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CSC3130 Mid-term Review

CSC3130 Mid-term Review. bridger.zhao@gmail.com. Department of Computer Science & Engineering. Scope: regular languages. DFA, NFA and conversion from NFA to DFA. Regular expression. Equivalence of DFA, NFA and regular expression. Closure properties of regular languages.

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CSC3130 Mid-term Review

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  1. CSC3130 Mid-term Review bridger.zhao@gmail.com Department of Computer Science & Engineering

  2. Scope: regular languages • DFA, NFA and conversion from NFA to DFA. • Regular expression. • Equivalence of DFA, NFA and regular expression. • Closure properties of regular languages. • Pumping lemma for regular languages. • DFA minimization algorithm, DFA equivalence. CSC3130 Tutorial One

  3. Scope: context free languages • Context free grammars. • Derivations, parse trees, and ambiguity. • Removal of ε and unit productions, Chomsky Normal Form. • The CYK parsing algorithm. • Pushdown automata. • Pumping lemma for context-free languages CSC3130 Tutorial One

  4. NFA to DFA & State minimization CSC3130 Tutorial One

  5. DFA Hacking • Let L be a regular language, L’={u|uv is in L and |u|=|v|}. That is, L’ is the first halves of strings in L. Prove that L’ is also regular. • Hints: Let M be the DFA for L, How do we characterize a middle state? CSC3130 Tutorial One

  6. Closure Properties • For regular languages: union, intersection, concatenation, Kleen closure, reverse, complement, etc. • For context free languages, exclude intersection and complement. • Usage: is L L regular? Is non palindromes regular? CSC3130 Tutorial One

  7. Context Free Language • S → |aaSbbb • CFG to PDA see 08L10,17/23 CSC3130 Tutorial One

  8. CNF • Consider the following simple grammar: • A  cA | a • B  ABC | b • C  c • How to convert this grammar to CNF? CSC3130 Tutorial One

  9. Conversion into CNF Step 1: Convert every production into either: A B1B2…Bn or A a e.g. A  bCDeF becomes: A  BCDEF B  b E  e CSC3130 Tutorial One

  10. Conversion into CNF Step 2: Convert production of the form A B1B2…Bn into A  C1C2 : e.g. A  BCDEF becomes: A  BX X  CY Y  DZ Z  EF CSC3130 Tutorial One

  11. Example • Convert the following CFG into Chomsky Normal Form: S  • S  ABBA • B  bCb • A  a • C  c CSC3130 Tutorial One

  12. Solution Step 1: B bCb becomes B DCD D b Step 2: S  ABBA becomes S AE E BF F BA B DCD becomes B DG G CD Result: S  S AE E BF F BA B DG G CD A  a C  c D b S  S  ABBA B  bCb A  a C  c CSC3130 Tutorial One

  13. Pumping Lemma for CFL • L = { ww | w is in {0,1}* } • Assume L is a CFL • Let n be the Pumping Lemma constant • Considerz = 0n1n0n1n, |z| = 4n > n • We want write z = uvwxy with |vwx|  n and |vx| 1 • There are several cases for uvwxy CSC3130 Tutorial One

  14. Pumping Lemma for CFL 0n1n0n1n • Case 1: vwx is within the first block0’s • v and x must contain 0’s • Let vx consist of k 0’s, where k > 0 • Choose i=0, uviwxiy=0n-k1n 0n1n • If 0n-k1n 0n1n in form of ww, the end of first w must lie in the second block of 0’s • Then, the two w’s are in the different pattern:0+1+0+ and 0+1+.So, 0n-k1n 0n1n L CSC3130 Tutorial One

  15. Pumping Lemma for CFL • Case 2: vwx spread over the first block of 0’s and first block of 1’s • vx has at least one 0’s or one 1’s • Let vx consist of k 0’s and p 1’s, where k0 ,p>0 or k>0,p 0 • Choose i=0, uviwxiy=0n-k1n-p 0n1n • If 0n-k1n-p 0n1n in form of ww, the end of first w must lie after the first block of 1’s • Then, the number of 1’s is different in two w. So, 0n-k1n-p 0n1n L 0n1n0n1n CSC3130 Tutorial One

  16. Pumping Lemma for CFL • Case 3: vwx is within the first block of 1’s • v and x must contain 1’s • Let vx consist of k 1’s, where k > 0 • Choose i=0, uviwxiy =0n1n-k 0n1n • If 0n1n-k 0n1n in form of ww, the end of first w must lie in the second block of 0’s • Then, the number of 1’s is different in two w. So, 0n1n-k 0n1n L. 0n1n0n1n CSC3130 Tutorial One

  17. Pumping Lemma – Example • Case 4: vwx spread over the first block of 1’s and second block of 0’s • vx has at least one 1’s or one 0’s. • Let vx consist of k 1’s and p 0’s, where k 0,p>0 or k>0,p 0 • Choose i=0, uviwxiy =0n1n-k 0n-p1n • For 0n1n-k 0n-p1n in form of ww, if the end of first w lie in the first block of 1’s, then the number of 0’s is different in two w. • if the end of first w lie in the second block of 0’s, then the number of 1’s is different in two w. So, 0n1n-k 0n-p1n L 0n1n0n1n CSC3130 Tutorial One

  18. Pumping Lemma for CFL • Other cases: vwx is within the second half of z • the argument is symmetric to the cases where vwx is within the first half of z • We can conclude that L = { ww | w is in {0,1}* } is not context free. 0n1n0n1n 0n1n0n1n 0n1n0n1n CSC3130 Tutorial One

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