1. CS 3240 – Chapter 4

2.  Closure Properties  Algorithms for Elementary Questions:  Is a given word, w, in L?  Is L empty, finite or infinite?  Are L 1 and L 2 the same set?  Detecting non-regular languages CS 3240 - Properties of Regular Languages2

3.  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 Languages3

4.  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 ✓ ) CS 3240 - Properties of Regular Languages4

5.  Proof from set theory:  L 1 ∩ L 2 = (L 1 ’ ∪ L 2 ’)’  Since complement and union are closed, intersection must be also! QED CS 3240 - Properties of Regular Languages5

6. 6 Note how the intersection is never shaded L 1 ’ ∪ L 2 ’ shades everything but where they overlap Therefore, (L 1 ’ ∪ L 2 ’)’ is the overlap (intersection)

7.  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 CS 3240 - Properties of Regular Languages7

8.  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? CS 3240 - Properties of Regular Languages8

9. 9 -x 1 x2x2 +x 3 a b a b a,b aaaa b b b b Double-a EVEN-EVEN

10. x i, y i ab x 1, y 1 x 2, y 3 x 1, y 2 x 2, y 3 x 3, y 1 x 1, y 4 x 1, y 2 x 2, y 4 x 1, y 1 x 3, y 1 x 3, y 3 x 3, y 2 x 1, y 4 x 2, y 2 x 1, y 3 x 2, y 4 x 3, y 2 x 1, y 3 x 3, y 3 x 3, y 1 x 3, y 4 x 3, y 2 x 3, y 4 x 3, y 1 x 2, y 2 x 3, y 4 x 1, y 1 x 1, y 3 x 2, y 1 x 1, y 4 x 3, y 4 x 3, y 2 x 3, y 3 x 2, y 1 x 3, y 3 x 1, y 2 For union: assign accepting states where any original x i or y i accept. For intersection: assign accepting states only where both original x i or y i accept simultaneously. No need to compute (L 1 ’ ∪ L 2 ’)’ ! For difference, assign accepting states where one accepts and the other does not.

11. 11 a b The resulting machine… a a a b b b a a a a a aa a b b b b b b b

12.  Given a word w, and a regular language, L, can we answer the question:  Is w ∊ L?  You tell me… CS 3240 - Properties of Regular Languages12

13.  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 CS 3240 - Properties of Regular Languages13

14.  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 CS 3240 - Properties of Regular Languages14

15.  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. CS 3240 - Properties of Regular Languages15

16.  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   CS 3240 - Properties of Regular Languages16

17.  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)… CS 3240 - Properties of Regular Languages17

18.  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 CS 3240 - Properties of Regular Languages18

19.  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 L 1 = L 2 iff (L 1 ∩ L 2 ’) ∪ (L 1 ’ ∩ L 2 ) = ∅ CS 3240 - Properties of Regular Languages19

20. CS 3240 - Properties of Regular Languages20

21. CS 3240 - Properties of Regular Languages21

22.  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 CS 3240 - Properties of Regular Languages22

23. CS 3240 - Properties of Regular Languages23

24. CS 3240 - Properties of Regular Languages24

25. CS 3240 - Properties of Regular Languages25

26.  Consider a n b n  ab is regular  ab + aabb = a n b n, 0 ≤ n ≤ 2, is regular  Any finite language is regular (why?)  But a n b n, n ≥ 0 is not regular (why not?)  How do we prove it’s not regular!?! CS 3240 - Properties of Regular Languages26

27.  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 a n b n where n is arbitrarily large  But intuition is often faulty. We need a proof! CS 3240 - Properties of Regular Languages27

28.  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) CS 3240 - Properties of Regular Languages28

29.  Consider a k b k, where k is greater than the number of states in a supposed DFA accepting all a n b n, 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 a k+im b k is also accepted!  This contradicts the existence of a DFA that accepts a n b n CS 3240 - Properties of Regular Languages29

30. CS 3240 - Properties of Regular Languages30 The first “revisit”

31.  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 CS 3240 - Properties of Regular Languages31

32.  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 CS 3240 - Properties of Regular Languages32

33.  Consider the string a p b p  It is in this language  It is long enough (≥ p in length)  Now let a p b p = xyz  Remember |xy| ≤ p  What can you conclude about y? CS 3240 - Properties of Regular Languages33

34.  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 CS 3240 - Properties of Regular Languages34

35.  a i b j, i > j  PALINDROME  w = w R (same backwards and forwards)  ww  Equal halves  PRIME (a m where m is prime)  SQUARE (a m where m is a perfect square) CS 3240 - Properties of Regular Languages35

36.  Strings with equal number of a’s and b’s  NOTPRIME CS 3240 - Properties of Regular Languages36

37.  NOTPRIME is pumpable!  Let y = the whole string (a km )  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 CS 3240 - Properties of Regular Languages37