Followup

1
Followup

• We know that if L1 and L2 are regular, then L1L2
  is regular.
• Let L L1L2 be a regular. Does this imply that
  L1 and L2 are regular?
• Consider L1aibi i?0, L2?

2
HW 4.1.17

• M(tail(L)) given M(L)
• Add new starting state
• Add transitions from new starting state to all
  other states on l

3
HW 4.1.25

• What the heck does min(L) mean?
• Consider L ab, abc
• min(L) ab
• Consider L(abcd)
• min(L) abc, not ac!
• Answer Sheet Take the transition graph of a dfa
  for L and delete all edges going out of any final
  state. Note that this works only if we start
  with a dfa.

4
5a Decision Problems for Regular Languages
4.2,4.3

• Decision Procedure for Regular Language
• Membership Tests for Empty, Finite, Infinite
  Language
• Equality of Regular Languages
• Pigeonhole Principle
• A Non-Regular Language
• The Pumping Lemma for Regular Languages
• The Language made up of All Possible Regular
  Expressions is NOT Regular (not in text)

5
Theorem 4.5 page 112

• Membership in Regular Language?
• To be a valid "Decision Procedure," it must end
  in finite length of time
• Feed word to dfa...
• This result is not so obvious for other classes
  of languages

6
Theorem 4.6 page 112

• Regular Language Empty?
• Reduce dfa to minimal number of states
• see if you can get from initial state to final
• Regular Language Finite?
• Minimal dfa
• see if any path from initial to final state
  includes a cycle

7
Theorem 4.7 page 112-3

• Equality of Regular Languages
• One proof in text
• Another would be to build minimal dfa for each.
• If the languages were equal, the dfa's would be
  identical (except for renaming of states).

8
Pigeonhole Principle

• If we put n objects (pigeons) into m boxes
  (holes), and if n gt m, then at least one box must
  have more than one object in it.

9
Proof on Non-Regularity

• Example 4.6 page 114
• anbn n?0 is not regular based on Pigeonhole
  Principle
• If it were regular (i.e. assume it is regular),
  it would be accepted by a "finite" state machine.
  This machine must therefore have a finite number
  of states. (duh!)
• Call that number m. Consider ambm.
• am must visit to m1 states (but only m in
  machine). Why?
• What are the pigeons? What are the pigeon holes?
• Therefore, at least one state is visited more
  than once in path from start to final state.
• We could repeat this loop or even eliminate it
  and still accept resulting word with more (or
  fewer) as than bs
• Contradiction!

10
Pumping Lemma for Regular Languages

• Theorem 4.8 pages 115-116
• Let L be an infinite regular language. Then there
  exists some positive integer m such that any w ?
  L with w ? m can be decomposed as
• w xyz,
• with
• xy ? m,
• and
• y ? 1,
• such that
• wi xyiz
• is also in L for all i 0, 1, 2, ...

11
Illustrations of "Pumping"

• Consider a machines accepting
• abc
• How many states?
• what are good choices for x, y, z?
• (aa)ba(bbaa)
• more possible w's, more choices for x, y, z

12
Proof of Pumping Lemma

• Consider dfa accepting word. It has a finite
  number of states, call this m.
• Consider a word w of length at least m.
• Machine accepting w must pass through w1
  states
• Apply Pigeonhole Principle
• Must pass through at least one state at least
  twice

13
About the Pumping Lemma

• Why its called a lemma
• Used to prove that languages are NOT regular, not
  to illustrate that they are regular.
• The joke We are working with the value of m,
  but if the language is not regular, no such m
  exists!

14
Selecting a Good w

• Consider L yyR y ? a,b
• w abba, w2 abbaabba, etc.
• doesn't work because w lt m
• How do we know this???
• w a2m am am, can be pumped, but doesn't prove
  anything
• should have tried ambmbmam, or ambbam

15
Pumping Lemma Proof for anbn

• ASS U ME that L is accepted by a dfa
• The dfa has some finite of states, call it m
• Consider w ambm. It can be decomposed into xyz,
  where y must ak for some kgt0, since xym.
  (i.e. xam-k-j, yak, zajbm for some j0)
• By P.L., all words of the form am(i-1)kbm
  i0 will also be accepted.
• Since am-kbm (i0 above) will be accepted, and is
  not in L, CONTRADICTION! (? assumption is false)

16
Applying the Pumping Lemma

• Exercise 4.3.3 page 122
• Let w (of pumping lemma) ambm
• What about L?
• Exercise 4.3.5(a)
• Let wap, where p is prime and pgtm1
• yak, for some kgt0, then xz ap-k.
• Consider xy(p-k)z y(p-k)xz ak(p-k)ap-k
  a(k1)(p-k)
• contains (k1)(p-k) a's which is a composite
  number
• Exercise 4.3.9(f)

17
Language of all Legal Regular Expressions is
not Regular

• Note that picking many words doesn't get us very
  far
• a?a?(a?b)?acan be raised to mth power and pumped
  
• Pick w (ma)m
• go through formal proof by trying to pick all
  possible x's, y's and z's
• What if parentheses were not part of Regular
  Expressions?
• nfa on next slide

18
nfa to accept Language of Regular Expressions
(less parentheses)

,?
a,b,?,?
a,b,?,?
NOTE ? and ? represent actual symbols of the
alphabet, not the empty word and empty set.
19
In-Class ExerciseExercise 4.3.10 page 123

• Work in groups of 2 or 3
• First 2 groups finished, put on board
• Answer on next slide (no peeking)