Properties of Regular Languages

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 languagesCS 3240 - Properties of Regular Languages 2

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

CS 3240 - Properties of Regular Languages 3

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✓)


Proof from set theory: L1 ∩ L2 = (L1’ ∪ L2’) ’ Since complement and union are closed,

intersection must be also! QED



• Note how the intersection is never shaded• L1’ ∪ L2’ shades everything but where they overlap• Therefore, (L1’ ∪ L2’) ’ is the overlap (intersection)

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


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?


9

-x1 x2 +x3

a

b

a

b a,b

1y 2y

3y 4y

aaaa

b

b

b

b

Double-a

EVEN-EVEN

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.

11

11a

b

The resulting machine…8

47

9 1 3 6

12 10 5 2

a

a

a

b bb

a

a

aa

a

aa

ab

b

b

b

bb

b

Given a word w, and a regular language, L, can we answer the question: Is w ∊ L?

You tell me…


A graph theory problem: Find a path from the start to a final state

in the associated FA Algorithm:

“mark” the start staterepeat:

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


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


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.


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


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)…


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-

1CS 3240 - Properties of Regular Languages 18

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) = ∅


Not all languages are regularWe 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


Consider anbn

ab is regularab + aabb = anbn, 0 ≤ n ≤ 2, is

regularAny finite language is regular (why?)But anbn, n ≥ 0 is not regular (why

not?)

How do we prove it’s not regular!?!CS 3240 - Properties of Regular Languages 26

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!


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)


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



The first “revisit”

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: |y| > 0 |xy| ≤ p xy*z ∈ L


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


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?


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”1. Because a pumped string falls “outside” the

language1.Must anticipate all possible partitions xyz


aibj, i > jPALINDROME

w = wR (same backwards and forwards)ww

Equal halvesPRIME (am where m is prime)SQUARE (am where m is a perfect

square)


Strings with equal number of a’s and b ’s

NOTPRIME


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 languagesCS 3240 - Properties of Regular Languages 37

Properties of Regular Languages

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