Ogden s Lemma. and Formal Languages. Automata Theory CS 573. The proof is similar but more fussy. than the proof of the PL4CFL.

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1 CS 573 Automata Theory and Formal Languages Professor Leslie Lander Lecture # 24 December 4, 2000 Ogden s Lemma (6.2) Let L be a CFL, then there is a constant n such that if z is a word in L with z > n and we mark any n or more of the positions of z as distinguished, then z = uvwxy such that v and x together have at least ONE distinguished position vwx has at most n distinguished positions uv i wx i y is in L for all i CS 573 L. Lander, 2000 Class 24-8 Ogden s Lemma Back to Chapter 6 No proof! The proof is similar but more fussy than the proof of the PL4CFL Example 6.3 Let L 4 = {a i b j c k : i, j, k all different} Choose z = a n b n+n! c n+2n! and make the positions of the a s (1 through n) the distinguished positions Write z = uvwxy v and x contain at least one a We could have: CS 573 L. Lander, 2000 Class 24-9

2 Using Ogden s lemma Either v or x could contain two different letters but then uv 2 wx 2 y would have symbols out of order (a s after b s or b s after c s) (1) So v can be non-empty and in the a s with x in the b s or (2) v can be non-empty and in the a s with x in the c s or (3) v and x can be in the a s (at least one of them non-empty) CS 573 L. Lander, 2000 Class Substitution Back to Chapter 2 Dealing with the 3 cases (1) Let q = n! / v (an integer since v < n) Consider uv 2q+1 wx 2q+1 y = a n+2n! b n+n!+2q x c n+2n! --the number of a s and c s are equal (2) With the same q consider uv q+1 wx q+1 y = a n+n! b n+n! c n+2n!+q x --the number of a s and b s are equal (3) Let q = n! / vx, then uv q+1 wx q+1 y = a n+n! b n+n! c n+2n! CS 573 L. Lander, 2000 Class Substitutions and homomorphisms Algebra is concerned with structures, Σ* is a monoid: (w 1 w 2 )w 3 = w 1 w 2 w 3 = w 1 (w 2 w 3 ) w ε = ε w = w Concatenation is a way to combine elements, it is an associative operation The empty string is the identity for this operation CS 573 L. Lander, 2000 Class 24-13

3 More algebra For languages L 1 and L 2 over another alphabet Γ [i.e. given elements L 1 and L 2 of P(Γ*)], we have a composition L 1 L 2 that satisfies: (L 1 L 2 )L 3 = L 1 L 2 L 3 = L 1 (L 2 L 3 ) = {w 1 w 2 w 3 : w k L k, k = 1, 2, 3} with the identity ε = {ε}: {ε}l 1 = L 1 So P(Γ*) is also a monoid under concatenation CS 573 L. Lander, 2000 Class Defined by the symbols in the alphabet Hence, for any string w = a 1 a 2...a n we have f (w) = f (a 1 a 2...a n ) = f (a 1 )f (a 2 )...f (a n ) so it is enough to define f on the individual symbols in Σ Note that each symbol in Σ is mapped to a language over Γ Structure-preserving maps in algebra are called homomorphisms CS 573 L. Lander, 2000 Class homomorphisms We can create a function from Σ* to P(Γ*) for some other alphabet Γ that respects the structures This function f : Σ* P(Γ*) must preserve concatenation and the identity: f (uv) = f (u)f (v) and f (ε) = ε It follows immediately that f (u 1 u 2...u n ) = f (u 1 )f (u 2 )...f (u n ) CS 573 L. Lander, 2000 Class Properties For any function f : Σ* P(Γ*) and language L Σ* we can define the language f (L) = U f (w) Γ* w L Next we call f : Σ* P(Γ*) a substitution if f is a homomorphism (i.e. f (uv) = f (u)f (v), f (ε) = ε ) and f (a) is a regular language for each a in Σ CS 573 L. Lander, 2000 Class 24-17

4 Regular expressions Theorem 3.4 If f is a substitution and L is regular, then f (L) is regular Proof: Let r be the reg. expr. for L, then f (r) is a well-defined reg. expr. (i) for a symbol a in Σ, a is a basic regular expression and f (a) is a regular language (by hypothesis) and therefore has a regular expression r a (ii) also f ( )= and f (ε)= ε so that all basic reg.expr s map to reg.expr s CS 573 L. Lander, 2000 Class Simpler homomorphisms We can simplify the type of homomorphism we consider Consider f : Σ* Γ* such that f (uv) = f (u)f (v), f (ε) = ε This is a homomorphism from Σ* to Γ* Note that for a single symbol a in Σ, f (a) may be a string in Γ*, not necessarily a single symbol CS 573 L. Lander, 2000 Class Regular Expression constructs Next r will be a composition of simpler regular expressions and all we need to show now is that f preserves these constructs: f (L 1 L 2 ) = f (L 1 ) f (L 2 ) f (L 1 L 2 ) = f (L 1 ) f (L 2 ) f (L*) = f (L)* Hence f (r 1 r 2 ) = f (r 1 ) f (r 2 ), f (r 1 +r 2 ) = f (r 1 )+f (r 2 ), f (r *) = f (r)* This is enough to complete the proof CS 573 L. Lander, 2000 Class Theorem 3.5 Theorem 3.5 If f : Σ* Γ* is a homomorphism L 1 is a regular language in Σ* and L 2 is a regular language in Γ*, then f (L 1 ) is a regular language in Γ* and f -1 (L 2 ) is a regular language in Σ* where f -1 (L 2 ) = {w Σ*: f (w) L 2 } Proof: the first part is a special case of the result for substitutions: each f (a) is a singleton, hence regular CS 573 L. Lander, 2000 Class 24-21

5 Proof continued Given a DFA for L 2 with transition function δ, we can define a DFA for f -1 (L 2 ) using the transition function δ new (q, a) = δ*(q, f (a)) Since f (a) may be a string, we need to use the extension of δ, which is defined for strings It is easy to show that we can use δ new to get a DFA for the inverse image f -1 (L 2 ) CS 573 L. Lander, 2000 Class Conclusion But h 2 (a) = 0, h 2 (b) = 1, h 3 (c) = 1, so h 2 {a n bc n-1 : n > 1} = {0 n 1 n : n > 1}, which certainly is not regular Hence {a n ba n : n > 1} cannot be regular CS 573 L. Lander, 2000 Class Application Define h 1 : (a+b+c)* (a + b)* by h 1 (a) = a, h 1 (b) = ba, h 1 (c) = a Define h 2 : (a+b+c)* (0 + 1)* by h 2 (a) = 0, h 2 (b) = 1, h 3 (c) = 1 Then h -1 1 ({a n ba n : n > 1}) a*bc* = {a n bc n-1 : n > 1} If {a n ba n : n > 1} were regular then {a n bc n-1 : n > 1} would be regular by the previous theorem, then h 2 {a n bc n-1 : n > 1} would be regular CS 573 L. Lander, 2000 Class Substitution and homomorphism in CFL s Theorem 6.2 and 6.3 A substitution maps a CFL to a CFL A homomorphism maps a CFL to a CFL The inverse image of a CFL under a homomorphism is a CFL In Chapter 11 we find rec. enum. languages are closed under sustitution but not recursive languages CS 573 L. Lander, 2000 Class 24-25

6 Another solution to Problem 3.1 (h) Suppose {xx R w : x, w (0+1) + } were regular, then {xx R w : x, w (0+1) + } (01) + (10) = {(01) n (10) m 1 s : m > n > 0, s > 0} would be regular But {0 n 1 m 2 s : m > n, s > 0} = h -1 {(01) n (10) m 1 s :m>n>0,s>0} where h(0) = 01, h(1) = 10, h(2) = 1 and {0 n 1 m 2 s : m > n, s > 0} is not regular by the simple PL4RL CS 573 L. Lander, 2000 Class Exercise in the book (3.4) Let L be regular, which below are regular? {a 1 a 3 a 5...a 2n-1 : a 1 a 2 a 3...a 2n L} {a 2 a 1 a 4 a 3... a 2n a 2n-1 : a 1 a 2 a 3...a 2n L} {x 2 x 1 : x 1 x 2 L} {x L : if xy L then y = ε} {x L : if x = yz and y L then y = x} {x Σ* : xy L for some y} {x : x R L} {x : xx R L} CS 573 L. Lander, 2000 Class Quotients Another construct that is important for regular language theory is the quotient Given languages L 1, L 2, write L 1 /L 2 for {x Σ* : xy L 1 for some y L 2 } Theorem 3.6 If L 1 is regular, then L 2 is regular CS 573 L. Lander, 2000 Class 24-27