Properties of Regular Languages

We will talk about 4.2 & 4.3:
Elementary Questions
about Regular Languages
Pumping Lemma ...
4.2&3: Pumping Lemma

       Standard Representations
         of Regular Languages
          Regular Languages


DFAs
   ...
4.2&3: Pumping Lemma


When we say:   We are given
               a Regular Language   L



We mean:   Language L is in a ...
Elementary Questions
       about
 Regular Languages



                       4
4.2&3: Pumping Lemma


             Membership Question
Question:   Given regular language L
            and string w
    ...
4.2&3: Pumping Lemma


For any w   * : For a DFA of   L
       w                           w
                   or


    w...
4.2&3: Pumping Lemma


Question:    Given regular language     L
             how can we check
             if L is empty:...
4.2&3: Pumping Lemma


DFA

      L



DFA

      L

                          8
4.2&3: Pumping Lemma


Question:   Given regular language   L
            how can we check
            if L is finite?



...
4.2&3: Pumping Lemma

DFA


      L is infinite



DFA


      L is finite

                           10
4.2&3: Pumping Lemma




Theorem 4.6 (p.112)
There is an algorithm for determining
whether a regular language is empty,
fi...
4.2&3: Pumping Lemma


Question: Given regular languages     L1 and L2
          how can we check if L1       L2 ?


Answe...
4.2&3: Pumping Lemma

          ( L1    L2 )     ( L1    L2 )


L1        L2           and        L1        L2

     L1   ...
4.2&3: Pumping Lemma



          ( L1      L2 )        ( L1    L2 )


L1        L2               or          L1        L2...
4.2&3: Pumping Lemma




Theorem 4.5     (p.112)

Membership algorithm.

Theorem 4.6     (p.112)


Algorithm for determini...
Examples
                                     4.2&3: Pumping Lemma




Question: Given regular languages   L1 and       L2...
Examples cont.
                                      4.2&3: Pumping Lemma




Question: Given regular language L
         ...
Existence Examples
                                      4.2&3: Pumping Lemma




Question: Given regular language L
     ...
Existence Examples cont.
                                             4.2&3: Pumping Lemma




Question: Given regular lan...
Non-regular languages


        4.2&4.3         20
4.2&3: Pumping Lemma

                            n n
                          {a b : n 0}
Non-regular languages
        ...
4.2&3: Pumping Lemma

How can we prove that a language     L
is not regular?


Prove that there is no DFA that accepts    ...
The Pigeonhole Principle




                           23
4.2&3: Pumping Lemma

 4 pigeons




3 pigeonholes




                               24
4.2&3: Pumping Lemma




A pigeonhole must
contain at least two pigeons




                                   25
4.2&3: Pumping Lemma




  n pigeons

                ...........



m pigeonholes                 n   m

       ............
4.2&3: Pumping Lemma



       The Pigeonhole Principle
n pigeons
m pigeonholes
                There is a pigeonhole
    ...
The Pigeonhole Principle
          and
         DFAs


                           28
4.2&3: Pumping Lemma




     DFA with   4 states

                             b
 b       b

q1   a   q2 b       q3   b  ...
4.2&3: Pumping Lemma

In walks of strings:
 a, aa, aab no state is repeated
    b, aba have states repeated
For any string...
4.2&3: Pumping Lemma


However,                aabb
in walks of                         At least one
                     ...
4.2&3: Pumping Lemma
                                       Magic
If string   w has length | w |   4:   number?


    The ...
In general, for any DFA:
                                           4.2&3: Pumping Lemma




 String   w has length       ...
4.2&3: Pumping Lemma

In other words for a string    w:

   a     transitions are pigeons

   q     states are pigeonholes...
The Pumping Lemma




                    35
4.2&3: Pumping Lemma




Take an infinite regular language L


     DFA that accepts   L


                               ...
4.2&3: Pumping Lemma




Take string   w with w L



     There is a walk with label   w :


                  .........
 ...
4.2&3: Pumping Lemma


If string   w has length | w |       m
                               number of states
            ...
4.2&3: Pumping Lemma




Let q be the first state repeated
in the walk




              ......       q   ......
         ...
4.2&3: Pumping Lemma




Write   w   xyz


                          y



                 ......   q   ......


         ...
4.2&3: Pumping Lemma

Observations:       length   |x y|       m    number
                                              o...
4.2&3: Pumping Lemma

Observation:   The string       xz
               is accepted



                            y



  ...
4.2&3: Pumping Lemma

Observation:   The string       xyyz
               is accepted



                            y



...
4.2&3: Pumping Lemma

Observation:   The string       xyyyz
               is accepted



                            y


...
i       4.2&3: Pumping Lemma

In General:   The string       xy z
              is accepted      i 0, 1, 2, ...


        ...
4.2&3: Pumping Lemma

In General:   xy z ∈ L
                    i
                                  i      0, 1, 2, ...

...
4.2&3: Pumping Lemma



In other words, we described:




The Pumping Lemma !!!




                                      ...
4.2&3: Pumping Lemma


           The Pumping Lemma
• Given a Infinite Regular language    L
• there exists an integer   m...
4.2&3: Pumping Lemma


Example of pumping lemma

    L M = { all strings without
            substring 001 }
1            ...
4.2&3: Pumping Lemma


  L M = { all strings without
          substring 001 }
   1                 0          0,1
       ...
4.2&3: Pumping Lemma

     The Pumping Lemma: L is reg. & inf.

• there exists an integer m                  Let m = 4
 • ...
4.2&3: Pumping Lemma




For any w L(M), |w| 4:
Then either w= 00...0, or w has 1s;
(1)w contains 0 only: let let x=00, y=...
4.2&3: Pumping Lemma


    Example of pumping lemma

   L M = { a nb m : n   2, m      2}


L(M) is infinite and regular, ...
4.2&3: Pumping Lemma


        L M = { anbm : n            2, m      2}
      The Pumping Lemma: L is reg. & inf.


• ther...
4.2&3: Pumping Lemma


    Example of pumping lemma

   L M = { uabv : u,v      {a,b}* }


L(M) is infinite and regular, s...
4.2&3: Pumping Lemma



       L M = { uabv : u,v                  {a,b}* }
      The Pumping Lemma: L is reg. & inf.


• ...
4.2&3: Pumping Lemma




          Discussion


 Section 4.2
 What is the Pumping lemma?
 Show a regular language satis...
Applications
         of
the Pumping Lemma



                    58
4.2&3: Pumping Lemma




Theorem: The language L     n n
                          {a b : n 0}
                 is not reg...
4.2&3: Pumping Lemma


                     n n
            L {a b : n 0}
      Since L is infinite
      L is regular  L ...
4.2&3: Pumping Lemma

          (The Pumping Lemma)
• Given a infinite regular language       L
 for any positive integer ...
n n                 4.2&3: Pumping Lemma

       L {a b : n 0}
Since L is infinite, if L is regular then L
satisfies the P...
4.2&3: Pumping Lemma

Any decomposition of w = ambm:
                          m      m
                w       a b       ...
4.2&3: Pumping Lemma
                        m    m
w         xyz          a b         y       k
                         ...
4.2&3: Pumping Lemma

               m k      ki      m         m ( i 1) k        m
   wi      a         a b              ...
4.2&3: Pumping Lemma




Therefore L does not satisfy the
pumping lemma.




Conclusion: L is not a regular language


   ...
4.2&3: Pumping Lemma




Non-regular languages           R
                        L {uu : u           *}


            Re...
4.2&3: Pumping Lemma




Theorem: The language
                  R
          L {uu : u        *}         {a, b}
          ...
4.2&3: Pumping Lemma

                     R
          L {ww : w {a, b}*}
      Since L is infinite
      L is regular  L ...
4.2&3: Pumping Lemma
                       R
                L {uu : u        *}
   Since L is infinite, if L is regular ...
4.2&3: Pumping Lemma


Any decomposition of w :
                m   m   m   m
         w    a b b a           xyz

it must...
4.2&3: Pumping Lemma
                            m       m   m   m
w        xyz           a b b a                       y ...
4.2&3: Pumping Lemma


            m k    ki       m   m   m           m ( i 1) k    m       m   m
 wi     a         a b b...
4.2&3: Pumping Lemma




Therefore: L does not satisfy the
pumping lemma.




Conclusion: L is not a regular language


  ...
Prove a language is not
regular by pumping lemma

   How to write a
    good proof?

                            75
4.2&3: Pumping Lemma
Show L={w: na(w) > nb(w)} is not
          regular.

w = ambm-1




Let y = ak

wi     L

     L is n...
4.2&3: Pumping Lemma


Non-regular languages
                        n l n l
                L {a b c          : n, l     ...
4.2&3: Pumping Lemma




Theorem: The language
                n l n l
          L {a b c         : n, l   0}
          is...
n l n l                 4.2&3: Pumping Lemma

      L {a b c        : n, l   0}


Assume for contradiction
that L is a reg...
n l n l                 4.2&3: Pumping Lemma

            L {a b c          : n, l   0}
   Since L is infinite, if L is re...
Any decomposition of w :
                                       4.2&3: Pumping Lemma



                  m m 2m
         ...
4.2&3: Pumping Lemma


                   m m 2m                        k
w= x yz           a b c                 y    a ,...
4.2&3: Pumping Lemma




Therefore: L does not satisfy the
pumping lemma.




Conclusion: L {a nbl c n l : n, l    0}
    ...
4.2&3: Pumping Lemma



      n l n l
L {a b c        : n, l     0} is not regular,
                     n l   k
how about...
4.2&3: Pumping Lemma




Non-regular languages           n!
                         L {a : n 0}


            Regular lan...
4.2&3: Pumping Lemma




Theorem: The language         n!
                        L {a : n 0}
                        is n...
4.2&3: Pumping Lemma

                        n!
            L {a : n 0}
   Since     L     is infinite
    L is regular  ...
4.2&3: Pumping Lemma
                    n!
            L {a : n 0}
Let   m be the integer in the Pumping Lemma


Pick a s...
4.2&3: Pumping Lemma

    Any decomposition of w :
                          m!
              w= a              xyz
From t...
4.2&3: Pumping Lemma
            m!                          k
xyz    a                          y    a , 1 k             ...
4.2&3: Pumping Lemma

            2          m! k
  w2     xy z      a          1 k   m

Since:            n!
          L ...
However:
                                                  4.2&3: Pumping Lemma


        m!    m! k         (m 1)!       ...
4.2&3: Pumping Lemma


What happen if m = 1?

If m =1, we choose w = a, with x= = z, and
y=a is the only decomposition for...
4.2&3: Pumping Lemma

                 n!
Show L {a : n              0} is not regular.
Since L is infinite, if L does not...
4.2&3: Pumping Lemma
                           n!
     Show L           {a : n 0} is not regular.
                     w ...
4.2&3: Pumping Lemma




Non-regular languages L     n l   k
                          {a b c : k        n l}



         ...
4.2&3: Pumping Lemma




Theorem: The language L {a nbl c k : k                 n l}
                                  is ...
4.2&3: Pumping Lemma
                   n l   k
            L {a b c : k     n l}
   Since L is infinite, if L is regular ...
4.2&3: Pumping Lemma

    Any decomposition of w :
                       m! ( m 1)!
              w=a c                 x...
4.2&3: Pumping Lemma

            m! ( m 1)!               k
xyz       a c                y      a , 1 k             m
   ...
4.2&3: Pumping Lemma

            m! ( m 1)!                 k
xyz       a c                   y     a , 1 k             m...
4.2&3: Pumping Lemma




 Therefore: L does not satisfy the
 pumping lemma.


Conclusion: L {a b c : k
                   ...
4.2&3: Pumping Lemma

        n!                2   6   24   120
Show {a : n 0} {a, a , a , a , a             , ...} not r...
4.2&3: Pumping Lemma
          n!                                 2      6     24     120
  L {a : n 0} {a, a , a , a , a ...
4.2&3: Pumping Lemma




 Consider the following two languages,
 determine which is regular.
           n l
  L1 {a b : n...
4.2&3: Pumping Lemma




 Consider the following two languages,
 determine which is regular.
           n l
  L1 {a b : n...
4.2&3: Pumping Lemma




Non-regular languages           n l
                        L {a b : n               l}


       ...
4.2&3: Pumping Lemma




Theorem: The language L              n l
                                  {a b : n             l...
4.2&3: Pumping Lemma


                       n l
              L {a b : n            l}
   Since     L     is infinite
  ...
4.2&3: Pumping Lemma

                    n l
                L {a b : n     l}
Let   m be the integer in the Pumping Lemm...
4.2&3: Pumping Lemma

    Any decomposition of w :
                      m! ( m 1)!
              w=a b                xyz...
4.2&3: Pumping Lemma

            m! ( m 1)!               k
xyz       a b                y      a , 1 k              m
  ...
4.2&3: Pumping Lemma

            m! ( m 1)!                 k
xyz       a b                   y     a , 1 k            m
...
4.2&3: Pumping Lemma




Therefore: L does not satisfy the
pumping lemma.


  Conclusion: L       n l
                   {...
4.2&3: Pumping Lemma




Theorem: The language L                      n l
                                          {a b :...
4.2&3: Pumping Lemma

                                   is not regular
               n l
     L {a b : n               l...
Similar problem:
                                                    4.2&3: Pumping Lemma




 Show L {a b c : k          ...
4.2&3: Pumping Lemma

   Prove or disprove:
If L1   L2 and L1 are regular, then L2 must be regular.

If L1 L2 is regular a...
4.2&3: Pumping Lemma



       Regular or Non-regular?

                     R
     L1 {uww v : u, v, w {a, b} }.


      ...
4.2&3: Pumping Lemma




  4.3 (p.122)
 3, 4, 5abcd, 6 ~9, 13, 14, 15abcf, 16, 17, 19,
 24, 26 and 16 (p.110)
Hand in: 4bc...
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Class6

  1. 1. Properties of Regular Languages We will talk about 4.2 & 4.3: Elementary Questions about Regular Languages Pumping Lemma for Regular Languages 1
  2. 2. 4.2&3: Pumping Lemma Standard Representations of Regular Languages Regular Languages DFAs Regular Grammars NFAs Regular Expressions 2
  3. 3. 4.2&3: Pumping Lemma When we say: We are given a Regular Language L We mean: Language L is in a standard representation 3
  4. 4. Elementary Questions about Regular Languages 4
  5. 5. 4.2&3: Pumping Lemma Membership Question Question: Given regular language L and string w how can we check if w L? Hint: Take a DFA that accepts L Answer: Take the DFA that accepts L and check if w is accepted 5
  6. 6. 4.2&3: Pumping Lemma For any w * : For a DFA of L w w or w L w L Answer: There is a unique path starting from q0 for w in such DFA. w L if it ends at a final state. Theorem 4.5 (p.112) 6
  7. 7. 4.2&3: Pumping Lemma Question: Given regular language L how can we check if L is empty: (L )? Answer: Take the DFA that accepts L Check if there is a path from the initial state to a final state 7
  8. 8. 4.2&3: Pumping Lemma DFA L DFA L 8
  9. 9. 4.2&3: Pumping Lemma Question: Given regular language L how can we check if L is finite? Answer: Take the DFA that accepts L Check if there is a walk with cycle from the initial state to a final state 9
  10. 10. 4.2&3: Pumping Lemma DFA L is infinite DFA L is finite 10
  11. 11. 4.2&3: Pumping Lemma Theorem 4.6 (p.112) There is an algorithm for determining whether a regular language is empty, finite, or infinite. 11
  12. 12. 4.2&3: Pumping Lemma Question: Given regular languages L1 and L2 how can we check if L1 L2 ? Answer: check if( L1 L2 ) ( L1 L2 ) Since L1and L2are regular, (L1 L2 ) (L1 L2 ) is 3 L also regular. From Thm. 4.6, we can check whether L3 is empty. If L3= then L1= L2 otherwise L1 L2 . Theorem 4.7 (p.112) 12
  13. 13. 4.2&3: Pumping Lemma ( L1 L2 ) ( L1 L2 ) L1 L2 and L1 L2 L1 L2 L L2 L1 L1 2 L1 L2 and L2 L1 L1 L2 13
  14. 14. 4.2&3: Pumping Lemma ( L1 L2 ) ( L1 L2 ) L1 L2 or L1 L2 L1 L2 L2 L1 L1 L2 or L2 L1 L1 L2 14
  15. 15. 4.2&3: Pumping Lemma Theorem 4.5 (p.112) Membership algorithm. Theorem 4.6 (p.112) Algorithm for determining whether a regular language is empty, finite, or infinite. Theorem 4.7 (p.112) Algorithm for checking whether 2 regular languages are the same. 15
  16. 16. Examples 4.2&3: Pumping Lemma Question: Given regular languages L1 and L2 how can we check if L1 L2 ? Question: Given regular languages L1 and L2 and any string w how can we check if w L1 L2 ? 16
  17. 17. Examples cont. 4.2&3: Pumping Lemma Question: Given regular language L how can we check if L? Definition: A language is said to be a palindrome language if L = LR. Given regular language L how can we check if L is a palindrome language. 17
  18. 18. Existence Examples 4.2&3: Pumping Lemma Question: Given regular language L how can we check if there exist any w L such that wR L ? Question: Given regular language L and string u how can we check if L contains any w such that u is a substring of it, i.e. w =v1u v2 for u, v1,v2 *? 18
  19. 19. Existence Examples cont. 4.2&3: Pumping Lemma Question: Given regular language L how can we check if there exist any w L such that |w| is an odd number ? Question: Given regular language L how can we check if L contains an infinite number of odd-number strings? You are able to do (4.2) hw# 1~15 except 10 p.113. 19
  20. 20. Non-regular languages 4.2&4.3 20
  21. 21. 4.2&3: Pumping Lemma n n {a b : n 0} Non-regular languages R {ww : w {a, b}*} Regular languages a *b b*c a b c ( a b) * etc... 21
  22. 22. 4.2&3: Pumping Lemma How can we prove that a language L is not regular? Prove that there is no DFA that accepts L Problem: this is not easy to prove Solution: the Pumping Lemma !!! 22
  23. 23. The Pigeonhole Principle 23
  24. 24. 4.2&3: Pumping Lemma 4 pigeons 3 pigeonholes 24
  25. 25. 4.2&3: Pumping Lemma A pigeonhole must contain at least two pigeons 25
  26. 26. 4.2&3: Pumping Lemma n pigeons ........... m pigeonholes n m ........... 26
  27. 27. 4.2&3: Pumping Lemma The Pigeonhole Principle n pigeons m pigeonholes There is a pigeonhole n m with at least 2 pigeons ........... 27
  28. 28. The Pigeonhole Principle and DFAs 28
  29. 29. 4.2&3: Pumping Lemma DFA with 4 states b b b q1 a q2 b q3 b q4 a a 29
  30. 30. 4.2&3: Pumping Lemma In walks of strings: a, aa, aab no state is repeated b, aba have states repeated For any string w, |w| 3, then its walk in this DFA may repeat the state(s) or does not repeat. b b b q1 a q2 a q3 b q4 a a 30
  31. 31. 4.2&3: Pumping Lemma However, aabb in walks of At least one bbaa state is repeated strings with length 4: abbabb abbbabbabb... b b b q1 a q2 a q3 b q4 a a 31
  32. 32. 4.2&3: Pumping Lemma Magic If string w has length | w | 4: number? The transitions of string w are more than the states of the DFA Thus, at least one state must be repeated b b b q1 a q2 a q3 b q4 a a 32
  33. 33. In general, for any DFA: 4.2&3: Pumping Lemma String w has length number of states A state q must be repeated in the walk of w walk of w ...... q ...... Repeated state 33
  34. 34. 4.2&3: Pumping Lemma In other words for a string w: a transitions are pigeons q states are pigeonholes walk of w ...... q ...... Repeated state 34
  35. 35. The Pumping Lemma 35
  36. 36. 4.2&3: Pumping Lemma Take an infinite regular language L DFA that accepts L m states 36
  37. 37. 4.2&3: Pumping Lemma Take string w with w L There is a walk with label w : ......... walk w 37
  38. 38. 4.2&3: Pumping Lemma If string w has length | w | m number of states of the DFA then, from the pigeonhole principle: a state q is repeated in the walk w ...... q ...... walk w 38
  39. 39. 4.2&3: Pumping Lemma Let q be the first state repeated in the walk ...... q ...... walk w 39
  40. 40. 4.2&3: Pumping Lemma Write w xyz y ...... q ...... x z 40
  41. 41. 4.2&3: Pumping Lemma Observations: length |x y| m number of states length | y| 1 of DFA y ...... q ...... x z 41
  42. 42. 4.2&3: Pumping Lemma Observation: The string xz is accepted y ...... q ...... x z 42
  43. 43. 4.2&3: Pumping Lemma Observation: The string xyyz is accepted y ...... q ...... x z 43
  44. 44. 4.2&3: Pumping Lemma Observation: The string xyyyz is accepted y ...... q ...... x z 44
  45. 45. i 4.2&3: Pumping Lemma In General: The string xy z is accepted i 0, 1, 2, ... y ...... q ...... x z 45
  46. 46. 4.2&3: Pumping Lemma In General: xy z ∈ L i i 0, 1, 2, ... The original language y ...... q ...... x z 46
  47. 47. 4.2&3: Pumping Lemma In other words, we described: The Pumping Lemma !!! 47
  48. 48. 4.2&3: Pumping Lemma The Pumping Lemma • Given a Infinite Regular language L • there exists an integer m • for any string w L with length | w | m • exists a decompsition of w: wxyz with |x y| m and | y | 1 • such that: wi L for all i 0, 1, 2, ... i xy z 48
  49. 49. 4.2&3: Pumping Lemma Example of pumping lemma L M = { all strings without substring 001 } 1 0 0,1 1 q0 q1 0 1 q3 q2 0 49
  50. 50. 4.2&3: Pumping Lemma L M = { all strings without substring 001 } 1 0 0,1 1 q0 q1 0 1 q3 q2 0 L(M) is infinite and regular, show it satisfies the pumping lemma. 50
  51. 51. 4.2&3: Pumping Lemma The Pumping Lemma: L is reg. & inf. • there exists an integer m Let m = 4 • for any string w L with length | w | m a decomposition w x y z with | x y | m and | y | 1 1 0 0,1 1 q0 q1 0 1 q3 q2 0 • such that: i for all xy z L i 0, 1, 2, ... 51
  52. 52. 4.2&3: Pumping Lemma For any w L(M), |w| 4: Then either w= 00...0, or w has 1s; (1)w contains 0 only: let let x=00, y=0 (or x= , y=0) and z= rest of the string (2)w contains one: w is in the form of (1+01)* if w stops at q0, or w is in the form of (1+01)*0 if w stops at q1, or w is in the form of (1+01)*000* if w stops at q2. Since w contains 1, the first 1 in w must occurs at the first or second symbols of w. For the first case, let x= , y=1, and z= rest of the string ; for the second case, let x= , y=01, and z= rest of the string. 52
  53. 53. 4.2&3: Pumping Lemma Example of pumping lemma L M = { a nb m : n 2, m 2} L(M) is infinite and regular, show it satisfies the pumping lemma. 53
  54. 54. 4.2&3: Pumping Lemma L M = { anbm : n 2, m 2} The Pumping Lemma: L is reg. & inf. • there exists an integer m Let m = 5 • for any string w L with length | w | m a decomposition w x y z with | x y | m and | y | 1 • such that: i for all xy z L i 0, 1, 2, ... 54
  55. 55. 4.2&3: Pumping Lemma Example of pumping lemma L M = { uabv : u,v {a,b}* } L(M) is infinite and regular, show it satisfies the pumping lemma. 55
  56. 56. 4.2&3: Pumping Lemma L M = { uabv : u,v {a,b}* } The Pumping Lemma: L is reg. & inf. • there exists an integer m Let m = 4 • for any string w L with length | w | m a decomposition w x y z with | x y | m and | y | 1 • such that: i for all xy z L i 0, 1, 2, ... 56
  57. 57. 4.2&3: Pumping Lemma Discussion  Section 4.2  What is the Pumping lemma?  Show a regular language satisfies the pumping lemma. 57
  58. 58. Applications of the Pumping Lemma 58
  59. 59. 4.2&3: Pumping Lemma Theorem: The language L n n {a b : n 0} is not regular Proof: Use the Pumping Lemma 59
  60. 60. 4.2&3: Pumping Lemma n n L {a b : n 0} Since L is infinite L is regular L satisfies P.L. (L satisfies P.L.) ( L is regular) (contrapositive) Goal is to show L does not satisfy the pumping lemma then L is not regulatr. by contradiction i.e. assume L is regular and show L does not satisfy the pumping lemma  contradiction L is not regular 60
  61. 61. 4.2&3: Pumping Lemma (The Pumping Lemma) • Given a infinite regular language L for any positive integer m • there exists an integer m we pick a string w L • for any string w L with length | w | m for any possible decomposition of w • exist a decompsition of w: w xyz with |x y| m and | y | 1 some i such that wi= xyiz L • such that: wi L for all i 0, 1, 2, ... i xy z 61
  62. 62. n n 4.2&3: Pumping Lemma L {a b : n 0} Since L is infinite, if L is regular then L satisfies the Pumping lemma. Let m be the positive integer in the Pumping lemma. Pick a string w such that: w L length | w| m m m We pick w a b 62
  63. 63. 4.2&3: Pumping Lemma Any decomposition of w = ambm: m m w a b xyz it must be that length | x y | m, | y | 1 m m m m xyz a b a...aa...aa...ab...b x y z Thus: k y a , 1 k m 63
  64. 64. 4.2&3: Pumping Lemma m m w xyz a b y k a , 1 k m i wi xy z L k i a...aa...aa...aa...aa...ab...b...b x y y ......y z m k ki m m ( i 1) k m wi a a b a b for all i 0, 1, 2, ... We need to find some i in order to have wi= xyiz L 64
  65. 65. 4.2&3: Pumping Lemma m k ki m m ( i 1) k m wi a a b a b for all i 0, 1, 2, ... We need to find some i in order to have wi= xyiz L Let i = 2 m ( 2 1) k m m k m w2 a b a b Since 1 k m , thus m k m n n w2 a b L {a b : n 0} 65
  66. 66. 4.2&3: Pumping Lemma Therefore L does not satisfy the pumping lemma. Conclusion: L is not a regular language 66
  67. 67. 4.2&3: Pumping Lemma Non-regular languages R L {uu : u *} Regular languages 67
  68. 68. 4.2&3: Pumping Lemma Theorem: The language R L {uu : u *} {a, b} is not regular Proof: Use the Pumping Lemma 68
  69. 69. 4.2&3: Pumping Lemma R L {ww : w {a, b}*} Since L is infinite L is regular L satisfies P.L. (L satisfies P.L.) ( L is regular) (contrapositive) Goal is to show L does not satisfy the pumping lemma then L is not regulatr. by contradiction i.e. assume L is regular and show L does not satisfy the pumping lemma  contradiction L is not regular 69
  70. 70. 4.2&3: Pumping Lemma R L {uu : u *} Since L is infinite, if L is regular then L satisfies the Pumping lemma. Let m be the positive integer in the Pumping lemma. Pick a string w such that: w L and length | w| m m m m m We pick w a b b a 70
  71. 71. 4.2&3: Pumping Lemma Any decomposition of w : m m m m w a b b a xyz it must be that length | x y | m, | y | 1 m m m m xyz a...aa...a...ab...bb...ba...a x y z k Thus: y a , 1 k m 71
  72. 72. 4.2&3: Pumping Lemma m m m m w xyz a b b a y k a , 1 k m i wi xy z L 2m m a...aa...aa...aa...a...ab...bb...ba...a x y y... y z m k ki m m m m ( i 1) k m m m wi a a b b a a b b a for all i = 0, 1, 2, 3, … We need to find some i in order to have wi= xyiz L 72
  73. 73. 4.2&3: Pumping Lemma m k ki m m m m ( i 1) k m m m wi a a b b a a b b a for all i = 0, 1, 2, 3, … We need to find some i in order to have wi= xyiz L Let i = 2 m ( 2 1) k m m m m k m m m w2 a b b a a b b a Since 1 k m , thus m k m m m R w2 a b b a L {uu : u {a, b}*} 73
  74. 74. 4.2&3: Pumping Lemma Therefore: L does not satisfy the pumping lemma. Conclusion: L is not a regular language 74
  75. 75. Prove a language is not regular by pumping lemma How to write a good proof? 75
  76. 76. 4.2&3: Pumping Lemma Show L={w: na(w) > nb(w)} is not regular. w = ambm-1 Let y = ak wi L L is not regular. 76
  77. 77. 4.2&3: Pumping Lemma Non-regular languages n l n l L {a b c : n, l 0} Regular languages 77
  78. 78. 4.2&3: Pumping Lemma Theorem: The language n l n l L {a b c : n, l 0} is not regular Proof: Use the Pumping Lemma 78
  79. 79. n l n l 4.2&3: Pumping Lemma L {a b c : n, l 0} Assume for contradiction that L is a regular language Since L is infinite we can apply the Pumping Lemma 79
  80. 80. n l n l 4.2&3: Pumping Lemma L {a b c : n, l 0} Since L is infinite, if L is regular then L satisfies the Pumping lemma. Let m be the positive integer in the Pumping lemma. Pick a string w such that: w L and length | w| m We pick w a b c m m 2m 80
  81. 81. Any decomposition of w : 4.2&3: Pumping Lemma m m 2m w=a b c xyz From the Pumping Lemma it must be that length | x y | m, | y | 1 m m 2m w = xyz a...aa...aa...ab...bc...cc...c x y z k Thus: y a for some k , 1 k m 81
  82. 82. 4.2&3: Pumping Lemma m m 2m k w= x yz a b c y a , k 1 i wi xy z L i 0, 1, 2, ... = am-k(ak)ibmc2m = am+(i-1)k bmc2m Let i = 0 0 m k m 2m w0 x y z a b c L since (m-k)+m = 2m- k 2m for k 1 82
  83. 83. 4.2&3: Pumping Lemma Therefore: L does not satisfy the pumping lemma. Conclusion: L {a nbl c n l : n, l 0} is not a regular language 83
  84. 84. 4.2&3: Pumping Lemma n l n l L {a b c : n, l 0} is not regular, n l k how about L1 {a b c : k n l} ? L1 must be non-regular otherwise …. 84
  85. 85. 4.2&3: Pumping Lemma Non-regular languages n! L {a : n 0} Regular languages 85
  86. 86. 4.2&3: Pumping Lemma Theorem: The language n! L {a : n 0} is not regular n! 1 2  (n 1) n Proof: Use the Pumping Lemma 86
  87. 87. 4.2&3: Pumping Lemma n! L {a : n 0} Since L is infinite L is regular L satisfies P.L. (L satisfies P.L.) ( L is regular) (contrapositive) Goal is to show L does not satisfy the pumping lemma thus L is not regular. by contradiction 87
  88. 88. 4.2&3: Pumping Lemma n! L {a : n 0} Let m be the integer in the Pumping Lemma Pick a string w such that: w L length | w| m m! We pick w a 88
  89. 89. 4.2&3: Pumping Lemma Any decomposition of w : m! w= a xyz From the Pumping Lemma it must be that length | x y | m, | y | 1 m m! m m! w = xyz a a...aa...aa...aa...aa...a x y z k Thus: y a for some k , 1 k m 89
  90. 90. 4.2&3: Pumping Lemma m! k xyz a y a , 1 k m m! k k i m! ( i 1) k wi a (a ) a i 0, 1, 2, ... Let i = 2 2 m! k w2 xy z a 1 k m 90
  91. 91. 4.2&3: Pumping Lemma 2 m! k w2 xy z a 1 k m Since: n! L {a : n 0} In order to show w2 L, we need to show m!+k is not a factorial number. By showing m! < m!+k < ( m+1)! It works only for all k = 1, 2, 3, …, m when m 2 ! 91
  92. 92. However: 4.2&3: Pumping Lemma m! m! k (m 1)! given m 1 m!< m!+k is trivial since k 1. To show m!+k < (m+1)! From (m+1)! = (m+1) m! = m! + m m! > m! + k 1 = m!+k since m k & m! > 1! =1 m! k p! for any p 92
  93. 93. 4.2&3: Pumping Lemma What happen if m = 1? If m =1, we choose w = a, with x= = z, and y=a is the only decomposition for w = a. Then wi = ai for all i = 0, 1, 2, … Clearly that when i= 3, w3 = a3 L={an! : n 0}. Now, put everything together: 93
  94. 94. 4.2&3: Pumping Lemma n! Show L {a : n 0} is not regular. Since L is infinite, if L does not satisfy pumping lemma then L is not regular. So the goal is to prove L does not satisfy pumping lemma. (it is proved by contradiction) For any given positive integer m in m pumping lemma, w L and |w| the if m =1, we choose w = a, then x = z = , and y=a is the only decomposition for w = a. Then wi = ai for all i = 0, 1, 2, … Clearly that when i= 3, A contradiction to w3 = a3 L={an! : n 0}. pumping lemma! 94
  95. 95. 4.2&3: Pumping Lemma n! Show L {a : n 0} is not regular. w L and |w| m If m > 1, we pick w = am!. Consider all possible decompositions of w = xyz with |xy| m, |y| 1, k y must be in the form of y a for some k , 1 k m m! k k i m! ( i 1) k wi a (a ) a for i= 0, 1, 2, … 2 m! k choose i = 2: w2 xy z a 1 k m From m! < m!+k m! + m < m! + m m! (since m!>1) = (m+1)! i.e. m! < m!+k < (m+1)!, m!+k is not a factorial number. w2 L, a contradiction. Thus L does not satisfy pumping lemma, and therefore L is not regular. 95
  96. 96. 4.2&3: Pumping Lemma Non-regular languages L n l k {a b c : k n l} Regular languages 96
  97. 97. 4.2&3: Pumping Lemma Theorem: The language L {a nbl c k : k n l} is not regular Proof: Use the Pumping Lemma Choose a w from L and, later,exist an i such that wi L What does wi L imply? 97
  98. 98. 4.2&3: Pumping Lemma n l k L {a b c : k n l} Since L is infinite, if L is regular then L satisfies the Pumping lemma. Let m be the positive integer in the Pumping lemma. Pick a string w such that: w L length | w| m We pick 99
  99. 99. 4.2&3: Pumping Lemma Any decomposition of w : m! ( m 1)! w=a c xyz From the Pumping Lemma it must be that length | x y | m, | y | 1 m! (m 1)! m! ( m 1)! w = xyz a c a...aa...aa...ac...cc...c x y z Thus: y a k for some k , 1 k m 100
  100. 100. 4.2&3: Pumping Lemma m! ( m 1)! k xyz a c y a , 1 k m m! k k i ( m 1)! m! ( i 1) k ( m 1)! wi a (a ) c a c i 0, 1, 2, ... How to choose i such that wi L? i.e. choose i such that m!+(i-1)k = (m+1)! m! m Is such i an integer? choose i 1 k 101
  101. 101. 4.2&3: Pumping Lemma m! ( m 1)! k xyz a c y a , 1 k m m! k k i ( m 1)! m! ( i 1) k ( m 1)! wi a (a ) c a c i 0, 1, 2, ... Since 1 k m, k | m! m! m when i 1 (: a positive integer) k m!+(i-1)k = (m+1)! which causes wi L A contradiction to pumping lemma! 102
  102. 102. 4.2&3: Pumping Lemma Therefore: L does not satisfy the pumping lemma. Conclusion: L {a b c : k n l k n l} is not a regular language Theorem: The language L {a nbl c k : k n l} is not regular 103
  103. 103. 4.2&3: Pumping Lemma n! 2 6 24 120 Show {a : n 0} {a, a , a , a , a , ...} not regular. Since L is infinite, if L does not satisfy pumping lemma then L is not regular. So the goal is to prove L does not satisfy pumping lemma. (it is proved by contradiction) For any given positive integer m in m pumping lemma, w L and |w| the if m =1, we choose w = a, then x = z = , and y=a is the only decomposition for w = a. Then wi = ai for all i = 0, 1, 2, … Clearly that when i= 3, A contradiction to w3 = a3 L={an! : n 0}. pumping lemma! 104
  104. 104. 4.2&3: Pumping Lemma n! 2 6 24 120 L {a : n 0} {a, a , a , a , a , ...} w L and |w| m If m > 1, we pick w = am!. Consider all possible decompositions of w = xyz with |xy| m, |y| 1, k y must be in the form of y a for some k , 1 k m m! k k i m! ( i 1) k wi a (a ) a for i= 0, 1, 2, … 2 m! k choose i = 2: w2 xy z a 1 k m From m! < m!+k m! + m < m! + m m! (since m!>1) = (m+1)! i.e. m! < m!+k < (m+1)!, m!+k is not a factorial number. w2 L, a contradiction. Thus L does not satisfy pumping lemma, and therefore L is not regular. 105
  105. 105. 4.2&3: Pumping Lemma  Consider the following two languages, determine which is regular. n l L1 {a b : n l} n l L2 {a b : n l} Both are non-regular. Which is easier to prove by pumping lemma? 106
  106. 106. 4.2&3: Pumping Lemma  Consider the following two languages, determine which is regular. n l L1 {a b : n l} n l L2 {a b : n l} Both are non-regular. Which is easier to prove by pumping lemma? 107
  107. 107. 4.2&3: Pumping Lemma Non-regular languages n l L {a b : n l} Regular languages 108
  108. 108. 4.2&3: Pumping Lemma Theorem: The language L n l {a b : n l} is not regular Proof: Use the Pumping Lemma The proof is similar to the proof in n l k L {a b c : k n l} 109
  109. 109. 4.2&3: Pumping Lemma n l L {a b : n l} Since L is infinite L is regular L satisfies P.L. (L satisfies P.L.) ( L is regular) (contrapositive) Goal is to show L does not satisfy the pumping lemma thus L is not regular. by contradiction 110
  110. 110. 4.2&3: Pumping Lemma n l L {a b : n l} Let m be the integer in the Pumping Lemma Pick a string w such that: w L length | w| m We pick w a b m! ( m 1)! 111
  111. 111. 4.2&3: Pumping Lemma Any decomposition of w : m! ( m 1)! w=a b xyz From the Pumping Lemma it must be that length | x y | m, | y | 1 m! (m 1)! m! m 1! w = xyz a b a...aa...aa...ab...bb...b x y z Thus: y a k for some k , 1 k m 112
  112. 112. 4.2&3: Pumping Lemma m! ( m 1)! k xyz a b y a , 1 k m m! k k i ( m 1)! m! ( i 1) k ( m 1)! wi a (a ) b a b i 0, 1, 2, ... How to choose i such that wi L? i.e. choose i such that m!+(i-1)k = (m+1)! m! m Is such i an integer? choose i 1 k 113
  113. 113. 4.2&3: Pumping Lemma m! ( m 1)! k xyz a b y a , 1 k m m! k k i ( m 1)! m! ( i 1) k ( m 1)! wi a (a ) b a b i 0, 1, 2, ... Since 1 k m, k | m! m! m when i 1 (: a positive integer) k m!+(i-1)k = (m+1)! which causes wi L A contradiction to pumping lemma! 114
  114. 114. 4.2&3: Pumping Lemma Therefore: L does not satisfy the pumping lemma. Conclusion: L n l {a b : n l} is not a regular language Theorem: The language L n l {a b : n l} is not regular 115
  115. 115. 4.2&3: Pumping Lemma Theorem: The language L n l {a b : n l} is not regular Proof: Another way to prove L is not regular by closure property of regular family. L is regular then L is also regular What is the complement of L? 116
  116. 116. 4.2&3: Pumping Lemma is not regular n l L {a b : n l} Proof: Assume L is regular then L is also regular Let R = L L(a*b*) an intersection of two regular languages, thus R is a regular language too. In fact, R = {anbn : n 0} which is not regular. A contradiction! Therefore, the assumption that L is regular is not true, i.e., L is not regular. 117
  117. 117. Similar problem: 4.2&3: Pumping Lemma Show L {a b c : k n l} is not regular. n l k Show L ={ w: na(w) nb(w) } is not regular. Is L* regular or non-regular? Prove or disprove: If L1 and L2 are non-regular, then L1 L2 must be non-regular. If L1 and L2 are non-regular, then L1 L2 must be non-regular. If L is regular then any subset of L must be regular too. If L is non-regular then any subset of L must be non-regular . 118
  118. 118. 4.2&3: Pumping Lemma Prove or disprove: If L1 L2 and L1 are regular, then L2 must be regular. If L1 L2 is regular and L1is finite, then L2 must be regular. The family of regular languages closed under finite number union. The family of regular languages closed under infinite number union. 119
  119. 119. 4.2&3: Pumping Lemma Regular or Non-regular? R L1 {uww v : u, v, w {a, b} }. R L2 {uww v : u, v, w {a, b} , | u | | v |}. // L1 is regular, but L2 is non-regular. 120
  120. 120. 4.2&3: Pumping Lemma 4.3 (p.122) 3, 4, 5abcd, 6 ~9, 13, 14, 15abcf, 16, 17, 19, 24, 26 and 16 (p.110) Hand in: 4bcf, 5bcd, 6, 13, 16, 24, 16(p.110). 121

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