ !
For difference, assign
accepting states where one
accepts and the other does
not.
The resulting machine…
a
11

a
a
a
a
b
a
1
9
b
8
3
b
b
b
b
b
12

b
a
a
10
a
7
6
b
b
a
5
b
2
a
4
a
11

Given a word w, and a regular language, L,
can we answer the question:
 Is w ∊ L?

You tell me…
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
A graph theory problem:
 Find a path from the start to a final state in the
associated FA

Algorithm:
“mark” the start state
repeat:
mark any state with an incoming edge from a previously
marked state
until an accepting state is marked or no new states were
marked at all
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
Attempt to convert the associated FA to a
regular expression
 By the state bypass and elimination algorithm

If you get a regular expression, then a string is
accepted
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

Suppose a minimal machine, M, for the
language L has p states
If M accepts any non-empty words at all, it
must accept one of length <= p
 Why?

So…
 Systematically try all possible strings in Σ* of
length 1 through p. If none are accepted, then no
non-empty strings at all are in L.
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
Convert its machine to a regular expression
It is infinite iff it has a star


Another way:


 A language is infinite if there is a cycle in an
accepting path
 A (tedious) graph theory problem

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Suppose L’s minimal machine, M, has p states
 Any path of length p has (or is) a cycle

 And any cycle must have or be a cycle of length p or less
 Because a state is revisited after at most p characters
So, infinite languages have a machine with at least
one cycle of length p or less in an accepting path*
 And all non-empty languages have a string of length
p or less (already showed that)…

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
Let m denote the length of a cycle in an accepting path
 We know m ≤ p

Let k be the length of a string in L such that k ≤ p
 There has to be one if the language is infinite!

Then strings of length k + im are accepted, i ≥ 0
 By traversing the cycle i times



But k + im ≤ p + ip = (i+1)p
So, there must be some i such that p ≤ k+im ≤ 2p
Procedure: Test all strings of length p through 2p-1
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

That is, are they the same set of strings?
Set-theoretic argument:
 Two sets are equal if their symmetric difference is
empty (denoted by A ∆ B or A ⊖ B)
 A∆B=A∪B–A∩B=A–B∪B–A

But A – B = A ∩ B’, and B – A = B ∩ A’

So L1 = L2 iff (L1 ∩ L2’) ∪ (L1’ ∩ L2) = ∅
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

Not all languages are regular
We need to recognize whether languages are
regular or not
 We don’t want to waste time using regular
language processing techniques where they don’t
apply
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
Consider anbn
ab is regular
ab + aabb = anbn, 0 ≤ n ≤ 2, is regular
Any finite language is regular (why?)
But anbn, n ≥ 0 is not regular (why not?)

How do we prove it’s not regular!?!




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
Finite Automata don’t have unlimited
counting capability
 They only have a fixed number of states

Intuitively, we see that an automaton can’t
keep track of counts for anbn where n is
arbitrarily large

But intuition is often faulty. We need a proof!
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

Any accepted string of length p (the number
of states) or greater forces a cycle in an
accepting path.
In other words, at least one state is visited a
second time
 And that “revisit” must happen within the first p
characters of the string
▪ Because that’s when the (p+1)th state is entered
 This could be any state (start, final, other)
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
Consider akbk, where k is greater than the number of
states in a supposed DFA accepting all anbn, n ≥ 0
 Before the first b is encountered, a state has been visited at least
twice (because there are more a’s than states)
 Suppose the length of the associated cycle is m
 Then the string ak+imbk is also accepted!

This contradicts the existence of a DFA that accepts anbn
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The first “revisit”
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
For every infinite regular language, L, there is
a number, p, such that for all strings, s, in L,
where |s| ≥ p, you can partition s into three
concatenated substrings, xyz, such that:
1. |y| > 0
2. |xy| ≤ p
3. xy*z ∈ L
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
You can only use the pumping lemma to show
that a language is not regular
 By showing it fails the “pumping” conditions of
infinite regular languages
 Note: Some non-regular languages pump!

The trick is to find a convenient string
 Usually the condition |xy| ≤ p is also key
 Sometimes pumping down (i = 0) is easiest
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
Consider the string apbp
 It is in this language
 It is long enough (≥ p in length)

Now let apbp = xyz
 Remember |xy| ≤ p
 What can you conclude about y?
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
You can treat proving a language non-regular
as a “game”:
1. You pick a string, s, in L, where |s| ≥ p
▪ You may pick any such string; choose wisely!
2. Opponent picks x, y, and z
▪ But must obey |xy| ≤ p and |y| > 0
3. You show it can’t be “pumped”
▪ Because a pumped string falls “outside” the language

Must anticipate all possible partitions xyz
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

aibj, i > j
PALINDROME
 w = wR (same backwards and forwards)

ww
 Equal halves


PRIME (am where m is prime)
SQUARE (am where m is a perfect square)
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

Strings with equal number of a’s and b’s
NOTPRIME
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


NOTPRIME is pumpable!
Let y = the whole string (akm)
The number of a’s will always be a multiple
of km, hence not prime
 Note: zero is not a prime number

This does not violate the pumping lemma
 The pumping lemma draws no conclusion about
non-regular languages
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