# l6 - Lecture 6 Finishing Regular expressions David Dill...

This preview shows pages 1–7. Sign up to view the full content.

Lecture 6: Finishing Regular expressions David Dill Department of Computer Science 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Outline More closure properties Minimization of DFAs. Decision problems on regular languages. 2
String Homomorphisms Start with function h : Σ 1 Σ 2 . Then extend it to strings: Base: h ( ǫ ) = ǫ Induction: h ( xa ) = h ( x ) h ( a ) . So, if h (0) = ab and h (1) = ǫ then h (0110) = abab h can also be applied to entire languages: h ( L ) = { h ( x ) | x L } . 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Closure under Homomorphisms Theorem: If L Σ 1 is regular and h : Σ 1 Σ 2 is a string homomorphism, then h ( L ) is regular. Construction: There is a regular expression R for L . Substitute h ( a ) for each occurrence of each a in R . The proof is by induction on the structure of regular expressions. In the induction cases, you have to prove things like h ( L ) = [ h ( L )] 4
Inverse Homomorphism h 1 ( L ) = { x | h ( x ) L } . Consider h that deletes “ c ”: h ( a ) = a , h ( b ) = b , h ( c ) = ǫ . Consider L = ( abc ) . What is h ( L ) ? What is h 1 [( ab ) ] ? Answer: c ( c ac bc ) not the same language . h 1 ( L ) is the largest language L such that h ( L ) = L . 5

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Closure under Inverse Homomorphism Theorem: If L Σ 2 is regular and h is a homomorphism from Σ 1 Σ 2 , then h 1 ( L ) is regular. Construction:
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 15

l6 - Lecture 6 Finishing Regular expressions David Dill...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online