Note: Text Cross-References in Parenthesis Are for 2Nd Ed

Chapter 4 Properties of Regular Languages 2/14/03 ==> 10/1/03Note: text cross-references in parenthesis are for 2nd ed.NOTE: The “prime” symbol (as in L’) means complement ClosureIf L1 and L2 are regular languages, Then so are L1L2, L1L2, L1L2, L1’, and L* Regular languages are closed under reversal. Homomorphism:Let  and  be two alphabets. h:   * is a homomorphism: h(ai) = u  *,  ai  extending to strings we have: w = a1a2...an h(w) = h(a1)h(a2)...h(an) Homomorphic image:If L is a language on , then homomorphic image of L is h(L) = { h(w) : w  L }, h = homomorphism Example 4.2, 4.3 Theorem 4.3Let h be a homomorphism.If L is a regular language, then its homomorphic image h(L) is also regular.==> Regular languages are closed under homomorphisms.CS373 Fall 2003 Chapter 4 N. Guydosh Page 1  Definition 4.2 Right QuotientLet L1 and L2 be languages on same .Then right quotient is defined as L1/L2 = {x: xy  L1, for some y  L2}==> “Set of all strings in L1 that have a suffix in L2 with this suffix removed” n m Example: L1 = {a b : n  1, m  0}  {ba} m L2 = {b : m  1} n m L1/L2 = {a b : n  1, m  0} aaabbbbb  L1  aaabb  L1/L2 bbb is suffix==> aaab  L1/L2==> aaa  L1/L2 Theorem 4.4If L1 and L2 are regular languages over , then L1/L2 is also regular==> Family of regular languages is closed under right quotient with a regular language.See examples 4.4 and 4.5 Elementary questions about regular languages (4.2) - or - Regular questions about elementary languages! Standard representationA regular language is given in standard representation (SR) if, and only if, it is described by:(1) Finite automaton, (2) Regular expression, or (3) Regular grammarInformal descriptions and set notation are excluded.CS373 Fall 2003 Chapter 4 N. Guydosh Page 2 Theorem 4.5Given a standard representation of a regular language L on  and any w  *, then  an algorithm determining whether or not w  L.==> Construct DFA for L and test w.Theorem 4.6 an algorithm for determining whether a regular language in standard representation form is empty, finite, or infinite.==> Construct DFA. If simple path (no edge or vertex repeated) from q0 ==> qf then non-empty.Question: Does any walk from q0 ==> qf implies  a simple path also? If q0 ==> qf path passes through vertex which is base of cycle then infiniteTheorem 4.7Given two regular languages L1 and L2 in standard representation form,  an algorithm to determine if/(if not) L1 = L2.==> Define regular language: L3 = (L1  L2’)  (L1’  L2) = L1  L2 ... “Exclusive OR” Since L3 is a regular language,  DFA M accepting L3.Use Theorem 4.6 to see if L3 is empty.L3 =  if, and only if, L1 = L2. (Why?)CS373 Fall 2003 Chapter 4 N. Guydosh Page 3 Identifying Non-Regular Languages (Sect 4.3) Pigeon hole principle (a target of denial) Put n objects into m boxes.If n > m, then at least one box must have more than one item in it. Can prove L is non-regular language by assuming contrary ==> denying pigeon hole property.Example 4.6 Prove L = {anbn : n  0} is non-regularAssume contrary: (Assume L is regular language)==>  DFA, M = (Q, {a,b}, , q0, F) i Consider *(q0, a ) i > 0==> infinite number of is and finite number of states. n m ==>  q  Q and some n, m (n  m) such that *(q0, a ) = q = *(q0, a ) due to pigeon hole principle. ... q would be the first state to repeat for a sufficiently long string of a’s. But since, by assumption (contrary), n n n M accepts a b , we have *(q, b ) = qf  F[Uses: *(q,w1w2) = *(*(q,w1),w2)]But also, m n m n n *(q0,a b ) = *(*(q0,a ),b ) = *(q, b ) = qf  FContradiction since m  n and ambn should be rejected. QEDCS373 Fall 2003 Chapter 4 N. Guydosh Page 4  “Pumping Lemma” (another object of denial) Application of pigeon hole principleTheorem 4.8Let L be an infinite regular language.Then, some integer m > 0 such that  w  L and |w|  m, we can decompose w as w = xyz with |xy|  m|y|  1 i such that wi = xy z, i  0 (pumped up w) is also in L.Proof: L is regular language ==>  DFA M with states q0, q1, q2, ..., qn , (n+1 states)Let w  L such that |w|  m = n + 1M goes through the following sequence of states as it processes w: q0, qi, qj, ..., qfSince |w|  n + 1,Number of states in walk = |w| + 1 > n + 1 See example below ... Number of states in walk > number of states==> at least 1 repeated state (pigeon hole principle)CS373 Fall 2003 Chapter 4 N. Guydosh Page 5 Example: w = abc number of states = |w|+1 = 3+1. = 4. Sequence looks like: q0 , qi , qj , ..., qr , ..., qr , ..., qf ------x y substrings x, y, z such that*(q0 , x) = qr*(qr , y) = qr*(qr ,z) = qf with |xy|  n + 1 and |y|  1 Note: Reason for |xy|  n + 1 =m: If qr is the first occurrence of a vertex which will be also the next repeated one, st Then it (1 qr) will occur no later than nth move since there are only n+1 states in going from q0 to qf . Since there can be no repetitions of states in the “y loop or segment”, and we need at least 1 move to complete the loop, it follows that number of moves in xy = |xy|  number of states = n+1 = m. Remember the example above relating number of moves (string length) with number of states. nd Thus *(q0 , xz) = qf skip the loops, and |xy|  n + 1 = m … see p. 116 (120, 2 ed) i and *(q0 , xy z) = qf i  0 any number of loops will work xyiz = pumped string is accepted QEDCS373 Fall 2003 Chapter 4 N. Guydosh Page 6 Observation: - If assuming a language L is a regular language (accepted by DFA)And pumping lemma is denied in the process, then L is not a regular language (contradiction)- We are guaranteed existence of m as well as the decomposition xyzButWe do not know what they are. - Cannot claim contradiction just because the pumping lemma is violated for some specific value of m or specific xyz … see middle p. 117 (bottom p. 120).- But: Pumping Lemma holds  w  L, |w|  m and all i.Thus If Pumping Lemma is violated for even one w or i, then L is not a regular language. Game Theory Approach to Applying Pumping Lemma1. Opponent picks m2. Given m, we pick w  L such that |w|  m (totally free to choose)3. Opponent chooses decomposition xyz with |xy|  m and |y|  1. Opponent makes optimal choice.) i 4. We try to pick some particular i such that the pumped string wi = xy z is not in L. … make no assumptions on what partition is – must guarantee violation for all possible partitions …If we can do this, we win.Else inconclusive (opponent wins).CS373 Fall 2003 Chapter 4 N. Guydosh Page 7

Note: Text Cross-References in Parenthesis Are for 2Nd Ed

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